ptate CoIIcse of ^Qvicnltnvt ^t Cornell Winiiittsiit^ Stdaca, i%. j^. Hiiirarp Cornell University Library QO 42.A7 Chemical calculations, 3 1924 002 967 382 The original of tliis book is in tlie Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924002967382 Chemical Calculations BY E. HAEMAN ASHLEY, Ph.D. ASSISTANT PROFESSOR OF CHEMISTRY IN THE UNIVERSITY OP MAINE ILLUSTRATED NEW YORK D. VAN NOSTRAND COMPANY 26 Park Place 1915 Copyright, 1915, BT D. VAN NOSTRAND COMPANY Stanbope jpresz H. GILSON COMPANY BOSTON, U.S.A. PREFACE In putting forth this small work, the author has in- tended to write a textbook on Chemical Calculations fol- lowing more closely than other books at present on the market, the needs of the student who wiU later find occupation in chemical laboratory work. For this reason, constant reference will be made to Chemists' Handbooks, the best of which, in the author's opinion, is "Van Nos- trand's Chemical Annual for 1913," edited by Prof. J. C. Olsen. Constant reference will be made to this work by means of footnotes, with the intention of lead- ing the student to the most convenient source of tabu- lated data, and to explain the method of utilizing the same. The purchase of the "Annual" will not be nec- essary, much as it may be desirable, in following the ex- planations or solving the problems. The problems are nimierous and of varying degrees of difficulty; the instructor is the best judge of the type of problem to assign a class or individual. Those of the more difficult sort may be given the more advanced student, thus holding the class together. Many of the problems conform to the type usually found in works of this natm-e and in the selection of -these, most of the books on this subject have been consulted. Some of the types are new to works of this kind; their inclusion is made in the hope of making the book more practical. The method of solving the problems is treated in the text, and in the earlier chapters typical problems are solved at the end of the reading matter just before the problems are stated. The answer or answers have been IV PREFACE attached to each problem rather than given in a list in the appendix, with the belief that it is more convenient for student and instructor. It may be urged that the inclusion of the answer with the problem may be a temp- tation offered the student to work for this answer. How- ever, a student who will do this, will always take the trouble of looking up the answer, even if it is at the end of the book. Some of the topics are discussed from two or more methods of approach in the belief that one of the methods may appeal to a student and be more easily comprehended by him than the others. If some of the topics are treated in too elementary or extended a fashion to suit some, it must be remembered that this subject is not simple to all students. The author does not anticipate adverse criti- cism on the score of the work being too elementary when taken as a whole. The treatment of ratios may be so full as to be tedious. This is for the reason that it is wished to lead the student away from proportions as generally expressed, i.e., x : y = a : b, but to have him at once set down his mathematical expression in a form for immediate solution. The algebraic method has been extensively employed and the treatment of indirect analysis and the calculation of indirect factors have been given considerable space, an amount of space not meas- ured by the importance of indirect analysis, but rather by the importance of the methods of chemical and alge- braic reasoning. Many problems have been solved in steps, after which it is shown how the same problem may be solved by one expression. The problems are more numerous than will be solved by one class, if small; two or more problems of the same nature are included so that in a large class there will be enough to go around. Furthermore, a large number of problems will give the l-JXJlJli AUJli instructor more choice, and there is less likelihood of their being solved by one class and handed on to the next in the nature of heirlooms. In the calculation of the problems, the use of loga- rithms is strongly recommended; as also is the slide rule. The latter is to be especially recommended, as the time allotted the subject may be devoted to theory, relegating the mechanical solution of the problems to a subordinate place. With the use of this device the subject may be thoroughly treated in. haK the usual time. The problems have been checked and rechecked, but notwithstanding this, errors will imdoubtedly creep in. The author will appreciate having his attention called to such. As five- and, in a few instances, six-place logarithms have been used, answers are often stated to a niunber of significant figures not justified by the number of signifi- cant figures in the data given. It was thought best to carry out the answer, as this involved no extra work and is valuable in inculcating accuracy. The author wishes to express his appreciation to Mr. Harry R. Lee, Superintendent, of the Virginia Electro- lytic Company for checking problems, reading the manuscript and giving much criticism and advice; to Mr. G. C. Merrill, formerly of the College of Montana, for the checking of a large number of problems. Profes- sors Ralph H. McKee and Charles W. Easley of the University of Maine have been generous with help and advice, as have many others of the author's colleagues. An expression of thanks is due Mr. Charies F. Guhlmann of the General Chemical Company for reading the chap- ter on mixed acids and offering very helpful advice for this section. jj jj j^ UNnrERSiTY OF Maine, Obono, Maine, Jan. 30, 1915. TABLE OF CONTENTS Page Preface iii-v CHAPTER I: Ratios 1-25 Decomposition of Mercuric Oxide — Ratios Involved in the Decomposition of Mercuric Oxide — Metathesis between Barium Chloride and Sodiimi Sulphate — r Fac- tors — Percentage Composition ,of Chemical Com- poimds — Equivalents — Successive Reactions — Fac- tors Other than Chemical — Problems. CHAPTER II: Approximate Numbers 26-31 Addition and Subtraction of Approximate Numbers — MultipUcatJon of Approximate Numbers — Division of Approximate Numbefs — Recapitulation. CHAPTER III: Interpolation 32-47 Functions — Graphic Representation of Functions — Change of Percentage Composition with Baum6 for Sulphuric Acid — Interpolation — Extrapolation — Problems. CHAPTER IV: Heat 4S-58 Fahrenheit Scale — Centigrade Scale — Reaumur Scale — Absolute Scale — Calorie — British Thermal Unit — Specific Heat — Heat of Fusion — Heat of Vaporization — Heat of Combustion — Conversion of the Centigrade into British System and Vice Versa — Problems. CHAPTER V: Specific Gravity 69-79 Definitions — Specific Gravity of Body Heavier than and Insoluble in Water — Specific Gravity of Sohd Substance Lighter than Water — Specific Gravity of vm CONTENTS Paqb Powders — Pyknomefcer Method for Liquids — Sinker Method for Liquids — Constant Weight Hydrometers — Constant Volume Hydrometers — Hydrometers — Baum6 Hydrometer — Conversion of Baume Readings into Specific Gravity and Vice Versa — Calculation of Density Determinations to Different Conditions of Temperature — Problems. CHAPTER VI : Gas Calculations 80-140 Boyle's Law — Charles' Law — Gas Thermometer — Law of Boyle and Charles Combined — Standard Con- ditions of Temperature and Pressure — Correction of Barometer for Temperature — Moist Gases — Gay- Lussac's . Law — Gas Analysis — ■ Avogadro's Law — ■ Atomic Weights and Molecular Weights — Vapor Density' — Graham's Law — Volume Occupied by a Gram Molecule of a Gas — Measurement of Vapor Density — Deviations from Gas Laws — Volume of Gas from Weight of Substance (Metric System) — Volume of Gas from Weight of Substance (English System) — Calculations of Weighings in Air to Value in Vacuo and Vice Versa — Standard Unit of Volu- metric Apparatus — Mohr's Liter — Dalton's Law of Partial Pressures — Problems. CHAPTER VH: Calculation of Atomic Weights AND Fokmulas 141-171 Atomic Weights — Valence — Combining Weights — Relationship between Atomic Weight, Valence and Combining Weight — Calculation of Atomic Weight from an Analysis — Law of Dulong and Petit — Molec- ular Weight by Elevation of Boiling Point and De- pression of Freezing Point — Molecular Weight of an Organic Acid — Molecular Weight of an Organic Base — Formula of a Compound Given Molecular Weight and Percentage Composition — Formula from Percentage Composition — Formulas of Minerals, Isomorphic Replacement — Problems. CONTENTS IX Page CHAPTER VIII: Gravimetric Analysis 173-201 Direct Gravimetric Analysis — Elimination of a Con- stituent from an Analysis — Factor Weights — • In- direct Analysis — Analysis of Oleum — Problems. CHAPTER IX: V<3Lumetric Analysis 202-251 Single Factor Solutions — Methods of Standardization — Equivalent Values of Single Factor Solutions — Factor Weights for Volimietric Solutions — Normal Solutions — Simplification of Calculations by the Use of Normal Solutions — Calculation of Normality — Factor to a Given Normality — • Volumetric Determi- nations using two Solutions — Back Titrations — Vol- umetric Analysis using Two Indicators — Volumetric Analysis of Oleum — Adjustment of Strength of Solu- tions — Problems. CHAPTER X: Use of Specific Gravity Tables and Acid Calculations ,. . . . 252-267 Calculation of Data in Specific Gravity Tables — Use of Specific Gravity Tables — Calculation of Mixed Acid — Strengthening of Mixed Acid by Use of Oleum — Problems. CHEMICAl CALCULATIONS CHAPTER I RATIOS Chemical calculations are based on the two following laws: Law of the Conservation of Mass. — In any system un- dergoing change, chemical or otherwise, no change in the mass of the system has ever been observed. Law of Constancy of Composition. — The masses of the elements taking part in a given chemical change always exhibit a definite and invariable ratio to each other. The Decomposition of Mercuric Oxide. — Let it be required to find the number of grams of oxygen which will be liberated by the decomposition of 1.7000 g. of mercuric oxide. The decomposition is represented by the equation: 2HgO = 2Hg + O2 parts by weight, 2 (216.6) 2 (200.6) 2 (16) In the atomic weight tables,^ 200.6 is given as the atomic weight of mercury and 16.00 as that of oxygen. The sym- bol of mercuric oxide, HgO, shows that one molecule of mercuric oxide contains one atomic weight of mercury ' The atomic weights are revised annually by an international committee. See "Van Nostrand's Chemical Annual," p. 1, third issue, 1913. Frequent reference will be made to this book, where the title will be abbreviated to Chem. Ann., always referring to the edition of 1913. 1 2 CHEMICAL CALCULATIONS and one atomic weight of oxygen, consequently the sum of these atomic weights (200.6 + 16.00), 216.60, represents the molecular weight of mercuric oxide.' Single atoms and . molecules are so small that they cannot be dealt with as such, but aggregations of atoms or molecules may be taken in the ratio of their atomic or molecular weights, in which case an equal number of atoms are dealt with. Numbers representing molecular or atomic weights are relative only, being based upon a convenient standard, for example, the weight of an atom of oxygen, the atomic weight of which is arbitrarily fixed at 16.00. In a complete reaction, when all the initial substances are adjusted in the ratio of the molecular or atomic weights, all the initial substances disappear and the products stand in relation to each other in the ratio of their molecular or atomic weights and nothing remains over. Ratios Involved in the Decomposition of Mercuric Oxide. — Returning to the problem, the ratio of oxygen to mercuric oxide is 2 (16) parts by weight of oxygen to *? ("] fi^ 2 (216.6) parts by weight of mercuric oxide, or ■ This is the fixed and invariable ratio which always holds between these two substances in this particular reaction. In other words, 2 (16) parts by weight of oxygen are evolved by the decomposition of 2 (216.6) parts by weight of mer- curic oxide. Then, if 2 (216.6) parts by weight of mercuric oxide yield 2 (16) parts by weight of oxygen, one part by weight of oxygen is given by , Jl ^ . parts by weight of 1 Tabulations of molecular weights are to be found in chemist's handbooks. In such a simple case as this, no time would be saved by so doing, but in many cases such tabulations are useful and much time saved and liability to error reduced. See Chem. Ann., pp. 39-47; 100-336. RATIOS 3 merciiric oxide, and as the unit of weight iinder considera- tion is the gram, the problem is solved by the expression: 2^(2166) ^ ^-^^^^ " ^-^^^^ ^- °^ °^Seii, yielded by 1.7000 g. mercuric oxide. Again, employing the same equation, how much mer- cmy will be obtained by the decomposition of 1.7000 g. of mercuric oxide? The quantity of mercury is now sought; consequently the ratio of mercury to mercuric oxide being 2 (200.6) , . • 1 • .u „ )^^ „ J. , and reasomng precisely as m the previous problem, the expression becomes: ^^§: X 1.7000 = 1.5744 g. of mercury, yield from 1.7000 g. mercuric oxide. In all, this reaction yields the following ratios: 2(16) {ay Oxygen to mercuric oxide, ^».^ o + 2(16 ) (6) Oxygen to mercury, 2(200 6) (c) Mercury to mercuric oxide, 2 (216.6) 2(16) 2(200.6)" 2 (200.6) 2(216.6)' ' It is evident that oTaffifiT' ^**'"' '^ equivalent to kto-o J but the equation as written is more accurate, chemically speaking, than the simpler equation, HgO = Hg + O, in which the oxygen is rep- resented as in the atomic state, a condition fulfilled only at the instant of decomposition. It is necessary for the student to bear this in mind, as in gas equations involving volumes it is a matter of importance. For calculations involving weights only, such an equation serves the purpose, but in the calculation it is a simple matter to write the correct equation, and in solving t6 cancel; i.e., /(1 6) ^ 16 ;? (216.6) 216.6' 4 CHEMICAL CALCULATIONS 2(200.6) (d) Mercury to oxygen, 2(16) / X T>T ■ -J ^ 2(216.6) (e) Mercuric oxide to oxygen, ^ .^^. — ■ -J X 2(216.6) (/) Mercunc oxide to mercury, , . • If X represents the units of weight in any system: (a)' , - X = weight of O2 produced by x units of HgO. (6)' .fi>f > ft\ ^ = weight of O2 equivalent to x units of Hg. (c)' - ^- „"J a: = weight of Hg produced by x units HgO. (d)' ^ " X = weight of Hg equivalent to x imits of O2. 2 (lb; (e) ' ^ , ' a; = weight of HgO produced by a; units of O2. 2 (lb; (/) ' L(xr)'gx a; = weight of HgO produced by x units of Hg. These six ratios are applied in problems typified by the following: (a)" How many grams of oxygen will be liberated by the decomposition of 3.000 g. of mercuric oxide? Ans. gwV) ^ ^-^^^ " ^-^^^^ ^• (6)" How many grams of oxygen are associated with 3.000 g. of mercury in mercuric oxide? Ans. 2^^^ X 3.000 = 0.2393 g. RATIOS 5 (c)" How many grams of mercury are produced by the decomposition of 3.000 g. of mercuric oxide? ^"^- Hl^ X ^-^OO = 2-7784 g. (d)" How many grams of mercury are associated with 3.000 g. of oxygen in mercuric oxide? Ans. ^^^^ X 3.000 = 37.613 g. (e)" How many grams of mercuric oxide must be de- composed to obtain 3.000 g. of oxygen? Ans. ^^^p X 3.000 = 40.613 g. (/)" How many grams of mercuric oxide must be de- composed to obtain 3.000 g. of mercury? Metathesis between Barium Chloride and Sodium Sulphate. — Again take the equation: BaCl2 + NazSO* = BaSOi -1- 2 NaCl. The molecular weights being the sum of the atomic weights, are: BaClg Na2S04 Ba = 137.37 Naa = 2 (23.00) = 46.00 201 = 2(35.46)= 70.92 S = 32.07 208.29 40 =4(16.00) = 64.00 BaS04 Ba = 137.37 Na S = 32.07 CI 40 = 4 (16.00) = 64.00 2 NaCl 142.07 = 23.00 = 35.46 2 (58.46) 233.44 6 CHEMICAL CALCULATIONS As before the ratios are: 208 29 a) Barium chloride to sodium sulphate: h) Barium chloride to barium sulphate: c) Barium chloride to sodium chloride: 142.07 208.29 233.44' 208.29 2 (58.46) 142 07 d) Sodium sulphate to barium chloride: " • 142.07 233.44" 142.07 2 (58.46) ■ 233.44 208.29" 233.44 142.07" 233.44 e) Sodium sulphate to barium sulphate: /) Sodium sulphate to sodium chloride: g) Barium sulphate to barium chloride: h) Barium sulphate to sodium sulphate: i) Barium sulphate to sodium chloride: , , j) Sodium chloride to barium chloride: " • k) Sodium chloride to sodium sulphate: A^ " • I) Sodium chloride to barium sulphate: -pr^nr-rr^ • 233.44 These twelve ratios cover all cases which will arise from this reaction. In the typical examples given below, these ratios are employed in solving problems. (a)' How many grams of barium chloride will be re- quired to react with (are equivalent to) 5.0000 g. of sodium sulphate? 208 2Q Ans. jg^ X 5.0000 = 7.3305 g. RATIOS 7 {by How many grams of barimn chloride are required for the precipitation of 5.0000 g. of barium sulphate? 208 2Q Ans. 2^1^ X 5.0000 = 4.4613 g. (c)' Required the number of grams of barium chloride equivalent to 5.0000 g. of sodium chloride. 208 2Q Ans. 2^^;^ X 5.0000 = 8.9074 g. {dy How many grams of sodium sulphate are necessary for the precipitation of the barium of 5.0000 g. of barimn chloride? 142 ny Ans. i^gi X 5.0000 = 3.4104 g. (e)' How many grams of sodium sulphate have been added to barium chloride if 5.0000 g. of barium sulphate are precipitated? Ans. 111^ X 5.0000 = 3.0430 g. (/)' How many grams of sodium sulphate are equiva- lent to 5.0000 g. of sodium chloride? 142 07 Ans. 2758^ ^ ^"^^^^ " ^"^^^^ ^■ (g)' How many grams of barium sulphate are precipi- tated by 5.0000 g. of barimn chloride? 9^^ 44. Ans. |g|| X 5.0000 = 5.6037 g. (h)' How many grams of barium sulphate are precipi- tated by 5.0000 g. of sodium sulphate? O'i'i 44 Ans. =g^ X 5.0000 = 8.2157 g. 8 CHEMICAL CALCULATIONS {i)' How many grams of barium sulphate are equiva- lent to 5.0000 g. of sodium chloride? Ans. 2^^ X ^-0000 = 9-9829 g. (jY How many grams of sodium chloride are formed when 5.0000 g. of barium chloride are added to a solution containing sodium sulphate? Ans. ^^^ X 6.0000 = 2.8067 g. (k)' How many grams of sodium chloride are equiva- lent to 5.0000 g. of sodium sulphate? Ans. ?^^^ X 5.0000 = 4.1149 g. (Z)' If 5.0000 g. jf barium sulphate are precipitated, how many grams of sodium chloride are formed? Ans. \ff 4^ X 5.0000 = 2.5043 g.^ ■ These examples are typical of the commonest and simplest chemical calculations. It is to be noted that there are three terms, of which two are expressed in similar units, the third being dissimi- lar. In these examples there are two terms expressing , atomic weights and one term in grams. To solve a problem of this type: Make a pbaction op the two terms in similab units, placing that term representing the magnitude sought over the line (making it THE NUMERATOR), THE OTHER TERM OF THE SAME KIND OP UNIT UNDER THE LINE (MAKING IT THE DENOMINATOR), MULTIPLY BY THE TERM OP THE DISSIMILAR UNIT. This rule of constructing ratios is, as stated, empirical, but a little reflection will show its validity. It is given in this form in the belief that it is more easily grasped and that a little practice will result in making application of it almost mechanical. It must be borne in mind that it is apphcable only when an increase in one term produces a corresponding increase in the other; in other words, in those cases only in which the two units going to make up the fraction vary directly. The great majority of chemical problem RATIOS 9 Factors. — The fractions composed of the terms of similar units are constant for the sahtie equation and when frequently employed it is advantageous to solve the ratio once for all. The values thus obtained are called factors.' In the reaction between bariimi chloride and sodiimi sul- phate, the fractions formed by the molecular weights yield the following constants: (a)" 5^^ = ttIS =1.4661: factor of BaCl2 to NaaSOi. (b)" S^ = IS^ =0.8923: factor of BaCU to BaSOi. r5a&U4 ZooAi (^)" ^1 = 2^) ='■''''■■ factorofBaCUoNaCl. (d)" §§^* = ^^ =0.6821: factor of Na^S04 to Baa. (e)" ^^^* = ^^ = 0.6086: factor of NaaSO* to BaS04. ^ BaS04 233.44 ^fy^l = ^ = 1.2151:factorofNa.S04toNaCl. ^^y ^S"* " Hit " ^■^^^^' ^^''*°'' °^ ^^^^' *° ^^^^" 00" ^^ = ^^ = 1.6431 : factor of BaS04 to Na2S04. ^ Na2S04 142.07 come under this head, the chief exceptions being in gas calculations, as for instance Boyle's law, which states that the volume of a gas varies inversely with the pressure. In such cases the rule is re^ versed so far as the composition of the fraction is concerned. This is treated in another place. 1 See Chem. Ann., pp. 10-36. 2 When chemical symbols are written in this manner, the frac- tion is considered as made up of the terms corresponding to the molecular weight of the symbol, ^^ gQ representing the fraction 208.29 142.07 ■ 10 CHEMICAL CALCULATIONS ^y ^l = 2^ =1.9966:factorof BaSO.toNaCl. (^■)" 'S = Wf^ ='■''''■■ factorof NaCltoBaCl. W'l^^ = ^^= 0.8230: factor to NaCltoNa^SO*. (0" 1^, = ^So'f? = 0.5009 :factorofNaCl to BaS04. BaSOJ 233.44 From an inspection of these ratios it will be noted that of the twelve, each has its reciprocal; (a)", „ „^ having ., . , . . ,w, Na2S04 ,, w, BaCU . , ,„ BaS04 Its reciprocal in {d) , -^^^ ; (6) , ^^^m (g) , -^^ ; and so on throughout. This then eliminates six of the twelve as superfluous, for knowing the value of the factor, the value of its reciprocal follows. To multiply by a number is equivalent to dividing by its reciprocal and vice versa. Consider the reciprocal factors (b)" and (g)". Problems involving these may be typified as: (6)" How many grams of barium chloride are required to produce X grams of barium sulphate? By employing factor (&)", the answer is obviously 0.8923 X g. (g)" How many grams of barium sulphate will be pre- cipitated by X grams of barium chloride? -Factor (gr)" gives the answer, 1.1207 X g. Supposing factor for (b)" only were known and problem (fif)"is to be solved. This is done by simply dividing by the reciprocal factor for (6)", the result being, Q-^ = 1.1207 X g., as before.1 ' When a factor is given and its reciprocal is desired, instead of dividing by the factor given, if logarithms are used and the loga- rithm of the factor is known, it is more convenient to merely add the cologarithm. RATIOS 11 Factors as given in chemists' handbooks are usually tabulated in four columns: Given, Sought, Factor and Logarithm of the factor. In problem (b)" barium sul- phate is given; barium chloride is sought and the factor is 0.8923. In problem (g)" these conditions are reversed; barium chloride is given and barium sulphate is sought; consequently the operation is reversed.' Percentage Composition of Chemical Compounds. — The law of constancy of composition, stating that every well-defined chemical compound always consists of the same elements united in invariable proportion by weight, furnishes a means of calculating the ratio composition of any compound and consequently the weight of any ele- ment or group of elements contained in any weight of that compound. The compound MgS04 • 7 II2O is com- posed of Mg 24.32 .„„_ , „ — = 0.0987 parts of MgSOi • 7 H2O 246.50 magnesium, per unit weight MgS04 • 7 H2O, ,.- c^ — -,T ^ = o^g rr. = 0.1301 parts of MgSOi • 7 H2O 246.50 sulphin: per unit weight MgSOi • 7 H2O. 110 _ 1106) _ 7140 narts o£ MgS04-7H20 - 246.50 " °-^^*" P^'*' "*• oxygen per unit weight MgS04 • 7 H2O, 1 See Chem. Ann., p. 13. In this tabulation, both factors are given, one being the reciprocal of the other. In the problem (6), barium chloride is "required," while barium sulphate is "found." An inspection of the arrangement of the tables wUl show when the column headed "a" is to be used, as the line giving the conditions of "required" and "found" are given on the line marked "A." The factor (g)" is found in column "B" as the line marked "B" fulfills the conditions of the problem. 12 CHEMICAL CALCULATIONS 14 H 14(1.008) MgS04 • 7 H2O 246.50 = 0.0572 parts of 1.0000 hydrogen per unit weight MgS04 • 7 H2O. Or, if magnesium sulphate is regarded as made up of the radicals, MgO -803-7 H2O, then MgO _ 40.32 246.50 80.07 MgS04-7H20 SO3 MgS04-7H20 7H2O 246.50 7 (18.016) MgSOi - 7 H2O ~ 246.50 = 0.1636 parts of magnesium oxide, = 0.3248 parts of sulphur trioxide, = 0.5116 parts of water. 1.0000 Equivalents. — If a compound contains the elements of a radical, even though it be impracticable actually to convert the compound into the desired radical, yet the amount of the radical which might theoretically be derived if the conversion were possible is readily calculated, though the equations may be lacking. Thus: what is the arsenic trioxide equivalent of potassium arsenite? Representing the equivalence by the sign =, K3ASO3— ^AszOs. But arsenic trioxide contains two atoms of arsenic in the molecule while potassium arsenite contains only one atom of arsenic; hence two molecules of potassium arsenite must be taken, 2 K3ASO3 = AS2O3 and the factor is AS2O3 197.92 2 K3ASO3 2 (240.22) = 0.4112. RATIOS 13 Successive Reactions. — When a compound is con- verted into some other substance by successive steps, and the amount of one of the initial compounds, needed to furnish a certain amoimt of a final product, is required, it is unnecessary to calculate the intermediary products. For example, the Solvay process for the manufacture of sodium carbonate depends upon the equations: NH3 + H2O + CO2 = NH4HCO3, NaCl + NH4HCO3 = NaHCOs + NH4CI, 2 NaHCOs = Na^COs + H2O + CO2. It is desired to know the amount of ammonia involved in the production of one ton of sodium carbonate. It is needless to calculate the amount of acid ammonium car- bonate and of acid sodium carbonate; it is necessary only to notice that one molecule of ammonia yields one mole- cule of acid ammonium carbonate which in turn yields one molecule of acid carbonate, while it requires two mole- cules of acid sodium carbonate to yield one molecule of sodimn carbonate. The equations must then be modified to 2 NH3 -I- 2 H2O -f 2 CO2 = 2 NH4HCO3, 2 NaCl + 2 NH4HCO3 = 2 NaHCOs + 2 NH4CI, 2 NaHCOs = Na^COs + H2O + CO2. The sodiiun carbonate depends upon acid sodimn carbon- ate for its production, the weight of which, requisite to produce one ton of sodium carbonate, being ^^^^^^' X 1 = X tons of NaHCOs; (1) Na^jCOs and X tons of acid sodimn carbonate require 2 NH4HCO3 2 NaHCOs X X = F tons of NH4HCO3; (2) 14 CHEMICAL CALCULATIONS and the production of Y tons of ammonium acid car- bonate call for 2NH3 X Y = Z tons of ammonia. (3) tuting in (2) 1 suit is 2 NH4HCO3 Substituting in (2) the value of X found in (1), the re- 2NH4HCO3 2NaHC03 y -, ^ y 2 NaHCOs NaaCOs ' and substituting in (3) this value of Y and canceling; 2 NH3 ^^^aiS€Q» - 2^feSG0, ^„ Then the number of tons of ammonia sought is 2NH3 therefore Z = ^/^J-^y X 1 = 0.3213 ton. 106.00 It is not necessary to go through all this; the above is offered in proof. These reactions are summed up as follows : 2 NH3 = 2 NH4HCO3 = 2 NaHCOs = NaaCOa; therefore 2 NH3 = Na2C03, and the ratio is determined. ^ ' Another method of arriving at the same result may be exem- pUfied by taking the three equations given above, canceUng hke molecules on opposite sides of the equality sign and adding: 1 1 2NH3 +!^H20 +\^C02 =-%^S^QQf^ 2 NaCl +-^ '?HiI-ICa =-2^faSGQff + 2 NH4CI 2 NH3 + 2 NaCl + H2O + CO2 = Na^iCOa + 2 NH4CI from which it is at once evident that 2 NH3 = NanCOs. RATIOS 15 Factors Other than Chemical. — Factors are of very general application, tables of lengths, weights and other measures coming mider this head. Thus, three feet make one yard. The factors are f = 3 for the conversion of yards into feet and its reciprocal J = 0.333 is the factor to convert feet into yards. Such examples might be in- stanced indefinitely. Take the calculation of factors for the conversion of the metric system of lengths into the EngHsh system.' Having given Metric system. 10 decimeters = 1 meter 100 centimeters = 1 meter 1000 millimeters = 1 meter 1000 meters = 1 kilometer English system. 12 inches = 1 foot 3 feet = 1 yard 5280 feet = 1 mile and also the fundamental relation, 1 meter = 39.37 inches. The factor for the conversion of meters into feet is ?^ = 3.2808, which means that 1 meter = 3.2808 feet. Its reciprocal, 19 1 313-^ = 0.30480, or ^^^ = 0.30m, is the factor for the conversion of feet into meters. In the same way, -^^ = 1.0936: factor of meters to yards. 36 QC 1 ) or nnoa = 0.91440: factor of yards to meters, then as 1 inch is i^^th of a foot and 1 foot = 0.30480 meters, 0.30480 ^ 0.02540: factor of inches to meters, 1 See Chem. Ann., pp. 469-480. 16 CHEMICAL CALCULATIONS and as, 1 meter = 100 centimeters, then 1 inch = 2.540 centimeters. So a table may be constructed : Given. Sought. Factor (or number). Logarithm. meters yards meters feet centimeters inches yards meters feet meters' inches centimeters 1.0936 0.9144 3.2808 0.3048 0.3937 2.5400 0.03886 1.96114 0.51598 1.48402 1.59517 0.40483 PROBLEMS I. Given the reaction: PbClz + K2Cr04 = PbCr04 + 2 KCI. 278.02 323.2 (a) What is the factor of lead chloride to lead chromate? (6) If 0.1784 g. of lead chromate are precipitated by an excess of potassium chromate from a solution containing lead chloride, how many grams of lead chloride were present? (c) How many grams of lead chromate are obtained from 1.000 g. of lead chloride? (d) How many grams of lead in 0.7325 g. of lead chromate? PbCl2 278.02 Ans. (a) (6) (c) PbCr04 323.1 0.1784 X 0.8605 1 = 0.8605. 0.1535 g. PbCU. 0.8605 Pb = 1.1622 g. PbCr04. 207.1 X 0.7325 = 0.4695 g. Pb. PbCr04 323.1 11. Hydrous sodium carbonate may be converted into the anhydrous salt by heat according to the equation: NajCOa • 10 H2O = NajCOa + 10 H2O. 286.16 106.00 180.16 (a) How many pounds of anhydrous sodium carbonate may be obtained from 15.000 pounds of the crystallized salt? (6) EATIOS 17 What is the factor of hydrous sodium carbonate to anhydrous sodium carbonate? (c) If 17.000 pounds of hydrous sodium carbonate axe converted into the anhydrous form, what is the loss in weight? Ans. (a) (6) NazCOs • 10 H2O NazCOa-lOHaO NajCOs ■■ ^^ X 15.000 = 5.5214 lbs. Na^COa. 286.16 106.00 = 2.6996. (c) The loss in weight is the water driven off. This problem may be solved in two ways: By using the factor found in (6), or by calculating the water directly. ^^~ = 6.2973 lbs. NajCOa remaining. Or 17.000 - 6.2973 = 10.703 lbs. water driven off. IOH2O 180.16 NazCOs . 10 H2O driven off. 286.16 X 17.000 = 10.703 lbs. water 1. (a) What is the factor for the conversion Mg2P207 — » P2O6? (b) How many grams of phosphoric anhydride are contained in 0.7256 g. of magnesium pyrophosphate? (c) What is the factor for the conversion of (NH4)3 PO4 • 12 MoOs to P2O5? (d) How many grams of phosphoric anhydride are equivalent to 0.1500 g. of ammonium phosphomolybdate? Ans. (a) 0.63793. 2. Verify the following factors: (6) 0.4629 g. (c) 0.03784. (d) 0.005677 g. Given. Sought. Fafitor (number). KaPtCls PbMoOi NO PbS04 Mn2P207 Pt MoOs NaNOa Pb MnO 0.40151 0.39226 2.8327 0.68311 0.49961 18 CHEMICAL CALCULATIONS 3. How many grams of chromic sulphide will be formed from 0.7182 g. of chromic oxide according to the equation: 2 CrjOs + 3 CS2 = 2 CrjSs + 3 CO2? Ans. 0.9460 g. 4. How many grams of tin must be treated with nitric acid to obtain 2 kilos of stannic oxide? Sn^SnOz, Ans. 1576 g. 5. How much charcoal is required to reduce 1.500 g. of arsenic trioxide according to the equation: AS2O3 + 3 C = 3 CO + 2 As? Ans. 0.2728 g. 6. Iodine is liberated according to the equation: 2KI + Cl2 = 2KCl + l2. (a) How many grams of iodine are liberated from 0.1726 g. of potassium iodide? (6) If 1.5370 g. of iodine are liberated, how many grams of potassium iodide were decomposed? Ans. (a) 0.1319 g. (6) 2.0105 g. 7. Oxygen is prepared from potassium chlorate according to the equation: 2 KCIO3 = 2 KCl + 3 O2. (a) What is the yield of oxygen when 7.0680 g. of potassiimi chlorate are decomposed? (6) How many grams of potassium chlorate must be decomposed to get 2.0000 g. of oxygen? Ans. (a) 2.7681 g. (6) 5.1067 g. 8. In the compound CaCOs • 3 Cas (P04)2, (a) how much phosphorus is contained in 5.000 g.? (6) How much phosphoric anhydride in the same amount? Ans. (a) 0.9031 g. (6) 2.067 g. 9. Sulphur dioxide may be produced by the reaction: Cu + 2 H2SO4 = CuSOi + 2 H2O + SO2. RATIOS 19 (a) How much copper and (b) how much of a 93.20% H2SO4 must be taken to obtain 64.00 g. of sulphur dioxide? Ans. (a) 63.50 g. (&) 210.3 g. 10. How much superphosphate can be made from one ton of calcium phosphate, 93.50% pure? The reaction is Caa (POOz + 2 H2SO4 = 2 CaSOi + CaH4(P04)2. Ans. 0.7056 ton. 11. Sulphuric acid is made according to the equation: 2 S + 3 O2 + 2 H2O = 2 H2SO4. (a) If brimstone containing 97.00% sulphur is used, how much sulphuric acid is obtained from one ton? (6) If pyrites con- taining 96.00% FeSa is used to furnish the sulphur, how many tons are required to yield a ton of sulphuric acid? Ans. (a) 2.9667 tons. (6) 0.6371 ton. 12. Sulphuric acid reacts with sodium hydroxide thus: H2SO4 + 2 NaOH = Na2S04 + 2 H2O. If 0.2073 g. of sulphuric acid are added to 0.1705 g. of sodium hydroxide, (a) how much sodium sulphate is formed and (6) which is left over, caustic or acid, and how much? Ans. 0.3003 g. 0.0014 g. NaOH. 13. (a) What is the percentage of manganese ia pure potassium permanganate (KMn04)? (6) If contaminated to an extent of 2.00% of impurities? Ans. (o) 34.76%. (6) 34.06%. 14. Dolomite is a double carbonate of calciimi and magne- sium which may in some cases be represented by CaCOs • MgCOs. (a) What are the percentages of magnesium carbonate and of calcium carbonate in such a sample? (6) How many pounds of magnesium carbonate and calcivuu carbonate in a ton of this 20 CHEMICAL CALCULATIONS dolomite? (c) How many tons of Epsom salts, MgSO* • 7 H2O, can be obtained from a ton of dolomite? Ans. (a) 54.27% CaCOs, 45.73% MgCOa. (&) 1085.4 lbs. CaCOs, 914.6 lbs. MgCOs. (c) 1.337 tons. 16. What are the percentages of the elements in ammonimn phosphomolybdate if it is (NH4)3P04 • 12 M0O3 • 3 H2O? Ans. N = 2.18%, = 35.63%, H = 0.93%, Mo = 59.65%, P = 1.61%. 16. Regarding ammonium phosphomolybdate as made up of the radicals NH3, H2O, P2O5 and M0O3, what is the percentage composition of these radicals ia the molecule? See Prob. 15. Ans. P2O5 = 3.69%, H20= 4.20%, NH3= 2.65%, M0O3 = 89.47%. 17. Potassium antimonyl tartrate (tartar emetic) corresponds to the formula K2H2 (C4H406)2 • Sb203. (a) What are the per- centages of the different elements in this compound? (6) What is the percentage of Sb203? (c) How many grams of tartar emetic must be taken to obtain 5.0000 g. of antimony? Ans. (a) K = 11.76%, H = 1.52%, C = 14.44%, = 36.11%, Sb = 36.17%. (b) 43.39%. (c) 13.8245 g. L8. 5.000 g. of arsenic (the metal) are oxidized and the oxide dissolved iu caustic potash. How many grams of potassium arsenite are formed? The reactions are 4As + 3 02 = 2As203, AS2O3 + 6 KOH = 2 K3ASO3 + 3 H2O. Ans. 16.026 g. RATIOS 21 19. (a) How many pounds of salt are required to make 1500 lbs. of salt cake (NaaSO*)? (6) How many pounds of Glauber's salt (Na2S04 • 10 H2O) will this amount of salt cake make? Reactions: 2 NaCl + H2SO4 = 2 HCl + NasSOi, NazSOi + 10 H2O = Na2S04 • 10 H2O. Ans. (a) 1234.4 lbs. (6) 3402.2 lbs. 20. How many grams of ammonium dichromate may be pre- pared from 500 g. of potassiiun dichromate according to the reactions: K2Cr207 + H2SO4 = K2SO4 + H2O + 2 CrOa, 2 CrOa + 2 NH3 + H2O = (NH4)2 CrzO,? Ans. 428.4 g. 21. Iodine may be obtained from potassium iodide according to the equations: NaCl + H2SO4 = NaHSOi + HCl 4 HCl + MnOj = MnCl2 + 2 H2O + CI2 CI2 + 2KI = 2KCl + l2. How much sulphuric acid must be taken to produce 5.000 g. of iodine? Ans. 7.728 g. 22. The Leblanc process for the manufacture of sodium car- bonate is 2 NaCl + H2SO4 = Na2S04 + 2 HCl, NajSOi + 2 C = Na2S + 2 CO2, NajS + CaCOs = Na2C03 + CaS. How many tons of sodium carbonate may be obtained from a ton of salt? Ans. 0.9066 ton. 23. From the equations: AICI3 + 3 NH4C2H3O2 = Al (C2H302)3 + 3 NHiCl, Al (C2H302)3 + H2O = Al (OH) (C2H302)2 + HC2H3O2, 2 Al (OH) (C2H302)2 -H 8 O2 = AI2O3 + 7 H2O + 8 CO2. (a) How many grams of almninum chloride are required to yield 0.3000 g. of aliuninmn oxide? (b) How many grams of 22 CHEMICAL CALCULATIONS aluminum oxide are obtained from 0.8300 g. of aluminum chloride? Ans. (a,) 0.7836 g. (6) 0.3177 g. 24. Chrome iron ore is CrjOs • FeO, and may be converted into potassium dichromate as follows : 4 FeO . CrjOs + 4 K2CO3 + 4 CaO + 7 O2 =4 K2Cr04 + 4 CaCrOi + 2 FezOs + 4 CO2. The calcium chromate is converted into potassium chromate, CaCr04 + K2SO4 = CaS04 + K2Cr04; and potassium dichromate is obt£^ined from potassium chromate, 2 KsCrOi + H2SO4 = K2SO4 + H2O + K2Cr207. How many tons of potassium dichromate can be obtained from a ton of chrome iron ore, if the conversion is complete and the ore is 92.00% FeO • CraOs? Ans. 1.2092 tons. 26. The barometer reads 30.00". What is the reading in the metric system?' Ans. 76.20 cm. 26. A piece of aluminum wire 200 mm. long weighs 0.1327 g. What length should be taken to make a centigram rider? Ans. 15.07 mm. 27. It is required to find the height of a can to hold one quart, the diameter of which must be 4.5", one half inch being allowed at the top for air space. Ans. 4.13". 28. In estimating the capacity of paint cans, they were filled to within one half inch of the top with water at 60° F. The weights of water were A, 4.169 lbs. C, 2.084 lbs. B, 1.042 lbs. D, 8.338 lbs. What were the capacities of these cans to within one-half inch of the top? (1 gallon = 231 cu. in.) Ans. A, 2 qts. B, 1 pt. C, 1 qt. D, 1 gal. ' The sign (') is used as an abbreviation for feet, (") for inches. RATIOS 23 29. It is desired to make a 50-cc. burette, graduated to tenths of a cubic centimeter, the graduations to be 2 mm. apart. What should be the diameter of the glass tube? Ans. 0.798 cm. 30. Eimer & Amend's catalogue gives the following data about platinum foil: "Platinum foil, medium, tAts" thick, one gram per square inch." Assuming the price of platinum to be $1.60 per gram, what would a cone for electrolysis cost, having a slant height of 4", diameter 3"? Ans. $30.16. 31. The Westinghouse handbook gives the following data relative to copper wire. Gage. Diam. (") Area sq. in. Lbs. per 1000 ft. Ft. per lb. 10 0.10381 0.0081532 31.37 31.38 If one foot = 30.480 em. and one pound = 453.59 g. : (a) What length in centimeters must be taken to weigh 1.500 g.? (6) What is the weight in pounds of 7.000 m.? Ans. (a) 3.213 cm. (6) 0.73191b. 32. Working up from the datum, one inch ■■ the following: 2.540 cm., verify Given. Sought. Factor. square centimeters square inches cubic centimeters cubic inches square meters square yards cubic meters cubic yards square inches square centimeters cubic inches cubic centimeters square yards square meters cubic yards cubic meters 0.1550 6.4516 0.06102 16.387 1.1960 0.8361 1.3079 0.7645 33. The internal diameter of a spherical glass bulb is 4.382" what are its contents in cubic cm.? Ans. 721.9 cc. 24 CHEMICAL CALCULATIONS 34. A Dumas flask for the determination of molecular weights measures 8.5" in inner diameter. Neglecting the volume of the drawn-out neck, what is its capacity in liters? Ans. 5.27 L. 36. Oil of vitriol is shipped in iron containers. A scow is to be fitted with two tanks running fore and aft. It is able to carry 200 tons of the acid. If the tanks are to be 60' long, what must be the diameter of the tanks to carry this acid? One cubic foot of oil of vitriol weighs 114.47 lbs. Ans. 6.09'. 36. (a) How many liters in a cubic foot? (6) How many liters in a gallon? (c) How many quarts in two liters? Ans. (a) 28.317 L. (6) 3.785 L. (c) 2.113 qts. 37. A circular piece of filter paper 15 cm. in diameter yields 0.00025 g. of ash. (a) What will the weight of ash be from a circular piece 5.5 cm. in diameter? (6) From a piece of the same kind of paper 3" square? Ans. (a) 0.00003 g. (6) 0.00008 g. 38. A tank measures 15.00' X 20.00' X 6.00'. How many liters wiU it hold? Ans. 50,969 L. 39. From the fundamental relation between meters and inches, (a) derive the ratio of square feet to square meters, (6) square meters to square yards, (c) If a wall has an area of 2057 square feet, how many square meters does it contain? (d) If cloth costs $2.50 per square meter, what wiU it cost per square yard? Ans. (a) 0.09290. (6) 1.1960. (c) 191.1 sq. m. id) $2.09. 40. (a) How many cubic inches in a liter? (6) How many cubic meters in an excavation 72' x 50' X 12'? Ans. (a) 61.025 cu. in. (6) 1223.2 cu. m. 41. Derive the value of a kilometer expressed in mUes from the fundamental unit; one meter = 39.37 inches. Ans. 1 km. = 0.62137 mi. RATIOS 25 42. A nautical mile is the length of one minute of the earth's circumference at the equator and is equal to 1.1527 statute or land nules. (a) What is the circumference of the earth at the equator? (6) What is the diameter of the earth at the equator in meters? (c) In kilometers? (d) How many feet in a nau- tical mile (knot)? Ans. (a) 24,898 mi. (6) 12,755,000 m. (c) 12,755 km. (d) 6086 ft. CHAPTER II APPROXIMATE NUMBERS An abstract number is accurate as stated, but when a number obtained by measurement represents some multiple of units obtained by measurement or manipula- tion, it is not absolutely accurate, but only relatively so. This error is called the experimental error or the error of measurement. For example, suppose a sub- stance is weighed on a balance sensitive to a milligram, and equihbrium is established with 1.628 g. as a counter- balance. Should this same weight be determined by a balance sensitive to tenths of a milligram, it cannot be predicted that it will weigh 1.6280 g.; in fact, it may weigh anything between 1.6275 and 1.6285 g. These figures are the extreme Umits of variance. The error in weighing on the balance sensitive to a milligram may be less than this; it can only be said with certainty that it is not more. There must be a figure beyond which there is uncertainty. Such numbers are approximate numbers. By convention, when a number expresses a multiple of some unit which has been experimentally determined, or derived, it is customary that the last figure given is the last figure known with any degree of certainty. In the example above, the last figure known is 8 in the thousandths place. The uncertainty is ±5 in the ten thousandths place or ±0.0005 g. This is the maximum apparent error. A number expressing some multiple of a unit, experi- mentally determined, or derived, always contains some 26 APPROXIMATE NUMBERS 27 error, which is great or small, according to the difficulty, or the dehcacy of the operation by which it is determined; hence there must be some figure in the result beyond which there is uncertainty. In atomic weight tables the element iodine will be foimd stated to five significant figures, while the majority of the atomic weights are given only to three or four significant figures. This is for the reason that this element, a very important one iij analyti- cal chemistry, has been very carefully studied, and also because of some of its properties, by taking advantage of which it (the atomic weight) may be determined with the degree of accuracy indicated. This is the convention, and strictly speaking is seldom accurately followed, as frequently results having an error in the third place are reported as far as the fourth. There are methods of abbreviated multiphcation and division, which eHminate superfluous figures, and indicate the extent to which the result may be relied upon in the use of approximate mmibers. Logarithms, however, do all this with the same degree of certainty, and with much less mental fatigue, especially when many extended oper- ations are to be performed. Logarithms themselves are only approximate nimibers, and when employed should have a degree of accuracy in their last significant figure equal to, or in excess of, the accuracy of the last significant figure in the approximate numbers involved. For this reason, four- and five-place tables are generally employed, with the balance of opinion in favor of the five-place tables for the usual calculations performed by the chemist. In chemists' handbooks the logarithms of analytical and other factors are given to five places in the mantissa, though many of the figures are reliable only to four places, yet quite as many are accurate to five significant figures. Also, to obtain four significant figures from four-place 28 CHEMICAL CALCULATIONS tables requires interpolation, while the five-place tables give four figures by direct reading and five by interpola- tion.i Addition and Subtraction of Approximate Numbers. — It is required to add the following atomic weights, assuming the maximum apparent errors, in order to obtain the molecular weight of alumimmi hydroxide, A1(0H)3. Al =27.1 3 =3X16.00 =48.00 3H = 3X 1.008 = 3.024 A1(0H)3 =78.124 The last figure (4) in the thousandths place is manifestly uncertain, for the reason that 27.1 has an error of ±5 in the hundredths place. Strictly speaking, the sum is accurate only to the extent of the tens place, and the result would read 78.1, but the underlined figure is re- tained for the reason that less error is caused by its reten- tion than by its rejection.^ ' Logarithms have been employed in the solution of problems in this book, except where the operation was very simple. In such cases the 10" Mannheim slide rule was used. The author would Strongly recommend this simple device for the checking of results in the laboratory. It cannot be used exclusively, as the accuracy of the rule is only to three figures in the higher numbers. The mantissa only of a logarithm may be used, neglecting the charac- teristic, the position of the decimal point being determined by in- spection. A chemists' slide rule, manufactured by Kruffel & Esser and devised by the author is now on the market. It is designed to faciUtate the computation of such problems as are found in this book and is recommended to students, as by its use much time is saved in calculating the answers to the problems. ^ This convention has not been strictly followed in the tabula- tion of molecular weights. See Chem. Ann., pp. 39-47; 100-336. One more figure is carried when the number is to be used in further computation. APPROXIMATE NUMBERS 29 Subtraction of approximate numbers leads to the same results. Multiplication of Approximate Numbers. — Multiply- ing the number 1.862, containing maximum apparent error, by 0.6257, the results obtained are: (a) By straight multiplication, assuming no errors, 1.862 X 0.6257 = 1.1650534. (6) By assuming the maximum apparent error acting in a negative (subtractive) direction, 1.8615 X 0.6257 = 1.16474055. (c) By assuming the maximum apparent error acting in a positive (additive) direction, 1.8625 X 0.6257 = 1.16536625. In practice, having a problem of this nature, it is not known whether the error is working in a negative direc- tion (decreasing) as in (&) or positive direction (increas- ing) as in (c). If there is any error at all it must be either positive or negative and it is not known which, nor is it known to what extent; it is only known that it is not more than ±5 in the ten thousandths place; consequently, the most probable result is obtained from (a) and is written 1.165, the last figure being uncertain as is seen by an inspection of the results of (b) and (c). Again, multiplying 57.2, containing maximum apparent error, by 15.27: (a) Assuming no error, 57.2 X 15.27 = 873.444. (6) Assuming the maximum apparent error as being negative, 57.15 X 15.27 = 872.6805. (c) Assuming the maximimi apparent error as being positive, 57.25 X 15.27 = 874.2075. 30 CHEMICAL CALCULATIONS The result as obtained from (a) is 873. It is evident that when a result is obtained by a multipUcation involving a unit multiple (approximate number) given to the degree of accuracy adopted by convention, the result may be depended upon only to the same number of significant figures as is contained in the approximate number. It often happens that both the numbers to be multiplied are approximate nimibers. Take the case of multiplying 572 and 5725 each containing maximum apparent errors. The result is 3274700 and is reliable only the third figure, for the reason that 572 contains only three significant figiures, hence the product is accurate to the same number of places only. Division of Approximate Numbers. — Take a case of division involving approximate numbers; 5724 contains maximum apparent error, 172 is absolute, (a) ^ = 33.2791 ■ ■ • ■ ^ = 0.0300489 . . . (6) ^^ = 33.2762 . . . : ^^^ = 0.0300515 . . . ^1794 ^ 172 (c) -^ = 33.2821 . . . ■.^^ = 0.0300462 . . . Again, if the number which has a maximum apparent error contains only three significant figures, as 152, while 1645 is absolute (contains no error), the following results are obtained: (a) ^ = 10.8223 . . . : ~ = 0.0924012 . . . ^ 152 1645 (6) J-ll = 10.8581 . . . : jg45 = 0.0920972 . . . APPROXIMATE NUMBERS 31 The limit of accuracy, evidently, is the same as in the mul- tiplication of approximate numbers. Recapitulation. — I. The sum obtained by adding approximate numbers is retained to one place beyond the indeterminate figure.' II. Subtraction follows the same rule as addition. III. The result obtained by multiplication involving approximate numbers is retained to the same number of significant figures as is contained in the approximate number having the smallest number of significant figures. IV. Division foUows the same rule as multiplication. It does not come within the province of this book to give instruction in the use of logarithms, but a few re- marks advocating their use are appropriate. A consider- ation of approximate numbers sho\ys that results obtained in the manipulation of such numbers are to be depended upon to a degree commensurate with the degree of accu- racy of the approximate numbers employed. In most chemical operations, readings are taken to foiu: or five significant figures. Thus a burette is read to four places with the last figure estimated. The average masses determined by the analytical balance seldom exceed six figures, and allowing for tare, the net weight seldom ex- ceeds five, and is generally accurate only to four figures. These approximate numbers, entering as factors into subsequent calculations, carry with them their inherent errors, with the consequence that most results are stated to four significant figures only. In an analysis the con- stituents are generally given to hundredths of a per cent. '■ One more figure is carried when the result is to be used in further computation; this applies also .for subtraction, multiplication and division. CHAPTER III INTERPOLATION Functions. — When two quantities are interdepend- ent, one is said to be a function of the other. The relation may be very simple, as between pounds and kilo- grams. An increase of weight in a body of x pounds will cause an increase of y kilograms, x being a function of y. Expressed mathematically, it is a; = hy. Conse- quently, knowing the ratio of x to y, x being given, y is readily determined. Let x be the number of pounds and y the number of kilograms change in weight of a system; to convert x poimds into kilograms it is only necessary to multiply by the factor k. In this instance, h = 0.45359; so the expression becomes x = 0.45359 y. This is dired variation, and by plotting the curve, measuring one of the units along the ordinate and the other along the abscissa, a straight line is obtained. Such curves are seldom plotted.^ Graphic Representation of Functions. — Any tabu- lation, when plotted, yields a curve, the simplicity of which depends upon the particular case under consider- ation. When two quantities vary directly the curve is a straight line; ^ when varying inversely, an hyperbola ' A tabulation of results obtained by solving the different values of one quantity in terms of the other, where the resulting curve is a straight line, is sometimes contained in handbooks for quick reference. ' When the equation is of the first degree, that is, when it does not contain either of the quantities raised to any power other than the first. 32 INTERPOLATION 38 10 Z5 o /• X PLATE I = .45359 y 1 1 ' 1 2 3 4 KILOGRAMS Plate I. 34 CHEMICAL CALCULATIONS results. Any value of either x or y, in terms of the other, is as accurately or more accurately obtained by using the factor in the equation under consideration as by consult- ing the curve. Plate I represents the curve of pounds equivalent to kilograms; any value of x in terms of y will fall on the line OA. 1.0 .9 ^ PLATE n ^ S ,7 Log.y X X y 2.6 H !•= o o ->A .5 .2 7 / / / / / • A b Id NUMBER PliA-CB II. Plate II represents the relation existing between a number and its logarithm; i.e., the curve fulfils the equa- tion log y = X. (In this case, to obtain values of y in terms of x involves extended mathematical calculations; rNTERPOLATION 35 consequently these values are determined and once for all tabulated.) Any value of x in terms of y, or vice versa, will fall on the curve. 20 30 ' 40 50, DEGREES BEAUME Plate III. Change of Percentage Composition with Baume for Sulphuric Acid. — Plate III shows the relation between the percentage composition and the strength in degrees 36 CHEMICAL CALCULATIONS Bauin6 of sulphuric acid.^ The "Baum6" will be, for_ the present, used as the weight of a unit volume.^ An increase in the Bauni6 of sulphuric acid is accompanied by an increase in the percentage of sulphuric acid (H2SO4) ; so if X represents the percentage of sulphuric acid (H2SO4), in a given sample, and y its Baum6, the relation be- tween these quantities is represented by the equation X = {f)y.' The distances along the axis OY represent the degrees Baum6 (° B6.) of sulphuric acid and those along OX the percentage composition; if one varied directly as the other a straight line would result. Let 5° B6. correspond to 5.28%H2S04 (point E), then 60° B6. would correspond to V- X 5.28 = 63.36% H2SO4 (point F'). Connecting these points the curve OA is obtained. OB is the true relationship between degrees Baum6 and the percentage sulphuric acid, determined experi- mentally; the curve continually diverging from the straight line OA. Owing to this divergence, acid of 60° B6. contains 77.67% H2SO4 (point F) instead of 63.36%, the percentage calculated on the assumption that the per cent H2SO4 varied directly as the Baum6; i.e., that the curve was the straight line OA. This differ- ence, 14.31%, is represented by the distance FF'. The divergence continually increases as the acid approaches 100% H2SO1; in fact, as the strength of the acid ap- proaches and passes 93.19% H2SO4, corresponding to 66° B6., a very shght change in the Baum^ strength ' The data has beeu taken from the tables of Ferguson and Talbot for Sulphuric Acid. See Chem. Ann., pp. 388-393. * The method of determining the Baum6 of a Uquid will be dis- cussed at greater length under Specific Gravity, p. 65. ^ The value of the functions {f)y in the equation x = {f)y in this case varies with the concentration. The value of x corresponding to y must be determined experimentally. IJNTJfiKPOLATION 37 causes a considerable change in the per cent H2SO4, the curve approaching perpendicularity. It is for this reason that the method is not employed for acids of a Baum^ strength greater than 66° B6. Interpolation. — Consider the curve OA included be- tween the points C and D'; being given: C = 48.00° Be. corresponding to 50.69% H2SO4 D' = 52.00° B6. corresponding to 54.91% H2SO4 difference = 4.00° Be. 4.22% H2SO4 It is required to find the per cent H2SO4 cprresponding to 50° B6. (point m') ; or, having the abscissa of the point m' lying on the Une OA, it is required to find its ordinate. The abscissa distance of CD' is 4.00; the difference between the abscissas of C" and D'. The ab- scissa distance of Cm' is the difference in Baum4 between C and m'; 50.00° B6. - 48.00° B6. = 2.00° B6. But the abscissa distance of CD' corresponds to the ordinate distance 4.22; hence the ordinate distance of Cm' is ?^ X 4.22 = 2.11. The ordinate of m' is then 50.69 + 4.00 2.11 = 52.80, the percentage of sulphuric acid correspond- ing to 50.00° Be., if these quantities varied according to the curve OA. This is interpolation. It can only be employed when the curve, between the Umits given, may be considered a straight line. The curve OB showing the relation between the Baum^ strength and the percentage strength of sul- phuric acid has been determined experimentally. It is found to depart considerably from a straight line, but between two points on the curve in close enough proximity the line is sensibly straight.^ This is shown in Plate IV, ' That is, the error is less than the errors of the other factors entering into consideration. 38 CHEMICAL CALCULATIONS 7c : ^ 1 J , 7 T_ i. J /Pi 4 T E" Z -^ / ^ V Z i s ■<:/, df hhlie L ■ ^ / -^ ^ , • '^v Z ^'^^^ 2 5"^ ^^ ?z ^, J 57 ^r il 11 S A o < Z Z ^ y^ y 4l'^ tt^'l ■^ - i = / J , ^ ^ Z ^F ^ ' " " - ^^ ^ - - - Z _j t'h - / m ^ z ^f/ ~ Z V ^ <.i l-en-- I -' _ Z^° : y z_ r^y ui 'Jit jS^ o : z 1 J ^> "■ I z 7 UJ J- , 7 7 : z; Z 7 - ' J. - 2 Z" -. J 7 ER - ' - ^' :^2 - ao Z Z 2i' ^ . ^ Z '^ - -, ^™' ^ Z 2 J. J. - -^ J --J - ^% It- ^4 2 fU' 50^' j'^ 40 45 6 , 65 60 Plate IV. an enlargement of a small section of Plate III in the region of CD'. Working with the ciuve OB, the following data being given: C = 48.00° B6. corresponding to 59.32% H2SO4 D = 52.00° B6. corresponding to 65.13 % H2SO4 difference = 4.00° B6. 5.81% H2SO4 IJNTjaKPOLATION 39 It is required to find the percentage of sulphuric acid corresponding to 50.00° B6., or the ordinate of m} 50.00 - 48.00 = 2.00 = the abscissa distance of Cm, 4.00 = the abscissa distance of CD, 5.81 = the ordinate distance of CD, then the ordinate distance of Cm is 150X5.81=2.91, consequently the ordinate of Cm, or the per cent of sul- phuric acid corresponding to m, is 59.32 + 2.91 = 62.23%. This value, obtained by interpolation, agrees fairly closely with the value found experimentally, 62.18%; the dis- crepancy is 0.05%. Once more, working with the curve OB, given E = 5.00° B6. corresponding to 5.28% H2SO4 F = 60.00° B6. corresponding to 77.67% H2SO4 difference = 55.00° B6. 72.39%, H2SO4 It is required to find the per cent of sulphuric acid corre- sponding to 50.00° B6. 50.00° — 5.00° = 45.00° = the excess of ° B6. of required acid over 5.00° B6., the ° BL below. 55.00° = the interval in ° B6. between the points given. 72.39% = the interval in per cent H2SO4 corresponding to an interval of 55.00° B6. ^^ X 72.39 = 59.23% H2SO4 corresponding to an 55.00 increase of 45.00° B6. 59.23-1-5.28=64.51% H2SO4 corresponding to 50.00° B6. 1 Note that the m' and m correspond to the same Baum& 40 CHEMICAL CALCULATIONS This corresponds to point m" . The difference between the per cent of sulphuric acid found by this interpola- tion, 64.51%, and the per cent of acid found by the previ- ous interpolation, 62.23%, is due to the wide separation of the points ifi) and {F) as against the relative proximity of the points (C) and (D) of the previous interpolation, between which intervals the determined curve OB is almost a straight hne. m" must fall on a point in a straight line connecting (E) and (i^). The difference, 64.51% - 62.23% = 2.29%, is due to the curvature of OB between the intervals given, and is represented by the distance mm" . From these considerations, it becomes apparent that to interpolate, the curve must be sensibly a straight line. To obtain this condition, when the curve as a whole deviates from a straight line, the tabular difference must be small enough, or the fixed points near enough, that the curve between the points given is sensibly straight. Con- sequently, the greater the curvature, the smaller must be the interval between the given points to make the inter- polation accmrate to the desired degree. Extrapolation. — Again, working with results experi- mentally determined (as represented by the curve OB), being given 40.00° B6. corresponding to 48.10% H2SO4 41.00° B6. corresponding to 49.47% H2SO4 difference = 1.00° B6. 1.37% H2SO4 It is required to find the percentage of sulphuric acid corresponding to 42.00° B6. 42.00° B^. does not lie between the points given, but it may be calculated if the curve is assumed to be sensibly straight between the intervals 40.00° B6. and 42.00° Be.; i.e., if within these limits the Baum6 is proportional to the percentage of INTERPOLATION 41 sulphuric acid without too much error. On this assump- tion the percentage of sulphuric acid corresponding to 42.00° B6. is 49.47 + 1.37 = 50.84% H2SO4. As a matter of experiment, 42.00° B6. corresponds to 50.87% H2SO4, hence an error has been made of 0.03% H2SO4 (50.87 - 50.84) which is relatively small. This is extrapolation and is resorted to with caution for a reason which will appear in the following. Given 65.75° B6. corresponding to 91.80% H2SO4 66.00° B6. corresponding to 93.19% H2SO4 difference = 0.25° B6. 1.39% H2SO4 It is required to find the percentage of sulphuric acid corresponding to 66.25° B^. On the same assumptions as above, the percentage of sulphuric acid corresponding to 66.25° Be. is 93.19 + 1.39 = 94.58% H2SO4. As a matter of experiment 66.25° B^. is found to corre- spond to 95.28% H2SO4, consequently an error of 0.70% H2SO4 (95.28 — 94.58) has been made by extrapolating notwithstanding the small interval taken (0.25° B6.). In other words, the curve is not sensibly a straight line between the intervals given (65.75° B6. and 66.25° B€.). This is shown in the large scale Plate V, page 42. PROBLEMS I. Given 60° B6} correspondiag to 77.67% H2SO4, 61° B6 corresponding to 79.43% H2SO4. 1 The Baum6 is given in the tables to but two significant figures though they are accurate to two places to the right of the decimal point, 60° B6. being acciu'ate to 60.00° B6. 42 CHEMICAL CALCULATIONS lOOr 98- 97- 96- o a: 395. a. 94- z ?93- 92- 91- 90 PLATE V Detail of plate P95.28 994.58 M 64 65 66 67 68 , DEGREES BEAUME Plate V. 69 70 INTERPOLATION 43 (o) If the BaumI of a sample of sulphuric acid is 60.85° Bd., what is the percentage of H2SO4? (6) What is the Bauin6 of 78.00% sulphuric acid? Am. The tabular difference in ° B6. is 1° B6. (61 - 60); the tabular difference of per cent H2SO4 is 1.76% (79.43% - 77.67%). The increment in ° B6. of the sample, above the given ° Bi., is 60.85° - 60.00° = 0.85° B6. Assuming that a small increase in Baum4 is accompanied by an increase, in the same ratio, of the percentage of H2SO4, the increase of the percentage of H2SO4 is ^ X 1.76 = 1.50%. The percentage of H2SO4 in the acid is 77.67% + 1.50% = 79.17%. (6) Analogously, 78.00 - 77.67 = 0.33, ^ X 1.00 = 0.19. 60.00 + 0.19 = 60.19° B6. corresponding to 78.00% H2SO4. II. Given 65.75° B6. corresponding to 91.80% H2SO4, 66.00° B6. corresponding to 93.19% H2SO4. If an acid is 66.05° B6., what is its per cent of H2SO4?- Ans. It may be assumed that the next interval will be in approximately the same ratio of increase as in the one previous. Tabular difference is 66.00° - 65.75° = 0.25° B6. Tabular difference is 93.19% - 91.80% = 1.39% H2SO4. 66.05° - 66.00° = 0.05° B6. 5::5|x 1.39 = 0.28% H2SO4. 0.25 93.19 + 0.28 = 93.47% H2SO4 corresponding to 66.05° B6. 44 CHEMICAL CALCULATIONS 43. Given 'B6. Per cent HjSO,. Determine 65 88.65 (a) Per cent HJ.SO4 of 65. 15° B6 acid 65i 89.55 (6) Per cent H2SO4 of 65.62° B6. acid 65J 90.60 (c) Per cent H2SO4 of 65.97° B6. acid 65J 91.80 (d) Per cent H2SO4 of 66.06° B6. acid 66 93.19 (e) °B6. of 89,00% acid (/) "36. of 91.00% acid (s) °B6. of 93.35% acid Ans. (a) 89.19%; (6) 91.18%; (c) 93.02%; (d) 93.52%,; (e) 65.09°; (/) 65.58°; (g) 66.03°. 44. A liquid of 63° B6. weighs 110.29 lbs. per cubic foot, 64° B6. weighs 111.65 lbs. per cubic foot. What is the weight per cubic foot of a liquid of 63.18° B6.? Ans. 110.53 lbs. 45. Sulphuric acid of 65.25° B^. corresponds to 96.10% oil of vitriol, 65.50° B6. corresponds to 97.22% oil of vitriol. If a sample has a Baum6 of 65.37°, what is the percentage of oil of vitriol? Ans. 96.64%. 46. The following is taken from the nitric acid tables of Lunge and Rey.^ Specific gravity. Per cent HNO3. 1.470 1.475 82.90 84.45 (a) What is the per cent of nitric acid corresponding to 1.472 specific gravity? (6) If a sample of nitric acid contains 83.15% HNOs, what is its specific gravity? Ans. (a) 83.52%. (6) 1.471 sp.gr. 1 See Chem. Ann., pp. 401-402. INTERPOLATION 45 47. Ferguson's tables of aqua ammonia furnish the follow- ing': Specific gravity. Per cent NHj. 0.8903 0.8889 31.85 32.34 What is the per cent of ammonia in a sample, if the specific gravity is 0.8895? (Notice that as the specific gravity decreases the per cent of ammonia increases. These two properties vary inversely.) Am. 32.13%. 48. Given the following: °Be. Per cent NH,. 28.00 28.25 33.32 33.81 What is the percentage of ammonia of a sample of 28! 18° B6.? Ans. 33.67%. 49. Given the following logarithms and the numbers corre- sponding: Number. Logarithm. 1021 1022 3.00903 3.00945 (a) What is the logarithm of 1021.6? (6) What is the number corresponding to the logarithm 3.00922? Ans. (a) 3.00928; (6) 1021.5. 60. The boiling point of pure water, accurately determined in Deer Lodge, Montana, Oct. 25, 1910, was found to be 95.56° C. Assuming the thermometer to be correct, what was the baro- metric pressure, given the following data? > See Chem. Ann., pp. 408-409. 46 CHEMICAL CALCULATIONS At 95.00° C. the pressure of aqueous vapor is 633.78 mm. Hg.* At 96.00° C. the pressure of aqueous vapor is 657.54 mm. Hg. Am. 647.09 mm. Hg. 51. In testing the accuracy of a thermometer, the following readings were taken: Barometric pressure, 652.70 mm. Hg. Reading of thermometer in melting ice, 0.00° C. Reading of thermometer in vapor of boUiag water, 95.62° C. Assuming the bore of the thermometer to be uniform and that the barometric pressure exerts no appreciable influence on the melting point of ice, what is the error of the thermometer at these two points? (Use the data given in Problem 50.) Ans. Correct at 0.00° C. 0.18° C. too low at 95.62° C. 62. The following readings were obtained on a Jolly balance: 'Load. Balance stretches to 0.040 g. 0.120 g. 22.0 cm. 26.2 cm. If the balance stands at 23.7 cm., what load is in the pan? Ans. 0.072 gr. 63. Being given the following data on hydrochloric acid: Sp. gr. 1.155 corresponds to 30.55% HCl. Sp. gr. 1.145 corresponds to 28.61% HCl. (a) Extrapolate the percentage of hydrochloric acid corre- sponding to 1.160 sp. gr. (6) 1.140 sp. gr. (c) The specific gravity of hydrochloric acid containing 31.25% HCl. (i) 28.14% HCl. Ans. (a) 31.52%; (6) 27.64%; (c) 1.159 sp. gr; (d) 1.143 sp. gr. ' See Chem. Ann., pp. 461-466; also pp. 70-71. INTERPOLATION 47 64. Under a pressure of 760 mm. hydrochloric acid (solution) gives a constant boiling point distillate of 20.242% HCl, while under 750 mm. the composition of the distillate is 20.266% HCl. What is the probable composition of a distillate collected under 765 mm. Ans. 20.230% HCl. CHAPTER IV HEAT Theee are four units of heat intensity: I. Fahrenheit (written ° F.) . — On the thermometer the melting point of water is marked 32°; the boiling point (b.p.) of water under standard atmospheric pressure (760 mm. Hg) is marked 212°. The distance between these points is divided into 180 (212 — 32) equal spaces, and the same divisions are extended above and below these points. The zero of this scale is 32° (intervals) below the melting point of ice. A change of heat intensity which will cause a change of volume of the substance, usually mercury, through one interval of the scale is a change of one degree Fahrenheit. This scale is used largely in English-speak- ing countries.^ II. Centigrade or Celcius (written ° C). On the thermometer the melting point (m.p.) of ice is marked 0°; ' The expansion of the liquid in the capillary tube is made a measure of heat intensity. The mercurial and other Hquid ther- mometers are based upon the assumption that a change in tempera- ture of any number of degrees is accompanied by a change in the volume of the liquid by a proportionate amount. Through small intervals this is true; the coefficient of expansion of the liquid is sensibly uniform. Mercury, being a liquid, does not exhibit a uniform expansion for a given range of temperature, at aU tem- peratures. Gases, however, more closely approximate a uniform coefficient of expansion, and for this reason the expansion coefficient of hydrogen has been adopted as the standard of temperature meas- urement. The hydrogen thermometer and the mercury thermometer agree closely in the moderate ranges. See Chem. Ann., pp. 539-541. 48 HEAT 49 the boiling point (b.p.) of water under standard conditions of pressure (760 nun.) is marked 100°. The interval is divided into 100 equal divisions, and the same divisions are extended above and below. This scale is used almost exclusively in scientific work and is common in most Con- tinental coimtries. III. Reaumur (written ° R.) . — On the thermometer the melting point of ice is marked 0° and the boiling point under standard conditions of pressure (760 mm.) is marked 80°; the interval is divided into 80 divisions. The grad- uations are extended above and below these points as with the previous scales. This scale is little used except on the Scandinavian Peninsula, in Russia and in some in- dustries. IV. Absolute (written ° A.). — This scale was derived from observations of the changes of volinne of gases ac- companying changes of temperature. The scale interval is the same as in the Centigrade system, but the zero is 273° below the melting point of ice. This scale is chiefly used in gas calculations. The calorie, also called the gram-calorie, is the amount of heat required to raise the temperature of one gram of water one degree Centigrade.^ This unit is very small and a larger unit, the big Calorie,^ also called the kilogram- calorie (written with a capital C), equal to 1000 gram- calories is often employed.' The British Thermal Unit (written B.T. U.) is the amount of heat required to raise the temperature of one 1 See Chem. Ann., p. 481. » Ibid., p. 481. ' The value of the calorie varies slightly according to the tempera- ture at which it is determined. The amount of heat required to raise one gram of water, at 0° C, one degree Centigrade is not the same as is required to raise the same amount of water at some other tempera^ ture, e.g., 15° C, one degree Centigrade. 50 CHEMICAL CALCULATIONS pound of water at 39.1° F. through one degree Fahren- heit.i Specific Heat. — The number of heat units absorbed by a gram of a substance when its temperature is raised through one degree Centigrade or given up when it falls through one degree Centigrade is called the specific heat of that substance.^ When heat passes from one body to another, the amount of heat energy (calories or B.T.U.) gained by the one body is equal to the loss of heat sustained by the other. The Heat of Fusion (H.F.) is the number of heat units required to change a gram of the substance from the soUd to the liquid state without any change of temper- ature. In the case of ice it requires 80 calories to change one gram of ice at 0° C. to water at the same temperature. The heat of fusion' of ice is therefore 80. The Heat of Vaporization (H.V.) is the quantity of heat required to change a gram of the substance from the liquid to the gaseous state without any change of temper- ature. In the case of water it is 536.6 calories per gram. ' See Chem. Ann., p. 481. ' The specific heat of a [substance is a ratio of the heat capacity of a unit weight of the substance in question and the heat capacity of the same weight of water which is taken as unity, and is independent of the units of heat measurement employed. The specific heat of magnesium is 0.250; the heat capacity of one pound of magnesium is 0.250 B.T.U.; of one gram of magnesiimi is 0.250 calories; of one kilo- gram of the same substance is 0.260 Calories or kilogram-calorie. For a tabulation of the specific heats of the elements, see Chem. Ann., pp. 4-9. The value of the specific heat of a substance varies with the temperature at which it is measured. In obtaining an- swers to problems, the specific heat will be assumed to be independ- ent of the temperature of determination. ' This term is synonymous with "Latent Heat of Fusion" used by many writers. The term "Latent" is tautologous in view of the definition of the Heat of Fusion. £LJ!jiV'l' 51 The Heat of Combustion of a substance may be given in the British or Centigrade systems.' In the Centigrade system it is the nimiber of large or small calories evolved from the combustion of a kilogram or a gram of the sub- stance. In the British system it is the number of B.T.U. evolved from the combustion of one pound of the sub- stance.^ Conversion of Centigrade into British System Units and Vice Versa. — It follows from the definitions of the different vmits of temperature,' that to convert ° F. to ° C, solving |(° F. - 32), gives ° C. ° C. to ° F., solving f ° C. + 32, gives ° F. ° F. to ° R., solving | (° F. - 32), gives ° R. ° R. to ° F., solving f ° R. + 32, gives ° F. ° C. to ° R., solving | C, gives ° R. ° R. to ■* C, solving | ° R. gives ° C. °C. to ° A., solving °C. + 273, gives" A. ° A. to ° C, solving ° A. - 273, gives ° C. To convert Fahrenheit or Reaumur to Absolute, first con- vert to Centigrade and then convert to Absolute. > See Chem. Ann., pp. 481, 502-511. ^ In American and British mechanical engineering practice, heata of combustion are generally given in B.T.U. This is not considered the best practice by metallurgical engineers, who prefer the kilo- gram-calorie by reason of its obvious advantages when used in con- junction with the metric system of weights and volumes. ' The interval between the melting point of ice and the boiling point of water is 180° for the Fahrenheit scale, 100° for the Centi- grade scale and 80° for the Reaumur scale. The zero of the Centi- grade and Reaumur scales corresponds to 32° in the Fahrenheit. Hence F.-32 ^ C. ^ R. 180 ~ 100 80' Solving yields the formulas given above. 52 CHEMICAL CALCULATIONS From definition, knowing lib. = 453.59 g. and 1° F. = | ° C. : 1 B.T.U. = 453.59 g. of water raised | ° C. = 453.59 X | = 252 cal. = 0.252 Cal. 1 Cal. = 1000 g. of water raised 1° C. = 1000 cal. = 1.000 Cal. Then the factor, given Calories to find B.T.U., is and the factor, being given B.T.U. to find Calories, is 0.252 as found above. The heats of combustion may be converted as follows: Knowing 1 kilogram = 2.2046 lbs., then if the combus- tion of a kilogram of substance produces X Calories, the fact may be expressed: 1 kilo. = X Cal. or 2.2046 lbs. =XCal.; then for 1 lb. = ^^ and by the factor for the conversion of Calories into B.T.U., found above, the combustion of one pound pro- duces X 3.96832 _ 2.2046 ~ 1-8000 ^< therefore the factor for the conversion of the heat of combustion of a substance expressed in Calories into the heat of combustion expressed in B.T.U. is 1.8. Con- versely, given the heat of combustion of a substance ex- pressed in B.T.U., the factor to convert into Calories is 1 1.8000 = 0.55556.: ' A distinction between B.T.U. and Cal. when used in expressing a definite amount of heat and when expressing a heat of combustion must not be overlooked, as these are two different kinds of quan- tities. HEAT 53 PROBLEMS I. Convert into Centigrade the following temperatures: (a) 80° F. (6) 28° R. (c) 250° A. Ans. (o) " C. = I (80 - 32) = | (48) = 26.66° C. (6) ° C. = f X 28 = 35° C. (c> ° C. = 250 - 273 = - 23° C. n. Convert into Fahrenheit the following temperatures: (o) 60.5° C. (6) 40° R. (c) 273° A. Ans. (a) ° F. = f X 60.5 + 32 = 108.9 + 32 = 140.9° F. (6) ° F. = f X 40 + 32 = 90 + 32 = 122° F. (c) ° C. = 273 - 273 = 0° C; ° F. = I X + 32 = 32° F. in. (a) How many calories must have been imparted to 500 g. of water if its temperature is raised from 12° C. to 26° C? * (6) How many British thermal units? (c) How many Calories? Ans. (a) 26° - 12° = 14° raise in temp. 500 X 14 = 7000 cal. (6) Factor to convert Calories into B.T.U. = 3.96832, 7.000 X 3.96832 = 27.778 B.T.U. (c) 7000 cal. = iUi = 7.000 Cal. IV. 200 g. of lead at 100° C. are plunged mto 200 g. of water at 0° C, and the resulting temperature is 3.17° C; what is the specific heat of lead? ^ Ans. 200 (3.17 - 0) = 634 cal., heat capacity of the lead, 200 (100 - 3.17) = 19,366 cal., heat capacity of the water; 1 Although the value of the calorie varies slightly according to the temperature at which it is defined, this variation will not be considered in the problems in this book. The calorie will be assumed to have a constant value independent of the temperature of defi- nition. ^ In calculating the specific heats, it will be assumed that the specific heat of the substance under consideration is independent of the temperature, within ordinary ranges. 54 CHEMICAL CALCULATIONS then as the specific heat is, by definition, the ratio of the heat capacity of the substance compared with the heat capacity of water, Or the problem may be solved: Let X = the specific heat of the lead, then 200 Z (100 - 3.17) = 200 (3.17 - 0); solving, X = 0.0327. V. If 10 g. of ice at 0° C. are brought into contact with 200 g. of water at 30° C, what is the temperature of the resulting mixture? Ans. Let X = final temperature. 80 X 10 = cal. used to melt the ice without any rise in temp, of the resulting water. 10 Z = cal. used in raising the temp, of this water from 0° C. to Z°. 200 (30 — Z) = cal. given up by the water to melt the ice, and to raise the resultant water to temp. Z°. Then as the heat absorbed is equal to the heat given up, (80 X 10) + lOZ = 200 (30 - Z), X = 24.8° C. VI. In the combustion of one gram of a substance, the tem- perature of 700 g. of water was raised 11.03° C. (a) How many Calories were liberated? (6) How many Calories per kilo- gram? (c) How many Calories per pound would the substance liberate? (d) How many British thermal units per pound? ,. 700 X 11.03 _ 7 _„, p , (6) 7.721 X 1000 = 7721 Cal. per kUo. (c) 1 lb. =453.59 g.; so, 7.721 X453.59 = 3502.2 Cal. per lb. (d) Factor to convert Cal. to B.T.U. = 3.96832, 3502.2 X 3.96832 = 13,898 B.T.U. per Ib.i '■ It wiU be noticed that in this problem, starting with the heat given by the combustion of one gram, we have determined the heat of combustion in the British system. HEAT 55 66. The melting point of yellow phosphorus is 44.2° C, and its boiling point is 290° C. What are the corresponding tempera- tures in Fahrenheit? Ans. m. "p. 111.6° F., b. p. 554° F. 66. Water is at its greatest density at 4° C. What is this temperature in Fahrenheit? Ans. 39.2° F. 67. The average factory temperature is assumed to be 60° F. To what does this correspond in Centigrade? Ans. 15.56° C. 58. Sir James Dewar reached a temperature of -264° C. by evaporating liquid hydrogen under diminished pressure. What is this temperature in Fahrenheit? Ans. —443.2° F. 69. Zero of the Absolute scale is -273° C. What is the cor- responding temperature in Fahrenheit? Ans. —459.4° F. 60. Benzaldehyde melts at 7.7° F. and boils at 355.82° F. Find the corresponding temperatures in Centigrade. Ans. m. p. -13.5° C, b. p. 179.9° C. 61. If the melting point of platinum is 1753° C, what is this in Reaumur? In Fahrenheit? Ans. 1402° R., 3187° F. 62. Caproic acid melts at —5.2° C. and boils at 205° C. What are the corresponding temperatures in Reaumur? Ans. m. p. —4.16° R., b. p. 164° R. 63. Carbon monoxide melts at — 168.8° R. and boils at — 152° R. What are the corresponding temperatures on the Fahrenheit scale? Ans. m. p. -347.8° F., b. p. -310° F. 64. If —271° C. is the lowest temperature so far obtained, what is this temperature in the Absolute scale? Ans. 2° A. 65. Given the following: (o) Hydrogen, critical temp. — 240°C. ; b.p. at atm. press., —251°C. (6) Carbondioxide, critical temp.31°C.; b.p.atatm. press., —80°C. (c) Alcohol, critical temp. 243° C; b. p. at atm. press., 78° C. 56 CHEMICAL CALCULATIONS What are these temperatures in the Absolute scale? Ans. (o) Grit. temp. 33° A., b. p. 21° A.; (6) Grit. temp. 304° A., b. p. 193° A.; (c) Grit. temp. 516° A., b. p. 351° A. 66. (a) At what point is the reading on the Gentigrade scale and the Reaumur scale the same? (6) Gentigrade and Fahrenheit? Ans. (a) 0°; (6) -40°. 67. 100 g. of water at 5° G. are mixed with 100 g. of water at 28° C. ' What is the temperature of the resultant mixture? Ans. 16.5° G. 68. What is the resultant tertiperature when 100 g. of water at 8° C. are mixed with 250 g. of water at 30° C.? Ans. 23.7° C. 69. On mixing the following quantities of water, 1 kilogram at 30° C. 1.5 kilograms at 25° G. 3 kilograms at 20° C. 4.5 kilograms at 10° G. what is the temperature of the resultant? Ans. 17.25° G. 70. What temperature wUl result from mixing 15 g. of ice at 0° G. with 300 g. of water at 30° G.? Ans. 24.8° G. 71. Equal weights of water and ice at 0° G. are mixed, with the result that the mixture had a temperature of 5° G. What was the temperature of the water? Ans. 90° G. 72. If 10 g. of steam at 100° G. are condensed in 100 g. of water at 15° C., what is the resultant temperature? ^ Ans. 71.5° C. ' Authorities differ in the values assigned to the heats of fusion and the heats of vaporization. This may be owing to different temperatiu-es being taken in defining the calorie. The value 80.0 calories will be taken as the heat of fusion of ice in obtaining the answers to the problems in this book. Also 536.6 calories will be taken as the heat of vaporization of water. HEAT 57 73. 150 g. of copper at 100° C. are mixed with 200 g. of water at 12° C. and the resulting temperature is 17.8° C. What is the specific heat of copper? Ans. 0.094. 74. 200 g. of mercury at 94° C. are mixed with 200 g. of water at 17° C. and the resulting mixture has a temperature of 19.49° C. What is the specific heat of mercury? Ans. 0.0334. 75. The specific heat of iron is 0.113. If 100 g. of water at 18° C. are mixed with 150 g. of iron at 65° C, what is the re- sultant temperature? Ans. 24.8° C. 76. When one gram-molecule of carbon combines with two gram-molecules of sulphur to form carbon disulphide, 19,000 calories are absorbed. How many British thermal units are absorbed? Ans. 75.39 B.T.U. 77. When two gram-molecules of hydrogen unite with one gram-molecule of oxygen to form water, 68,300 calories are evolved. To how many British thermal units does this cor- respond? Ans. 271 B.T.U. 78. In the reaction CaO -I- 2 HCl = CaClj -|- H2O, 183.45 British thermal units are evolved. To how many Calo- ries does this correspond? Ans. 46.23 Cal. 79. The reaction NH3 -1- HCl = NH4CI liberates 166.2/ B.T.U. How many Calories does this corre- spond to? Ans. 41.9 Cal. 80. By the combustion of a certain mass of coal 200 g. of water had its temperature increased from 10° C. to 18.5° C. How many Calories were absorbed? Ans. 1.7 Cal. 81. If the combustion of 7 g. of a substance raises the tem- perature of 500 g. of water from 9° C. to 17.4° C, how many calories are yielded per gram? Ans. 600 cal. 82. The heat of combustion of Black Mountain coal is 8333 Calories. What is this in the British system? Ans. 15,000 B.T.U. 58 CHEMICAL CALCULATIONS 83. The heat of combustion of a sample of coke was found to be 8006 Calories. Convert this into British thermal units. Ans. 14,411 B.T.U. 84. The heat of combustion of a sample of pine was found to be 9153 British thermal units. What is the value in Calories? Ans. 5085 Cal. 85. Convert into Calories, 18,718 British thermal units, the heat of combustion of a sample of petroleum. Ans. 10,399 Cal. 86. The coefficient of expansion of a gas is j^-y of its volume at 0° C. for each degree Centigrade of change. What is the coefficient of expansion for a degree Fahrenheit? Ans. jjj. 87. One gram of a sample of coal raises 700 cc. of water from 6.72° C. to 18.26° C. What is its heat of combustion in British thermal units? Ans. 14,540 B.T.U. 88. The combustion of two drams of a substance caused the temperature of 9 ounces of water to be raised 15.43° F. (a) What is the heat of combustion in the British system? (&) In Calories? (16 drams = 1 oz. avoirdupois.) Ares, (a) 1111 B.T.U.; (6) 617.2 Cal. CHAPTER V SPECIFIC GRAVITY Fob the principles underlying this subject, the student is referred to textbooks on Physics. The density of a body is defined as the weight of a unit volume, that is J ., , mass density^ = — ^ > volume from which it follows that mass = density X volume and mass volume = density The relative density, or specific gravity, of a substance, is the ratio of its mass, referred to the mass of an equal volume of some substance taken as standard. The standard substance for solids and liquids is pure water taken at its maximum density (4° C. or 39.2° F.). Gases have such a low specific gravity referred to water that the weight of a unit volume of a gas is usually referred to the weight of the same unit volume, say either a liter of ' The density of a body is the weight of a unit volume. The specific gravity, or the synonymous term, relative density, is the ratio of the density of the body in question, referred to the density of some substance which is taken as unity. In the metric system, the standard employed is water at 4° C. Hence, in this system, the density is numerically equal to the specific gravity (or the relative density). The formulas given are appUcable only when the units of measurement are in the metric system. 59 60 CHEMICAL CALCULATIONS hydrogen, ^ oxygen, or of air. The specific gravity of gases will be taken up under Gas Calculations. Specific Gravity of Body Heavier than and Insoluble in Water. — The relative density, or specific gravity, is determined according to the following formula, let Wi = weight of the body in air,^ Ww = weight of an equal volume of water at 4° C, rel. dens, (or sp. gr.) = =- • yy w According to the principle of Archimedes,^ let Wb = weight of the body in water at 4° C, then W^ = Wb- Wi', in which Wb — Wb is the loss in weight in water of a substance heavier than water {i.e., the body is wholly immersed). Hence rel. dens. = .fj^ ^j^-r > Wb — Wb so, to obtain the relative density (or specific gravity), it is only necessary to weigh the substance in air and in water and solve by the formula. To obtain the relative density when the body is weighed in water at some other temperature than 4° C, say f C, then Wb rel. dens. = ^^^ — ^j^ X rel. dens, of water at t° C. Wb—Wb ' Strictly, this weight should be corrected to weight in vacuo. The reduction of the weight in air to the weight in vacuo is treated on pages 102-104 of this book. For a fuller discussion, see Chem. Ann., pp. 519-530. 2 "A body wholly or partly immersed in a fluid (liquid or gas) is buoyed up with a force which is equal to the weight of the volume of the fluid which the body displaces,'' A Text-Book of Physics, by Duff. SPECIFIC GRAVITY 61 The Specific Gravity of a Solid Substance Lighter than Water is obtained by means of a sinker attached to the soUd; the combination being of a relative density greater than water so that it will sink. Let Ws = weight of the sinker in water, TF's+6 = weight of the sinker with body attached in water. Then reLdens. = ^^^^^,_^,^^^ . In practice such a determination is most easily carried out by weighing the light substance in air and then weigh- ing the hght substance in air suspended from the hook with the sinker attached to it by a slender thread (the sinker being immersed in water) and taking the weight of this combination. Finally, the weight of both the substance and sinker in water is taken. Let SWh + Ws'] = weight of body in air and sinker in water, W'a+6 = weight of body and sinker, both in water. Then rel. dens. = ^^^^^,\_^,^^^ The Specific Gravity of Powders is determined by using a specific gravity bottle, flask or pyknometer. The empty bottle is weigh-ed, then filled with water and weighed again. This gives the weight of water held by the bottle. The bottle is emptied and dried and a con- venient amount of powder introduced when the bottle and powder are weighed. This gives the weight of the powder. The bottle containing the powder is now filled to the mark and weighed again. Let Wh = weight of powder taken, Wm+i = weight of flask full of water, Wh+f+w = weight of powder, flask and water. Wh Then reLdens. = :™^ jjj^ 1 — tSt \* Wh — \Wh+f+w — Ww+f} 62 CHEMICAL CALCULATIONS Pyknometer Method for Liquids. — The relative den- sity of liquids cannot be obtained by the means outlined above. An obvious method is to compare the weight of a given volume of the liquid in question with the weight of the same volume of water; account being always taken of the temperatures of the two liquids. Then, letting Wi = weight of a given volume of liquid, Ww = weight of the same volume of water, Wi rel. dens. = ™r- • This method is used when extremely accurate results are desired. Sinker Method for Liquids. — The principle of Archi- medes states that a body is buoyed up by an amoimt equal to the weight of liquid displaced. Knowing the amount of this buoyancy in water and in the liquid the specific gravity of which is to be found, the specific gravity of the liquid can be calculated. The buoyancies are measured by the losses in weight of a solid body in the liquid of unknown density and in water respectively. Any sohd (usually a metal) of a specific gravity such that it will sink in both liquids and is not acted upon by either of the liquids will suffice. Let Wb = weight of the solid body in air, Wb = weight of the solid body in water, Wb" = weight of the solid body in the liquid whose density is to be determined. Wb - Wb" Then rel. dens. Wb - Wb' Constant Weight Hydrometers. — Another method which is based upon the principle of Archimedes is that a body lighter than a liquid in which it is immersed SPECIFIC GRAVITY 63 sinks until it displaces an amount of liquid equal to its own weight. Let A = area of cross section of a cylindrical floating body, h = depth to which the cylinder sinks in watery Ww = weight of water displaced. In the metric system, the unit of weight, the gram, being the weight of one cubic centimeter of water (at 4° C), then if A and h are measured in square and linear centi- meters respectively. Ah represents a volume of water dis- placed which is numerically equal to the weight in grams of water displaced. That is Wu, = Ah. If the same cylinder is next placed in a liquid of a differ- ent density, it will sink to such a depth that it displaces an amount of liquid equal to its own weight. Let d = the density of this liquid, h' = the depth to which the cylinder sinks in the liquid, Wi = the weight of liquid displaced. The liquid displaced is Wi = Ah'd. Then as equal weights of the liquids are displaced (this is the method of the constant weight hydrometer), Ww = Wi, and as the density of water is unity, then Ah'd = Ah, and d = p- In words, the depth to which a cylindrical floating body will sink is inversely proportional to the density of the liquid in which it is immersed. This is the principle of the hydrometer, which is extensively used commercially. Constant Volume Hydrometers. — Still another method is the use of the principle of flotation, differing 64 CHEMICAL CALCULATIONS from the former method in that the volume of hydrom- eter immersed is constant (that is, that it is immersed to the same depth in the different liquids), the weight of the liquid ^displaced varying and serving as a measure of the specific gravity. The instrument employed is called the Nicholson's Hydrometer. Let Wh+g — weight of hydrometer plus the weight in grams necessary to be added for it to sink to the mark in water, Wh+i = weight of hydrometer plus the weight neces- sary to be added for it to sink to the mark in the liquid of unknown density. The weight of the volume of water displaced by the hy- drometer when immersed to the mark is Wh+w', the weight of the same volume of liquid displaced by the hydrometer when immersed to the mark is Wh+i- This gives the weights of equal volumes, hence ' rel. dens. = ^^■ Hydrometers are instruments of glass, weighted at their lower end, the upper end terminating in a thin cylindrical tube. This thin cylindrical tube is graduated, not into lengths, but is calibrated so as to read specific gravity directly.^ As the density of a liquid changes with the temperature, the liquid should always be at the tem- perature at which the hydrometer was calibrated or proper correction made. The usual temperature is 60° F. or 15.5° C. The Baume Hydrometer is an instrument constructed as a specific gravity hydrometer, but divided into degrees ' This same principle is involved in the Westphal's or Mohr balance. 2 See Chem. Ann., pp. 524-525. SPECIFIC GRAVITY 65 Bavun6.^ These divisions are arbitrary and unscientific, but owing to conservatism and their established use in factories manufacturing heavy chemicals, they are still retained. As in specific gravity determinations, the density of water is taken as unity, but the Baum^ scale differs from the specific gravity scale in that, as the den- sity of a liquid decreases below that of water the degrees Baum6 increase. For liquids heavier than water an in- crease in specific gravity is accompanied by an increase in degrees Baum6; consequently it is necessary to know whether the degrees Baum^ apply to a liquid heavier or lighter than water.^ Conversion of Baume Readings into Specific Gravity and Vice Versa. — The formulas showing the relations between specific gravity and degrees Baimi6 for liquids heavier than water are (temp. 60° F.) sp. gr. 145 For liquids lighter than water, ^ 140 - 130 X sp. gr. _ 140 _ ^^^ sp. gr. sp. gr. 140 ^P- S""- ~ 130 + °B6.»' 1 See Chem. Ann., pp. 525-526. ' For a tabulation of equivalent degrees Baum§ and specific gravity for liquids heavier than water, Ibid., pp. 379-383. For Uquids lighter than water, Ibid., pp. 383-387. ^ The Baum6 scale as devised by Baum6 was graduated in the following manner: The depth to which a hydrometer sank when immersed in a solution made by dissolving 15 parts of pure salt in 85 parts of distilled water was marked 15, temperature 12.5° C. The point to which it sank in pure water was marked 0. The in- 66 CHEMICAL CALCULATIONS Calculation of Density Determinations to Different Conditions. — In purely scientific calculations, water is taken as standard at 4° C, while in commercial labora- tories the standard is often in the neighborhood of 15° C. ; consequently specific gravities determined by these standards will not agree. As the temperature of water in- creases from 4° C, it expands. The weight being con- stant, with increase of volmne, the specific gravity is lowered. In the case of water this increase of volume with rise of temperature is not uniform, but has been determined with great care. Knowing the relative den- sity of water at various temperatures, the volume of a gram is obtained by taking the reciprocal of the density. The expansion of liquids and solids being appreciable, conditions should always be given with densities. This is very important in the case of liquids and differences 15° are appreciable with solids. D ^fo C. is to be interpreted ID to mean that the temperature of the substance at . the time of the determination of its density was 15° C. and 17 5° that the unit of density is water at 15° C; D ' C. means that the temperature of the substance was 17.5° C. terval was divided into 15 equal parts and the same graduations extended. This is for use with Uquids heavier than water. The hydrometer for liquids lighter than water was marked at the point to which it sank in a ten per cent solution of salt and the point to which it sank in pure water was marked 10. The interval was divided into ten equal spaces and the same distances extended. These solutions will not give the exact relations between specific gravity and degrees Baum6 cited in the formulas above. The makers of instruments produced so many so-called Baumfi hydrometers, no two of which agreed, that the Manufacturing Chemists' Association adopted as the BaumS scale one that will agree with the formulas given above, which is now regarded as standard. SPECIFIC GRAVITY 67 and is referred to water at 4° C} To convert densities of substances taken under one set of conditions into equivalent densities of the same substance under other conditions requires a knowledge of the coefficients of ex- pansion (cubical) of the substance under consideration and water, or what comes to the same thing, the variations in density of the substance and water with change of temperature. For water this has been carefully deter- mined and may be obtained from tables;^ in the case of other liquids it must be experimentally determined for each individual substance. x° x° To convert D -5 C. to Z) -5- C. requires only a knowl- y s edge of the densities of water at y° C. and z° C. Thus, let _ T)^° (^ _ weight of imit volume of the body at x° C. _ x ~ y° ' weight of unit volume of water at y° C. y _ a;°p _ weight of unit volume of the body at a:° C. ^x ~ ^ ■ ~ weight of unit volume of water at z° C. z d = -, x = dy, y d' = -, X = d'z, z dy = d'z, d = ~d', d' = ^d. y z 1 Specific gravities may be corrected for the values of the mass factor in vacuo (see Tables of Lunge and Isler, Chem. Ann., pp. 394-396) or this correction may be neglected (see Tables of Fergu- son and Talbot, Ibid., pp. 388-393). For corrections for weighing in air, see pp. 102-104 of this book. For examples of specific gravi- ties given imder specified conditions, see, for example. Aluminum, Chem. Ann., p. 100; Antimony Pentachloride, p. 106, etc. 2 For changes in the density and volume of water with change of temperature, see Chem. Ann., pp. 457-460. 68 CHEMICAL CALCULATIONS To convert D ^ C. into -D -5 C. requires a knowledge of the variation in density of the body under consideration with changes of temperature.^ Thus, let , _ ^ j/° p _ weight of unit volume of the body at y° C. _y x° ' weight of unit volume of water at x° C. x ji_j)^(^ _ weight of unit volume of the body at z° C. _ z x° ' weight of unit volume of water at x° C. x d = -, X = dy, X = d'z, X X dy = d'z, d = -d', d' =y-d. y z PROBLEMS I. A sample of bismuth weighed 14.738 g. in air and 13.235 g. in water, (o) What is the density of the bismuth? (6) What is the weight of a cube of bismuth, 2 cm. on an edge? (c) How many cubic centimeters in a kilogram of bismuth? Wh Ans. (a) Rel. dens. = == — ^rrrr Wb — Wb 14.738 - 13.235 = 1.503 g. loss of weight in water. 14.738 „„„. -^^ = 9.805 sp.gr. (6) Mass = rel. dens. X vol. Mass = 9.805 X (2)^ = 78.445 g. (c) Vol. = ^^5^- sp. gr. ^ For corrections in the case of sulphuric acid, see Chem. Ann., p. 391. OJTJliVylllHJ ii±ta.VlTY 69 n. A sample of cork weighed 2.140 g. in air. A silver sinker (specific gravity, 10.53) of 10.000 g. was employed, the combina- tion of sinker and cork, in water, weighing 2.274 g. Find the specific gravity of the cork. Am. Sp. gr. = ^ The sinker will displace a volume of water equal to its own volume. The weight of this water will be equal to the loss of weight of the sinker, when weighed in water. ^^ ; = 0.9497 cc. = 0.9497 g. 10.53 10.00 - 0.9497 == 9.0503 g., weight in water of sinker. Substituting ia the formula, 2.14 ^ 2.14 ^ 2.14 + 9.0503 - 2.274 8.9163 ^^' ^' III. A block of pine weighed 6.431 g. in air. With a sinker attached to the block by a fine thread, the sinker being in water and the block in air, the combination weighed 18.530 g.; the combination of both sinker and block in water weighed 7.635 g. Find the specific gravity of the block of pine. Wh 6.431 6.431 18.53 - 7.635 10.895 = 0.5903. IV. Find the specific gravity of a sample of sand, from the following data: Weight of sand taken, 4.655 g.; weight of bottle full of water, 80.04 g.; weight of bottle containing sand and filled up with water, 82.755 g. Ans. Sp. gr. = =- j— — r , Wb — (Wb+f+w — Ww+f) 4.655 4.655 ^ ^ ggg 4.655 - (82.755 - 80.04) 1.94 70 CHEMICAL CALCULATIONS V. A platinum ball weighed 42.96 g. in air, 40.96 g. in water, 39.548 g. in sulphuric acid and 41.264 g. in naphtha, (a) Find the specific gravity of the sulphuric acid, (6) of the naphtha and (c) of the platinum.^ Am. (a) Sp. gr - ^'' ~ ^^" Wt- Wb' 42.96 - 39.548 _ 3.412 42.96 - 40.96 2.00 1.706. ,,, 42.96 - 41.264 1.696 „„.„ (^) 42.96 - 40.96 = 2:00" = °-^'^- (,) 42.96 ^42^^ ^ ' 42.96 - 40.96 2.00 VI. (a) Convert specific gravity 1.7957 into degrees Bauml. (6) Convert 65.25° Bauni6 (heavier than water) into specific gravity, (c) Convert specific gravity 0.7692 into degrees Baum^. (d) Convert 51° Bauni6 (lighter than water) into specific gravity. Ans. Substituting in the proper formulas : (a) ° B6. = 145 - -^z = 145 - 80.75 = 64.25°. , . o-DA ^ 140 - 130 X 0.7692 ^ 140 - 100 ^ 40' ^ ^2° ^"^ ■ 0.7692 0.7692 0.7692 ' Vn. A liquid shows a specific gravity, D f^^^ C, of 1.3182. 15.56 20° The same liquid shows a specific gravity, Z) z-r-rz-„ C, of 1.3142. 15.56 ' For a tabulation of the specific gravities of elements and chemical compounds, see Chem. Ann., pp. 100-337. Si-JiiUlD'KJ GRAVITY 71 1 17 ro What is the specific gravity, D -~ C? Density of water at 15.56° C. = 0.999040; at 17.5° C. = 0.998713.' Ans. To change D |||-I C. (= d) to Z) ^°C. ( = d'). y° = 15.56°, then y = 0.999040, z° = 17.5°, then z = 0.998713. ,, _^, 0.999040 „,.,,„„ , „.„„ n 15.56°^ To change D M! C. to i) g|| C. ^ ti ^- = i-^i«2 ^ii^^=l:!^ difference = 0.0040 20° — 15.56° = 4.44°, change in temperature, ^^^ = 0.000901, difference per ° C. 4.44 17.5 — 5.56 = 1.94, difference in temperatiu'e, 1.94 X 0.000901 = 0.0017. 1.3186 - 0.0017 = 1.3169 specific gravity D ^^ C. 17.5 89. A. steel sphere of 1.90 cm. diameter weighed 28.25 g. Knowing that 1 cc. of water at 4° C. weighs one gram, what is the relative density of the steel sphere? Ans. 7.866. 90. A block of wood, 7.49 X 7.46 X 3.78 cm., weighs 152.7 g. What is its specific gravity? Ans. 0.723. 91. The largest diamond found weighed 3200 carats. If in the form of a cube, what is the length of an edge? (One carat = 205 milligrams; sp. gr. of diamond, 3.530.) Ans. 5.706 cm. 92. A footnote in "Pepys' Diary" mentions a stone shot weighing 770 lbs., which fired by the Turks in 1807, struck • See Chem. Ann., pp. 457-458. 72 CHEMICAL CALCULATIONS H.M.S. "Lion." If spherical and composed of granite (sp. gr., 2.5), what was its diameter? (One cubic foot of water weighs 62.4 lbs.) Ans. 25.35 in. " 93. Linseed oil has a specific gravity of 0.930. What will it weigh per gallon? (One gallon =231 cubic inches; one cubic foot of water = 62.37 lbs.) Ans. 7.758 lbs. 94. A drum has a capacity of 4 cubic feet. How many pounds of ammonia of 0.8917 specific gravity will it. hold? (Take the weight of one cubic foot of water as 62.37 lbs.) Ans. 222.5 lbs. 95. What is the weight of 15 cubic feet of oil of vitriol, the specific gravity being 1.8354? Ans. 1717 lbs. 96. What is the volume of 100 lbs. of hydrochloric acid of 1.2003 specific gravity? Ans. 1.335 cu. ft. 97. A casting of iron weighs 1000 kUograms. Taking the specific gravity of iron as 7.23, what is its volume? Ans. 138 L. 98. A platinum wire 7.25 cm. long weighs 1.0762 g. The specific gravity of platinum is 21.48. Find the diameter of the wire. Ans. 0.938 mm. 99. 0.0203 g. of gold (specific gravity = 19.32) was plated on a cubical brass weight 1.5 cm. on an edge. What is the thickness of the gold? Ans. 0.00078 mm. 100. What is the radius of a steel sphere (specific gravity, 7.81) equal in weight to a brass sphere (specific gravity, 8.40) 1.5 cm. radius? Ans. 1.54 cm. 101. Faraday estimated that the ductility of gold was so high that the gold in four English sovereigns could be drawn into a wire long enough to surround the earth. The weight of a sovereign is 7.988 g., and it contains 91.66% gold. If a quadrant of the earth is 10,000,857 meters, what is the thick- ness of the wire? (Specific gravity of gold, 19.3.) Ans. 0.0002198 mm. 102. A casting of iron is suspected of having internal cavities. In air it weighs 170.42 g.; in water 145.60 g. The specific SPECIFIC GRAVITY 73 gravity of cast iron is 7.23. Has the casting any cavities, and if so, what is their volume? Ans. 1.25 cc. 103. In obtaining the specific gravity of a sample of heavy spar (BaSOO, the following weights were obtained: Weight in air, 5.127 g.; weight in water, 3.969 g. What is the relative density of the sample? Ans. 4.427. 104. In obtaining the specific gravity of a brass weight, the following readings were obtained : Weight in air, 116.62 g. ; weight in water, 102.81 g. ; temperature of the water, 20° C. (One gram of water expands to 1.001773 cc. at 20° C.) What is the specific gravity of the weight? Ans. 8.430. 105. Find the weight of a cubic foot of water at 60° F., knowing that the relative density of water at 60° F. is 0.999050. Ans. 62.363 lbs. 106. Calculate the relative density of a block, given : Weight of block alone in air 152 . 7 g. Weight of block alone in air, and sinker in water 218.5 g. Weight of block and sinker, in water 9 . 5 g. Ans. 0.7308. 107. Find the relative density of gutta-percha from the following data: Weight of gutta-percha in air, 4.152 g.; weight of sinker in air, 10.450 g.; weight of sinker in water, 7.546 g.; weight of gutta-percha and sinker, in water, 7.405 g. Ans. 0.967. 108. A sample of willow weighed in air 3.820 g. A sinker of lead (specific gravity, 11.4) of a volume of 1.632 cc. was employed, the combination weighing in water 14.26 g. What is the specific gravity of the willow? Ans. 0.5843. 109. Determine the relative density of a block of wood from the following data: Weight of block in air, 3.750 g.; weight of lead sinker, 10.000 g.; weight of lead smker and block under water, 8.315 g. (Specific gravity of lead, 11.34.) 4ns. 0.824. 110. A platinum ball weighs 19.278 g. in air, 18.382 g. in water and 17.643 g. in siilphuric acid, (a) What is the 74 CHEMICAL CALCULATIONS specific gravity of the platinum ball, and (6) of the sulphuric acid? Am. (a) 21.52. (6) 1.8248. 111. A specific gravity flask holds 83.327 g. of alcohol of specific gravity 0.8164; 155.79 g. of sulphuric acid, and 120.44 g. of potassium hydrate solution, (a) Determine the specific gravity of th^ sulphm-ic acid, and (6) of the potassium hydrate solution. Ans. (a) 1.526. (6) 1.180. 112. A piece of glass weighed 5.236 g. in air and its specific gravity was 3.256. It weighed 3.702 g. in a solution of am- monia. Find the specific gravity of the ammonia. Ans. 0.9539. 113. A cylinder sank 54.40 cm. when immersed in gasoline, and 39.85 cm., in water. What is the relative density of the gasoline? Ans. 0.7325. 114. A cylinder was immersed in water at 4° C, and was marked 1.000 at the depth to which it sank. It was then immersed in a liquid of 1.2083 specific gravity, and the depth to which it sank was marked 1.250. The distance between these marks was divided into 25 equal spaces. When the cylinder was placed in a third liquid, it sank to the 1.150 mark; what is the specific gravity of this liquid? Ans. 1.125. 115. One side of a U-tube is fiUed with glycerine, the other with mercury (relative density, 13.6). If 17.4 cm. of mercury balance 187.8 cm. of glycerine, what is the specific gravity of the glycerine? Ans. 1.26. 116. A cylinder, when immersed to a certain depth in water, weighed 37.93 g. WTien immersed to the same depth in gasoline it weighed 27.55 g. What is the relative density of the gasoline? Ans. 0.7263. 117. Find the specific gravity of the liquid from the following: Weight of specific gravity bottle 40.327 g. Weight of specific gravity bottle and water. . 143.252 g. Weight of specific gravity bottle and liquid . . 108 . 779 g. Ans. 0.665. SPECIFIC GRAVITY 75 118. Bunsen gives the following data; from it calculate the relative density of calcium. Weight of empty bottle 13 . 64 g. Weight of bottle filled with naphtha 20 . 275 g. Weight of bottle partly filled with naphtha. . . . 16.650 g. Weight of bottle partly filled with naphtha and calcium 19 . 150 g. Weight of bottle fuU of naphtha and calcium . . 21 . 576 g. Relative density of the naphtha . 758 g. Ans. 1.58. 119. A sample of bronze is made up of 31.50% zinc, 3.00% tin and 65.50% copper. What is its specific gravity, supposing no change in volume occurred in alloying? (Specific gravities: zinc, 7.142; copper, 8.93; tin, 7.29.) Ans. 8.226. 120. A piece of brass weighed 9.0410 g. in water at 4° C. and 10.2621 g. in air. The specific gravity of copper is 8.930 and of zinc is 7.142. What is the per cent of copper and zinc, supposing these two metals, only, are present, and no change of volume took place in alloying? Ans. 75.00% Cu, 25.00% Zn. 121. An alloy of gold (sp. gr., 19.32) and silver (sp. gr., 10.53) has a specific gravity of 13.6312. Assuming that no change of volume occurs in alloying, what is its percentage composition (a) by volume? (6) By weight? Ans. (o) 35.28% Au, 64.72% Ag. (6) 50.00% Au, 50.00% Ag. 122. If the volume of the moon is iVth and its mass Ast that of the earth, (a) what is the density of the moon compared to the earth? (6) If the relative density of the earth is 5.53, what is the relative density of the moon? Ans. (a) 0.605. (6) 3.34. 123. An amalgam consisting of 60.34% mercury (specific gravity, 13.59) and of 39.66% gold (specific gravity, 19.3) shows 76 CHEMICAL CALCULATIONS a specific gravity of 15.47. What is the contraction in volume that has taken place in the formation of a kilogram of the amalgam? Ans. 0.31 cc. 124. Lupton states that an alloy of 50.00% by weight of platinmn (specific gravity, 21.5) and 50.00% by weight of cop- per (specific gravity, 9.00) has the same color and density as gold (specific gravity, 19.5). What is the contraction in mak- ing 50 cc. of the alloy? Ans. 26.84 cc. 125. Find the degrees Baum^ corresponding to the following specific gravities: (a) 1.8354; (6) 1.5263; (c) 1.2205; (d) 0.8963; (e) 0.9315. Ans. (a) 66° B6. (6) 50° B6. (c) 26.2° Bd. (d) 26.2° Be. (e) 20.3° B6. 12G. Find the specific gravities corresponding to the follow- ing: Liquids heavier than water; (a) 65.75° B6.; (6) 30.6° BL Liquids lighter than water; (c) 20.85° B6. ; (d) 26.92° B6. Ans. (a) 1.8297 sp. gr. (6) 1.2675 sp.gr. (c) 0.9281 sp. gr. id) 0.8922 sp. gr. 127. The allowance for temperature, of 13% to 26% nitric acid, is 0.00029 specific gravity, for each degree Fahrenheit, (a) Given a sample of acid of specific gravity 1.1154 at 60° F., what is its specific gravity at 45° F.? (6) At 78° F. (c) What is the weight of 3.4 cubic feet of this acid at 80° F. (d) What weight of this acid will occupy 10 cubic feet, at 42° F.? (e) What is the volume in cubic feet of 100 lbs. of this acid at 60° F.? (/) At 73° F.? (One cubic foot of water at 60° F. weighs 62.37 pounds.) Ans. (a) 1.1197 sp. gr. (6) 1.1102 sp.gr. (c) 235.3 lbs. (d) 698.9 lbs. (e) 1.437 cu. ft. (/) 1.442 cu. ft. SPECIFIC GRAVITY 77 128. An acid of a certain concentration was found to have a specific gravity of 1.5281 at 56° F., and a specific gravity of 1.5209 at 72° F. (a) What is the average expansion per de- gree F. in cubic centimeters? (6) What is the change per degree F. of the specific gravity? (c) Change of the strength Baurn^ per degree F.? (d) What is the specific gravity of this acid at 60° F.? (e) What is the Baumd strength of this acid at 60° F. (/) Assuming the changes of specific gravity and of the Baumd strength to vary uniformly with the temperature, what is the specific gravity of the acid at 50° F.? (g) What is the strength Baum6 at 80° F.? Ans. (o) 0.000296. (6) 0.00045. (c) 0.02812. id) 1.5263. (e) 50.00° B^. if) 1.5308. (?) 49.44°. 129. 60° F. is the temperature at which degrees Baum^ are tabulated. An acid of a certain concentration changes 0.0235° B6. for each degree change of temperatiure (Fahrenheit), (a) If the Baumd at 42° F. of a sample of this acid is 66.46° Bd., what is the Baiun6 at the temperature of tabulation? (6) What would be the Baum^ of this acid at 73° F.? (c) If at 60° F. the per cent of acid correspondiag to 66° B6. is 93.19%, and 65.75° B6. corresponds to 91.80% acid, what is the per cent acid in this sample? Ans. (o) 66.04° B6. (6) 65.73° B6. (c) 93.41%. 130. A sample of sulphuric acid is 65.25° B^. at 60° F. How many pounds per cubic foot does it weigh? Ans. 113.40 lbs. 131. What must be the diameter of a drum, to hold 400 lbs. of 26° B6. ammonia (ammonia water), length of drum to be 2.5 feet? (Ammonia is lighter than water.) Ans. 1.91 ft. 132. Accurate volumetric analysis requires that correction be made for changes of volume of standard solutions* with change of temperature. A solution was standardized at 72° F. 78 CHEMICAL CALCULATIONS This solution showed a specific gravity of 1.0277 at 84° F., and 1.0378 at 40° F. (a) What is the expansion per unit volume per degree Fahrenheit? (6) If a determination was made with this solution at 55° F., using 98.00 cc, what correction must be made to find what the volume would be at 72° F., the tempera- ture at which it was standardized? (c) What is the volume corrected to 72° F.? Ans. (a) 0.000225. (6) +0.37CC. (c) 98.37 cc. 15° 133. The specific gravity, D — C, of a 70% mixture of alco- hol and water (by weight) is 0.87187. The specific gravity, 20° D -75" C., is 0.86766 for the same mixture, (a) Calculate 17 5° 15° D — fr- C. and (6) D — r C. for this liquid. Density of water 4 15 at 15° C. = 0.999126. Am. (a) 0.86976. (6) 0.87264. 134. The specific gravity, D i^^ C, of 50.87% H2SO4 is 15.56 1.4078. (o) What is the specific gravity, D i^5_ C? (6) 15° D — C? Sulphuric acid of this concentration decreases 0.000738 D i§^ C. for each ° C. Density of water at 15.56° C. 15.56 = 0.999040. Ans. (a) 1.4065. (6) 1.4069. 136. Nitric acid containing 50.32% HNO3 has a specific grav- 15 56° ity) ^ W-" „o C., of 1.3182. The same acid shows a specific 15.56 20° gravity, D _,. .„„ C, of 1.3134. What is the specific gravity, 15.56 D ^ C? Density of water at 15.56° C. = 0.999040; at 17.0 17.5° C. = 0.998713. Ans. 1.3165. SPECIFIC GRAVITY 79 136. The Bauin^, D ||4S C, of 77.67% H2SO4 is 60.00°. 15.00 20° This same acid shows a Baum6, D , , .„- C, of 59.79°. What la.oD is the Baum6, D ^ C? Am. 59.97° Bd. 137. Calculate the difference in specific gravity in the tables of Ferguson and Talbot and those of Lunge and Isler on an acid of 50.00% H2SO4, being given the following: Ferguson and Talbot 49A7% H2SO4 = 1.3942, D Jf^C. lO.OD 50.87% H2SO4 = 1.4078, D ^^° " 15.5 Lunge and Isler 50.11% H2SO4 = 1.4000, D^U 50.63% H2SO4 = 1.4050, D ^ C. Density of water at 15.56° C. = 0.999040. 3O4 = 0.000738 D ^ Ans. L. and I., 0.0023 sp. gr. lower. Correction for 50% H2SO4 = 0.000738 D if^ C. for each ° C. 15.60 V v P' = p' PV = P'V'} CHAPTER VI GAS CALCULATIONS Boyle's Law. — The temperature remaining constant, the volume of a gas varies inversely as the pressure to which it is subjected. Let V be the volume of a gas under a pressure P, and let V be some other volume of the same quantity of the gas and P' its corresponding pressure. The analytical expression of this law is or Charles' Law. — The pressure remaining constant, the volume of a gas varies directly as the absolute temper- ature to which it is subjected. Let V be the volume of a gas at a temperature T, and let V be some other volume of the same quantity of the gas and T' its corresponding temperature. Then the analytical expression of this law is V T — = -=-.2 v r Since 0° C. corresponds to 273° A., the law of Charles may be stated: The pressure remaining constant, a true ' P'F' = fc, a constant, therefore on plotting the changes of a given volume of a gas under varying pressure or temperature an hyperbola results. * Note that T and T' are in the absolute temperature scale. 80 GAS CALCULATIONS 81 gas expands or contracts ^f ^ of its volume at 0° C, for each degree Centigrade rise or fall in temperature.^ Furthermore, the volume remaining constant, the pressure on a gas varies directly as the absolute temper- ature. Let P be the pressure of a gas at temperature T and let P' be some other pressm-e on the same quantity of the gas and T' its corresponding temperature. Then the analytical expression of this fact is P^T_ p, j,r The gas thermometer is based upon this law. The standard degree of temperature is a temperature interval such as will cause the pressure on a confined gas to change x^7 of that change in pressure which is shown by a true gas between the temperature of melting ice and the temperature of water boiling under standard pressure. Thus, the pressure exerted by a gas is used as a means of measm-ing temperature, and is employed in the hydrogen thermometer, in which the volume is kept constant and differences of pressure caused by different temperatures are measured. This unit has been chosen for the reason that the expansion coefficient of hydrogen is very uni- form over wide ranges of temperature, a property of all gases in a condition far removed from their liquefaction point. Mercury, being a liquid, does not expand with this regularity with increase of temperature, although at ' jjj may be expressed as a decimal, when it becomes 0.003663. Every gas has its own coefficients of expansion, one imder constant pressm-e, which is the coefficient usually required for gas calculations, the other under constant volume, a coefficient of a different numerical value. For air the coefficient has been found to be 0.0036706 (under constant pressure) and the decimal value 0.00367 is usually employed for every gas where extreme accuracy is not required. See Chem. Ann., p. 73. 82 CHEMICAL CALCULATIONS ordinary temperatures the difference between a temper- ature reading with a hydrogen thermometer and with a mercury thermometer is slight.^ Laws of Boyle and Charles Combined. — The laws of Boyle and Charles are readily combined in one expression. Considering a given weight of gas, Boyle's law is Poc— , hence P = K^f^' in which K is a constant. The law of Charles is PxT, hence P = K'T, in which K' is a constant. Combining these expressions:^ „ r PV P = r y, or ^ = r, in which r is a constant for the same quantity of gas. From this it follows that the same mass of gas under the conditions P', V and T' gives P'V __ «. hence PV _ P'V T r 1 Gases only conform to these generalizations approximately. In general, the fiirther a gas is removed from its liquefaction tem- perature, the more closely does it obey the gas laws. An "ideal" gas ia a hypothetical gas which is supposed to exactly obey the gas laws. A distinction between gases and vapors is sometimes made, assigning to the formed' a condition of temperature above its critical temperature, and to the latter a condition of temperature below its critical temperature. ' If the volume varies inversely as the pressure and directly as the absolute temperature, then the product of the volume into the pressure is equal to the absolute temperature into a constant. When a gram molecular volume (G.M.V.) is under consideration, this is usually expressed, PV = BT. GAS CALCULATIONS 83 Knowing five of these quantities, the sixth may be ob- tained by solving the equation. It is more to be recom- mended, however, that in solving gas equations the logic of the case should serve as a guide than that a formula should be used unthinkingly. For example, a volume of a gas equal to 7 at a temperature T is to be changed to temperature T', pressure remaining constant. Knowing that a gas expands with a rise and contracts with a fall of temperature, a ratio may be made employing T and T' which is a proper or an improper fraction according as the gas contracts or expands. This ratio is multiplied by V to obtain V. The same may be said for changes of volume of gases with change of pressure, temperature remaining constant. Let V be the volmne of a gas at pressure P, which is to be changed to pressure P'. The new volume V wUl be found by using P and P' as terms of a ratio, recollecting that if the new condition of the gas is to be a smaller volmne than V, the ratio must be in the form of a proper fraction; if larger, an improper fraction. This fraction is multiplied by V to obtain V. Standard Conditions of Pressure and Temperature. — As changes of temperature and pressm-e exert so consider- able an influence upon the volume of a gas and conse- quently upon the weight of a imit volume, standard conditions have been adopted, which are: 0° C. as stand- ard temperature, and a pressure equal to that exerted by a column of mercury 76 cm. high when at a temper- ature of 0° C.^ The measurement of temperature has already been discussed; it remains to take up the measure- ment of pressure. Pressure is usually expressed as so many units of weight per square unit of area. Atmos- ' See Chem. Ann., p. 3. 84 CHEMICAL CALCULATIONS pheric pressure is measured by the height of a column of mercury which will balance this downward pressure of the atmosphere. Then if the height of the mercury column is taken to be standard at 76 cm. at a temper- ature of 0° C. and the specific gravity of mercmy is 13.596 at this temperature, the pressure per square cen- timeter is 13.596 X 76 = 1033.3 g. As atmospheric pres- sure is always measured by a mercury barometer or a barometer standardized against a mercury barometer, it is only necessary to indicate the height of the column of mercury. Correction of Barometer for Temperature. — Mercury expands and contracts with rise and fall of temperature; consequently its specific gravity increases and decreases. So, merely measuring the height of the column of mercury is not an exact measurement of the pressure. As 0° C. is the standard of temperature of a gas, so also is this same temperature taken at which the height of the mercury column is standard. For example, at 0° C. the specific gravity of mercury is 13.596, then the pressure of 76 cm. of mercury at' this temperature is 76 X 13.596 = 1033.296 g. per square centimeter. At 15° C. the specific gravity of mercury is 13.560, then the pressure of 76 cm. of mercury at this temperature is 76 X 13.560 = 1030.56 g. per square centimeter. The height of the column of mercury is usually measured by graduations on the glass, or the glass tube is mounted in a brass jacket which carries the graduations. If the expansion of the mercury and of the substance carrying the graduations were the same, no correction for temperature in reading the barom- eter would be necessary; because, as the mercury ex- panded, though its specific gravity would be lowered, the material carrying the graduations would expand by an equal amount and these expansions would neutralize each GAS CALCULATIONS 85 other, as the graduations would register a greater length than the true length.* As an expansion of the mercury is accompanied by an expansion of the material carrying the graduations, it is only necessary to determine the difference between the coefficient of expansion of the mercury and its containing tube. This is the apparent expansion of the mercury, and for measuring this height on a glass tube 0.00017 of the column at 0° C. for each degree departure from this temperature is the correction. For a brass scale it is 0.00016. Hence 1 ± ai is the length of the apparent column as compared with the column at 0° C. In this expression a is the apparent coefficient of expansion and t is the nmnber of degrees from 0° C. Then if P is the observed height of the barometer, the corrected reading ^^ 1 ± at Moist Gases. — Volumes of gases are often measured over liquids which may or may not exert an appreciable vapor pressure. The vapor pressure of a saturated vapor depends upon the temperature only, and is independent of the pressure or the presence or absence of an inert gas. If a sufficient amoimt of a volatile liquid is introduced into a Torricellian vacumn above the mercury in a barom- eter or into a barometer tube containing a gas, the height of the mercury column will be depressed by an amoimt which is independent of all conditions except the temper- ature. If, then, the volmne of a confined gas is measured ' For the very accurate reading of some forms of the mercury barometer, capillarity must also be taken into accoimt. This cor- rection win not be considered here as it varies with the form of the instrument and the diameter of the tube. This correction is usu- ally suppUed by the manufacturer with each instrument. 2 For a fuller discussion, see Chem. Ann., pp. 542-543. 86 CHEMICAL CALCULATIONS over a volatile liquid such as water, the volume will appear greater than the volume of the same amount of dry gas by an amount corresponding to the vapor pressure of the water (if this be the liquid .employed) ^ at that temper- ature. If this vapor pressure were a constant quantity or increased regularly with rise in temperature, it would be a simple matter to correct for it; but such not being the case, the vapor pressures corresponding to various temperatures are determined experimentally and tabu- lated.^ Let Pg+„ = the pressure of the moist gas, Pw = the pressure of the water vapor, Vg+^ = the volume of the moist gas, Vg = the volume of the dry gas. Then Pg+w — Pw is the pressure of the dry gas, the volume of which is ^g + w When measuring a gas over mercury, whether moist or not, a common procedure is to bring the mercury to the same level inside and outside the tube, the pressure of the gas being that indicated by the barometer. It may not be convenient to do this; then, to measure the pressure of a gas confined in this way, the height of the mercury in the tube must be subtracted from the baro- metric reading.^ ' Every liquid has a vapor pressure which is peculiar to that liquid. To determine the variation of vapor pressure of a liquid with change of temperature is a matter of experiment. ^ For the values for water, see Chem. Ann., pp. 461-466. ^ If the gas is measured over some liquid other than mercury and the level inside and outside the tube is not the same, the height of the liquid must be reduced to the equivalent height of a column of mercury. This necessitates a knowledge of the specific gravities of the mercury and the liquid. GAS CALCULATIONS 87 Let Pb = the pressure indicated by the barometer, Ph = the height of raercury in the tube. Then the pressure of the moist gas is Pb — Ph and the pressure of the dry gas is {Pb — Ph) — Pw, or Pb — {Ph + PJ), and the volume of the dry gas is y _ Pb — {Ph + Pw) V a — p J^ g+w — p y g+w Gay-Lussac's Law of Combination by Volume. — In a chemical combination of gases, producing a gas, there is always a simple ratio between the volumes of the factors (the gases entering into the reaction) and the volumes of the products (the gas or gases resulting). Chemical equations represent not only combination by weight, but also combination by volume. From the equation • Tables are constructed for reducing the volume of a gas, moist or dry, to the volume which the gas would occupy dry at standard conditions (0° C. and 760 mm.). For example, 300 cc. of a gas is measured moist at 18° C. and 765 mm. What is the volume, dry, at 0° C. and 670 mm.? Using the decimal coefficient, the expres- sion to solve is Vg = i + (ig x 0.00367) ^ 760 " ^ ^°°- '^^^ logarithm of the factor 7«nfi i n 00367 1) ^^^ *^^ values of Pw for different temperatures are given in the table. To solve by use of the table: ^°g 760(l+0.003670 ^"^^°^- ='''''' Ten. aq. vap. for 18° C. = 15.38 mm.; 765 - 15.38 = 749.62; log = 2.87484 log 300 = 2.47712 sum of logs = 2.44337 antilog = 277.57 This considerably simplifies the calculation. See Chem. Ann., pp. 70-71. 88 CHEMICAL CALCULATIONS 2H2O = 2H2 + O2 (a) 36.032 4.032 32 lb) 2 vols. 2 vols. 1 vol. the subscripts (a) give the decomposition by weight, while subscripts (6) indicate the volumes entering into the reaction. It is to be noted that in writing equations from which calculations involving volimies are to be made, care must be taken to represent the molecular condition of the gases. When this is done the coefficients of the symbols indicate the volumes of factors and products. Gas Analysis. — Gay-Lussac's law finds constant ap- plication in gas analysis. Given a mixture of hydrogen, carbon monoxide and methane; the mixture may be analyzed by exploding the whole with an excess of oxygen and noting the contraction. The resulting carbon dioxide is next absorbed in potassium hydroxide and the contrac- tion again noted. The reactions taking place are: (a) 2H2 + 02 = 2H2O 2 vols. 1vol. vol. (b) 2 CO + 02 = 2CO2 2 vols. 1vol. 2 vols. (c) CH4 + 202 = CO2 -1- 2H20 1vol. . 2 vols. 1vol. vol. In regard to the contraction caused by oxidation,- the following is to be noted: Reaction (a) shows that two volumes of water vapor and one volume of oxygen, in all three volumes, disappear in the oxidation of two volimies of hydrogen (the water vapor being condensed to a Kquid, the volume of which may be neglected, being approximately only tyVi? of the volume it occupies as steam). Then, if x represents the volume of hydrogen originally present, | x is the amount of contraction due to the oxidation of hydrogen. GAS CALCULATIONS 89 Reaction (6) shows that where originally there were three volumes of gas present before the explosion, two volumes of carbon monoxide and one volume of oxygen, after oxidation there remain only two volumes of carbon dioxide, a net loss of one volume. If y represents the volume of carbon monoxide origiually present, ^ y repre- sents the contraction due to the oxidation of carbon monoxide. Reaction (c) shows . that where originally there were three volumes, one volmne of methane and two volumes of oxygen, after explosion there remains but one volume, the loss in this case being two volumes. As this contrac- tion was caused by one volume of methane, then, if z represents the volume of methane originally present, the contraction due to methane is 2 z. After exploding the mixtiu:e, the resulting carbon dioxide is absorbed by potassium hydroxide, the reaction being CO2 + 2 KOH = K2CO3 + H2O. Reaction (a) produces no carbon dioxide, hence no part of the carbon dioxide absorbed can be ascribed to hydrogen. » Reaction (6) shows, that carbon dioxide to the extent of two volumes will be absorbed by the potassimn hydrox- ide, and as these two volumes of carbon dioxide resulted from two volumes of carbon monoxide, it is evident that, if 2/ be the volume of carbon monoxide originally present, the same volume will be absorbed. Reaction (c) shows that one volume of carbon dioxide is absorbed which was produced from one volume of methane. If 2 represents the volume of methane origi- nally present, the contraction, due to the absorption of carbon dioxide produced from z volumes of methane, will be z. 90 CHEMICAL CALCULATIONS To put these facts in algebraic form, let A = volume of mixture taken, B = volimie of contraction after explosion, C = voliune of carbon dioxide absorbed, X = volimie of hydrogen, y = volume of carbon monoxide, 2 = volume of methane, (o) A = x + y + z. (b) B = ix + ^y + 2z. (c) C = y + z. (a) A = X + y + z. (c) C ^y + z. (d) A — C = X. (by subtraction.) Substituting the value of (d) in (a), clearing (6) of frac- tions and solving: (e) A = A-C + y + z. (b) 2B = ZA -3C + y + 42. A-2B=-2A + 2C-3z. (by subtraction.) ,„ -3A + 2B + 2C (/) z= 3 Substituting (/) and (d) in (a) and solving for y: A=A-C + y + :^^A±IB±2C, 3A = 3A-3C + 3y-3A + 2B + 2C. 3A -2B + C y= — 3 Avogadro's Law. — Equal volumes of gases, elemen- tary or compound, at the same temperature and pressure, contain the same number of molecules. This law gives a method of determining molecular weights. Let n be the number of moleciiles of gas A, and GAS CALCULATIONS 91 m the weight of each molecule, then the total weight w of this gas is w = nm. Also let m' be the weight of each molecule of another gas B of the same volume and under the same condi- tions of temperature and pressure, then the weight w' of this gas is w' = nm'; the ratio of the weight w to w' is w _ nm _ w' nm' ' consequently, w_ _ m w' m' or the ratio of the molecular weights of two gases is the ratio of the weights of equal volmnes imder the same con- ditions. So if 5 is the gas taken as standard, the molecular weight m of the gas A is w, or the weight of a unit volimie of the gas A referred to the weight of the same volume in the same condition as the gas B. Atomic Weights and Molecular Weights. — Hydro- gen was the gas formerly taken as the basis of atomic weights; but lately an ideal gas of a weight ^V of oxygen gas has come into almost universal acceptance. For practical purposes air is often used in determining molec- ular weights. From what has been stated it is evident that these gases bear the same relation to the determina- tion of molecular weights as water bears to the deter- mination of specific gravity. As one cubic centimeter of water at 4° C. is taken as the unit weight of a unit volume in specific gravity determinations, so in molecular weight determinations, which are analogous, the unit 92 CHEMICAL CALCULATIONS of weight is the weight, und^r standard conditions, of one liter of hydrogen, ^ of the weight of oxygen gas, or, lastly, the weight of one liter of air. Atomic weights are now based on oxygen as 16, and as a molecule of oxygen is composed of two atoms, the molec- ular weight of oxygen is 32, so the unit in atomic weights is ^V of the weight of oxygen gas."^ One liter of oxygen, at sea level, latitude 45°, under a pressure of 760 mm. of mercury and at 0° C. weighs 1.4290 g. Hence one liter of an imaginary gas having a unit atomic weight would weigh, per liter, 1.4290 „„.,„.„ g2 = 0.044656 g. Consequently, if m is the molecular weight of any gas, 0.044656 m = weight of a liter under standard conditions, and, conversely, given the weight w of a liter of gas under standard conditions, the molecular weight m is w m = 0.044§56 Instead of dividing the weight of a liter of the gas by 0.044656 it may be multiplied by the reciprocal 22.393. In general terms the relation between the weight w oi a. liter of gas and its molecular weight m is vn w = 22 X 1.4290 = 0.044656 m g. Vapor Density. — Relative density being defined as the ratio of the weight of a \mit volume of the substance imder consideration referred to the weight of the same ' Most of the common elementary gases have two atoms in the molecule. Examples: CI2, H2, N2, etc. See Chem. Ann., p. 3. GAS CALCULATIONS 93 volume of another substance taken as standard/ the for- mula for relative density is Density to standard substance X _ wt. of unit vol. of substance wt. of imit vol. of X The standard substance for soUds and liquids is water at 4° C» ; for gases Density tp ^ oxygen^ _ wt. of one liter of gas wt. of one liter of ideal gas, ^ O2 (0.044656) ' _ ., , . wt. of one liter of the gas Density to air = ^t. of one liter of air (1.2926) ' Density to hydrogen _ wt. of one liter of the gas wt. of one liter of hydrogen (0.089873) ' hence it is evident that given the specific gravity (vapor density) of a gas to hydrogen, to obtain the molecular weight m, divide the weight w) of a liter by 2.016, the molecular weight of hydrogen: 0.089873 r^r^AAnon j-u t ^ = 0.044580; therefore m = 2.016 "•"--""". '-^ — 0.04450 Air weighs 1.2926 g. per liter under standard conditions; considering it to be a chemical compound it has a molecu- lar weight referred to ^ oxygen of 1.2926 0.044656 = 28.946. 1 When the metric system is employed, the specific gravity and the density of a gas is mmierically the same if the miit of volume is the hter. As aheady mentioned, the terms relative density and specific gravity are sjnionymous; when deaUng with gases the term vapor density is frequently employed. * The specific gravity to ^^ oxygen is the molecular weight. 94 CHEMICAL CALCULATIONS Consequently, given the specific gravity ^ of a gas referred to air, the molecular weight m is m = gX 28.946, and, conversely, given the molecular weight, its specific gravity is m 9 = 28.946 The density of a gas (pressure remaining constant) is inversely proportional to the absolute temperature, that is, D' T' also the density of a gas varies directly as the pressure (temperature remaining constant), or D' P'' Graham's Law. — The volumes of two gases which diffuse in equal times under the same conditions are in- versely proportional to the square roots of their densities. Effusion, or the passage of gases through a minute orifice in a thin plate, obeys the same law as diffusion, which is the passage of gases through a porous diaphragm. The analytical expression of this law is V2 Vd[' °'' V2' D,' in which Vi is the volume of a gas of density Di which diffuses or effuses in the same time as a volume V2 of another gas of density D2. Bunsen found that the ratios of the times of diffusion (or effusion) of two gases which will diffuse (or effuse) the GAS CALCULATIONS 95 same volume is directly proportional to the square roots of their densities; that is, in which Di is the density of a gas that requires time Ti to diffuse (or effuse) the same volume as the gas of density D2 which requires time T2. Volume Occupied by a Gram Molecule of a Gas. — A gram molecule of a substance will be defined as that weight of the substance, in grams, numerically equal to its molecular weight. For example, the molecular weight of oxygen is 32, hence a gram molecule of oxygen is 32 g. Given the following: Substance. Molecular weight. (0) cu (6) HCl (c) CO2 70.92 36.468 44.00 to find the volmne occupied by one gram molecule of each of these substances. (a) 70.92 X 0.044656 = 3.167 g. per liter, 70.92 3.167 or m one expression 70.92 70.92 X 0.044656 (&) Similarly, 36.468 36.468X0.044656 (c) In the same way, 44.00 44.00 X 0.044656 = 22.393 liters per gram molecule; = 22.393 liters per gram molecule. = 22.393 liters per gram molecule. = 22.393 liters per gram molecule. 96 CHEMICAL CALCULATIONS A consideration of Avogadro's law would have led to these results, for since all equal volumes of gases under the same conditions contain the same number of mole- cules, and as this fact serves as a means of determining molecular weights, it follows that if weights of gases are taken according to their molecular weights, these gases must occupy the same volume. These facts give an alternative method for calculating the molecular weights of gases, being given the weight of a specified volume. As one gram molecule of a gas occupies 22.4 liters ' at standard conditions of temper- ature and pressure, the molecular weight of a gas is that weight, in grams, which will occupy this volume. To calculate the molecular weight of propylene: 298.5 cc. of propylene at 18° C. and 768 mm. weigh 0.5315 g. Re- ducing this volume to 0° C. and 760 mm., il X ?lx 298-5 = 283 cc. Then, the molecular weight being the weight of 22.4 liters, 22,400 , 283 X 0.5315 = 42.07. Measurement of Vapor Density. There are three gen- eral methods for obtaining vapor density. These are the methods of Dumas, Victor Meyer and Hofmann. Dumas' Method is applicable to gases and liquids which may be vaporized. In this method, a bulb of about 200 cc. capacity is provided with a narrow neck through ' The law of Charles states that gases have the same coefficient of thermal expansion. This is not strictly true; although the co- eflScients are nearly equal, they differ to a slight extent. This being the case, all gases at 0° C. and 760 mm. will not occupy ex- actly the same volume. 22.4 Uters may be taken as an approxi- mate value in the majority of cases. This volimie is termed the gram molecular volimie (G.M.V.). GAS CALCULATIONS 97 which the material is introduced and which is sealed when full of the gas to be determined. To make a vapor den- sity determination, a small quantity of the Kquid to be investigated is introduced or the gas is passed in, displac- ing the air, after ascertaining the weight of the open bulb in air. If used for a hquid the bulb is placed in a water or oil bath which is about 25° C. above the boiling point of the liquid introduced, which is thereby vaporized and escapes through the neck. As soon as no more vapor issues from the bulb, the neck is sealed off by a blowpipe and the temperature of the bath is taken. After cooling, the bulb is again weighed and then the neck is broken off tmder water, which is drawn into the flask, filling it completely, except for a small bubble. Filling the bulb and weighing the water it holds gives the capacity, the barometer reading giving the pressure of the gas. If D is the vapor density of the substance, W the weight of the vapor in the bulb at the temperature of the bath and at atmospheric pressure,^ and W the weight of the air which the bulb would hold at the same temperature, then the vapor density compared with air is W D = — • W The Method of -Victor Meyer is not applicable to gases, but only to substances which volatilize without decom- position, and consists in vaporizing a known weight of substance at a temperature considerably above its boil- ing point in an apparatus of special form. The vapor displaces its own voliune of an inert gas, which is col- lected (usually over water) and its volume measured, noting the temperature and the height of the barometer. ' Correction should be made for the buoyant effect of weighing in air. This will be treated subsequently. 98 CHEMICAL CALCULATIONS The volume of air is corrected to standard conditions, dry. Knowing the weight of substance taken and the volume which it occupies in the form of vapor, the vapor, the density, or what is more often required, the molecular weight, is easily calculated. Suppose 0.1910 g. of sub- stance gave 60.50 cc. of air at 11° C. and 752 mm. To calculate the molecular weight (vapor density to -^ oxy- gen). As the air was measured over water, the volume must be calculated to standard conditions, dry. The tension of aqueous vapor at 11° C. is 9.81 mm. 273 ,, 752 - 9.81 ,, ^^ ^„ ^^ „„ 284 ^ — 760 ^ " ^^' The weight of vapor which would be required to fill 22.4 liters (the molecular weight) is 5^ X 0.1910 = 75.34. 56.79 Hofmann's Method is used for substances which may be volatilized undecomposed and consists in vaporizing a known weight of substance at a constant known temper- ature in vacuo. This is usually done over mercury. A graduated glass tube closed at one end is filled with mer- cury and inverted in a mercury trough. The height of the mercury is noted and a weighed amount of substance in a small bottle is introduced into the tube by inserting it under the mercury up into the open end of the tube. The tube is surrounded by a jacket through which circu- lates steam or vapor from some high boiling point sub- stance. The temperatiu-e maintained in the jacket is considerably higher than the boiling point of the sub- stance examined. This high temperature drives the stopper out of the bottle, vaporizes the substance and depresses the column of mercury, the height of which is GAS CALCULATIONS 99 noted.^ From the position of the mercury, the volume and the pressure of the vapor inside the tube is deter- mined. The temperature of the vapor is read. This will be the temperature of the jacket. Suppose 0.0648 g. of substance yields 65.30 cc. of vapor at 100° C, the barometer reading 752 mm., the mercury in the tube standing 480 mm. above the level of the mercury in the trough. To calculate the molecular weight of this sub- stance (neglecting corrections for the expansion of the mercury, its vapor tension and its specific gravity). J^ X 760 X 65.30 = 17.11 cc. at 0° C. and 760 mm. The weight of 22.4 liters, or the molecular weight, is 22,400 17.11 X 0.0648 = 84.86. Deviations from the Gas Laws. — Up to the present, calculations have been made assuming the absolute truth of the laws of Charles and Boyle. These laws are only approximations, for, as the critical points of a gas are approached or the gas is near a change of state, the laws lose accuracy. A gas far removed from its liquefaction point acts very closely in accordance with the laws of Charles and Boyle; the farther removed, the closer its compliance with these laws. It can only be said then, that if the gas under consideration acted as an ideal gas (an ideal gas is an imaginary gas which obeys the laws of Charles and Boyle) the predictions would be 1 By this method, the gas may be obtained mider reduced pres- sure which may allow of the examination of substances which de- compose on heating to their boiUng point under higher pressures. Further, the diminishing of the pressure by lowering the boiling point allows of a lower temperature being employed for vaporiza- tion. 100 CHEMICAL CALCULATIONS correct.' As an instance, carbon disulphide has a molec- ular weight of 76.12, while its specific gravity to ^V oxy- gen is 76.4. The volume occupied by a gram molecule of this gas is 76.4 X 0.044656 = ^^'^^ ^*^''^- Also, mercuric chloride having a molecular weight of 271.52, in gaseous condition shows a specific gravity to ^ oxygen of 283.5. The volume occupied by a gram molecule is then 271.52 283.5 X 0.044656 = 21.47 liters. This last case is a considerable divergence from the theoretical and it will be noted that in this case the vol- ume of gas is taken of a substance which at ordinary temperatures is a solid. Water at 0° C. and 76 cm. pressure is ordinarily in the solid condition, but at ele- vated temperatures becomes gaseous. The same is true of mercury and other substances which at ordinary temperatures do not exist as gases. It may be predicted, however, that if they were in the gaseous state they would approximate the volume calculated from the gas equa- tions. Thus, if steam, H2O, with a molecular weight of I81OI6 existed as a gas at 76 cm. and 0° C, one gram molecule would occupy 22.4 liters. Also, the molecular weight of mercury is 200.6 (the atom and the molecule of mercury are the same) ; then 200.6 g. of mercury vapor • It is for this reason that some of the answers given to the problems do not agree exactly with data obtained from chemists' handbooks, it often happening that the specific gravity calculated and observed difiei* sUghtly. The reason for this is that molecular weights are often determined by gravimetric methods which in many cases are more accm'ate than those just described. See Chem. Ann., p. 3. EXPR. AX: GAS CALCULATIONS 101 if it existed as a gas at standard conditions would occupy 22.4 liters. On this basis it is possible to calculate the volume of mercury vapor and the volume of any weight of it at any temperature at which it is a gas. One gram of mercury vapor at standard (if it were a gas) would occupy .^^ X 22.4 = 0.1117 hters, and at 400° C. the volume of this amount would be 1^ X 0.1117 = 0.2753 liters, and the voliune of the vapor at one-half atmospheric pressure, ^ X 0.2753 = 0.5506 liters = 550.6 cc. To restate the problem: What is the volume of 1.000 g. of mercury vapor at 400° C. and under one-half atmos- pheric pressure. Solving, by one expression, 1.000 ^ 673 ^ 760 ^ __ . . _... ... 200:6 X 273 X 380 X 22-* = ^-^^^^ ^'^^''- This is a fair approximation to the truth, as the mercury would be in the vapor state, the boiling point of mercury under atmospheric pressure being 357.33° C. Volume of Gas from Weight of Substance (in Metric System). — The foregoing yields a method for calculating the volumes of a gas resulting from a given weight of a reacting substance. Take the reaction: 2 KCIO3 = 2 KCl + 3 O2 (a)- Parts by weight (g.) 2 (122.56) = 2 (74.56) + 3 (32) (&) Parte by vol.' 2 vols. 2 vols. 3 vols. = 3 (22.4) liters. ' Potassium chlorate and potassium chloride are not gases, but if they were, the volume 'relations of (6) would hold if they corre- sponded to the formulas assigned to them. 102 CHEMICAL CALCULATIONS Volume of Gas from Weight of Substance (in Ounces). — By a coincidence approximately the same relations exist between the volume of a gas in liters and its weight expressed in grams and the volume of a gas expressed in cubic feet and its weight expressed in ounces; for one cubic foot = 28.32 liters and one ounce = 28.35 g. For example, one gram molecule of oxygen has been shown to occupy 22.4 liters. To calculate how many cubic feet in 32 ounces of oxygen, 1 oz. = 28.35 g., 1 cu. ft. = 28.32 L.; therefore 32 oz. will occupy 22.4 X 28.35 = 635.04 L., and 635.04 28.32 = 22.42 cu. ft. Consequently, gas calculations being rough in most cases, it is possible to calculate chemical equations involving gases to units in the Enghsh system with the same facility as with the metric system. Defining an ounce molecule as that weight in ounces corresponding to the molecular weight, the volume of an ounce molecule of a gas under standard conditions is 22.4 cubic feet.^ Calculation of Weighings in Air to Values in Vacuo and Vice Versa. — The principle of Archimedes as stated is that a substance immersed in a liquid is buoyed up (loses weight) by an amoimt equal to the weight of the liquid displaced. If for the term "liquid," the term "fluid" is substituted, the law has a more general applica- tion, for gases come under the head of fluids. Commonly, weighings are carried out in the atmosphere; consequently, the weight of an object shown by the ordi- ' If the English system is used in all the units, it is better to calculate the temperature to Absolute. The other units, such aa inches of mercury, may be retained. GAS CALCULATIONS 103 nary balance may not be the true weight, as the object weighed has displaced a certain amount of air, and con- sequently has lost weight by an amount equal to the weight of the air displaced. If, however, the weights used to counterbalance the object are of the same volume they have likewise lost weight to the same extent; conse- quently, the mass shown by the coimterpoising weights is the true mass of the object weighed. If, on the other hand, the substance weighed and the weights used to counterpoise are of different volimies, the mass may be more or less than the true weight. If the volume of the substance is greater than that of the weights counter- balancing it, the substance has lost weight, and the apparent value is smaller than the real value; if the weights occupy a greater volume than the object weighed they have lost weight to a greater extent than the object and the apparent weight of the object is larger than the real value. Let W be the true weight in vacuo of the body W of a density D. The body then occupies a volume yy Let W be the apparent weight of the standard grams necessary for counterpoise and D' the density of the standard grams. The volume of the standard grams is Yj. If d be the weight of a imit volume of air at the W time of weighing, then the weights have lost -jy '^ ^^^ ^^^ W loss of weight of the object is j^d. Then the condition of equilibrium is W W W-^d^M'-'^rd. 1 The weights are supposed to be corrected so that in vacuo they have this value. 104 CHEMICAL CALCULATIONS Solving for W and W gives also 1-- 1-- D' D W = W' ~; W = W ^; 1 -— 1 - — D D' and since d is small in comparison with D and D', by division the above expressions give W = F'+ W'd(^^-^^ iW'^W+Wd (^ -^} The latter form is more convenient for general use as W'dljT — jy], or Wdijy — r-j may be solved once for all and tabulated for bodies of differing densities as brass weights are almost universally used.^ Standard (United^ States) Unit of Volumetric Appa- ratus. — The metric system was designed to connect the units of length and mass. The gram was intended to be the mass of a cube of water, at 4° C, of one centi- meter edge. Or, the kilogram was to be the mass of a cube of water, at 4° C, of one decimeter edge. This relationship was very closely approximated, but the error is appreciable, a kilogram of water being found to con- tain 1000.027 cc. This discrepancy is very small and causes no inconvenience, but because of it, the units of capacity for volumetric apparatus have been changed. The liter, defined as the volume occupied by a quantity of pure water at 4° C, having a mass of one kilogram, and the one-thousandth part of the liter, called the milliliter ' See Stewart and Gee, "Elementary Practical Physics," Vol. 1, p. 90; also Treadwell, "Analytical Chemistry," Vol. 11, p. 14, 1912. GAS CALCULATIONS 105 or cubic centimeter, are employed as units of capacity .* As 4° C. is an inconvenient temperature to work at, volumetric apparatus is calibrated to contain, or deliver, a specified volume at some other temperature. In this coimtry 20° C. has been adopted by the Bureau of Stand- ards. At whatever temperature the apparatus is caU- brated to contain or deliver, this temperature should be marked on the instrument. Volumetric instruments are often caUbrated by weigh- ing into them the calculated amount of water at a definite temperature which the instriunent should contain, or de- liver, to be correct. As the weighing is done in air, cor- rection must be made for the buoyant effect of air. The water to be weighed may or may not be at the temperature at which the instriunent is calibra,ted to be correct, and allowance must be made for the expansion of the water and the glass. The cubical coefficient of expansion of the glass in volumetric apparatus is taken as 0.000025 for each degree centigrade. Suppose the temperature at which the instrument is calibrated to be correct is 20° C, the expres- sion for the volume of water the instrument is to contain, or deliver, at other temperatures is V = v + vX 0.000025 (20 - f), in which V is the volume at 20° C, v the volume at t° C, the temperature (<°) at which the water and the instru- ment are at the time of calibrating. Thus, to calculate the weight, W, of water, which should be weighed into a flask at f C. such that the meniscus shall be at the place to be marked to show a correct vol- ume at 20° C. Let V be the volume it is desired to have the flask contain at 20° C. and d the density of the water 1 Circular of the Bureau of Standards, No. 9; also Sutton, Vol- umetric Analysis, d. 24, 1911. 106 CHEMICAL CALCULATIONS at f C. Let 6 be the buoyancy constant/ t the temper- ature of the flask and water at the time of weighing. Then the number of grams W, which should be placed on the other pan (the flask being tared by a similar flask and beads, foil, etc.), is W = Vd + Vhd+VX 0.000025 (20 - 0, in which d is the density of water at the temperature t. The Mohr Liter. — Mohr proposed as a unit of vol- ume that volume occupied by one gram of water at 17.5° C. weighed in air with brass weights. In English-speaking countries 60° F. or 15.5° C. is commonly used as also is 15° C. In standard flasks,> burettes, etc., the temper- ature at which it was caUbrated is always marked on standard instruments.^ To calculate, for example, the true capacity of a Mohr's cubic centimeter: Air at standard weighs 0.0012926 g. per cc; then at 17.5° C, the barometer being at standard: ' The buoyancy constant for a body of a specified specific grav- ity is obtained by solving (^ ( rj ~ 77/ J for correcting weighings made in air to obtain values in vacuo and conversely i\YP~n\ ^°^ converting weights in vacuo to weights in air. This latter quantity is intended as the value of (b) in the above. ' "Measures marked, e.g., '25 cc. 15° C should contain or de- liver, as the case may be, 25 true cc. when the instrument is at the temperature of 15° C. On the other hand, a flask marked ' 1000 g. 16° C should, of course, contain 1000 g. of distilled water at the temperature of 16° C, i.e., a Mohr's liter. Vessels graduated accord- ing to Mohr's system should bear the word 'gramme' or the letters 'grm' together with the temperature. It should be noted that the German Kaiserliche Normal-Eichungs Kommission no longer em- ploys Mohr's unit." Sutton, Volumetric Analysis, p. 24. As to the last sentence, the United States Bureau of Standards could also be included. GAS CALCULATIONS 107 1 + 0.00367 X 17.5 ^ 0-0012926 = 0.0012146 g. per cc. One gram of water at 17.5° C. = 1.001289 cc, then 1.001289 X 0.0012146 = 0.0012162 g. air displaced by the water. 0-7 = 0.11905 cc. air displaced by the weights. 0.11905 X 0.0012146 = 0.0001446 g. air displaced by the weights. 0.0012162 - 0.0001446 = 0.0010716 g. loss of weight of the water. 1.0000000 + 0.0010716 = 1.0010716 g. true weight of 1 cc. of water under these conditions equals 1.001072 true cc. 1.001289 - 1.000000 = 0.001289 cc. expansion of the water due to change from 4° C. to 17.5° C. 1.001072 + 0.001289 = 1.00236 true cc. in a cubic centi- meter as recommended by Mohr. Then a Mohr's liter is 1002.36 true cc, or is 2.36 cc larger than a true liter. Dalton's Law of Partial Pressures. — The pressure exerted by a mixture of gases, or of gases and vapors, or vapors, which do not react chemically, is equal to the sum of the pressures which each would exert if it alone occupied the whole space occupied by the mixture. For simplicity a gas saturated with water vapor will be considered. Let Pg° = pressiu-e of a dry gas corresponding to a vol- ume Vg° and a temperature T°, Pg' = pressure of the same gas corresponding to a volume Vg' and a temperature T'. By the laws of Boyle and Charles: V ° P ' T° Ia = £x V- _. (-1) -tr I -p o '^ rpi \^) 108 CHEMICAL CALCULATIONS Let this gas be at Y J , PJ and T' and let it be saturated with water vapor. Let y'g+w = volume of the mixture after saturation, P'g+w = pressure of the gas after saturation, P'„ = pressure of aqueous vapor at 7". Then, by Dalton's law, P'„+«, = P'g + P'„; whence P'g = P'g+y, - P'„. Substituting in (1) y o p/ p/ /TT( g+w Cg J- Now let the gas be at Pg°, Va° and T°, and let it be sat- urated with water vapor. Let V°g+w = voliune of the mixture after saturation, P°g+w = pressure of the gas after saturation, P„° = pressure of aqueous vapor at T°. Then, as before, P%+„ = Pg° + P„°; whence P," = P°„+„ - P„°. Substituting these values in (2) 7o p/ p / rpo y g+w * g+w ^ w ^ The weight of a moist gas is the weight of the dry gas plus the weight of the water vapor. Let Wg+w = weight of Vg° volumes of moist gas at a pressure P°g+w and a temperature T°, Wg = weight of Vg° volumes of dry gas at a pressure Pg° and a temperature T°, ' The disoiission given on page 85, correcting the volume of a moist gas to the volume which the gas would occupy if dry, is a single case of Dalton's law. For values of the tension of aqueous vapor, see Chem. Ann., pp. 461-466. Mercury also has a small vapor tension; see Chem. Ann., p. 468. GAS CALCULATIONS 109 Ww = weight of Vg° volumes of water vapor at a pressure P„° and a temperature T°, A = weight of a unit volume of dry gas at Pg° and T°, B = weight of a unit volume of water vapor at Pg° and T°. Then p/ p / rpo W — ""''"' " V V A '' p o ^ rpl ^ ■^• Since the weight of a gas is directly proportional to its pressure and inversely proportional to its temperature, and p / mo W = — V — V B rr w p o ^N m, /^ ^, then Wg + y, = Wg+ W^. For similar reasons p/ p / rpo p / rpo Vy g+w ~ p~d X j^ X A -j- -p-Q X fpT X •£>. By means of this formula, the weight of a moist gas may be calculated for any temperature and pressure. For a specified gas, the formula may be considerably shortened. Take, for example, air saturated with water vapor. The density of aqueous vapor compared to air is 2g'Q43 = 0.62247 = g , approximately. Let A = weightof aunitvolumeof dryairatPg°andT°, B = weight of a imit volume of water vapor at Pg° and T°, Wa+w = weight of a unit volume of saturated air at Pg° and T°. Then B = iA. 110 CHEMICAL CALCULATIONS Then the weight of a unit volume of saturated air at P' and T is p/ p / mo p / rpo fK a+ui — p"o ^ rpi ^ -^ \ p I ^ rpi '^ % ■^i ■pi 3 J3 ' W° o+ui — p~d ^ rpr ■^ ■"■• But as the weight of a hter of air is given at 0° C. and 760 mm., then Pg° = 760, T° = 273 and A = 1.2926 g. Substituting, P' 3X3 I 970 Use of this formula is as follows: It is desired to know the weight Wa+m of a liter of air saturated with moisture at 15° C. {T') under a pressure of 754 mm. (P'b+„). The weight of a Kter of a gas is given under standard con- ditions, i.e., temperature 273° A., and pressure 760 mm. Of air this weight is 1.2926 g. The tension of aqueous vapor PJ at 15° C. is 12.76 mm." Substituting in the formula, • The fraction f is applicable only to air. For any other gas the requisite fraction could be worked out in the same manner as given for air. 2 The value 12.76 mm. is given to this degree of accuracy in the tables. When subtracted from the barometric reading given in millimeters, the result would give a reading which is apparently accurate to hundredths of a millimeter, and to be significant," would require that the barometric reading was accurate to hundredths of a rmHimeter, which is seldom the case. In the solution of prob- lems, however, the vapor pressures have been subtracted from the barometric readings as if these were as accurate as the tabulated values of the vapor pressures. GAS CALCULATIONS 111 Again, it is required to find the weight of dry oxygen contained in a Uter of oxygen saturated with moisture at 17° C. and under a pressure of 750 mm. (ten. aq. vap. at 17° C. = 14.45 mm.). ^'^m^ ^^° 760^^^ ^ ^^ ^ 0.044656 = 1.3019 g. O2, TF" = g^ X -^ X 18.016 X 0.044656 = 0.0144 g. H2O vapor, W = 1.3019 + 0.0144 = 1.3163 g. A gas need not be saturated; a knowledge of the degree of satiu-ation will suffice, from which the tension of the aqueous vapor may be calciilated. In the case of mois- ture present in the atmosphere this may be readily deter- mined by taking the dew point. The dew point is that temperature at which the air would be saturated if it contained the amotmt of water which is present at the time and place of determination. For example, it is necessary to know the aqueous pressure exerted in the laboratory at a certain time. Suppose the dew point is found to be 15° C. On consulting the table of aqueous tensions ^ the tension of aqueous vapor at 15° C. is found to be 12.73 mm. The relative himiidity is the ratio be- tween the amount of aqueous vapor present in the air at a certain time and place, and the amoimt which would need to be present to saturate the air at the temperature of the air at the time of taking. As an example: The dew point at a certain time and place was found to be 10° C. The temperature of the atmosphere was 20° C. The tension of saturated aqueous vapor at 10° C. is 9.18 mm.; at 20° C. it is 17.41 mm.; then the relative humidity is q 1 o itS = 0-5273 = 53%. 1 See Chem. Ann., pp. 461-467. 112 CHEMICAL CALCULATIONS PROBLEMS I. 200 cc. of a gas are at a pressure of 752 mm.^ and 15° C: (a) What is the volume under a pressure of 770 mm., tempera- ture remaining constant? (6) What is the volume if the tem- perature is lowered to 10° C, pressure remaining constant? (c) What is the volume if changed from 752 mm. and 15° C. to 770 mm. and 10° C? Ans. (a) III X 200 = 195.33 cc. (6) 15°C.=273+15=288°A.; 10° C. =273+10 = 283° A. ?^ X 200 = 196.53 cc. Zoo II. A barometer graduated on a glass scale reads 763.4 mm. at 19.5° C. (a) What is the reading corrected to 0° C? (6) If the corrected height of a barometer with a brass scale is 764.7 mm., what does the barometer read at 22° C? (c) If a barom- eter with a glass scale reads 754.3 mm. at — 10 C°., what is the height corrected to standard? , , , 763.4 763.4 _„„ „ ^'^^^ ^"^ 1 + 19.5 X 0.00017 = l-:003315 = ^^°-^'^°^- (*) l + 22jo.00016 = ^^^-^- X = 764.7 X (1 + 22 X 0.00016) = 767.39 mm. , , 754.3 „.- „ ^'^ 1 - 10 X 0.00017 = ^^^-^"^- 1 Unless otherwise stated, barometer readings in all problems are supposed to be corrected to values at 0° C. Barometer readings will be supposed to be correct as far as the second place past the decimal point even if given only to the decimal point. Thermom- eter readings will be considered correct to the second decimal place. GAS CALCtTLATIONS 113 m. A gas at 750 mm. and 12° C, measured moist, has a volmne of 325 cc. (a) What is its volume, dry, under the same conditions? (jb) Dry, at standard? (c) 160 cc. of a gas are measured moist at 15° C, the barometer (corrected) reads 743 mm. The mercury in the tube stands 150 mm. above the trough; what is the volume of the gas, dry, at standard? Ans. (a) Tension of aqueous vapor at 12° C. = 10.48 mm. 750r_iM8 X 325 = m^ X 325 = 320.46 cc. 7oO 750 (c) Tension of aqueous vapor at 15° C. = 12.73 mm. Using the decimal coefficient for the change of volume due to change in temperature,' 743 -(150 +12.73) ^ 1 ^ 160 = 1158 cc 760 ^ 1 + 0.00367 X 15 ^ ^^" "^'^ ""■ IV. How many cubic centimeters of nitrogen gas, at stand- ard, can be obtained from a liter of ammonia gas, at 15° C. and 780 mm.? Ans. 2NH3 = N2 + 3H2, 2 vol. 1 vol. 3 vol. V. A gaseous mixture contains nitrous oxide (N2O), nitric oxide (NO) and nitrogen. Calculate the composition of the mixture from the following : Volume of mixture taken 51 .00 cc. Volimie after adding hydrogen 108 . 30 cc. ' Where considerable accuracy is desired, the particular decimal coefficient of expansion of the gas under consideration should be used. See Chem. Ann., p. 73. Unless otherwise stated, the factor used in obtaining the answers to problems will be j^j. 114 CHEMICAL CALCULATIONS Volume after explosion 51.00 cc. Volume after adding oxygen 76.50 cc. Volmne after explosion 57.60 cc. Ans. Let x = cc. N2O, y = cc. NO, 3 = cc. N2. H2 added = 108.3 - 51.00 = 57.30 cc. Contraction = 76.50 - 57.60 =? 18.90 cc. f X 18.90 = 12.60 cc, H2 added in excess. 57.30 — 12.60 = 44.70 cc. is the required amount of Hj nec- essary for the first explosion. The reactions are: N20-l-H2 = N2 + H20, X + X = X + 2 NO + H2 = N2 + H2O, 2y + y = y + 0. Then the contraction due to the H2 explosion is (1) x + y = 44.70 cc. After the first explosion, if no H2 had been added in excess, the volume would be z cc. N2 originally present, x cc. N2 from the N2O, and | cc. of N2 from the NO; this is 51.00 - 12.60 = 38.40 cc. Hence (2) x + l + z = 38.40 cc. Then (3) a; + 2/ + g = 51.00 cc. (2) 2x + y +2z = 76.80 cc. (2) 2x + y + 2z = 76.80 cc. (1) x + y = 44.70 cc. (4) x + z = 25.80 cc. X +2z = 32.10 cc. (4) X + z = 25.80 cc. 2 = 6.30 cc. Substituting in (4) gives x = 19.50 cc. Substituting these values in (3) gives y = 25.20 cc. VI. (a) A liter of sulphur dioxide at standard weighs 2.9266 g. Calculate its molecular weight. (6) The molecular weight of GAS CALCULATIONS 115 acetylene is 26.016; what is the weight of 250 co. of the gas at 18° C. and 757 mm.? (c) If the specific gravity of hydrogen selenide referred to air is 2.806, what is its weight per liter? (d) Its molecular weight? Ans. (o) »7m1^L = 65.5; also, 2.9266 X 22.393 = 65.5. 0.044000 (6) 26.016 X 0.044656 = 1.1617 g. per L. at standard, ~ X 1.1617 = 1.0899 g. per L. at 18° C. and 760 mm. ^ X 1.0899 = 1.0856 g. per L. at 18° C. and 757 mm. ^ X 1.0856 = 0.2714 g. in 250 cc. at 18° C. and 757 mm. or in one expression, 26.016 X 0.044656 X |f X || X ^ = 0.2714 g. (c) 2.806 X 1.2926 = 3.627 g. per L. (d) 2.806 X 28.943 = 81.21. Vn. (a) If 30.82 cc. of oxygen (density to air, 1.1055) effuses through a small orifice in 55 seconds, what volume of hydrogen (density to air, 0.06965) will effuse in the same time under the same conditions? (6) What volume of sulphur dioxide will effuse through a small orifice in the same time as 83 cc. of ammonia? (c) 150 cc. of air effuses in the same time as €3.82 cc. of bromine. What is the molecular weight of the bromine? Ans. (o) Substituting in the formula, Fi = V"^^' ..=v/- (ja82)!jm055^^22.79cc. 0.06965 116 CHEMICAL CALCULATIONS (6) The densities of sulphur dioxide and ammonia are in the ratio of their molecular weights; then, as before, y^^Jmmm^ 42.80 CO. ' 64.06 (c) Substituting in the proper formula: A = ^Ig^g^y = 5.525, density to air, 5.525 X 28.943 = 159.89. Vm. (a) What volume of oxygen at 18° C. and 754 mm. is liberated by 1.763 g. of potassium chlorate when completely decomposed? (6) How much sulphuric acid must be taken to obtain 5.5 cubic feet of hydrogen at 17° C. and 762 mm. by acting on a metal? Ans. (o) 2KCIO3 = 2KC1 + 3O2, 2 (122.56) g. 3 (22.4) L. 1763 X ii x!4-?X 3 (22.4)= 0.5193 L. 2 (122.56) 273 754 (6) H2SO4 + M" =M"S04 + H2 98.09 oz. 22.4 cu. ft. 2II ^ 1 X 715 X '^-'^ = 2'-^^ ''■ = '-'^ ''''■ IX. (a) A mass of alumimun (density, 2.583) weighed in air at 18° C. and 742 mm. showed an apparent weight of 149.2350 g., brass weights (density, 8.4) being used. What is its weight in vacuo? (b) A mass of platinum (density, 21.48) weighed in slji at 15° C. and 765 mm., against brass weights, showed an apparent weight of 89.4130 g. Find its weight in vacuo. Ans. , , 149.2350 ^ 273 ^ 742 ^ „ „„,„„„« («^ -2:583- >< 291 X 760 ><0-°°^'^2^ = 0.0684 g. lost by aluminum. GAS CALCULATIONS 117 = 0.0210 g. lost by weights. 0.0684 - 0.0210 = 0.0474 g. difference in weights of air displaced. 149.2350 + 0.0740 = 149.2824 g., weight in vacuo. (6) Using the formula W=W' + W'd (k-h), d = III X 11^ X 0.0012926 = 0.0012333, W = 89.4130 + 89.4130 X 0.0012333 (-^ -], \21.48 8.4/' = 89.4130 + 89.4130 X 0.0012333 (0.04656 - 0.11905), = 89.4130 + 89.4130 X 0.0012333 (- 0.07249), = 89.4130 - 0.0080 = 89.4050 g. in vacuo. X. A flask which is to be marked to contain one liter at 20° C. (United States Standard) is counterpoised on a balance with a similar flask, finishing the coimterpoising with sand foil, etc. How many grams of water must be weighed into it to give the position on the neck which should be marked? The conditions are: ' Temperature of air in the balance 22° C. Temperature of the water to be used 22° C. Temperature of the flask 22° C. Barometric pressure 765 mm. Saturation of air in the balance 50% Vapor pressure of water at 22° C 19 . 66 mm. Weight of a liter of dry air at 0° C. and 760 mm. 1 . 2926 g. Coefficient of expansion of air . 00367 Density of weights 8.4 Density of water at 20° C 0.997797 Cubical coefficient of expansion of the glass. . . 0.000025 Ans. Weight of one cubic centimeter of air under these conditions 1 765 - 0.50 XIX 19.66 1 + (22 X 0.00367) 760 X 0.0012926 = 0.0011981 g. 118 CHEMICAL CALCULATIONS The weight of one liter of water in air under these conditions is: W = 1000 X 0.997797 + 1000 X 0.997797 X 0.0011981/-^ -^;g^). \.8.4 = 997.797 + 997.797 X 0.0011981 X (- 0.883159) = 997.797 - 1.056, = 996.741 g. Correcting for the expansion of the glass gives 1000 X 0.000025 (20 - 22) = - 0-05; and the amount to weigh in is 996.74 - 0.05 = 996.69 g. 137. One liter of a gas is under a pressure of 780 mm. TVhat will be its volume at standard pressure (760 mm.), the tejnpera- ture remaiaing constant? , Ans. J026.3 cc. 138. A gas of 300 cc. is under standard pressure^ What will be its volume at 784 mm., the temperature reniaimng constant? - Ans7 290.8 cc. 139. Five cubic feet of a gas are under a pressure of 27.3" of mercury. What is its volume at 29.9", the temperature remaining constant? Ans. 4.565 cu. ft. 140. A gas occupies a volume of 450 cc. under a pressure of 780 mm. The temperature remaining constant, what pressure must be applied to reduce the volume to 400 cc? Ans. 877.5 mm. 141. A gas occupying a volume of one liter, under standard pressure, is expanded to 1200 cc. The temperature remaining constant, by how many millimeters must the pressure have been diminished? Ans. 126.7 mm. 142. A gas of 200 cc. is at 15.7° C. Find its volume at 0° C. (at standard), the pressure remaining constant? Ans. 189.12 cc. 143. One liter of a gas is measured at — 15° C. What is its volume at 15° C, the pressure remaining constant? Ans. 1116.3 cc. GAS CALCULATIONS 119 144. A gas measured 150 cc, at 17.5° C, and on account of a change in temperature, the pressure remaining constant, the volume decreased to 125 cc. What is the new temperature? Am. - 30.9° C. 146. The pressure on a confined gas, at 15° C, was 792 nam. If the pressure later registered 820 mm., what is the tempera- ture then, the volume remaining unchanged? Ans. 25.17° C. 146. A liter of gas at standard conditions has its tempera- ture raised to 15° C. What must be the pressure on the gas, if the volume of gas is unaltered? Ans. 801.7 mm. 147. 183 cc. of a gas are at standard; the pressure is then raised to 792 mm. What is the temperature, the volume remaining constant? Ans. 11.5° C. 148. 250 cc. of a gas are at a temperature of 15° C. What is the volume of the gas at 0° C, the pressure remaining con- stant? Ans. 237 cc. 149. The pressiu-e on a certain volume of hydrogen is 730 mm. at the temperature of melting ice. The volume remaining con- stant, what is the temperature at a pressinre of 750 mm.? Ans. 7.48° C. 150. Given 250 cc. of a gas under a pressure of 765 mm. and at a temperature of 15° C, what is the volume corrected to standard conditions (0° C. and 760 mm.)? Ans. 238.5 cc. 161. A gas measures 300 cc. at standard conditions. What is its volume at 755 mm. and 17.5° C? Ans. 321.3 cc. 152. Under standard conditions of temperature and pressure, a gas measiu'es one liter. What is its volume at 17.7° C. and 748 mm.? Ans. 1082 cc. 163. A gas of 275 cc. under standard conditions changes its volume to 300 cc, when it is under a pressure of 754 mm. What is its temperature? Ans. 22.47° C. 164. 50 cc. of a gas at 780 mm. and at 10° C. changes its volume to 48 cc. under a pressure of 792 mm. What is the temperature at this pressure and volume? Ans. 2.87° C. 120 CHEMICAL CALCULATIONS 155. A gas is at a pressure of 748 mm. and a temperature of 12° C, when its volume is 200 cc. What must be the pressure of the gas if its volume is 178 cc, at a temperature of 0° C? Ans. 805.1 mm. 156. A volume of gas is confined at 0° C. and 760 mm. pres- sure, (a) What is this pressure in inches of mercury and (6) in pounds per square inch? (Sp. gr. of Hg = 13.6; 1" = 2.54 cm.; 1 sq. in. = 6.4516 sq. cm.) Ans. (a) 29.92". (6) 14.701 lbs. per sq. in. 157. A barometer with a glass scale reads 752.6 mm. at 15° C. What is the barometer reading corrected to 0° C? Ans. 750.68 mm. 158. A barometer with a brass scale shows a pressure 768.5 mm. at 18° C. What is the barometer reading corrected to 0° C? Ans. 766.28 mm. 159. The reading of a barometer, with a glass scale, at —5° C. is 753.2 mm. What is the reading corrected to 0° C? Ans. 753.84 mm. 160. A brass scale barometer reads 769.2 mm. at 15° F. What is the reading corrected to 0° C? Ans. 770.36 mm. 161. What must be the reading on a barometer with a glass scale, at 15° C, that the pressure at 0° C. may indicate 760 mm.? Ans. 761.9 mm. 162. If sufficient water is placed in a dry gas, at 15° C. and 753.8 mm. to thoroughly saturate it, what would be the pressure after saturation, the temperature remaining constant? (Ten. aq. vap. at 15° C. = 12.73 mm.) Ans. 766.53 mm. 163. If the atmosphere is saturated with water vapor at 14° C. and 758 mm., what percentage, by volume, of water vapor does it contain? (Ten. aq. vap. at 14° C. = 11.94 mm.) Ans. 1.58%. 164. A gas, measured moist, has a volume of one liter at 17.5° C, under a pressure of 758.9 mm. What is its volume, dry, under standard conditions? (Ten. aq. vap. at 17.5° C. = 14.91 mm.) Ans. 919.9 cc. GAS CALCULATIONS 121 166. 300 cc. of a gas are measured moist over water at 15° C. and under a pressure of 765 mm. (a) What would be its vol- ume, dry, at this temperature? (6) What is its volume, dry, under standard conditions? (Ten. aq. vap. at 15° C. = 15.38 mm.) Ans. (a) 295.01 cc. (b) 281.48 cc. 166. A certain reaction produces 22.4 L. of a gas, measured at standard conditions, (a) What volume would it occupy, moist, at 18° C. and at standard pressure? (6) At 18° C. and 770 mm.? (Ten. aq. vap. at 18° C. = 15.38 mm.) Ans. (a) 24.36 L. (6) 24.04 L. 167. 500 cc. of nitrogen are measured over water at 17° C, the barometer reading 750 mm. If the water stood 180 mm. in the tube, what is the volume of the nitrogen, dry, at standard? (Ten. aq. vap. at 17° C. = 14.91 mm.; sp. gr. Hg = 13.6.) Ans. 447.07 cc. 168. 180.5 cc. of air, saturated with moisture, at 18° C, are measured over mercury, the barometer reading 620.3 mm. The mercury stood 52 mm. in the tube. Find the volume of the air, dry, at standard conditions. (Ten. aq. vap. at 18° C. = 15.38 mm.) Ans. 123.7 cc. 169. 203 cc. of chlorine gas, at standard conditions, are nec- essary to decompose a certain amoimt of hydrobromic acid gas, also at standard conditions. What is the volume of hydro- bromic acid gas? Ans. 406 cc. 170. If to a mixture of 100 cc. of nitrogen and of 200 cc. of oxygen, 500 cc. of hydrogen are added, and the mixture ex- ploded : (o) What is the resultant volume if the water is allowed to condense? (6) What is the resultant volume if the water stays in the gaseous state? Ans. (a) 200 cc. (6) 600 cc. 171. Consider the following reactions: (1) 2 HBr -I- CI2 = 2 HCl -|- Br^, (2) 4 HBr + O2 = 2 H2O -I- 2 Brj. 122 CHEMICAL CALCULATIONS (a) If the volume of hydrobromic acid of (1) is one liter, how many liters of chlorine at the same conditions are necessary to liberate aU the bromine? (6) In equation (2) how many vol- lunes of oxygen of the same temperature and pressure are called for to liberate all the bromine? Ans. (a) 0.5 L. (6) 0.25 L. 172. (a) How many cubic centimeters of nitrogen are con- tained in a liter of ammonia gas? (6) How many cubic centi- meters of hydrogen? Ans. (a) 500 cc. (b) 1500 cc. 173. One liter of hydrogen and 400 cc. of oxygen at 760 mm. and 15° C. are exploded in a confined space. What is the vol- ume of the resultant gas at 170° C. and 760 mm. pressure? Ans. 1538.2 cc. 174. What is the ratio of the volumes of hydrogen to chlo- rine involved when hydrochloric acid is formed? (6) The ratio of the volume of chlorine to the volume of hydrochloric acid produced? (c) What is the ratio of the sum of the initial volumes (factors) to the resultant volume (products)? Ans. (a) 1/1. (6) 1/2. (c) 1/1. 175. One volume of ammonia gas is decomposed into its elements, (a) What is the volume of the resultant gases? (6) What is the ratio of the volume of the ammonia gas decom- posed to the volume of the nitrogen produced? (c) What is the ratio of the volume of the nitrogen to hydrogen produced? Ans. (a) 2V. (6) 2/1. (c) 1/3. 176. 1200 cc. of chlorine and 1500 cc. of hydrogen at the same temperature and pressure are combined to form hydro- chloric acid gas. (a) What volume of hydrochloric acid gas is formed? (6) Which gas is in excess and by how vs^y cubic GAS CALCULATIONS 123 centimeters? (c) What is the total volume of the gases after explosion? Am. (a) 2400 cc. (&) 300 cc. Hj. (c) 2700 cc. 177. Assuming air to be \ oxygen: (a) How many liters 'of air are required to burn one Uter of methane at the same tem- perature and pressure as the air? (6) If the pressure on the methane is two atmospheres and the air is one atmosphere, what volume of air is called for? (c) If the pressure on both is the same but the temperature of the air is 20° C. and the methane 0° C. what volume of air is called for? Ans. (a) 10 L. (6) 20 L. (c) 10.73 L. 178. 100 cc. of nitrous oxide (N2O) at standard are decomposed into its elements and yields 150 cc. of a mixture of nitrogen and oxygen. 125 cc. of hydrogen at standard are now introduced and the mixture exploded, and the resultant volume found to be 125 cc. at standard. Show that the symbol must be N2O. 179. 300 cc. of hydrogen and 100 cc. of nitrogen both at standard are combined to form ammonia gas in a closed system. If the volume of the resultant gas is 150 cc. and the pressure is unchanged, what is the temperature? Ans. —68.2° C. 180. 400 cc. of ammonia gas at standard are decomposed and the temperature reduced to standard. If the volume of the system is imchanged, what is the pressure on the resultant gas? Ans. 1520 mm. 181. 161.53 cc. of hydrogen at 18° C. and 754 mm. are united with 160.35 cc. of chlorine at 15° C. and 750 mm. What is the volume of the resultant gas at standard? Ans. 300 cc. 182. A liter of hydrogen at 39° C. is exploded with a liter of oxygen at 332.5 mm. The temperature is brought to 0°C. What volume of which gas is in excess? Ans. No excess. 183. 50 cc. of carbon monoxide are combined with oxygen in excess. After combination 60 cc. of gas remains. What was the volume of oxygen added? Ans. 35 cc. 124 CHEMICAL CALCULATIONS 184. 100 cc. of a mixture containing carbon monoxide and hydrogen are ejcploded with an excess of oxygen and 60 cc. of gas remain. Of this 60 cc, 40 cc. are absorbed by potassium hydroxide (the CO2). What was the volimie of the carbon monoxide, the hydrogen and the oxygen added? Ahs. 70 cc. O2, 40 cc. CO, 60 cc. H2. 185. If 50 cc. of ammonia gas at 15° C. and 766 mm. are de- composed into its elements, and 50 cc. of oxygen at 12° C. and 753 mm. are introduced, and the mixtiu'e exploded, what vol- umes of what gases remain at a temperature of 10° C. and 749 mm., disregarding the volume of water vapor formed? Ans. 25.12 cc. N2, 12.23 cc. O2. 186. How many hters of oxygen at 10° C. and 765 mm. might be obtained by the decomposition of a liter of nitrogen pent- oxide (NzOb) at 35° C. and 748 mm.? Ans. 2.246 L. 187. Given a mixture of hydrogen, carbon monoxide and nitrogen; calculate its percentage composition from the follow- ing: Volume of mixture taken 25.00 cc. Volume of oxygen added 10.00 cc. Volume after explosion 22 . 50 cc. Volume after CO2 absorption 12 . 50 cc. Ans. 20% H2, 40% CO, 40% N2. 188. Calculate the percentages of hydrogen, carbon monox- ide and propane from the following: Volume of sample taken 25 . 00 cc. Volume after adding oxygen 100 . 00 cc. Volimie after explosion 55. 50 co. Volume after CO2 absorption 17 . 50 cc. Ans. 28% H2, 32% CO, 40% CjHs. GAS CALCULATIONS 125 189. Calculate the .composition of a gaseous mixture of carbon monoxide and acetylene from the following data : Volume of mixture taken 20.00 cc. Volume of oxygen added 60 . 00 cc. Voliune after explosion 52 . 00 cc. Volume of oxygen left after KOH treatment. . . 24.00 cc. Ans. 8.00 cc. C2H2, 12.00 cc. CO. 190. Calculate the composition of a mixture of carbon mon- oxide and methane from the following : Volume of mixture taken 20 . 00 cc. Volume of oxygen added 40 . 00 cc. Volume after explosion 47 . 00 cc. Ans. 2.00 cc. CH4, 18.00 cc. CO. 191. A gaseous mixture of oxygen, nitrous oxide (N2O) and nitrogen is to be analyzed from the following: Volume of mixture taken 20.00 cc. Volume of hydrogen added 30.00 cc. Volume after explosion 18.00 cc. Volume of oxygen added 10.00 cc. Volume after explosion 19.00 cc. Ans. 8.00 cc. O2, 8.00 cc. N2O, 4.00 cc. N2. 192. By means of the formulas developed on page 90, cal- culate the percentage composition of a mixture of hydrogen, carbon monoxide and methane, from the following: Volume of mixture taken 50.00 cc. Volume of oxygen added 75.00 cc. Volume after explosion 60 . 00 cc. Volume after absorption of CO2 20. 00 cc. Ans. 20% H2, 40% CO, 40% CH4. 126 CHEMICAL CALCULATIONS 193. What is the weight of 300 cc. of nitric oxide (NO) under standard conditions? Am. 0.40204 g. 194. What is the weight at standard of a liter of methane, the molecular weight of methane being 16.032? Am. 0.7159 g. 196. A liter of a gas at standard conditions weighed 3.627 g. What is its molecular weight? Ans. 81.22. 196. 963 cc. of a gas at 754.3 mm. and at 17° C. weigh 1.368 g. What is its molecular weight? Ans. 34.05. 197. Nitrogen has a molecular weight of 28.02. (a) What is the weight of a liter at standard conditions? (6) What is its specific gravity referred to -^ oxygen? (c) To hydrogen? Am. (a) 1.251 g. (b) 28.02. (c) 13.90. 198. Bromine gas has a specific gravity, referred to air, of 5.524. What is its molecular weight? Ans. 159.9. 199. The density of hydrochloric acid referred to air is 1.2595. (a) What is its density referred to hydrogen? (6) To oxygen? Am. (a) 18.115. (6) 1.1393. 200. The density of nitrogen at a certain pressure is 0.9802, referred to air at standard, (a) What is the pressure upon it? (6) What is its density at 15° C. and, 760 mm. compared to air at standard? (c) If the density compared to air at standard is 0.9502, what is the temperature, (c?) What is the density of nitrogen at 755 mm. and 150° C. compared to air at standard? Ans. (a) 769.5 mm. (6) 0.9177. (c) 5.1° C. id) 0.6207. 201. A liter of ammonia gas under standard conditions weighed 0.7630 g. What is its specific gravity to hydrogen? Ans. 8.4885. GAS CALCULATIONS 127 202. A certain volume of chlorine weighs 0.6333 g.; it the same volume of silicon tetrafluoride weighs 0.9326 g., what is the molecular weight of the latter, taking the molecular weight of chlorine as 70.92? Am. 104.44. 203. What is the volume at standard conditions of 1.008 g. of hydrogen? (&) Of 2.016 g.? Am. (a) 11.2 L. (6) 22.4 L. 204. The molecular weight of carbon monoxide is 28.00. What is its specific gravity to air? Ans. 0.9674. 205. What is the weight of 275 cc. of carbon oxysulphide (COS) at 15.5° C. and 768 mm.? Am. 0.7054 g. 206. (a) What is the weight of 300 cc. of chlorine at stand- ard conditions if its molecular weight is 70.92? (6) What is its specific gravity to 51 oxygen? (c) To hydrogen? (d) To air? (e) To water? Am. (o) 0.9501 g. (6) 70.92. (c) 35.24. (d) 2.450. (e) 0.003167. 207. A bulb known to hold exactly 500 cc, when filled with hydrogen at standard weighed (with all corrections) 30.04494 g. ; when filled with ethane under the same conditions it weighed 30.67105 g. What is the molecular weight of ethane? Am. 30.05. 208. If the molecular weight of nitrous oxide is 44.02, at what pressure would 4 grams of this gas at 15° C. occupy a volume of one Uter? Am. 1631 mm. 209. Calculate the vapor density of carbon tetrachloride, com- pared with hydrogen, from the following readings obtained hy the Hofmann method: Weight of substance taken . 2832 g. Volume of vapor obtained 91 .50 cc. Temperature 99.50° C. Barometer 746.9 mm. Height of mercury in tube 283 . 4 mm. Weight of a liter of hydrogen 0.089873 g. Am. 77.05. 128 CHEMICAL CALCULATIONS 210. Calculate the molecular weight of an organic substance from the following readings obtained by Hof mann's method : Weight of substance taken . 0835 g. Volume of vapor obtained 81 . 40 cc. Temperature 100° C. Barometer 752 . 5 mm. Height of mercury in tube 413.5 mm. Ans. 70.38. 211. What is the vapor density, compared with air as unity, of iodine, yielding the following data when determined accord- ing to the method of Victor Meyer? Weight of iodine taken . 1735 g. Volume of air displaced 17 . 40 cc. Temperature 16° C. Barometer 722 . 3 mm. Tension of aqueous vapor at 16° C 13 . 57 mm. Ans. 8.757. 212. Victor Meyer's method yields the following results for carbon bisulphide. From them calculate the molecular weight. Weight of carbon bisulphide taken 0.0825 g. Volume of air displaced 27 . 34 cc. Temperature 16.5° C. Barometer 718 mm. Tension of aqueous vapor at 16.5° C 14.00 mm. Ans. 77.38. 213. If air has a density of 0.0012926 referred to water: (a) What is its density at 754 mm.? (6) At 754 mm. and 38° C? (c) At 770 mm. and -40° C? Ans. (a) 0.0012824. (6) 0.0011257. (c) 0.0015344. 214. If the barometer at a certain place varied between 740 mm. and 785 mm. and the temperature varies between 40° C. and —40° C. : (a) What is the maximum variation in the density of the atmosphere if these variations work in conjunc- tion? (6) What is the difference in the weight of a liter of air at GAS CALCULATIONS 129 the greatest apd the least density? (Density of the atmosphere = 0.0012926.) Ans. (a) 0.0004665. (6) 0.4665 g. 215. A certain volume of silicon fluoride weighed 3.4962 g. The same volume of air under the same conditions weighed 0.9694 g. What is the molecular weight of the sUicon fluoride? Ans. 104.4. 216. What is the specific gravity to air of hydrogen sulphide, the molecular weight of which is 34.086? Ans. 1.178. 217. The density of a gas is 0.8988 at 15° C. and 749 mm. What is its density at standard? Ans. 0.9621. 218. (a) At what temperature wiU one gram of hydrogen occupy a liter at 760 mm.? (6) At 750 mm.? Ans. (a) - 248.42° C. (6) - 248.79° C. 219. 283.5 cc. of a gas at 18° C. and 757 mm. weigh 0.51264 g. What is the weight of a Uter at standard? Ans. 1.9351 g. 220. What is the weight at 20° C. and 767 mm. of 263 cc. of propylene (CaHe)? Ans. 0.4644 g. 221. If 283.5 cc. of hydrogen selenide (HzSe) weigh 0.9497 g. at 17.4° C. and 747 mm., what is its molecular weight? Ans. 81.19. 222. If 54.62 cc. of oxygen (density to hydrogen, 15.88) pass through a minute opening in the same time as 58.33 cc. of nitrogen: (a) What is the density of nitrogen to hydrogen? (b) Its molecular weight? Ans. (a) 13.93. (6) 28.08. 223. 60.00 cc. of air effuse in the same time as 53.46 cc. of hydrochloric acid. What is the molecular weight of the hydro- chloric acid? Ans. 36.45. 224. 76.42 cc. of sulphur dioxide pass through a minute orifice in the same time as 108.1 cc. of oxygen. What is the molecular weight of the sulphxu- dioxide? Ans. 64.03. 225. It required 75.2 seconds for a certain volume of hydro- gen to pass through a minute orifice and 414.6 seconds for the 130 CHEMICAL CALCULATIONS same quantity of carbon oxysulphide. (a) What is the den- sity of the carbon oxysulphide to hydrogen? (b) Its molecular weight? Ans. (o) 30.4. (6) 61.28. 226. It required 105.6 seconds for a certain volume of air to effuse and 39.3 seconds for the same quantity of heUimi. (a) What is the density of helium to air? (6) Its molecular weight? Ans. (a) 0.1385. (6) 4.009. 227. It required 87.2 seconds for a certain volume of oxygen to pass through a minute orifice and 127.1 seconds for the same volimie of boron fluoride. What is the molecular weight of the boron fluoride? Ans. 67.98. 228. How many grams of sulphuric acid must act on a metal to produce 1200 cc. of hydrogen at 0° C. and 760 mm., all of the hydrogen of the acid being liberated? Ans. 5.256 g. 229. If 15 g. of iron react with sulphuric acid thus: Fe + H2SO4 = FeS04 + Ha, how many liters of hydrogen are liberated at standard? Ans. 6.017 L. 230. In the reaction: CaCOs + 2 HCl = CaCl2 + CO2 + H2O. (a) How many grams of calcium carbonate are required to pro- duce 1700 cc. of carbon dioxide at 780 mm. and 17° C? (6) If 860 cc. of carbon dioxide were liberated at 754 mm. and 20° C, how many grams of calcium carbonate were acted upon? Ans. (a) 7.3376 g. (6) 3.5515 g. 231. Lackawanna coal corresponds to the composition: Per cent Volatile combustible matter 5.00 Carbon 84.00 ' Ash 11.00 Considering the volatile combustible matter to be methane, what volume of air at standard, ^ of which is oxygen, is neces- sary to burn completely a kilogram? Ans. 8538.7 L. GAS CALCULATIONS 131 232. Consider the coal of problem just preceding. If the heat of combustion of methane burned to carbon dioxide and water is 13,344 Calories and of carbon burned to carbon dioxide is 8080 Calories, what is the heat of combustion of this coal? Am. 7454.4 Cal. 233. From the equations: NaCl + H2SO4 = NaHSOi + HCl 4 HCl + MnOj = MnCU + 2 H2O + CI2. (o) What volume of chlorine measured under standard con- ditions will be produced by 10.000 g. of sodium chloride? (6) What volume of chlorine measured moist will be produced? (Ten. aq. vap. at 0° C. = 4.6 mm.) Ans. (o) 0.9579 L. (6) 0.9637 L. 234. Consider the same reactions as in the previous problem, (a) One liter of chlorine measured at 764 mm. (the barometer not being corrected for temperature: brass scale) and 15° C, moist, indicates how much sodium chloride when the correc- tions are made? (6) How much sodium chloride is indicated if the readings are assumed to be correct as stated? (c) How many grams of sodium chloride are indicated if corrections were made for the pressure of water vapor and the tempera- ture only? (Pressure of water vapor at 15° C. = 12.73 mm. : apparent expansion of Hg = 0.00017.) Ans. (a) 9.757 g. (6) 9.948 g. (c) 9.782 g. 235. If ten grams of water vapor occupy a volume of one liter, the pressure being 20 atmospheres, what is the temperature? Ans. 166.1° C. 236. Water gas is produced according to the equation: C + H2O = CO + H2. (o) What volimie of carbon monoxide and hydrogen will be produced from a kilogram of water? (6) What volume of carbon monoxide and hydrogen will be produced from a kilo- gram of carbon? (c) What volume of air is necessary for the 132 CHEMICAL CALCULATIONS complete combustion of the gas produced in (a)? (Air = J oxygen.) Am. (a) 2486.7 L. (6) 3732.2 L. (c) 6216.7 L. 237. A balloon has a capacity of 100,000 liters. How many kilograms of (a) sulphuric acid and (6) iron are necessary to fill the balloon with hydrogen at 20° C. and 760 mm.? Ans. (a) 408 kg. H2SO4. (6) 232.3 kg. Fe. 238. How many liters of air at standard are required for the combustion of one kilogram of a coal of the following compo- sition? Per cent Carbon 65.72 Hydrogen 9.03 Nitrogen 0. 72 Oxygen 4.78 Ash 19.75 Ans. 8475 L. 239. Given a natural gas of the following composition by weight: Per cent Hydrogen 1 . 42 Methane 94. 16 Ethylene 0.30 Carbon dioxide . 27 Carbon monoxide . 65 Oxygen 0.30 Nitrogen 2. 80 Hydrogen sulphide 0. 18 How much air is necessary for the combustion of one kilogram? Ans. 13,598 L. 240. A sample of calcium carbonate when treated with an acid evolved 200 cc. of carbon dioxide at 18° C. and 754 mm. How much calcium carbonate does this amount indicate? Ans. 0.8316 g. GAS CALCULATIONS 133 241. From the reaction, 2Mg+02 = 2MgO, how many grams of magnesiimi can be bm-ned in a globe con- taining 3489 cc. of oxygen at 19.5° C. and 768 mm.? Am. 7.145 g. 242. (o) What volmne of nitrogen at 10° C. and 624 mm. is evolved from 3 liters of nitric oxide at 18° C. and 790 mm. when passed over red hot copper? (6) How many grams of copper oxide are formed? The reaction is 2NO-|-2Cu = 2CuO + N2. Ans. (a) 1846.8 cc. (6) 10.392 g. 243. (a) What volmne of hydrogen at 17° C. and 748 mm. is required to unite with one gram of chlorine to form hydro- chloric acid? (6) What voliune of hydrogen under the same conditions but moist? (Ten. aq. vap. at 17° C. = 14.45 mm.) Ans. (a) 0.3409 L. (6) 0.3476 L. 244. How many cubic feet of hydrogen at 62° F. and 29.4" will be yielded by the action of sulphuric acid on 10 lbs. of iron? Ans. 69.475 cu. ft. 245. 5 cubic feet of carbon dioxide measured at 70° F. and 30.7" were evolved by the action of hydrochloric acid on cal- cium carbonate. How much calcium carbonate was decom- posed? Ans. 1.326 lbs. 246. From the equation, NH4N03 = N20 4-2HaO, (o) how many cubic feet of N2O at standard will be produced by 15 lbs. of NH4NO3? (6) What is the resultant volume from the reaction at 30" and 300° C? Ans. (a) 67.16 cu. ft. (6) 422.87 cu. ft. 247. A gas tank contains 200,000 cubic feet of dry coal gas at 30" and 70° F. If 40% by volume of this is methane, how majiy pounds of methane are contained in the tank? Ans. 3322 lbs. 134 CHEMICAL CALCULATIONS 248. How many cubic feet of oxygen and hydrogen together at standard result from the decomposition of I lb. of water? Ans. 7.46 cu. ft. 249. Six cubic feet of sulphur dioxide are required, (a) How many ounces of sulphuric acid and (6) copper are required? The reaction is Cu + 2 H2SO4 = CuSOi + 2 H2O + SO2. Ans. (a) 52.54 oz. (6) 17.03 oz. 250. How many cubic feet of air at standard are required to bum a ton of 2000 lbs. of coal gas of the following composition by weight? Per cent Hydrogen 45 . 85 Methane 39.26 Ethylene 5. 17 Carbon dioxide 0. 82 Carbon monoxide 4. 78 Oxygen . 41 Nitrogen 3.71 Ans. 605,515 cu. ft. 261. If the reaction for the decomposition of oxalic acid- proceeded as follows: 2 (C2H2O4 • 2 H2O) = HCOOH + 5 H2O + 2 CO2 + CO, and the products were brought to standard conditions, (a) what is the volume of the gases (neglecting the volume of water vapor present) if 2 oz. of oxalic acid are decomposed? ^ (6) If the temperature is now raised to 140° C, what is the volume of the gases formed, pressure remaining at standard? (c) If the formic acid decomposes at 160° C. as follows: HCOOH = CO2 + H2, what is the total volume of the gases formed at 200° C? Ans. (a) 0.5332 cu. ft. (6) 2.420 cu. ft. (c) 3.072 cu. ft. ' Water freezes at 0° C. and boils at 100° C; formic acid melts at 8.6° C. and boils at 100.8° C. GAS CALCULATIONS 135 262. A flask weighs 139.8460 g. in air with brass weights (density, 8.4). The temperature is 18° C, the barometer 757 mm. The density of the flask is 3.45. What is the weight in vacuot (Use the decimal coefficient of expansion of a gas. 1 CO. air at standard weighs 0.0012926 g.) Ans. 139.8748 g. 263. A flask and water are counterpoised by 1008.2600 g., the weights being of brass (density, 8.4). The temperature is 18° C, the barometer 634 mm. Regarding the flask and water as having a density of 1, what is the weight in vacuo? Ans. 1009.1586 g. 264. It is desired to weigh out an amount of platinum (sp. gr., 21.5) in air at 17° C. and 764 mm. so that its true weight in vacuo should be 200.0000 g. Brass weights are used (den- sity, 8.4). What weight should be put on the other pan? (Use the decimal coefficient of expansion.) Ans. 200.01774 g. 266. It is desired to weigh' out 576.4213 g. to be true in vacuo. Brass weights (density, 8.4) are xised, the temperatin-e is 19° C, the barometer 764 mm. The density of the body to be weighed is 2.637; what weight must be put on the other side? (Use the decimal coefficient of expansion of a gas.) Ans. 576.2391 g. 266. It is required to get a weight of water that will be true in vacuo. Brass weights are used (density, 8.4), the density of the water is 0.999050, the weight required is 100.2037 g., the temperature of the air is 18° C. and the barometer is 763 mm. What weight must be put on the other pan? (Use the decimal coefficient of expansion of a gas.) Ans. 100.0961 g. 267. A flask of glass (density, 3.4) > weighing 203.8050 g. in air is filled with water at 17° C. when it weighs 1397.4370 g. in air, the temperature being 17° C, the barometer 756 mm.; brass weights are used (density, 8.4). What is the weight in vacuo of the combination? (Use the decimal coefficient of expansion of a gas. Density of water at 17° C. = 0.9988.) Ans. 1398.75 g. 1 Take into account the weight of air displaced by the glass of the flask. 136 CHEMICAL CALCULATIONS 258. (a) Calculate Wdi—, — j:]^ for a cubic centimeter of water weighed ia air under the following conditions: Temperatyre of the air 15° C. Temperature of the water 15° C. Barometric pressure 770 mm. Saturation of aqueous vapor in the air ..... . 50% Vapor pressure of water vapor at 15° C 12 . 728 mm. Coefficient of expansion of air . 00367 Weight of a liter of dry air at 0° C. and 760 mm. 1 . 2926 g. Density of water at 15° C 0.999126 Density of weights 8.4 Coefficient of cubical expansion of glass 0.000025 (jb) Using this last calculated quantity, calculate the appar- ent weight, in air, of 250 cc. of water under these conditions, (c) If a flask marked 250 cc. at 20° C. (United States Stand- ard) gives an apparent weight of water in air under the condi- tions enumerated above of 249.508 g., what is the error of the flask? Ans. (a) 0.0010903. (6) 249.508 g. (c) 0.031 cc. 269. Calculate the weight of a liter of water in air under the following conditions : Temperatin-e of the air 20° C. Temperature of the water 20° C. Barometric pressure 760 mm. Saturation of aqueous vapor in the air 50% Weight of a liter of dry air at 0° C. and 760 mm. 1 . 2926 g. Vapor pressure of water at 20° C 17. 406 mm. Density of water at 20° C 0.998230 Density of weights 8.4 Coefficient of expansion of air 0.00367 Ans. 997.173 g. 260. A flask calibrated to contain 500 cc. at 20° C. (United States Standard) was foimd to hold 499.149 g. of water when fiUed to the mark. The flask was tared with a similar flask, » See p. 104. GAS CALCULATIONS 137 finishing the tare with glass beads, foil, etc. Given the follow- ing conditions, what is the error of the flask? Temperature of water and flask 17° C. Temperature of air 17° C. Barometric pressure 760 mm. Saturation of aqueous vapor 50% Tension of aqueous vapor at 17° C 14.45 mm. Weight of a liter of dry air at 0° C. and 760 mm. 1 . 2926 g. Coefficient of expansion of air 0.00367 Density of water at 17° C 0.998801 Density of weights 8.4 Cubical coefficient of expansion of glass . 000025 Ans. +0.15 cc. 261. A flask is to be calibrated to hold 2000 ce. at 20° C. (United States Standard). It is tared by a similar flask, fin- ishing the tare with glass beads, foil, etc. How many grams of water should be weighed into the flask under the following conditions? Temperature of water and flask 17° C. Temperature of air 17° C. Barometric pressure 752 mm. Saturation of aqueous vapor ia the air 38% Vapor tension of water at 17° C 14 . 45 mm. Coefficient of expansion of air 0.00367 Weight of a liter of air at 0° C. and 760 mm. 1 . 2926 g. Density of water at 17° C 0.998801 Density of weights 8.4 Cubical coefficient of expansion of glass 0.000025 Ans. 1995.64 g. 262. (a) What is the volume of one gram of water at 15° C, weighed in air with brass weights (density, 8.4), the tempera- ture of the air being 15° C, pressure of the atmosphere 760 mm. (6) What is the volume of a kilogram of water under these conditions? (Use the decimal coefficient of expansion of a gas. Density of water at 15° C. = 0.999126: weight of one cubic centimeter of air at standard = 0.0012926 g.) Ans. (a) 1.001954 cc, (6) 1001.954 cc. 138 CHEMICAL CALCULATIONS 263. A flask calibrated to hold one Mohr's liter at 15° C. was counterpoised by another similar flask and sand. Filling up to the mark with water at 15° C. and 760 mm. required 999.8730 g. What is^the error? Ans. —0.127 cc. 264. (a) What is the volume of one gram of water at 60° F., weighed in air with brass weights (density, 8.4), the temperature of the air being 60° F., the barometer standing at 760 mm. (6) What is the weight of a kilogram of water under these con- ditions? (Use the decimal coeflicient of expansion of a gas. Density of water at 60° F. = 0.999050.) Ans. (a) 1.002037 cc. (6) 1002.037 cc. 266. If air is satiu-ated with water vapor at 20° C. and 760 mm.: (a) What per cent of its volume is aqueous vapor? (6) Taking air to have a molecular weight of 28.943, what is the weight of a liter of saturated air at this temperature? (c) What is its density referred to water? (Ten. aq. vap. at 20° C. = 17.41 mm.) Ans. (a) 2.291%. (6) 1.194 g. (c) 0.001194. 266. The temperature of a room is 18° C, the dew point is 15° C. What is the humidity? (Ten. aq. vap. at 18° C. = 15.38 mm.; at 15° C. = 12.73 mm.) Ans. 83%. 267. (a) What is the mass of nitrogen in a hter of nitrogen measured moist at 18° C. and 763 mm.? (6) What is the mass of the water vapor disseminated throughout the hter? (c) What is the mass of the moist nitrogen? (Ten. aq. vap. at 18° C. = 15.38 mm.) Ans. (a) 1.1548 g. (6) 0.1527 g. (c) 1.3075 g. 268. (a) What is the weight of a liter of water vapor satu- rated at 15° C. and 760 mm.? (6) What is its density compared to water at 4° C. (Ten. aq. vap. at 15° C. = 12.73 mm.) Ans. (a) 0.01277 g. (6) 0.00001277 g. ' GAS CALCULATIONS 139 269. (a) What is the mass of a liter of saturated air at 15° C. and 730 mm.? (6) What is the percentage by weight of air and moisture in air under these conditions? (Use the decimal co- eificient of expansion of a gas. Ten. aq. yap. at 15° C. = 12.73 mm.) Ans. (a) 1.1690 g. (6) 98.91% air. 1.09% H2O. 270. The hiunidity at 25° C. is 83%; the barometer stands at 765 mm. K the temperature were to suddenly drop to 15° C. and the barometer to 755 mm., how much moisture would be precipitated per cubic meter? (Ten. aq. vap. at 15° C. = 12.73 mm.; at 25° C. = 23.55 mm.) Ans. 1.669 g. 271. How many grams of water wUl evaporate into a five liter space at 20° C. and 760 mm.? (First calculate the. density of the aqueous vapor at 20° C. to water at 4° C; ten. aq. vap. at 20° C. = 17.41 mm.) " Ans. 0.0860 g. 272. (a) What weight of water vapor is present ia a cubic meter of air saturated at 20° C? (6) If the relative humidity is 60%? (c) If the air has a relative humidity of 80% at 20° C. and should be cooled to 10° C, what weight of water would be precipitated per cubic meter? (Density of aqueous vapor at 20° C. = 0.0000172; at 10° C. = 0.0000093.) Ans. (o) 17.2 g. (6) 10.32 g. (c) 4.46 g. 273. What is the weight of a liter of air at 18° C. and 762 mm., the dew point being 15° C? (Ten. aq. vap. at 15° C. = 12.73 mm.) Ans. 1.2050 g. 274. Calculate the weight of a liter of air 45% saturated with water vapor at 765 mm. and 18° C. (Tension of aqueous vapor at 18° C. = 15.38 mm. Weight of a liter of dry air at 760 mm. and 0° C. = 1.2926 g.) Ans. 1.2165 g. 276. Calculate the weight of a liter of carbon dioxide at 20° C. and 772 mm., the gas being 50% saturated with water 140 CHEMICAL CALCULATIONS vapor. (Tension of aqueous vapor at 20° C. = 17.41 nun. Weight of a liter of dry carbon dioxide at 0° C. and 760 mm. = 1.9768 g.) Ans. 1.8585 g. 276. Using the symbols of page 109, deduce the approximate formula Wg+y, = ^''+"' ~ ^•^^'"' X ^ X 1.9768 for the cal- culation of the volume of moist carbon dioxide at any tem- perature and pressure to the volume dry at 0° C. and 760 mm. (Weight of a Uter of dry carbon dioxide at 0° C. and 760 mm. = 1.9768 g.) CHAPTER VII CALCULATION OF ATOMIC WEIGHTS AND FORMULAS Atomic Weight. — The atomic weight of an element is, the ratio of the weight of an atom of that element to the weight of an atof" of some other element taken as standard. This stanuc^d is oxygen which is arbitrarily given an atomic weight of 16. Hydrogen, the hghtest of the elements, was formerly taken as unity, but this imit has been abandoned and oxygen has supplanted it, for the reason that most of the elements form binary and other compounds with oxygen which are stable at ordi- nary temperatures, while the same is not true of hydrogen. Again, it happens that when oxygen is taken as 16, more of the atomic weights of the elements can be expressed in whole numbers than is the case when the hydrogen stand- ard is employed, and also there are more compounds of importance to the chemist, containing oxygen, than hydrogen. Valence. — Valence is the power of an atom of an element to combine with a certain number of atoms of another element, taking the valence of hydrogen as unity. Valence is a numerical property of an element by virtue of which a definite number of atoms of another element are held in combination. Combining Weight. — The combining weight of an element is the weight of that element which will combine with, replace or be replaced by, 16 parts by weight of oxygen. 141 142 CHEMICAL CALCULATIONS Radicals or ions may be considered to have valence. Thus, sulphuric acid has two replaceable hydrogens, so the radical SO4 is divalent: the hydroxyl radical, OH, combines with one atom of hydrogen to form water, one atom of sodium to form sodium hydroxide and, conse- quently, has a valence of one. Relationship between Atomic Weight, Valence and Combining Weight. — As oxygen has a valence of two, the following equation shows the relationship between the atomic weight, the combining weight and the valence of elements A = -V in which A is the atomic weight, C the combining weight ^ and V the valence. Calculation of Atomic Weight from an Analysis. Simple Cases. — For example, mercury and oxygen have each a valence of two, the compound mercuric oxide is HgO, which by analysis gives: 92.61% mercury and 7.39% oxygen. Then X grams of mercuric oxide' contain 0.926 X g. of mercury and 0.074 X g. of oxygen. The equivalent of mercury according to definition is 0.9261Z^_ 0.9261 ,,,^ _.„. 0:0739XX1^ = 0:0739 ^l^ = 200-6- ' Combining weights may be referred to other elements than oxygen. If x represents the valence of the element to which the (J combining weight is referred, then the formula becomes A = — V. The term "equivalent" is often employed and by it is meant that weight of an element which will combine with, replace or be re- placed by 8 parts by weight of oxygen. With this definition the formula becomes A = CV. This latter is frequently used, but the formula given in the text is employed for the reason that it refers to oxygen as 16, which is the arbitrary standard to which all atomic and molecular weights are referred. ATOMIC WEIGHTS AND FOHMULAS 143 Substituting in the equation, the atomic weight is A=?f?X2 = 200.6. Again, iron in its trivalent form is combined with oxy- gen yielding an oxide which on analysis shows 30.06% oxygen, 69.94% iron Then, as before, the equivalent of the iron is 0.3006 ^^^~ ^^■'^'^' and the atomic weight of iron is A = ?^ X 3 = 55.84. Law of Dulong and Petit. — The formula A= ^ V^ supposes a knowledge of the valence of the element the atomic weight of which is to be calculated. The com- bining weight is found by an analysis of a compoimd, preferably a compound of the element with oxygen, if such compound be stable. In the case of compounds of elements which cannot be vaporized without decom- position, a knowledge of the valence of the element must be determined by other means than through molecular weights determined by vapor density methods. For this the law of Dulong and Petit is employed. This law states that the product of the atomic weight of an element into its specific heat (in the solid state) is a constant. The product is called the atomic heat and its value approx- imates 6.4.2 jf y, represents the atomic weight and S the specific heat, the law may be sjnnbolized: Atomic heat = 8w = 6.4. 1 P. 142. 2 Exceptions occur, notably in the cases of carbon, silicon, beryl- lium and boron. 144 CHEMICAL CALCULATIONS Values of the atomic weights so obtained are rough approximations, but they serve to indicate which of a series of values must be taken as the atomic weight. This is shown in the following: The specific heat of iron is 0.1162; ' and analysis of ferric oxide gives Per cent Iron 69.94 Oxygen 30.06 100.00 The problem is to find the atomic weight and the valence of iron in this compound. The combining weight of iron from these figures is 69.94 30.06 X 16.00 = 37.227. Then as oxygen has a valence of two, the lowest possible value for the atomic weight of iron is half the combining weight, or 18.6135; FeaO being a compound in which iron would show a minimum atomic weight as then its valence would be one. Without taking into consider- ation any other compounds of iron, the atomic weight of iron might be any whole multiple of this value, namely, 18.6132, 37.227, 55.8405, 74.454, etc. But from the law of Dulong and Petit the atomic weight of iron must approximate 6.4 ^= 01162 = ^^-°^- This shows 55.84 to be the atomic weight of iron. From Q the formula A = -^ F, the valence of iron in ferric oxide must be 55.84 = ^^7; F= 3. * See Chem. Ann., p. 6. ATOMIC WEIGHTS AND FORMULAS 145 Calculation of Atomic Weights from Analysis. More Complex Cases. — The above are examples of the simplest cases. A more complex example wiU now be considered. Arsenious oxide (AS2O3) is converted into silver arsenate (AgsAsO^: 1.5000 g. of arsenious oxide yield 7.0096 g. of silver arsenate. Knowing the symbols and the atomic weights of all the elements except the atomic weight of arsenic, which is to be determined, the conversion may be represented AS2O3 — » 2Ag3As04 (o) 1.5000 g. 7.0096 g. (6) 48 +2 X 775.28 + 2 X The subscripts (a) show the ratios by weight, (6) the ratios by molecular weights in which X is the atomic weight of the arsenic. These two ratios are equivalent: 1.5000 ^ 48 + 2Z 7.0096 775.28 + 2 Z' and solving for X gives X = 75.00 as the atomic weight of arsenic. Again, consider the equation, NaCl + AgNOa = AgCl + NaNOa, and suppose it were determined that 10.0000 g. of sodium chloride require 18.4535 g. of silver dissolved in nitric acid for complete precipitation. To calculate the atomic weight of sodium: According to the statement, 10.0000 g.NaCl = 18.4535 g. Ag, the amount of chlorine in the silver chloride is the amount of chlorine in the sodium chloride, which is and the weight of sodiiun in the sodium chloride is 10.0000 - 6.0657 = 3.9343 g. Na; 146 CHEMICAL CALCULATIONS consequently, the atomic weight of sodium is 11^x35.46 = 23.00. The molecular weight of a substance is the sum of the atomic weights of the atoms entering into the molecule. When a substance may be vaporized without decom- position, the molecular weight is obtained as given in the section devoted to gas densities. Molecular Weight by Elevation of Boiling Point and Depression of Freezing Point. — In addition to the method of measuring vapor densities, there are two methods conmionly employed with non-electrolytes (i.e., substances which do not ionize appreciably upon going into solution). These may be determined, in general, by measuring the depression of the freezing point and the elevation of the boiling point, due to the introduction of a solute. The depression of the freezing point of a given amount of solvent is the same for a gram-molecule of substance, and, in general, is independent of the composition of the substance if no ionization takes place. The amount of this depression depends upon the nature of the solvent. Let M be the molecular weight of the dissolved substance, &> its weight in grams, W the weight of the solvent in grams, A the observed change of temperature and K a constant depending upon the nature of the solvent. The molecu- lar weight is given by the formula ^ = ^A^- Using the symbols with the same meaning, as above, for the determination of molecular weights from the raising of the boiling point, the same formula applies except that K is a different constant and A represents ATOMIC WEIGHTS AND FORMULAS 147 the elevation of the boiling point of the solvent due to the introduction of the solute. B^low are a few values ioTK: Solvent. K Dep. o£ F. P. K Elev. of B. P. Water 1850 5000 510 2610 2160 3590 Ethyl ether Chloroform Molecular Weight of an Organic Acid from Analysis of the Silver Salt. — A method for the determination of molecular weights particularly applicable to organic acids is illustrated as follows: * One gram of silver ace- tate on analysis is found to contain 0.6464 g. of silver. As silver is univalent, then one hydrogen must have been replaced for each acidic hydrogen ion in the acetic acid, which being a monobasic acid has one replaceable hydro- gen ion. It is evident that in this compound there are 1.0000 - 0.6464 = 0.3536 g. of substance other than silver; the molecular weight of the radical is 0.3536 0.6464 X 107.88 = 59.015. The weight of the acetic acid, then, must be 59.015 + 1.008 = 60.02. Molecular Weight of an Organic Base from Analysis of the Hydrochloro Platinic Acid Salt. — The molecular weight of an organic base may be determined by weigh- 1 Strictly speaking, this method presupposes a knowledge of the number of replaceable hydrogen atoms and in reaUty only gives the molecular weight of the empirical formula of the acid. 148 CHEMICAL CALCULATIONS ing the hydrochloro platinic acid salt of the base, igniting, and weighing the residue of platinum after ignition.^ The formation of the salt may be symbolized (letting B represent the organic base) as follows: 2 B + HaPtCle = BsHaPtCle. One gram of methyl amine hydrochloro platinic acid leaves 0.4134 g. of platinum after ignition. To cal- culate the molecular weight of methyl amine, let x represent the molecular weight of methyl amine, then 1.0000 ^ 2x + HaPtCls _ 2x + 409.98 0.4134" Pt ~ 195.2 Solving gives x = 31.08. Formula of Compound, Given Molecular Weight and Percentage Composition. — The molecular formula of a compound is easily determined, knowing the molecular weight and the percentage composition. Take, for exam- ple, barium dichromate, given the following: Molecular weight 353.57 Barium 38.85% Chromium 29.47% Oxygen 31.68% The proportion of barium in the molecule is (atomic weight = 137.37) i05:00 ^ ^^^-^^ " ^^^■^^' i37!37 " ^' of chromium (atomic weight = 52.1), ^5:^X353 57-104 2 ^^ - 2- 100.00 ^ ^^^-^^ ~ ^"*-^' ^2T " ^' of oxygen (atomic weight = 16.00), ^3X353.57 = 112,1^2 = 7. Consequently, the sjnnbol is BaCraO?. 1 This method presupposes the organic base to be mono acidic. ATOMIC WEIGHTS AND FORMULAS 149 Formula from Percentage Composition. — Formulas may be empirical or molecular. The empirical formula is the simplest formula that can be assigned the substance and, in general, is only used in the absence of knowledge of the molecular formula. The molecular formula shows the number of atoms in the molecule and consequently in- dicates the molecular weight, that is, that weight of sub- stance which in the gaseous state would occupy 22.4 liters. To calculate the empirical formula of a compound, let A = symbol of one of the elements present in a com- poimd; B = symbol of another element present in the same compound; C = symbol of stUl another element present in the same compound. X = subscript of A in the compound under consideration; y = subscript ofB in the compound under consideration; z = subscript of C in the compound under consideration. a = atomic weight of the element A ; b = atomic weight of the element B; c = atomic weight of the element C. AxByCz = molecular symbol of the compound under con- sideration. Then the molecular weight of the compound is ax+ by+ cz. Let this be represented by M. Then %A=gxiOO; %B = gxiOO; % C = g X 100. Then %A : %B: %C = g X 100 : ^X 100 : f X 100, Hence = ax -.by : cz. %A,%B,%C = x:y :z. 150 CHEMICAL CALCULATIONS Then in which if is a constant. Multiples of K are chosen which will give x, y and z integral values. To apply the above, suppose a substan.ee analyzes Per cent Carbon 40.0 Hydrogen 6.6 Oxygen 53 . 4 Then if x, y and z represent the subscripts of carbon, hydrogen and oxygen respectively, y= -p = 6.6 2- ^g -6.6 As there cannot be less than one atom of each element present, it is evident that the ratio of the atomic weights of carbon, hydrogen and oxygen in this compound must be one atomic weight of carbon, one atomic weight of oxygen and two atomic weights of hydrogen, and the empirical or simplest formula is CH2O, corresponding to a molecular weight of 30.^ Such a compound is known and is formaldehyde (oxymethylene). The compound might, however, be (CH20)3„ and in the absence of knowl- edge of the value of x no further information may be '■ The value of X in this example is x-^ . Multiplying the values obtained for x, y and z by this quantity gives integral values for x, y and z of 1, 2 and 1 respectively. A value for K need not be de- termined as wiU be seen later. It is not necessary to have the composition of the compound expressed in percentages. For instance, the same result is obtained ATOMIC WEIGHTS AND FORMULAS 151 conveyed than is given in the empirical formula. There is a compomid yielding the same percentage composition which shows a molecular weight of 60 and corresponds to the molecular symbol C2H4O2 (acetic acid). In this case the value of x is two. Another compound, which on analysis has the same composition and has a molecular weight of 90, corresponds to the formula CsHeOs and is lactic acid. Finally, there is still another compound of molecular weight 180, being grape sugar, CeHi^Oe, in which the value of x is six. When the molecular weight of a compomid is known the molecular formula may be calculated: when this is lacking, given the percentage composition of a compound, in the absence of other knowl- edge, the empiric formula only may be determined. A complex case will now be considered. A compound analyzes: Per cent Carbon 49.97 Hydrogen 6.46 Nitrogen 17. 95 Oxygen - 25.62 100.00 from the following analysis of this same compound, 2.6000 g. of substance being taken: Carbon 1.0400 g. Hydrogen 0.1716 g. Carbon 1.3884 g. 2.6000 g. Then C = 1:0400 ^„ 08667 = 1, La H = S =017024 = 2, O = M|^ = 0.08679 = 1, lo as before. 152 CHEMICAL CALCULATIONS To calculate its empiric formula. As before, C = ^^ = 4.164, _ 17.95 _ ^ - 14:01 - ^•^^^' 16 The compound, then, corresponds to the formula C4.164 H6.406Ni,28iOi,6oi. But Sb formula must express the ratio of the atoms in simple whole numbers, the problem now presented being to find this ratio. It is apparent that the nitrogen atoms are present in the smallest number; then taking nitrogen as unity as in (a) : (a) (6) 4 164 4 164 ^ = r:il = ^-250^ 0MJ5 = 6.501 „ 6.406 _^„^ 6.406 ,... ^ = EMI = ^-^^ 0:6405 = ^°-°° --T 1.281 ^ nr^n 1.281 _ r\f\r\ ^ = Tm = 1-0^° o:64r5 = ^-ooo L601 _ imi _ " - 1281 ~ ^-^^^ 0.6504 " ^•^"" 1 The ratios in (o) assume nitrogen to be present to the extent of one atomic weight. With this assumption it is seen that the other elements are not present as whole but as fractional atomic weights, which, according to the atomic theory, cannot be. The ratios of (6), (c) and (d) may be arrived at by multiplying these ratios by 2, 3 and 4 respectively. Finally, a series of ratios ap- proximating whole numbers is reached. id) ^■'^^ -13 002- 0.32025 - ^^-^^ - = 13 ^•^^® -20 004- 0.32025 ^"•"^* ■ = 20 0.32025= ^•««« = = 4 ±''l = 5.000 = = 5 ATOMIC WEIGHTS AND FORMULAS 153 (C) C= iiM -9752 ^ 0.4270 ~ ^-^^^ 6.406 _ ^ - 04270 -^^•^'^ N = oWo = ^-«o« _ 1.601 _ ""0.4270""^-^^" 0.32025 From (a) the formula would be CssbHsNiOi^s. This ratio is not in whole numbers. Determining the ratio on the assumption that two nitrogens are present (by- dividing the nitrogen ratio, 1.281, by ^, giving the quotient 0.6405 and dividing through by this), the formula C6.6H10 N2O2.B is obtained (6). Taking nitrogen as being present to the extent of three atoms, the ratio as found in (c) is C9.76H15N3O3.76. Assimiing fom" nitrogen atoms to be present, the ratio numbers of (d) yield Ci3H2oN406. The case cited has been carried to the end and this method must yield results finally, but only on the assumption that the percentage composition as given is correct. That this ratio would finally have been reached might have been seen by an inspection of the ratios of (a). These ratios (a), it will be observed, are all evenly divisible by 0.25; consequently, dividing the ratios through by this nimiber would have led to the result, C13H20N4O6, by inspection, obviating the necessity of calculating (6), (c) and (d)} Such a short cut is not always apparent at a ^ The process amounts to finding the approximate lowest com- mon multiple of the subscripts. Knowing the lowest common multiple, by dividing the subscripts by it, the numbers so obtained 154 CHEMICAL CALCULATIONS glance, in which event the operations must be continued till approximately whole numbers are obtained.' Most acids, bases and salts may be regarded as being composed of two oxides.^ Na2S04 may be considered as made up of the basic oxide NaaO and the acidic oxide SO3; as Na20 • SO3. The analyses of many substances report the constituents present as oxides; particularly is this so in mineral analysis. These radicals or oxides may be treated in the same way as elements. A substance shows: Per cent Potassium oxide (K2O) 9 . 93 Aluminum oxide (AI2O3) 10 . 77 Sulphur trioxide (SO3) 33. 72 Hydrogen oxide (H2O) 45 . 56 yield the subscripts desired. As analyses contain unavoidable errors, the true mathematical lowest common multiple will not serve, hence it is best to proceed by some such method as outlined. Just where to stop depends upon the degree of accuracy of the analysis. 1 The sUde rule is of great assistance in calculations of this nature, as when set to show a ratio between two numbers, all num- bers coinciding are in the same ratio. ^ Halogen salts, sulphides, and general salts of acids which do not contain oxygen, may not be so considered. When an analysis shows acidic and basic oxides and also a halogen or halogens re- ported as such, the analysis, no matter how carefully done, even if theoretically perfect, wUl total more than one hundred per cent. This is for the reason that the metal combined with the halogen is calculated to the oxide where in fact it was present as a halogen salt. Hence, if the metal is calculated to an oxide and the halogen reported as such, the analysis will show too great a total by an amoimt equal to the oxygen calculated to be present but which was not present. Such an analysis will be too high by an amount equal to "the oxy- gen value of the halogen" or g^j X % Af , in which H is the molec- ular weight of the halogen and % M represents the per cent of the halogen reported. ATOMIC WEIGHTS AND FORMULAS 155 The calculation of its formula is K.0 = ?f| = 0.1054 : 0.1054 0.1054 ' ^^^^ = I^ = 0-10^* = 0.1054 0.1054 ' SOa = 1^=0.4211: 0.4211 . TTTT^ = 4, 80.07 ■ 0.1054 45.56 _ ^ . 2.529 ^^^-ISioIB-^-^^^ • 01054 = 24, and the compound is K2O • AI2O3 • 4 SO3 • 24 H2O, or it may be written: K2O • SO3 : AI2O3 • 3 SO3 : 24 H2O AI2 (804)3 • K2SO4 • 24 H2O, or KAl (804)2 • 12 H2O. In the problem above, the analysis is supposed to be correct as given. In practice, the analysis of a substance will contain errors, large or small, according to the accuracy of the methods employed and the difficulties encountered in the determination of the various constituents. Where the constituents are present in about equal amounts and all the constituents are determined with the same degree of accuracy, it is easier to find the ratios of the atoms or radicals by the method already given. In many cases, one or more constituents are present to but a small extent, or the methods by which they were determined do not lend themselves to the same degree of nicety of deter- mination as with some of the other constituents present. In the calculation of formulas many things must be taken into accoimt and it is largely a matter of judgment as to the maimer of calculation. These facts, the method of preparation or the manner of formation of the compound, its similarity to other well-known compoimds of the same type, the methods employed in the analysis: all these must be taken into account, leaving the determination 156 CHEMICAL CALCULATIONS of the formula of a compound to a large degree a matter of judgment. Suppose a substance on analysis yields . Pel cent Manganese 24. 50 Hydrogen 1.02 Arsenic 45.07 Oxygen (by diff.) 29.41 and let it be assumed that the arsenic is determined with the greatest degree of accuracy. Then 1 (19 ^^ = in = «-602- = ?^ = 1.838, and the compound approximates Mno.446Aso.6oiOi,838Ho.902. It is required to find the ratio of simple whole numbers which approximates these ratios. By inspection it will be noted that the ratio of manganese to hydrogen is approximately 1 : 2. Also it can be seen that the ratio of hydrogen to arsenic is nearly 3 : 2. Then it is evident that if these numbers representing the ratios found by analysis are divided by 0.15 that the ratios of manganese, hydrogen and arsenic fall ihto approximately even num- bers. This would indicate the arsenic to be present to the extent of four atoms (fT-rr ~ ^j. Then, proceeding on this assumption: ATOMIC WEIGHTS AND FORMULAS 157 446 Mn = -~ = 2.97 = 3 very closely, H = ' = 6.01 = 6 very closely, . 0.602 .„, . , , As = = 4.01 = 4 very closely, O = ~~ = 12.2 = 12 very closely. Consequently, the compound is Mn3H6As40i2 or MnjHe (AS03)4. Formulas of Minerals. Isomorphic Replacement. — A definite crystalline structure generally indicates purity and, consequently, definite chemical composition. Mit- scherlich propoimded the law that: "Substances which are analogous chemical compoimds have the same crystal- line form"; in other words, they are isomorphous. To a large extent this is true, but there are so many exceptions that the statement cannot be called a law. The "alums" consist of sulphates of a trivalent and a monovalent base with twelve molecules of water, all crystallizing isomor- phously. By manipulation a crystal may be grown -to contain one or more almns. Such crystals will not show definite chemical composition, but are mixtures, and may be regarded as a single alum, parts of which have been replaced by an analogous substance. This is isomorphous replacement. Strict rules for isomorphous replacement cannot be given. In general, it may be said that metals and metallic oxides of the same structure can replace each other and the same is true of the non-metals and acidic oxides. Thus AI2O3 and FejOa are mutually replaceable as are the oxides FeO, MgO, CaO, MnO, etc., these latter being similar in being oxides of divalent metals. In like manner 158 CHEMICAL CALCULATIONS the alkalies, NH4, Li, Na, K, etc., and their oxides (NH4)20, Li20, Na20, K2O, etc., may replace each other in a mineral. This is not all — CaO, for example, may be replaced by K2O or another alkali oxide, as also, in some instances, FeO may be replaced by AI2.O3. This leaves the matter to a large extent one of judgment: for more complete informa- tion the student is referred to works on mineralogy. The more common cases follow the rule of compounds of like structure replacing each other. Mosander examined a specimen of Menaccanite and found it to be composed of Per cent TiOj 46.92 FenOs 10.74 FeO 37.86 MnO 2.73 MgO 1.14 99.39 The percentages of the metals and oxygen are ^^ = ^ X 46.92 = 28.18% Ti and 18.74% 0, ■pip KCC CK ^ = ^X 37.86 = 29.44% Fe and 8.42% 0, ra^?^^^-^^^^-^^^''^'' ^^'^ 0.61% 0, ^ = Si ^ 1-1^ = "^-^^^ ^^ ^^^ _^% 0' Total 31.45% 0. ' These calculations are very easily performed by the use of the Bhde rule, as the error of the rule is, on the average, less than the error of analysis, etc. For the same reason approximate atomic weights may be used. ATOMIC WEIGHTS AND FORMULAS 159 If the metals replaced iron, the percentages replaced are Ti " ^St ^ ^^--^^ " 32.71% Ti as Fe, Mn = 54yx 2-12= 2.16%MnasFe, Fp ^^i S"! Mi = iS >< ^-^^ = _l:i^^° ^S as Fe, 36.45 + 36.95 = 73.40% total Fe. „ 73.40 , ^, , 1.314 „ ^ ^^ = 5-5:85 = 1-^1* ■ 0-J57 = 2-««' O = tro = 1-976 :J-|g= 2.99 = 3 approx. This mineral, considered as an oxide of iron,' approximates very closely to the formula Fe203 (hematite), part of the iron being replaced by other metals, chiefly titanium, making it (TiFe)203. The mineral as it analyzes is (Ti, Fe, Mn, Mg)203. An inspection of the analysis would show that there are no distinctively acid oxides present, which would suggest the method of calculation adopted. ' The metals could just as well be calculated on the supposition that they replaced titanium: Ti 48.1 Fe 55.85 j;i ^ 48.1 Mn 54.93 Ti ^ 48.1 M'g 24.32 X 36.95 = 31.82% Fe as Ti, X 2.12 = 1.86%JV[n as Ti, X 0.69 = 1.37% Mg as Ti, 35.05 + 28.18 = 63.23% total Ti. Ti = " = 1.315 = 2 approximately, O = ?^ = 1.966 = 3 approximately, 16 and the mineral becomes TijOs, or more generally M2O3, as before. 160 CHEMICAL CALCULATIONS Given the following analysis by Pisani; to determine the formula of the mineral. Per cent SiOj 45.95 AI2O3 0.85 FeO 8.91 MnO 10.20 ZnO 10.15 CaO 21.55 MgO 3.61 Loss on ignition 0.31 101.53 This mineral is obviously a siHcate. As a silicate may be regarded as a combination of basic and acidic oxides, the caleium oxide being present in the greatest amount, the metals will be calculated to calcium oxide. Disre- garding the loss on ignition: §^ = ^ X 8.91 = 6.96% CaO = FeO, CaO 56.09^^^20= 8.07% CaO ^ MnO, MnO 70.93 CaO ^ 56.09 ZnO 81.37 CaO 56.09 X 10.15 = 7.00% CaO = ZnO, X 3.61 = 5.02% CaO = MgO, MgO 40.32 21.55% CaO present, 48.60% CaO total. Calculating the AI2O3 into equivalent Si02, as it is prob- ably present as an aluminate replacing silica:^ ' On the supposition that the mineral contains the alumina combined as an aluminate and the sUica as a silicate, then the aluminate can be represented as 3 M"0 • AI2O3 and the silicate as M"0 • SiOz) whence it follows that as AUOa requires three M"0 and the SiOj one, hence AljOa = 3 SiOj. ATOMIC "WEIGHTS AND FORMULAS 161 m = li?X«-«^= 1.50%SiO. = Al.O, 45.95 % SiOa present, 47.45% SiOa total. _ 47.45 _ 0.788 _ ^'"' - 603" - "•^^^' 0788-^' „ „ 48.60 _ eRA 0.866 , , , ... CaU = „ = 0.866; _ „ = 1.1 = 1 approximately. The mineral is of the type CaO • Si02 or CaSiOa or R° SiOg in which R" is chiefly Ca partially replaced by Mn, Zn, Fe", Mg, with a smaU amount of SiOa replaced by AI2O3, and is (Ca, Mn, Zn, Fe, Mg)Si03: (pyroxene).i This latter formula gives no indication as to the amounts of the metals present and does not show the alumina re- placement. • The sum of the percentages of Si02 and CaO total 96.05%. The mineral as analyzed shows 0.31% loss on ignition. Deducting this leaves the analysis footing up to 101.56 — 0.31 = 101.25%. Calculating the percentage on this basis gives (see p. 176) : 47.45, 96.05 48.60 96.05 X 101.25 = 50.02% SiOj, X 101.25 = 51.23% CaO, 101.25% total. Then _ 50.02 _ 0.830 _ ^'"^ ~ 603 ~ "•^^"' 0.830 ~ ^' CaO = g^l = 0.915; 5|g = ^ = j approximately, and the ratio is as given before. 162 CHEMICAL CALCULATIONS PROBLEMS 277. The atom of oxygen is 1.1420 times as heavy as the atom of nitrogen, (a) Taking the atomic weight of oxygen to be 16, what is the atomic weight of nitrogen? (6) If oxygen were 100, what is the atomic weight of nitrogen? Ans. (a) 14.01. (6) 87.56. 278. When oxygen is taken as 16, the atomic weight of iodine is 126.92 and hydrogen is 1.008. What is the atomic weight of iodine in a system taking hydrogen as unity? Ans. 125.91. 279. Erdmann and Marchand heated mercuric oxide: 2HgO = 2Hg + 02. From 118.3938 g. of mercuric oxide, 109.6308 g. of mercury were obtained. Calculate the atomic weight of mercury. Ans. 200.17. 280. The atomic weight of lead was determined by Stas by converting lead into the nitrate: 100.00 g. of lead yielding 159.9703 g. of the nitrate. What is the atomic weight of lead from these data? Ans. 206.80. 281. Berzelius found that 2.265 g. of ferric oxide (FejOj) are formed from 1.586 g. of iron. Determine the atomic weight of iron from these data. Ans. 56.059. 282. Barium sulphate contains 58.85% barium, 13.73% sulphur and 27.41% oxygen, (a) What is the atomic weight of barium when oxygen is 16 and sulphur is 32.07? (6) What is the atomic weight of sulphur when barium is 137.41 and oxygen 16? Ans. (a) 137.41. (6) 32.07. 283. The analysis of chromous chloride (CrCl2) according to P61egot is 57.50% chlorine and 42.50% chromium. What is the atomic weight of chromium? Ans. 52.419. 284. Lead sulphate on analysis yields 68.31% lead, 10.58% sulphur and 21.11% oxygen. What is the atomic weight of lead? Ans. 207.1. ATOMIC WEIGHTS AND FORMULAS 163 285. According to Berzelius, litharge (PbO) contains 7.1724% ojcygen. What is the atomic weight of lead? Ans. 207.08. 286. Calculate the atomic weight of silver from the follow- ing figures determined by Stas. 53.1958 g. of silver form 92.6042 g. of silver bromide. Ans. 107.88. 287. Dumas passed hydrogen gas over heated copper oxide. The copper oxide lost 59.789 g.; the water formed weighed 67.282 g. Calculate the atomic weight of hydrogen. Ans. 1.0026. 288. Working with the reaction, 2Ag + Cl2 = 2AgCl, Stas found that 91.4620 g. of silver produced 121.4993 g. of silver chloride. Calculate the atomic weight of chlorine. Ans. 35.428. 289. According to the reactions, 2AuCl3 + 3SO2 + 6H2O = 2 Au + 6HC1 + 3H2SO4 3 H2SO4 + 3 BaCU = 3 BaSO, + 6 HCl, Levol found that 1000 parts of gold in the form of the chloride produce an amoimt of sulphuric acid that will precipitate 1728 parts of barium sulphate. Calculate the atomic weight of gold. Ans. 202.64. 290. Chromic oxide (CrjOs) contains 68.42% chromium and 31.58% oxygen. What is the atomic weight of chromium? Ans. 52.00. 291. Erdmann and Marchand determined the atomic weight of mercury by heating the sulphide with copper, HgS + Cu = CuS + Hg, and found 177.1664 g. of mercuric sulphide gave 152.745 g. of mercury. Calculate the atomic weight of mercury. Ans. 200.59. 292. The percentage composition of platinum chloride is Per cent Platinum 57.92 Chlorine 42.08 100.00 164 CHEMICAL CALCULATIONS Taking 35.46 as the atomic weight of chlorine and 0.0323 aa the specific heat of platinum, what is the atomic weight and the valence of platinum? Ans. At. wt. = 195.20. Valence = 4. 293. The percentage composition of ferrous oxide is Per cent Iron 77.73 Oxygen 22.27 100.00 The specific heat of iron is 0.1162. What is the atomic weight and the valence of iron in this compound? Ans. At. wt. = 55.84. Valence = 2. 294. Antimony forms compounds with oxygen of the follow- ing compositions: (a) (6) (c) Per cent Per cent Per cent Antimony 83.35 78.98 75.03 Oxygen 16.65 21.02 24.97 100.00 100.00 100.00 The specific heat of antimony is 0.0495. What is the atomic weight of antimony and its valence in the compounds (a), (6) and (c)? Ans. At. wt. = 120.2. Valence =(a) 3. (6) 4. (c) 5. j296. Siewert found that 36.865 parts of chromic chloride gave 100 parts of silver chloride: CrCls + 3 AgNOs = Cr(N03)3 + 3 AgCl. Calculate the atomic weight of chromium from these data. Ans. 52.15. 296. According to Seubert, when platinum tetrachloride (PtCU) is heated in hydrogen, hydrochloric acid is formed. This he precipitated with silver nitrate and weighed as the ATOMIC WEIGHTS AND FORMULAS 165 chloride of silver. 17.4139 g. of silver chloride were found to be equivalent to 5.9242 g. of platinum. The reactions are PtCl4 + 2 Hj = Pt + 4 HCl HCl + AgNOs = HNO3 + AgCl. Calculate the atomic weight of platinum. Ans. 195.05. 297. Berzelius determined the atomic weight of platinum by converting the platinum into platinic chloride (PtCU) and finding the amount of potassium platinic chloride (KaPtCle) formed, PtCl, + 2KCl = K2PtCl,. 24.735 g. of potassium platinic chloride were formed from 10.000 g. of platinum. Calculate the atomic weight of platinum. Ans. 197.46. 298. Berlin found that 100.000 g. of lead nitrate gave 97.576 g. of lead chromate, Pb(N03)2 + KaCrOi = PbCrOi + 2 KNO3. Calculate the atomic weight of chromium. Ans. 51.99. 299. Calculate the atomic weight of aluminum from the figures given by MaUet: 8.2144 g. of ammonium alum, Al2(S04)3-(NH4)2S04-24H20 gave 0.9258 g. of aluminum oxide (AI2O3). Ans. 27.113. 300. Mallet foimd that 6.9617 g. of aluminum bromide re- quired 8.4429 g. of dissolved silver for precipitation: AlBrs + 3 AgN03 = Al (N03)3 + 3 AgBr. Calculate the atomic weight of aluminum. Ans. 27.093. 301. Stas found that when he added 7.25682 g. of potassium chloride to 10.51995 g. of silver dissolved in nitric acid that 0.01940 g. of silver remained in solution. The reaction being KCl + AgNOs = AgCl + KNO3, calculate the atomic weight of potassium. Ans. 39.09. 302. Calculate the atomic weight of carbon in each case from the following figures of Rendtenbacher and Liebig: (a) 28.803 g. 166 CHEMICAL CALCULATIONS silver acetate (CHsCOOAg) yielded 18.612 g. silver. (6) 16.220 g. silver tartrate (CiHjOsAgj) yielded 9.6175 g. silver, (c) 25.898 g. silver malate (CJijOiJigd yielded 16.059 g. silver. Am. (a) 12.024. (6) 12.02. (c) 12.041. 303. Marignac made the following determinations for the cal- culation of the atomic weight of nitrogen: (a) From 314.894 g. of silver nitrate he obtained 200.000 g. of silver. (6) 14.1100 g. of silver nitrate required 6.1910 g. of potassium chloride for complete precipitation, (c) 10.339 g. of silver dissolved in nitric acid required 5.1200 g. of ammonium chloride for com- plete precipitation. Determine the atomic weight of nitrogen in each case and (d) the mean atomic weight of nitrogen from these determinations. Ans. (a) 13.974. (6) 14.055. (c) 13.932. id) 13.987. 304. 3.000 g. of iodine dissolved in 90 g! of ethyl ether raised the boiling point 0.283° C. What is its molecular weight? Ans. 254. 305. If 1.32 g. of formic acid dissolved in 100 g. of water lowers the freezing point 0.53° C, what is its molecular weight? Ans. 46.1. 306. 10.000 g. of naphthalene dissolved iu 100 g. of benzol lowers the freezing point 3.90° C. What is the molecular weight? Ans. 128. 307. 1.67 g. of benzoic acid dissolved in 70 g. of ether raised the boihng point 0.422° C. Calculate the molecular weight. Ans. 122. 308. 0.200 g. of alcohol dissolved in 100 g. of benzol lowered the freezing point 0.216° C. What is the molecular weight of the alcohol? Ans. 46.3. ATOMIC WEIGHTS AND FORMULAS 167 309. 1.000 g. of naphthalene dissolved in 50 g. of chloroform raised the boiling point 0.56° C. What is its molecular weight? Am. 128. 310. Calculate the molecular weight of nitrosodiphenyl amine if 3.58 g. of the substance dissolved in 150 g. of benzol lowers the freezing point 0.601° C. Am. 198.5. 311. Tartaric acid has two acidic hydrogen ions. When both these are replaced by silver the silver salt contains 59.31% silver. What is the molecular weight of tartaric acid assuming it contains no water of crystaUization? Ans. 150.046. 312. Silver acetate has one replaceable hydrogen ion. 1.0000 g. of silver produces 1.5475 g. of silver acetate. What is the molecular weight of acetic acid? Ans. 60.073. 313. The silver salt of fumaric acid (dibasic) contains 65.43% silver. What is the molecular weight of fumaric acid? Am. 116.016. 314. 2.0000 g. of the hydrochloro platinic acid salt of dimethyl aniline 3rield 0.5967 g. of platinum. What is the molecular weight of dimethyl aniline? Ans. 122.13. 315. 1.0000 g. of the hydrochloro platinic acid salt of pyri- dine jdelds 0.3436 g. of platinum on ignition. Calculate the molecular weight of pyridine. Ans. 79.08. 316. What is the molecular weight of ethyl amine, if 3.0000 g. of the hydrochloro platinic acid salt yield 1.1708 g. of platinum on ignition? Ans. 45.10. 317. An analysis of arsenic pentoxide shows 65.2% arsenic and 34.8% oxygen. Its molecular weight is 230. What is the formula? Ans. AS2O5. 318. Arsenic trichloride analyzes 41.4% arsenic and 58.6% chlorine. Its molecular weight is 181. What is the formula? Ans. AsCls. 319. Carbon trichloride shows a molecular weight of 237. Analysis shows it to contain 89.9% chlorine and 10.1% carbon. What is its molecular formula? Am. C2CI0. 168 CHEMICAL CALCULATIONS 320. Caxbanil contains Per cent Carbon 70.6 Hydrogen , 4.2 Nitrogen ' 11 .8 Oxygen 13.4 The molecular weight is 119. What is the formula? Ans. C7H5NO. 321. Phenyl cyanide has a molecular weight of 103. Its composition is Per cent Carbon 81.5 Hydrogen 4.9 Nitrogen 13. 6 What is the formula? Ans. C7H5N. 322. Phenyl disulphide has a molecular weight of 218. Analysis shows 70.6% CeHs and 29.4% sulphur. What is the formula? Ans. (06115)282. 323. What is the empiric formula of a compound analyzing: Per cent Phosphorus 3.99 Bromine 82.30 Chlorine; 13.70 Ans. PBrjCls. 324. What is the empiric formula of a compound of the com- position: Per cent Molybdenum 46.01 Oxygen 11 . 50 Chlorine 42.49 Ans. M02O3CI6. 326. If basic antimonyl sulphate has the composition Per cent SbO 40. 5 SO4 14.2 Sb 35.5 OH 10.0 what is the formula? Am. (SbO)2 Sd • Sbz (OH)*. ATOMIC WEIGHTS AND FORMULAS 169 326. Find the empiric formula of a compound of • Per cent Nickel 26. 12 Hydrogen 0.90 Arsenic 44 . 51 Oxygen 28.48 Am. NisHe (AsOa)*. 327. Find the empiric formula of a compound of Per cent Chromic oxide (CrjOs) 15. 9 Sulphur trioxide (SOs) 33.5 Ammonium oxide (NH4)20 5.4 Hydrogen oxide (H2O) 45 . 2 Am. Crj (804)3 • (NH4)2S04 • 24 H2O. 328. What is the formula of the following; lead being most accurately determined? Per cent Lead 83.61 Sulphur 9. 52 Chlorme 7.36 Am. Pb^SjCla. 329. Platinum is most accurately determined in the follow- ing. What is the formula? Per cent Lithium 2.91 Platinum 36.84 Chlorine 40. 10 Water •. 20.60 Am. LizPtCle • 6 H2O. 330. Calculate the simplest formula, considering arsenic oxide as the most accurately determined, of Per cent Arsenic oxide (A^Ob) 37.91 Calcium oxide (CaO) 18.28 Ammonium oxide (NH4)20 8.70 Water of crystallization 35. 23 Am. CaNHiAsOi . 6 H2O. 170 CHEMICAL CALCULATIONS 331. Calculate the empirical formula of the following sub- stance, considering calcium as being most accurately .deter- mined. Per cent Calcium 39.01 Phosphorus 19 . 85 Oxygen (by diflf.) 41 . 14 Ans. Ca3P208. 332. Find the formula of a substance analyzing Per cent Potassium 11.47 Antimonyl (SbO) 40.80 Carbon 14.65 Hydrogen 1 . 19 Oxygen 29.08 Water of crystallization 2. 60 considering the antimonyl most accurately determined. Ans. Ks (SbO) 2 CsHsOiz • H2O. 333. Given the following as the composition of Hypersthene, what is its formula? Per cent SiOj 54. 20 FeO.'. 21.70 MgO 24.10 100.00 Ans. (Mg, Fe) SiOa. 334. A sample of Smaltite examined by Hofmann gave the following analysis (disregard the Bi and S). What is its formula? Per cent As 70.37 Co .' 13.95 Ni 1.79 Fe 11.71 Cu 1.39 S 0.66 Bi 0.01 99.88 Ans. (Co,Ni,Fe,Cu) Asz. ATOMIC WEIGHTS AND FORMULAS 171 336. A sample of Wad analyzed: Per cent MnOa 59.94 MnO 6.58 ZnO 21 . 70 FejOs 0. 25 HaO 11.58 100.05 Disregarding the iron, what is the formula? Ans. 2 MnOz • (Zn.Mn) • 2 H2O. 336. What is the formula of a sample of DiaUage analyzed by Von Rath, and found to be (disregarding the water) : Per cent Si02 53.60 AI2O3 1.99 FeO 8.95 MnO 0.28 MgO 13.08 CaO 21.06 H2O 0.86 99.82 Ans. (Ca,Fe,Al,Mn,Mg) SiOs. 337. Show that the mineral analyzed below by Strtiber (Gastaldite) approximates the formula RsAUSisOa?. Per cent Si02 58.55 AI2O3 21.40 FeO 9.04 MgO 3.92 CaO 2.03 NajO 4.77 99.71 338. Deduce a formula for natrolite which was analyzed by Harrington with the following results: 172 CHEMICAL CALCULATIONS Per cent SiOj 47.09 AlaOs 26.99 NafiO 16.46 K2O 0.01 H2O 9.80 100.35 Am. NajO • AlA ■ 3 SiOj • 2 H2O. 339. What is the formula of Caberite which, according to Sachs, analyzes as foUows: Per cent AsjOs 40.45 NiO 26. 97 FeO 1.10 MgO 6.16 H2O 25.26 99.94 Ans. (Ni,Fe,Mg) ASO4 • 8 H2O. 340. Wolframite, according to Finlayson, gave on analysis the foUowing, what is its formula? Per cent WO3 76.24 FeO 16.39 MnO 6,05 CaO 1.05 MgO 0.11 99.84 Ans. (Fe,Mn,Ca,Mg)W04. CHAPTER VIII GRAVIMETRIC ANALYSIS In a previous section, the amount of an element or radical obtainable from a given weight of a substance entering into a reaction was considered. It was postu- lated that the substances were pure, and only on this assumption are such calculations possible. Direct Gravimetric Analysis. — Gravimetric analysis consists of weighing an element or compound of known composition and from this weight calculating the amount of a constituent present in the original substance. As gravimetric analyses are reported as percentages of constituents determined, it is necessary to know the weight of substance taken for analysis. Gravimetric analysis presupposes a knowledge of the elements or radicals present in the substance to be analyzed; in other words, it is preceded by a qualitative examination when the component elements or radicals are imknown. As an example: A salt known to contain sodium and chlorine is to be examined as to its piu-ity. One gram of the substance is weighed out and after proper precau- tions the chlorine is precipitated as silver chloride accord- ing to the reaction NaCl + AgNOa = AgCl + NaNOa, the silver chloride in pure state being found to weigh 2.4382 g. The weight of chlorine present in the silver chloride is 173 174 CHEMICAL CALCULATIONS Si = lW4X2-4^«2 = 0.6032 g. CI, and the percentage of chlorine is, one gram of substance being taken, 5^^X100 = 60.32% CI, or in one equation ""' ''-'^XJ^X 100 = 60.32% CI. AgCl 143.34 ^ 1.0000 The sodium of the sodium chloride is converted into pure sodium sulphate and weighed. The weight is found to be 1.2083 g. Then, as there are two atoms of sodium in sodium sulphate, 2Na^Na2S04, the per cent of sodium present is, 2Na _ 46.00 1.2083 _ N^;S04 ~ 142:07 X 1.0000 ^ ^"" ~ "^^•^^^'' ^^' and the percentage accounted for is 60.32% + 39.12% = 99.44%. From these figures the compound is found to be NaCl. The impurities are present to the extent of 100.00% — 99.44% = 0.56%. A complete quantitative analysis would require the nature and the amount of these impu- rities. If it had been known in the beginning that the substance was sodium chloride, the amount of which in the sample is the end desired, the same could have ' An expression such as this is intended to mean that the fraction immediately following the equality sign has as its numerator and denominator the molecular weights of the symbols of the fraction. The equality sign cannot be interpreted as the equality sign of mathematics, but is meant only in this Hmited sense. The value of the fraction is the factor to use when AgCl is weighed, found or given and CI is sought or required. See Chem. Ann., pp. 10-36. GRAVIMETRIC ANALYSIS 175 been calculated from the amount of silver chloride ob- tained, NaCl^ 58^ 2,4382 _ AgCl 143.34 ^ 1.0000 ^ ^"" ~ ^^'^^ ^'^ ^ ^^^• In the same way the percentage of a radical present may be calculated. Suppose one gram of a substance known to be a sulphate yields 1.2318 g. of barium sulphate and that it is desired to know the per cent of sulphuric anhy- dride present. This is '°' «^-«'xJ-:g^X 100 = 42.25% S03. BaSOi 233.44 '" 1.0000 When a reaction produces a gas, the volume may be measured and reduced to standard conditions and the weight of the same calculated. Suppose 0.5000 g. of calcite is treated with sulphuric acid, CaCOa + H2SO4 = CaSOi + CO2 + H2O, and that the carbon dioxide measured over water at 15° C. and 762 mm. is 118.2 cc. To determine the per cent of calcium carbonate, assuming that there are no other car- bonates present : The volume of dry gas at standard (the tension of aqueous vapor at 15° C. being 12.73 mm.) is SI ^ ^^^ 760^'^^ ^ ^-^^^^ = ^-^^^^^ ^- ^^^' the weight of which is ^^^V' ^ ^^-^^ " ^-^^^^ ^- *^^'- This indicates the presence of 176 CHEMICAL CALCULATIONS or better, proceeding to the final result by one expression, 273 762 - 12.73 0.1182 100.07 „ „q 288 >^ 760 >< "22X^0:5000 >^^"^"^^-^"^''^^^"'- Elimination of a Constituent from the Analysis. — Substances may contain water, water of crystallization, water mechanically held in the sample, oil and other impurities. When such substances are analyzed the amount of such material is determined; but when report- ing, the analysis is often desired on the dry basis, or as if no such substances were present. In this case the con- stituents are reported with the same relative accuracy as was attained in the actual examination, excluding the substance it is desired to eliminate. A compound was analyzed and found to contain: Per cent Moisture 1 . 47 Sodium oxide 57. 42 Carbonic anhydride 40.65 99.54 The analysis reported on the dry basis is 99.54^-^47 X ^^-^^ = ^^-^^^^ ^^^' 99.5f-L47 ^ *°-^^ = ^®'^« ^°^' 99.54% Factor Weights. — In the calculation of percentage compositions, a ratio is usually employed, such ratios being known as factors. If, then, a weight of substance is taken for analysis numerically equal to the factor which must be employed, the percentage composition is ascer- tained directly and is numerically equal to the mass of the substance separated and weighed; due cognizance GRAVIMETRIC ANALYSIS 177 being taken of the position of the decimal point. The factor for the deternaination of sulphur equivalent to C! on fiy barium sulphate is g^ = 23O4 = «-^^^^«- '^^^" if X units of mass of barium sulphate are produced from a quantity of material numerically equal to the factor 0.13738, the percentage of sulphur is 0.13738 X ol^ X 100 = 100 Z. The per cent of sulphur in the sample is given by the weight of the barium sulphate produced, the decimal point in this case being moved two points to the right. Weights numerically equal to the factor or to some simple multiple or submultiple of the factor which is to be em- ployed are called factor weights. Indirect Analysis. — Of indirect analysis there are so many different cases that all will not be treated individu- ally. A few typ^s will be considered. Case I. — In which an equivalent amount of some other element or radical is determined. In the reac- tion AS2O3 + 2 H2O + 2 12 = 4 HI -I- AssOb, the amoimts of iodine necessary for the oxidation of arse- nious oxide to arsenic oxide being known, the amount of arsenious oxide or arsenic present is readily calculated. 0.1014 g. of a sample containing arsenious oxide required 0.2582 g. of iodine to oxidize to arsenic oxide. The per- centage of arsenious oxide is AsaOs _ 198^ 0^2582 _ TT ~ 507.68 ^ 0.1014 ^ ^"" ~ 99.^7/0 AS2U3, 178 CHEMICAL CALCULATIONS or if the percentage of the metal arsenic had been required, M^ 149^ 0,2582 4 1 507.68 ^ 0.1014 ^ ^^ ~ ^^-^^ /o As. Ca&e II. — Simple cases in which mixtures are weighed and an element or radical in it subsequently determined. Given the following: Amount of substance taken 5 . 0000 g. Which yields 1.0000 g. AI2O3 + Fe^Os Which in turn yields . 1795 g. Fe to calculate the percentages of iron and aluminum in the sample. The percentage of iron is |i^ X 100 = 3.59% Fe. The amoimt of Fe203 equivalent to 0.1795 g. of Fe is ^' = Iiri X ^-1^^^ = ^-^^^^ S. Fe.03, then the amount of AI2O3 which was present in 1.0000 g. of the mixed oxides must have been 1.000 — 0.2566 = 0.7434 g. AI2O3. The per cent of alimiinum is 2A1__ M^ V 0-7434 AI2O3 ~ 102.2 ^ 5.0000 ^ ^"" - 7.897o Ai. Case III. — Simple cases in which mixtures are weighed and the mixture subsequently converted into a single compound. Given: Weight of substance taken 1 . 6000 g. Weight of mixture of AgBr + AgCl obtained. . 1 . 1060 g. Weight of AgCl produced from the mixture ' . . . . 9563 g. to calculate the percentages of bromine and chlorine present in the original sample. 1 The whole mixture AgCl + AgBr is converted into AgCl: X AgCl + V AgBr + I CI2 = (X + y) AgCl + 1 Br^. GRAVIMETRIC ANALYSIS 179 It is evident that an atom of chlorine has replaced an atom of bromine causing a loss of weight of 1.1060 — 0.9563 = 0.1497 g. The loss in molecular weight corre- sponding to this is AgBr - AgCl = 187.80 - 143.34 = 44.46, or what amounts to the same, the difference of the atomic weights of bromine and chlorine : 79.92 — 35.46 = 44.46. The amount of bromine present which would cause this loss of weight is Rr ' 7Q Q2 and the per cent of bromine in the substance taken is 0.2691 1.5000 X 100 = 17.94% Br. The amount of silver bromide equivalent to the bromine, i.e., the amount that must have been present in the mix- ture, is ^^ = ff X 0.2691 = 0.6323 g. AgBr, or what amounts to the same, BF^l = lf X«-^^9^ = «-«^23g.AgBr. The weight of silver chloride present in the mixture is 1.1060 - 0.6323 = 0.4737 g. The per cent of chlorine in the substance taken is CI 35.46^04737 ^,,^,_3^^^C, AgCl 143.34 "^ 1.5000 ' The fraction .= ™ is meant to show that the fraction just after the equality sign is made up of numbers equal to the molec- ular weights of the elements symbohzed and that the operation has been carried out. In this case, Br 70.92 70.92 Br - CI ~ 70.92- 36.46 44.64' 180 CHEMICAL CALCULATIONS To make this calculation algebraically, let X = weight of sUver chloride present in the mixture, y = weight of silver bromide present in the mixture, a = weight of mixed silver salts, h = weight of mixed silver salts converted into silver chloride, / = factor for the conversion of AgBr — > AgCl _ AgCl _ 143.34 _ - AiBr- 187:80 -^■^^^2^- Then x+ y = a, x+fy = b, eliminating x by subtracting and solving for y, a — b substituting the value of /, «-^ «-^ =4.2239 (a-6), " 1 - 0.76325 0.23675 X = a — y. Applying to the problem given: x+ y = 1.1060 X + 0.76325 y = 0.9563 _ 1.1060 - 0.9563 ^ 1 - 0.76325 = 0.6323 g. AgBr. Knowing that the reciprocal of 1 — 0.76235 is 4.2239, it is easier to proceed immediately to the result: y = 4.2239 (1.1060 - 0.9563) = 0.6323 g. AgBr, X = 1.1060 - 0.6323 = 0.4737 g. AgCl, and the percentages of chlorine and bromine in the sample taken are: GRAVIMETRIC ANALYSIS 181 CI ^^■«x?41xioo.7.8i%a. AgCl 143.34 ^ 1.5000 Instead of calculating the chlorine and bromine from the weights of the silver chloride and silver bromide, the amounts of chlorine and bromine can be determined directly, as, for instance, the factor of . ^ being " > A.gijr loV.oU the weight of bromine (z) is 70 no ^ = Sx*-2239(a-6) = 1.7975 (a - b). Multiplying factors are given in chemists' handbooks, the above being a method by which they are calculated.' To calculate the factor for the chlorine: x+fy= b, x+ y = a, b-fa . substituting the values of /, fe- 0.76325 a X = ______ =4.2239(6-0.76325 a) = 4.22396 - 3.2239 a. ine in silver = 0.24738, 0.23675 The factor to determine chlorine in silver chloride being CI 35.46 AgCl 143.34 then, if w is the weight of chlorine, w = 0.24738 (4.2239 6 - 3.2239 a) = 1.0449 6 - 0.79754 a. » This same factor could have been obtained in the method pre- Br 79 92 ceding the algebraic method: ^^ _ ^ = ^^q = 1-7975. 182 CHEMICAL CALCULATIONS . Case IV. — Complex cases in which a known weight of the mixture is subsequently converted into a single element or a radical. Given: Weight of KCl + NaCl in mixture 1.0000 g. Weight of CI in the mixture subsequently determined. . 0.5411 g. it is required to determine the weights of sodium chloride and potassium chloride present in the mixture. Several methods will be given. Method (a). — If the mixture of KCl + NaCl had been pure NaCl the amoimt of chlorine would be CI 35.46 ^ ^ QQQQ ^ 0.60657 g. CI; NaCl 58.46 if it had been pure KCl the amount of chlorine would be j^i = ^g X 1.0000 = 0.47559 g. CI. The difference between the chlorine content if the mixture had been pure sodium chloride or pure potassium chloride is 0.60657 - 0.47559 = 0.13098 g. The observed differ- ence due to the presence of potassium chloride in the mix- ture is 0.60657 - 0.5411 = 0.06547 g. The amount of potassium chloride in the mixture must be, therefore, 0.06547, 0.13098 X 1.0000 = 0.4999 g. KCl. Then the weight of sodium chloride present in the mix- ture is 1 . 0000 - 0.4999 = 0.5001 g. Knowing the amounts of sodium chloride and potassiiun chloride, also the weight taken, the percentages are calculated in the usual way. Method (b). — -If the mixture were pure potassium chloride, the weight of the mixture equivalent to 0.5411 g. of chlorine is KCl 74.56,, „.,,, , ,,^„ -Cr = 35146 ><0-^*ll = l-13^8s. GRAVIMETEIC ANALYSIS 183 1.1378 - 1.0000 = 0.1378 g., excess due to the presence of sodium chloride. The amount of sodium chloride cor- responding to this difference is NaCl _ 58.46 _ 58.46 ^ „ t^ts KCl - NaCl 74.56 - 58.46 ~ 16.10 ^ = 0.5004 g. NaCl, 1.0000 - 0.5004 = 0.4996 g. KCl. Method (c). — Let X = weight of KCl in mixture, y = weight of NaCl in mixture, a = weight of mixture, b = weight of CI in mixture, / = factor to determine the amoimt of CI in KCl = 0.47559, q = factor to determine the amount of CI in NaCl = 0.60657. Then x+ y = a, fx + qy = b. Substituting the values for the particular problem under consideration: x+ 2/= 1.0000 0.47559 X +0.60657 y = 0.541 1 a; = 0.4999 =g. KCl 2/ = 1.0000 -0.4999 = 0.5001 =g. NaCl. Another problem will be cited but slightly different from the foregoing. A sample containing potassiimi iodide and potassium bromide is to be examined for per- centage composition. One gram of the sample yields 1.5370 g. of mixed silver bromide and silver iodide. This mixture is now converted into silver chloride and weighs 1.1191 g. Although this latter compound contains none of the elements sought, these can be readily calculated. 184 CHEMICAL CALCULATIONS If the mixture weighed had been pure silver iodide, the amount of chloride would have been ^^ = ill X 1-^3^0 = 0-^3«^s-^^^^- If the mixture had been pure silver bromide, the amount of silver chloride would have been ^r = i^X 1-^^70 = 1-17^1 ^•^^^^- 1.1191 - 0.9383 = 0.1808 g., excess of silver chloride be- yond what it would be if it were pure silver iodide, the ex- cess being due to the presence of silver bromide. 1.1731 — 0.9383 = 0.2348 g., the difference in weight between what the silver chloride would weigh if derived from pure silver iodide or pure silver bromide. The weight of silver bromide present amounts to 1808 ^^ X 1.5370 = 1.1835 g.AgBr, and 1.5370 - 1.1835 = 0.3535 g.Agl. The percentages of the potassium bromide and potassium iodide in the original sample are: KBr 119.02, 1.1835,, ,„„ ^r n,.r» Tr-n AiB? = 187^0 X roOOO X 100 = 75.00% KBr, KI 166.02 0.3535 Ail = 234:80 X LOOOO X ^^ ' 25-00% KI. Algebraically, the factors being AgI-^AgCl = ^ = 0.61047, AgBr-»AgCl = |g = 0.76325; then, if Solving, GRAVIMETRIC ANALYSIS 185 X = weight of AgBr, y = weight of Agl, x+ y = 1.5370, 0.76325 X + 0.61047 y = 1.1191. X = 1.1835 g. AgBr, y = 1.5370 - 1.1835 = 0.3535 g. Agl. The percentages are calculated as given before. The calculation of a factor in this case is as follows: let X = grams AgBr present in the mixed salt, y = grams Agl present in the mixed salt, a = grams mixed salt, b = grams AgCl obtained from the mixed salt, / = factor for the conversion: AgBr— ♦AgCl, _AgCl_143^_ ~ AgBr - 187.80 ~ ";^'''^^^' k = factor for the conversion: Agl— »AgCl; AgCl _ 143.34 Then Agl 234.80 = 0.61047. x+ V = a, fx + ky = b, Ja y-f-k Substituting the values of / and k, y = 4.9959 a - 6.5454 b. If it is desired to know the amount of iodine in the mixed salts, the factor of iodine equivalent to silver io.dide being T^ = sItIk = 0.54054, then if g is the weight of iodine Agl 234.80 present, q = 0.54054 (4.9959 a - 6.5454 6) = 2.7005 a - 3.5381 6. 186 CHEMICAL CALCULATIONS The above is a general method.^ Having the value of y for any mixed salt of silver bromide and silver iodide, a factor such as q may be calculated for any desired element, radical or compoimd known to be present in the original sample. Case V. — In which a mixture containing two salts having an element or radical, in common is converted into another mixture, also containing an element or radical in common. Given, one gram of a mixture of sodium chloride and potassium chloride; the mixture being con- verted by treatment with sulphuric acid into a mixture of sodium sulphate and potassium sulphate, the weight of this latter mixture being 1.18600 g. It is required to calculate the weights of sodium chloride and potassium chloride present in the original mixture. If the sub- stance had been pure sodium chloride, the weight of so- dium sulphate obtainable would be ^; = iSS X 1-00000 = 1.21510 g. Na^SO. If it had been pure potassium chloride, the weight obtained from one gram would be ^§1 = iSlI ^ 1-00000 =1.16866 g.KaSO^. The difference between these weights is 1.21510 - 1.16866 = 0.04644 g. The difference between the weight found and the weight which would have been obtained if the substance had been pure sodium chloride is 1.18600 - 1.16866 = 0.01734 g. ' For a tabtilation of indirect factors, see Chem. Ann., pp. 37-38. GRAVIMETEIC ANALYSIS 187 Therefore, the amount of sodium chloride present in thp mixture of chlorides is qq^qI^ X 1.00000 = 0.37338 g. NaCl. By the same line of reasoning, the potassium chloride is 1.21510 - 1.18600 = 0.02910. ^^5??1? X 1.00000 = 0.62662 g. KCl, 0.04644 or, by difference, 1.00000 - 0.37388 = 0.62662 g. KCl. To calculate algebraically, let x = weight of NaCl, then 1 — a; = weight of KCl. m = factor ^^^J = 1.2151044, n = factor s^Fpi = 1.168656. 1.215104 a; + 1.168656 (1 - x) = 1.18600, 1.215104 X + 1.168656 - 1.168656 a; = 1.18600, 0.046448 X = 0.01734, ^ = o™ = «-^^^^«^-^^^i- 1.00000 - 0.37338 = 0.62662 g. KCl. To calculate the factor, let X = weight of sodium chloride, y = weight of potassium chloride, a = weight of mixed sulphates, b = weight of mixed chlorides, m = factor ^^' = 1.2151044, 2 NaCl n = factor _ L-p.^ = 1.168656. 188 CHEMICAL CALCULATIONS X = 1.215104 a; + 1.168656 (6 - a;) = a, 1.215104 a; + 1.168656 6 - 1.168656 a; = a, 0.046448 a; = a - 1.168656 6, a- 1.1686566 = 21.5293 (a- 1.168656 6) 0.046448 = 21.5293 a -25.1604 6. 1.168656 y + 1.215104 (6 - y) = a, 1.168656 2/+ 1.215104 6 - 1.215104?/ = a, -0.046448 y=- 1.215104 b + a, y = ''"Sms" ' = 21.5293 (1.215104 6 - a) = 26.1604 6 -21.5293 a. To calculate this factor in another manner, let X = weight of sodium chloride, y = weight of potassium chloride, a = weight of mixed sulphates, 6 = weight of mixed chlorides, m = factor ^^^^' = 1.2151044, 2 NaCl n = factor L. „| = 1.168656. Factor for sodium chloride: Factor for potassiimi chloride: a; + 2/ = 6, mx -\-ny = a, nx + ny = nb, mx + ny = a, {m — n) X = a — nb, a — nb X = m X = m — n a — m -' n 6. ^+ y = i, mx + ny = a, mx + my = mb, mx -\- ny = a, {m — n)y = mb — a, mb — a y m — n m m — n 6- m — n GRAVIMETRIC ANALYSIS 189 Substituting the values of m and n, 1 1.16866 , 0.046448 0.046448 a; = 21.5293 a -25.1604 6. _ 1.215104 1 ^ 0.046448" 0.046448 ' 2/ = 26.16046 -21.5293 a. An inspection of these factors shows that an error made in weighing the mixed chlorides or mixed sulphates is multipUed more than twenty times in calculating the results. This should serve as a warning against using an indirect method of analysis when a direct method may be made to serve. Such methods of indirect analysis, beauti- ful as they may appear on paper, are to be avoided, ex- cept, possibly, in those cases in which the multiplying factor is small. The errors are less when approximately equal weights of the two substances are present. They may become very large when one of the substances is present to a much larger extent than the other. A table of calculations is given on page 190 showing the errors introduced by only one milligram error in weighing the mixed chlorides or sulphates. From an inspection of this table, it is clear that this particular method, at least, is little short of useless, except in exceptional cases and when the weights of the two chlorides are about equal. 190 CHEMICAL CALCULATIONS •* o • o o t-l OS ■g CM M cq lo OS CO 3 O IC -^^ CO 1-H C0 (N iffl CM "3 e : ^ ^ CD ^ CO i-H g) • O CM t-- d c 00 i O 00 lO O TP i-H o i> (>i s ^ s o t> c; CM »o -^ »o 05 CX3 as T— 1 rH •3^. 00 00 00 CD CO CD o o o 00 00 00 ■* tH •* cc CO n 'S a « IS a c 1-1 th cq O O .-H M 03 Tfl a O^ 03 o^ l-H T— I T— ( r- 1^ t~ ?— 1 rH 1— I CM (N C § § § D o o g g 2 o sS- rH iH 1— 1 1-H i-H rH tH T-H rH o o o o o o o o o §=1 o o o o o o ^ M lO JO lO r-H 1— ( rH 03 CT> a> ^ M o o o o d o d d d o 8 3 III 8 S o i ?M H 2 iO lO »o OS O^ 05 tH l-t T-( odd odd d d d Ji -Ji : ^ :Ji : Ji :Ji : .SP -.SP " M ._hp . .SP '.SP • '3 :'3 : 'S **a) ! '33 -'3 m llll ^ m ^ M • .S;2.a 2 .'E .M .'C . j3 '"'S'"©. t-i O h Q, fci o fci a t. o hjii- M-^ bD-y* tn)3 boa S'o s s H C3 H m ^ 1-1 tj 1-1 13 _;t-ip^>-it3 •rtTji-H'S g+l+JJ g + g^ + S g+S+J '•io'i-o'a ■■i-sa-s's ■■s's'a's s Jg^g^ sg^g^ ■ 2 g fl g jn S— S— ' jn a— S-" jg n— jh— ' H H W 1 EH H W 1 H W H 1 GRAVIMETRIC ANALYSIS 191 Mixture of Sulphuric Acid and Sulphuric Anhydride (Oleum). — Another case of mixtures is the acids and their anhydrides. The commonest of these is "oleum" or fuming sulphuric acid, being a mixture of H2SO4 and varying amounts of SOs.^ In commerce an olemn is spoken of as being 30% oleum, for example, when it con- tains 30% by weight of free sulphur trioxide, the remainder being sulphuric acid, the per cent of free sulphur trioxide serving to designate the per cent oleum. Suppose it is required to know the total per cent of sul- phiir trioxide in a sample of oleum. Let X = per cent oleimi, i.e., the per cent uncombined SOs, 100 — X = per cent sulphuric acid, then as the factor „ „^ = „„' = 0.8163, if y is the ii2bv/4 ao.uy per cent total SO3, y = 0.8163 (100 - x) + x. Or being given the per cent total SO3, to find the per cent imcombined SO3, i.e., the per cent oleinn, from the equa- tion above is obtained x = 5.4436 (y - 81.63)." Examples. — (a) What is the percentage total SO3 in 30% oleum? (6) A sample of oleum analyzes 85.31% SO3. What per cent oleum is it? (a) y = 0.8163 (100 - 30) -|- 30 = 87.14% SO3. (6) X = 5.4436 (85.31 - 81.63) = 20.03% oleum. 1 As SO3 has a great affinity for water, oleum can contain no H2O as water. 2 For a table of percentages of oleum corresponding to percent- ages total SO3, see Chem. Ann., pp. 397-398. This table may be calculated by the method given. 192 CHEMICAL CALCULATIONS PROBLEMS 341. 0.2017 g. of a substance containing potassium yields 0.7092 g. of potassium platinic chloride. What is the per cent of potassium? Am. 56.56%. 342. A substance known to be largely NaHS04 • H2O is ex- amined as to its purity; 0.3062 g. of it produces 0.5093 g. of barium sulphate. What is the per cent of acid sodium sul- phate? ' Ans. 98.40%. 343. Calcium fluoride was analyzed by precipitating the fluorine as barium sUicofluoride (BaSiFe) and the calcium was estimated as the oxide. 0.5042 g. of the fluoride gave 0.6002 g. of barium silicofluoride and 0.3630 g. of calcium oxide. What is the per cent composition according to these figures? Ans. 48.52% F. 51.45% Ca. 344. A substance is suspected of being phosphorus pent- oxide (P2O6). 0.5043 g. of the substance dissolved in water gave 0.7835 g. of magnesium pyrophosphate (Mg2P207). Was the substance phosphorus pentoxide? Ans. Yes, 99.10%. 345. A sample of a substance is known to be an oxide of arsenic. 0.2504 g. gave 0.3912 g. of magnesium pjnroarsenate (Mg2As207). Is the substance arsenic oxide (AS2O6) or arsenious oxide (AS2O3)? Ans. 99.57% AS2O3. 346. A sample containing barium, chlorine and water of crystallization was analyzed and the following figures obtained; Weight of sample taken 1 . 0000 g. Weight after heating (water driven off) . 8522 g. Weight of barium sulphate obtained 0.9594 g. Weight of silver chloride obtained 1 . 1735 g. (a) Calculate the percentage of each constituent; (6) the formula of the substance and (c) the per cent errors on the assumption that the sample was pure. Ans. (a) 14.78% H2O; 56.46% Ba; 29.03% CI. (6) BaCl2-2H20. (c) H2O, -1- 0.03%; CI, ± 0; Ba, + 0.23%. GRAVIMETRIC ANALYSIS 193 347. Potassium chlorate was analyzed as follows: the oxy- gen determined by loss of weight on heating; one gram lost 0.3900 g. The chlorine determined by precipitation as silver chloride; one gram producing 1.1594 g. silver chloride. The potassium was weighed as potassium sulphate; one gram giving 0.7098 g. of potassium sulphate, (a) Find the per cent com- position from these figures and (6) the formula. Am. (a) 39.00% 0. 28.68% CI. 31.85% K. (6) KClOs. 348. A sample is analyzed for sodium oxide (Na20) and carbonic anhydride (CO2). 1.3417 g. of sodium sulphate are obtained from one gram of substance. The same amount pro- duces 229 cc. of carbonic anhydride measured moist at 17° C. and 757 mm. (o) What per cent of these constituents are present and (6) what is the formula of the compound? (Ten. aq. vapor at 17° C. = 14.45 mm.) Ans. (a) 58.55% Na20. 41.37% CO2. (6) Na^COa. 349. One gram of a solution of hydrogen dioxide is boiled liberating oxygen 2 H2O2 = 2 H2O + O2. 25.7 cc. of oxygen are produced measured moist at 758 mm. and 15° C. What is the per cent of hydrogen dioxide in the solution? (Ten. aq. vapor at 15° C. = 12.73 mm.) Ans. 3.69%. 360. A coal analyzes as follows: Per cent Volatile combustible matter including sulphur. . . 18. 10 Fixed carbon 74.52 Sulphur 0. 60 Ash 6.65 Water 0.73 100.60 What are these percentages on the dry basis? Ans. 18.23%; 75.06%; 0.61%; 6.70%. 194 CHEMICAL CALCULATIONS 351. A coke analyzes: Per cent Volatile combustible matter 1 . 58 Fixed carbon 88. 87 Sulphur 1. 18 Ash ■.. 8.99 Water 1.92 102.54 What is its composition on the dry basis? Ans. 1.61%; 90.56%; 1.20%; 9.16%. 362. The extracted pigment of a paint analyzes : Per cent White lead 30.42 Zinc oxide 61 . 52 Asbestine 7. 23 Unextracted oil 1.06 100.23 Calculate the percentages present if all the oil had been extracted. Am. 30.74%; 62.18%; 7.31%. 363. A sample of phosphate rock analyzes: Per cent Calcium phosphate 96 . 02 Ferric oxide 1.13 Aluminum oxide . 60 Insoluble 1 . 24 Moisture 0.73 99790 Calculate the percentages on the dry basis. Ans. 96.91%; 1.14%; 0.60%; 1.25%. 364. On the dry basis, a sample of coal analyzes: Per cent Volatile combustible 21 . 06 Fixed carbon 71 . 80 Ash 7.14 100.00 If the moisture present in the coal was 2.49%, what is the analysis on the wet basis? Ans. 20.54% 70.01% 6.96% 2.49% 100.00% GRAVIMETRIC ANALYSIS 195 355. A sample of coal was dried at 70° C, at which tempera^ ture a loss of 6.03% water was obtained. The partially dried sample was analyzed and showed: Per cent Volatile material 8 . 23 Fixed carbon 65.90 Afih 23.80 Moisture 2.07 lOOTOO What are the percentages on the original sample? Am. 7.73% 61.93% 22.36% 7.78% 100.00% 356. The pigment of a certain paint has the composition: Per cent (a) White lead 33.30 (6) Zinc oxide 66.70 100.00 The vehicle of this paint consists of Per cent (c) Linseed oil 32.40 (d) Japan dryer 13 . 60 (e) Asphaltum spirits -. 54.00 100.00 The pigment and vehicle are mixed to form the paint in the proportions: Per cent Pigment 63.00 Vehicle 37.00 100.00 Calculate the percentage composition of this paint. Am. (a) 20.98%. (6) 42.02%. (c) 11.99%. (d) 5.03%. (e) 19.98%. 196 ' CHEMICAL CALCULATIONS 357. A paint has the following composition: Per cent (o) White lead 40.00 (6) Zinc oxide 23.00 (c) Linseed oil 15 . 00 (d) Japan dryer 5 . 00 (e) Asphaltum spirits 17.00 100.00 (a) What is the percentage composition of the pigment? (6) What is the percenta;ge composition of the vehicle? Ans. (a) (a) 63.49%. (6) 36.51%. (6) (c) 40.55%. (d) 13.51%. (e) 45.95%. 358. Phosphoric anhydride (PsOs) is to be weighed as silver phosphate (Ag3P04). What is the factor weight to take in order that the weight of sUver phosphate obtained multiplied by 10 shall give the per cent of phosphoric anhydride? Ans. 1.6968 g. 359. The percentage of phosphoric anhydride (P2O5) is to be determined as magnesimn pyrophosphate (Mg2P207). What weight in grams must be taken in order that the weight of the pyrophosphate multiplied by 100 shall give the percentage of phosphoric anhydride? Ans. 0.6379 g. 360. Bicarbonate of soda (NaHCOs) is to be examined as to its purity. The sodium is converted into sodium sulphate and weighed as such. What weight of bicarbonate must be taken in order that the weight of sodium sulphate obtained multiplied by 100 shall give the per cent of bicarbonate? Ans. 1.1827 g. 361. Manganese dioxide (MnOj) is to be determined as tri- manganese tetroxide (Mn304). What is the factor weight in order that the weight of tetroxide in grams multiplied by 10 shall give the percentage of manganese dioxide? Ans. 11.3987 g. GRAVIMETRIC ANALYSIS 197 362. Sodium ammonium phosphate (NaNH4HP04 • 4 HjO) is to be analyzed. The phosphorus is precipitated as magnesium pyrophosphate (Mg2P207). What is the factor weight in order that the weight of the pyrophosphate in centigrams shall show the percentage of sodium ammonium phosphate? Ans. 1.8781 g. 363. What is the per cent of sulphur dioxide in a sample of sodium sulphite when 0.1063 g. of the same requires 0.1615 g. of iodine to oxidize it according to the reaction: NajSOa + 12 + H2O = NajSOi + 2 HI? Ans. 35.95%. 364. One gram of a sample containing an iodide is treated with chlorine, 2MI + Cl2 = 2MCl + l2, and the iodine liberated is treated with sodium thiosulphate, 2 NaaSsOs + I2 = 2 Nal + NajSiOe, 0.9213 g. of the thiosulphate being required. What is the per cent of iodine in the sample? Ans. 73.94%. 365. 1.5000 g. of a sample give 0.6214 g. of mixed strontium and calcium sulphates. From this mixture 0.4120 g. of calcium carbonate are obtained. Assuming calcium chloride and stron- tium chloride to have been present as such in the original sample, calculate the percentages of these compounds. Ans. 30.46% CaCl^. 3.50% SrCl2. 366. From the following calculate the percentages of iron and aluminum : Weight of substance taken .... 1 .0043 g. Which yields 0.2716 g. AI2O3 + FcjOs Which in turn yields 0.0726 g. Fe. Ans. 7.23% Fe. 8.86% Al. 367. One gram of sample gives 0.1485 g. of mixed calcium and strontiiun carbonates, from which mixtm'e 0.0957 g. of 198 CHEMICAL CALCULATIONS strontium sulphate are obtained. What are the percentages of calcium oxide and strontium oxide? Ans. 5.40% SrO. 4.01% CaO. 368. Calculate the percentages of sodium and potassium, given the following data: Weight of substance taken 2 . 0000 g. Weight of NaCl + KCl obtained.' 0. 6437 g. Weight of KzPtCle obtained from mixture 0. 6514 g. Ans. 5.24% K. 8.92% Na. 369. A substance contains sodium and potassium. It gives 36.42% of mixed sodium and potassium sulphates. From the mixed sulphates is obtained a weight of potassium platinic chloride (KaPtCU) that amounts to 47.20% of the weight of sample taken. What are the percentages of sodium and potas- sium oxide? Ans. 9.15% K2O. 8.51% NaaO. 370. 1.5000 g. of a substance containing bromine and chlo- rine yield 1.0000 g. of mixed silver chloride and silver bromide. The mixture is treated with chlorine, converting the silver bro- mide into silver chloride, the loss of weight due to this change being 0.1367 g. What are the percentages of chlorine and bromine? Ans. 16.38% Br. 6.97% CI. 371. What are the percentages of chlorine and iodine in a substance, 2.0000 g. of which yield 1.5230 g. of mixed silver chlo- ride and iodide, which mixture when converted into the chloride yields 1.1090 g. of silver chloride? Ans. 28.73% I. 15.69% CI. 372. A sample composed of potassium chlorate (KCIO3) and potassium bromate (KBrOs) is examined by the indirect method. One gram was taken from which 1.0474 g. of mixed silver chlo- ride and bromide were obtained. From this mixture 1.0138 g. of silver chloride were obtained. What are the percentages GRAVIMETRIC ANALYSIS 199 of potassium chlorate and potassium bromate in the original sample? Ans. 12.62% KBrOa. 77.42% KClOa. 373. If X — weight of AgCl present, y = weight of Agl present, o = weight of mixture of AgCl + Agl, 6 = weight of AgCl obtained from (a) : (a) Calculate the factor which multiplying (a — h) gives the weight of silver iodide, and (6) the factor that multiplying (o — 6) gives the weight of iodine. Atis. (a) 2.5672 {a — b). (b) 1.3877 (a -6). 374. Taking the factors of problem 373, (a) what are the amounts of silver chloride and silver iodide when the mixed salts weigh 1.8320 g. and the mixed salts converted into the chloride weigh 1.5232 g.? (6) What is the weight of iodine present? Ans. (a) 0.7928 g. Agl. 1.0392 g. AgCl. (6) 0.4285 g. I. 375. What is the factor to be used which multiplying the loss of weight caused by converting a mixture of silver chloride and silver bromide into silver chloride will give the amount of bro- mine present? Ans. 1.7976. 376. What is the factor to be used which multiplying the loss of weight caused by converting a mixture of silver chloride and silver iodide into silver chloride will give the weight of iodine? Ans. 1.3877. 377. One gram of a mixture of silver iodide and silver bfo- mide is converted into silver chloride, yielding 0.6487 g. of the same. What are the weights of iodine and bromine in the mix- ture? Am. 0.1065 g. Br. 0.4053 g. I. 378. 1.0721 g. of a mixture of calcium carbonate and stron- tium carbonate yield 1.4217 g. of a mixture of strontium sul- 200 CHEMICAL CALCULATIONS phate and calcium sulphate. How much calcium carbonate and strontium carbonate were contained in the mixture? Ans. 0.7574 g. CaCOa. 0.3148 g. SrCOa. 379. 0.5541 g. of a mixture of potassium iodide and bromide yield 0.3396 g. of potassium sulphate. How much iodine and bromine in the mixture? Ans. 0.2436 g. I. 0.1581 g. Br. 380. A mixture of potassium sulphate and sodium sulphate weighs 1.4304 g. From this mixture 2.1364 g. of barium sul- phate are obtained, (a) What are the weights of potassium sulphate and sodium sulphate in the mixture? (6) If 1.5000 g. of the substance were taken from which this mixture was ob- tained, what are percentages of sodium and potassiimi in the original sample? Ans. (o) 0.7257 g. NasSOi. 0.7047 g. K2SO4. (6) 15.67% Na. 21.08% K. 381. Calculate the factors that applied to the previous prob- lem (380) will give the weights of potassium and sodium. These are to be multiplying factors. Ans. K = 2.4278 a - 1.4775 6, Na= 1.0662 6 -1.4280 a. 382. One gram of a mixture consisting of silver chloride and silver bromide is found to contain 0.6635 g. of silver, (o) How much bromine and (6) how much chlorine does it contain? Ans. (a) 0.2128 g. Br. (6) 0.1237 g. CI. 383. Calculate factors which when multiplying the weight of the mixed salt of problem 382 and the weight of silver obtained from the mixtiure will give the weights of bromine and chlorine present. Ans. CI = 1.3884 b - 0.79753 a. Br = 1.7975 a - 2.3883 b. 384. A mixture of barium and strontium carbonates weighing 1.2601 g. contains 0.3192 g. of carbon dioxide, (a) What are GRAVIMETRIC ANALYSIS 201 the weights of barium carbonate and strontium carbonate in the mixture? (6) What are the weights of barium and stron- tium in the mixture? Ans. (o) 0.7504 g. BaCOa. 0.5097 g. SrCOa. (6) 0.5223 g.Ba. 0.3025 g. Sr. 385. Calculate the factors that applied to a problem such as 384 will give the weights of barium and strontium. Am. Ba = 2.7611 o - 9.2661 6. Sr = 7.9025 6 - 1.7617 a. 386. A sample of oleum weighing 3.1402 g. contains 2.7350 g. of sulphur trioxide. What per cent oleum is it? Ans. 29.78% oleum. 387. What is the per cent total sulphur trioxide in 25% oleum? Ans. 86.22%. 388. What is the per cent total sulphur trioxide in 15% oleum? Ans. 84.38%. 389. If acetic anhydride (CHaCO)! will take up any water present to form acetic acid, what is the amount of uncombined acetic anhydride present in a sample of mixed acetic acid and acetic anhydride showing 87.72% anhydride present? Ans. 18.13%. CHAPTER IX VOLUMETRIC ANALYSIS "VoLUMETKic analysis or quantitive chemical analysis by measure in the case of liquids and solids . . . depends upon the following conditions for its successful practice: 1. A solution of the reagent, the chemical value of which is accurately known, caUed the 'standard solu- tion. ' 2. A graduated vessel from which portions of it may be accvu-ately delivered, called the 'burette.' 3. The decomposition ^ produced by the standard solu- tion with any given substance must either in itself or by an indicator be such, that its termination is unmistakable to the eye, and thereby the quantity of the substance with which it has combined accurately calculated." ^ Single-factor Solutions. — The standard solution may be made to any desired strength. If a solution is to be frequently used for the determination of a substance, for example iron, it is convenient to adjust its strength so that one cubic centimeter shows the presence of 0.01, 0.001 or 0.0001 g. of iron, in which case the determination of the weight of iron in grams present in the sample is obtained directly by the burette reading. If it is de- sired to make up a solution of silver nitrate, one cubic centimeter of which is to be equivalent to 0.001 g. of chlorine, and the weight of silver nitrate which must be ' "Reaction" might be a better word. ^ Sutton, "Volumetric Analysis." 202 VOLUMETRIC ANALYSIS 203 contained in a liter of the solution is to be calculated, the required weight can be found readily. The reaction is M'Cl + AgNOa = M'NOs + AgCi; The weight of chlorine which is to be indicated by one liter is 0.001 X 1000. = 1.0000 g. The amount of silver nitrate necessary is ^•=ffxl. = 4.™0,. It commonly occurs that a substance with which it is desired to make a standard solution cannot be weighed out in a state of purity; hence it is necessary to standard- ize the solution after it is made up. The methods of doing this are many: typical cases will be considered. Methods of Standardization. — I. By weighing out the amount required and making up to the desired vol- ume. This can be done only when the substance is in a state of purity and the weighing can be accurately per- formed and presents no great difficulties in manipulation. Among such substances may be mentioned sodimn car- bonate, arsenious oxide and sodium oxalate. II. By causing a solution to react with a weighed amount of a chemically pure substance and measuring the volume of the solution necessary to do this. 0.21 194 g. of anhydrous sodium carbonate require 40.00 cc. of a solution of hydrochloric acid to completely neutralize it. The strength of the hydrochloric acid per cubic centi- meter is 2HC1 2 (36.47) ,0.21194 „„„„„^„ tt^, ~ X .„„„ = 0.003646 g. HCl per cc. NaiiCOs 106.00 ^ 40.00 III. By analyzing gravimetrically a measured portion of the solution. If 30.00 cc. of the same solution of 204 CHEMICAL CALCULATIONS hydrochloric acid as above yield 0.4299 g. of silver chloride, the content of hydrochloric acid per cubic centi- meter is HCl 36.47 ,0.4299 „„„„„,„ „^i Aici = 143:34 X 3000- = o-oo^^^ ^- ^^^ P^^ ''■ IV. By comparing a measured volume of the solution with a measured volume of another solution the strength of which is known and with which it reacts. If 45.00 cc. of the same hydrochloric acid solution' react with 45.20 cc. of a solution of sodium hydroxide, each cubic centimeter of which contains 0.003982 g. of sodium hydroxide, the strength of the hydrochloric acid solution is HCl 36.47^45.20X0.003982 „„„oRAfi wm rX TTTSK = 0.003646 g. HCl per cc. NaOH 40.01 45.00 Equivalent Values of Single-factor Solutions. — A solution may be made up to a strength to correspond to a simple factor, as, for instance, the solution of silver nitrate mentioned above, each cubic centimeter of which corresponds to a milligram of chlorine. When such a solution is to be used for another determination the value of the solution per cubic centimeter must be calculated for the new substance. Potassium permanganate is a reagent used in a nmnber of different volumetric determinations. Its value de- pends upon its oxidizing property. Iron is determined by it according to the equation 2 KMn04 + 10 FeS04 + 8 H2SO4 = K2SO4 + 2 MnSO* + 5re2(S04)3 + 8H20; oxalic acid, 2 KMn04 + 5 H2C2O4 + 3 H2SO4 = K2SO4 + 2 MnSO* + IOCO2 + 8H2O; VOLUMETRIC ANALYSIS 205 calcium when in the form of the oxalate, 2 KMnOi + 5 CaCaO* + 8 H2SO4 = K2SO4 + 2 MnSO* + 5 CaS04 + 10 COa + 8 H^O; manganese, 2 KMnOi + 3 MnSO^ + 2 H2O = K2SO4 + 2 H2SO4 + 5Mn02. There are many other similar reactions. If one cubic centimeter of the potassium permanganate solution repre- sents 0.001 g. of iron, then as 2 KMn04 = 10 Fe, and 2 KMn04 = 5 H2C2O4, it follows that 5 H2C2O4 = 10 Fe: the amoimt of oxalic acid equivalent to one cubic centi- meter of this permanganate solution is: ^§^'= imm^ ^-^^^ =0.000806 g. H2C2O4 per cc. In like manner, 2 KMn04 = 10 Fe; 2 KMn04 = 5 Ca; then 5 Ca = 10 Fe, and the amount of calcium indicated by each cubic centimeter of the permanganate solution is Finally, as 2 KMn04 = 10 Fe and 2 KMnO* = 3 Mn, it follows that 3 Mn = 10 Fe,and the value of the perman- ganate solution in terms of manganese is Factor Weights for Volumetric Solutions. — Factor weights ' are as applicable to volumetric as to gravimetric analysis. Acetic acid is to be determined by titration with sodium hydroxide, one cubic centimeter of which contains 0.04006 g. What weight of acetic acid should be taken » See p. 176. 206 CHEMICAL CALCULATIONS such that the burette reading in cubic centimeters may show the percentage of acetic acid directly? One cubic centimeter of the sodium hydroxide solution is equivalent to ^§^ = j^ X 0.4006 = 0.06003 g. HC2H3O2 per cc. Supposing the acetic acid were 100% pure, the mmiber of cubic centimeters required would be 100. This corre- sponds to 100 X 0.06003 = 6.003 g. of acetic acid. If this weight of acetic acid is weighed out and titrated with the sodium hydroxide under consideration, the burette reading in cubic centimeters will give the percentage of acetic acid present. Thus if 85.00 cc. of sodium hydroxide are required for neutralization, the acetic acid contains 85.00% HC2H3O2 as is readily seen by the following: 85.00 X0;06003 ,,,„„ s^ nno/ wn wn g-™ X 100 = 85.00% HC2H3O2. Instead of reading the percentage of acetic acid directly from the burette, it may be desirable to adjust the amount of acid taken, so that the burette reading multiplied by two gives the percentage of acetic acid.^ If half of 6.003 g. (= 3.0015 g.), that is, fifty times the factor, is weighed out and titrated, the burette reading multiplied by two gives the percentage of acetic acid. If 42.50 cc. are required for titration, the percentage of acetic acid is 2 X 42.50 = 85.00%, as is proved: 42.50 X 0.06003 3.0015 X 100 = 85.00% HC2H3O2 ' This would be the case if a 50 cc. burette were used, as it would obviate the trouble of filling the burette a second time if the per- centage exceeds 50%. VOLUMETRIC ANALYSIS 207 To put this in general terms; if the factor weight taken is m times the factor, the reading ^ must be multiplied by n, m and n being so chosen that their product equals 100. Thus, if the factor weight taken is five times the factor, the reading must be multiplied by twenty. Normal Solutions. — "A normal solution of a reagent is one which contains, in a liter, that proportion of its molecular weight in grams which corresponds to one gram (gram atom) of available hydrogen or its equivalent." ^ Submultiples of the normal solution are used and are designated half or seminormal (N/2 or 0.5 N), fifth or quintinormal (N/5 or 0.2 N), tenth or decinormal (N/10 or 0.1 N), hundredth or centinormal (N/100 or 0.01 N), etc. According to definition, the following weights of acids, bases or radicals contained in a liter are normal solutions: Acetic acid (one available H ion) = ^— = 60.032 g. ' The reading may be in cubic centimeters, as in volumetric anal- ysis, or in grams or some multiple of the gram, as in gravimetric analysis. * Sutton, "Volumetric Analysis." The words "gram atom" in parenthesis have been inserted by the -author, as in his opinion Sutton so intended. As the examples given by Sutton show, the nor- mal solution contains 1.008 g. of available hydrogen and not 1.0000 g., as his definition would lead one to suppose. His definition would be exact if hydrogen were taken as unity in the determination of atomic weights. Such, however, is not the case, the basis being oxygen = 16 which makes hydrogen = 1.008. ' The expression is intended to represent a fraction, the numer- ical value of which is obtained by giving to the numerator a number equal to the molecular weight of the symbol. For a Ust of acids and bases and the values of normal solutions of the same, with the indicators to use in each case, see Chem. Ann., pp. 52-53. The solubilities of many substances are too small to permit of the prep- aration of a solution of normal strength. The values given indicate the weight of substance which would be present in a normal solu- tion, solubiUty permitting. Such cases are illustrated by calcium 208 CHEMICAL CALCULATIONS Ammonia Ammonium hydroxide Hydrochloric acid Nitric acid Barium hydroxide Calcium hydroxide Sulphuric acid Oxalic acid Phosphoric acid' Phosphoric acid* (one available NH3 ion) } (one available OH ion) J (one available H ion) (one available H ion) I (two available OH ions) [ (two available OH ions) (two available H ions) (two available H ions) } (one available H ion) = [ (two available H ions) = NH3 1 NHiOH 1 HCl 1 HNOa 1 Ba(OH), 2 Ca(0H)2 2 H^SO^ 2 C2H2O4 H3PO4 1 H3PO4 = 17.064 g. = 35.08 g. = 36.47 g. = 63.048 g. = 85.71 g. = 37.06 g. = 49.04 g. = 45.01 g. = 98.024 g. = 49.01 g. Following is a list of a few oxidizing agents.' As one oxygen ion is equivalent to two hydrogen ions, the list will be evident. and barium hydroxides, which on account of their limited soIubUity must be used in fractional normal solutions. Because of its low solubiUty, calcimn hydroxide is not used in volumetric operations. ' As these solutions are used in neutralization, their value de- pends upon the acidity shown. When methyl orange is used as the indicator, the same shows color change when one hydrogen is replaced. * When phenol-phthalein is used as the indicator, the color change takes place when two hydrogens are replaced. ' For a list of volumetric oxidizing and reducing agents with the values of a normal in each case, see Chem. Ann., pp. 54-55. For a Ust of precipitation agents and data on the strengths of normal solutions of the same, see Chem. Ann., p. 56. VOLUMETRIC ANALYSIS 209 1 Iodine (liberates or is = 1 H ion) = j = 127.92 g. 1 Barium peroxide (liberates 1 O = 2 H ions) = — ^ = 84.7 g. 1 Potassium di-) ,,., , „^ „„. . K2Cr207 ...o , I (liberates 3 O = 6 H ions) = — -: — = 49.08 g. 1 Potassium per- ) ,,., . ., „ .„. ,, KMnOi r,-, an ^ ^ \ (liberates 2i O = 5 H ions)' = = — = 31.63 g. manganate ) 5 1 Arsenious oxide (takes up 2 O = 4 H ions) = -' = 49.5 g. In order to determine the amount of substance which must be present to form a solution of a required normaUty, the particular reaction in which the solution takes part must be known. Thus, in an acid solution, two molecules of potassium permanganate liberate five atoms of oxygen. Permanganic acid is HMnO*; the anhydride would be 2 HMn04 =. MnzO; + H2O. In acid solution, the decomposition may be abbreviated to Mn2O7 = 2Mn0 + 5O. Then as 5 = 10 H, the weight of salt requisite in a liter of a normal solution when used in an acid condition is 2JQ^ = ^^ = 1^ = 31.63 g. KMnO. 10 5 5 In an alkaline solution the essential features may be repre- sented by 2Mn207 = 2Mn02 + 3 0. Hence, when potassium permanganate is to be used in an alkaline solution, the weight necessary for a liter of a nor- mal solution is 2KMnO. ^ KMn04 ^ 15|15 ^ ^g^^^g ^ ^^^^^^ 6 3 3 1 That is, two molecules of KMnO^ yield five oxygens in an acid solution. 2 Potassium permanganate is rarely titrated in an alkaline or neutral solution. An example is the titration of manganese. See 210 CHEMICAL CALCULATIONS The calculation of the normal quantity of a salt is not always such a simple matter. The normal quantity of hydrochloric acid being 36.47, in the reaction NH4OH + HCl = NH4CI + H2O, the weight of ammonium chloride equivalent to this weight of hydrochloric acid, and, consequently, the normal quantity of ammonium chloride in this reaction, is NH4CI 53.502 ,, ^„ ,., ^^ .„„ HCr = -36^ X ^^-^^ = ^^-'^^ S- In like manner the amount of ammonium sulphate is obtained from the equation 2 NH4OH + H2SO4 = (NH4)2S04 + 2 H2O; consequently, the normal weight (the weight in grams per Uter, the normal weight of sulphuric acid being 49.04 g.) is (NH4)2S04 132.15 ^ . . . . „. _„_ H2SO4 = -98:09- ^ ^^-^ = ^^-^^^S- Simplification of Calculations by the Use of Normal Solutions. — Knowing the reaction, when normal quan- tities are used, the weights of the substances involved may be foimd by inspection. Defining oxidation as the process of adding oxygen to, or the removal of hydrogen from, or raising the valence of, the substance oxidized, and reduction as the reverse, the normal weights of com- pounds that are neither acids nor bases, but will react with oxidizing or reducing compounds, are readily deter- mined: The oxidation of iron raises its valence from two to three, a gain of one in valence, which is equivalent to reaction, Prob. 391. For other examples, see phosphoric acid with different indicators, p. 208. When a normal solution of potassium permanganate is spoken of, the solution equivalent to five oxygens is understood unless specifically quaUfied. VOLXIMETRIC ANALYSIS 211 one hydrogen. Hence, the normal amount of iron, i.e., the amount of iron shown by a liter of any normal oxidiz- ing or reducing reagent used for its determination, is 55.84 — ^ — = 55.84 g. Antimonious salts when acted upon by certain oxidizing agents are converted into antimonic salts, i.e., antimony is changed from the trivalent to the pentavalent condition, a gain in valence of two; hence, the weight of antimony shown by a normal weight of oxidizing or reducing substances is Sb 120.2 „_ ,. _ = -_ = 60.10 g. Normal solutions are of such strength that equal volumes are exactly equivalent; thus, one normal cubic centimeter of an acid exactly neutralizes one normal cubic centimeter of a base, producing a neutral solution.' Knowing the normality of a solution, its value in terms of any constituent with which it reacts can be calculated easily, and can often be done by inspection. If 30.00 cc. of an N/10 solution of sodium hydroxide are used to determine hydrochloric acid in a 2.0000 g. sample, the amount of hydrochloric acid present is (1.00 cc. N/10 HCl = Ioo^o = «-oo^«4^^-HCi) 30.00 X 0.003647 = 0.10941 g.; ' There is another system of concentrations. This is the Molar system and is used chiefly in physical chemistry. A molar solution contains one gram molecule of solute per liter of solution. Under this definition a molar solution of hydrochloric acid contains 36.468 g. HCl per liter as does also a normal solution. A molar solution of sulphuric acid contains 98.086 g. H2SO4 per liter, while a normal solution contains 49.043 g. So it is seen that a molar and a normal solution may or may not be equivalent. 212 CHEMICAL CALCULATIONS and as 2.0000 g. of sample were taken, the percentage of hydrochloric acid is 30.00 X 0.003647 ^ ^.„ . .-^ „p, 2:0000 ^ ^^ = ^-^^^^ ^^^- If the 30.00 cc. had been used in the determination of sulphuric acid, the same amount being taken (1.00 cc. 1 XT Qpi N/10 HaS04 = looo^io = 0.004904 g. H2SO4), the result would have been ^"•"%>^,y^»^ X 100 = 7.36% H.SO. Again, potassium permanganate is used to determine iron, calcium in the form of oxalate, and also nitrous acid. 45.00 cc. of a 0.1 N^ solution of potassium permanganate are required to bring about the reactions. The amounts present can be readily determined without recourse to stoichiometric ratios in the form of a reaction. Iron is oxidized by permanganate, the valence being increased by one. If all the iron were present in the ferrous con- dition, then as the amount of iron shown by one cubic centimeter of 0.1 N potassium permanganate is iximxro = '■'''''' ^■' the total amount of iron is 45.00 X 0.005584 = 0.2513 g. Fe. The calcium is separated as the oxalate which on treat- ment with sulphuric acid liberates one molecule of oxalic acid for each atom of calcium present. The oxalic acid is oxidized by the permanganate, one atom of oxygen being taken up for each molecule of oxalic acid present. Then, as one oxygen is equivalent to two hydrogens, the • The normality of a solution may as well be expressed as a decimal as a fraction; in fact, the decimal system is to be preferred. VOLUMETRIC ANALYSIS 213 amount of calcium equivalent to a cubic centimeter of a tenth normal oxidizing solution is 2-xOTxlO = 0-0020035 g. The amount of calcium present is 45.00 X 0.0020035 = 0.090158 g. Ca. Potassiiun permanganate oxidizes nitrous acid to nitric acid, a gain of one atom of oxygen which is equivalent to two hydrogens. The amount of nitrous acid equivalent to a cubic centimeter of a decinormal solution is 2 xWkl0 = 0002^^1 S- The amount of nitrous acid present is 45.00 X 0.002351 = 0.1058 g. HNO2. Calculation of Normality. Factor to a Given Nor- mality. — Thus far solutions have been considered as being adjusted to the normality indicated. This is a matter of considerable difficulty and it is a general prac- tice to calculate the strength of solutions by methods already given, not attempting to have the normahty more than approximate, the exact strength, however, always being known. A solution shows on standardization 0.02458 g. of sulphuric acid per cubic centimeter. A nor- mal solution of this acid contains 0.049045 g. per cubic centimeter; then the normality of this solution is 0.02458 _o5oigj.^ 0.049045 ~"-^0^^^- This shows the solution to be very nearly half normal, being a little strong. The factor to exactly half normal is 0-5012 _ 0.5000 - ^OO^- 214 CHEMICAL CALCULATIONS The use of this factor is to calculate the number of cubic centimeters used in an analysis to an equivalent number of cubic centimeters of exactly half normal, that the amount of substance indicated by it may be readily deter- mined as previously given. For example, 30.25 cc. of this sulphuric acid are used in determining sodium hydrox- ide. The number of cubic centimeters of exactly N/2 sulphuric acid equivalent to this number of cubic centi- meters is 30.25 X 1.002 = 30.31 cc. The amount of sodium hydroxide present is 2x'm0 ^ ^^-^^ = 0.6064 g. NaOH. It is often desirable to know the number of cubic centimeters of a solution of given normality equivalent to a given number of cubic centimeters of another solu- tion of different normality. This number is inversely proportional to the normality of the solutions under consideration; the greater the normality, the fewer the number of cubic centimeters necessary to bring about the reaction. It is at once apparent that it will require twice as many cubic centimeters of a half normal solution (N/2 or 0.5 N) to accomplish the same amount of chemical change as with a normal solution (N, N/1 or IN). Nor- malities expressed in the decimal system (0.5 N, 1 N) readily lend themselves to calculations of this natm-e. ' With a little practice and familiarity, the use of solutions stand- ardized in the normal system wiU become so easy that in large part the calculations may be performed mentally. An operator knows that a normal solution of any acid indicates 0.04001 grams of sodium hydroxide per cubic centimeter, knowing the molecular weight of sodium hydroxide to be 40.01, and a half normal solution being of haJf the value of a normal solution, the value 0.020005 g. of sodium hydroxide indicated by an N/2 solution of any acid is obtained by inspection. VOLUMETRIC ANALYSIS 215 Thus, how many cubic centimeters of a 0.1045 N solution are equivalent to 35.00 cc. of a 0.1003 N solution? The nimiber of cubic centimeters required being inversely proportional to the normalities or strengths of the solu- tions gives ^4S?1 X 35.00 = 33.59 cc. of 0.1045 N sol. = 35.00 cc. 0.1045 of 0.1003 N sol. This may be expressed in algebraic form. Let. V = number of cubic centimeters requisite for a cer- tain reaction, N' = normahty of the solution corresponding to V cubic centimeters, V" = number of cubic centimeters of a different strength requisite for the same reaction, N" = normality of the solution corresponding to V" cubic centimeters. Then V'N' = V"N". Taking the same example; substituting gives V X 0.1045 = 35.00 X 0.1003, V = ^4?S X 35.00 = 33.59 cc. 0.1045 To state another problem bearing on this same subject. 11.00 cc. of a 0.5200 N solution of hydrochloric acid are required to neutralize 50.00 cc. of a solution of barium hydroxide. What is the normality of the barium hydrox- ide solution? The normality being inversely proportional to the number of cubic centimeters requisite: 216 CHEMICAL CALCULATIONS ^J^ X 0.5200 = 0.1 144 N, or substituting in the formula gives 50.00 X N' = 11.00 X 0.5200, N' = ^^ X 0.5200 = 0.1144 N. Again, how many cubic centimeters of a solution whose factor to N/10 is 0.9924 are equivalent to 18.00 cc. of a solution whose factor to N/2 is 1.005? The number of cubic centimeters of N/10 solution equivalent to a solu- tion whose factor to N/10 is 1.005, is 18.00 X 1.005 = 18.09 cc. This number of cubic centimeters of N/10 solution is equivalent to 18.09 X 5 = 90.45 cc. of N/2 solution. Then the number of cubic centimeters of a solution whose factor to N/2 is 0.9924 equivalent to 90.45 cc. N/2 solution is S X ^0-4^ = 91-14 cc. Solving in another manner: The normality of the first solution, expressed decimally, is 0.1 X 0.9924 = 0.09924 N. The normality of the second solution, similarly expressed, is 0.5 X 1.005 = 0.5025 N. Then 0.5025 0.09924 X 18.00 = 91. 14 cc. Volumetric Determinations Using Two Solutions. Back Titrations. — Many volumetric analyses are car- ried out by the use of two solutions. These methods generally consist in adding an excess of the active reagent and determining this excess by means of a suitable second solution with which the first reacts. It is nec- essary to know the relative strengths of the two solutions employed so that the amount of the solution used in pro- VOLUMETRIC ANALYSIS 217 ducing the desired reaction may be ascertained. When the two solutions are exactly equivalent, cubic centimeter to cubic centimeter, subtraction of the volumes used gives the amount of the first solution necessary to bring about the desired reaction. A sample of calcium carbonate is to be analyzed, phenol-phthalein being used as the indicator, which is affected by the carbonic acid liberated when the calcium carbonate is treated with an acid; hence, the end point in this titration would occur too soon, were it attempted to titrate directly to a finish with sulphuric acid alone. A solution of sulphuric acid of exactly 0.1 N strength and a solution of exactly 0.1 N sodium hydroxide are em- ployed. 0.2000 g. of the sample are weighed out, water added, and treated with 45.00 cc. of the 0.1 N sulphuric acid and the solution boiled to expel the carbon dioxide. The solution is now acid with sulphuric acid, 6.45 cc. of 0.1 N sodium hydroxide being required to produce neutrality. The amount of sulphuric acid used in de- composing the calcivun carbonate is evidently 45.00 — 6.45 = 38.55 cc. of 0.1 N solution. Assuming that there are no other carbonates present, the equivalent of one cubic centimeter of 0.1 N solution of sulphuric acid in terms of calcium carbonate being ^ = 0.0050035 g., the percentage of calcium carbonate in the sample taken is 38.55 X 0.0050035 0.2000 X 100 = 96.44% CaCOa. Calcium acetate, Ca(C2H302)2 • H2O, is to be examined for the percentage of the same present in a sample weigh- ing 3.7246 g. The calcium acetate is digested with an excess of syrupy phosphoric acid, the acetic acid liberated being distilled into 50.00 cc. of 1.005 N sodium hydroxide 218 CHEMICAL CALCULATIONS solution. The excess of sodium hydroxide required 55.25 cc. of 0.2012 N sulphuric acid for neutralization. 50.00 X 1.005 = 50.25 cc. of 1 N NaOH solution taken, 2012 K^^ X 55.25 = 55.58 cc. of 0.2 N H2SO4 for back titration, 55.58 cc. of 0.2 N H2SO4 s 0.2 X 55.58 = 11.12 cc. of 1 N H2SO4, 50.25 - 11.12 = 39.13 cc. of 1 N NaOH consumed. Then, as CaCCaHaOa)^ • H2O = 2 HC2H3O2 = 2 NaOH, the amount of calcium acetate indicated by one cubic centi- meter of normal sodium hydroxide being Ca(C2H302)2 • H2O _ riQcnRf; ^ 2^^1060 0.088065 g., the per cent of calcium acetate is 39.13 X 0.088065 3.7246 X 100 = 92.52% Ca(C2H302)2 • H2O. Titration of "Mixed Acid." — A mixture of sub- stances may be analyzed volumetrically. A sample of "mixed acid" contains sulphuric acid, nitric acid and lower oxides of nitrogen, chiefly N2O3, which result from mixing the two acids. A sample is taken, treated with water and evaporated on the steam bath, the operation being repeated twice, which treatment leaves only sulphuric acid which is titrated with standard alkali solution. A second sample is titrated with standard alkali for total acidity and a third sample with potassium permanganate for nitrous anhydride. VOLUMETRIC ANALYSIS 219 D(da for determination of mixed acid : For total acid : Weight of sample taken = 4.6514 g. NaOH to neutralize = 74.10 cc. 1.00 cc. NaOH sol. = 0.052109 g. H2SO4. Per cent total acid as H2SO4 is 74.10 X 0.052109 ,, ,„„ „„ „, ^ ^^ggj^ X 100= 83.01%. For sulphuric acid (HNO3 + HNO2 being driven off by evaporation) : Weight of acid taken = 8.8872 g. NaOH to neutralize = 72.20 cc. 1.00 cc. NaOH sol. = 0.052109 g. H2SO4. Per cent sulphuric acid is 72.20 Xa052109 ^,„„.^^ 33^^ For nitrous anhydride: Weight of sample taken = 17.000 g. 0.1 N KMn04 to oxidize N2O3 to N2O6 = 8.60 cc. 1.00 cc. of 0.1 N KMn04 = 0.001900 g. N2O3. Per cent nitrous anhydride is ?:5«^^^ = 0.096%= 0.10%. To calculate the composition of the mixed acid: 83.01 - 42.33 = 40.68% HNO3 + HNO2 as H2SO4. The amount of acidity as nitric acid is 2 HNO3 2 (63.02) H2SO4 98.09 as HNO3. X 40.68 = 52.27% HNO3 + HNO2 220 CHEMICAL CALCULATIONS The equivalent of N2O3 in HNO3 is The amount of nitric acid present is 52.27 - 0.16 = 52.11% HNO3. From these figures the analysis of the mixed acid is Per cent Sulphuric acid 42 . 33 Nitric acid 52. 11 Nitrogen trioxide . 10 Water (by difference) 5.46 100.00 Volumetric Analysis Using Two Indicators. — Mix- tures may also be analyzed by taking advantage of the conduct of different salts toward indicators.* For exam- ple, phenol-phthalein reacts alkaline to sodium carbonate, neutral to bicarbonate, and acid to carbonic acid." If a mixture of sodium hydroxide and sodium carbonate is treated with a known amount of hydrochloric acid till the solution reacts just acid to phenol-phthalein, the following will be the reaction: (1) X NaOH + y NajCO, + (a; + y) HCl = y NaHCOa + XH2O + (x + 2/)NaCl. If the titration is continued, boiling out the carbonic acid, the solution will react neutral to methyl orange when the bicarbonate is decomposed: (2) y NaHCOa + y HCl = y NaCl + 2/ H2O + j/ CO2. ' See Chem. Ann., p. 51. 2 Phenol-phthalein as an indicator is used to show the presence of minute amounts of strong bases or weak acids. Methyl orange indicates the presence of minute amounts of weak bases and strong acids. For strong acids and strong bases either wiU serve. Methyl red is an indicator which acts similarly to methyl orange and pos- sesses advantages over the latter, being nearly colorless in a basic and a strong red in an acid solution. VOLUMETRIC ANALYSIS 221 Then, if p represents the total hydrochloric acid used, and q the hydrochloric acid used to convert the sodium hydrox- ide and the sodium carbonate into sodium chloride and sodium bicarbonate, x + 2t/ = p x-V y = q y = p-q X = 2q — p. To illustrate: one-half gram of a sample composed of sodium hydroxide and sodium carbonate requires 62.50 cc. N/10 hydrochloric acid to neutral reaction with phenol- phthalein. The titration is continued; 100.00 cc. of the same acid are required for complete neutrality (methyl orange). The last addition of hydrochloric acid is a meas- ure of the sodium carbonate present, for all the sodium hydroxide was neutralized with the first addition of acid. 100.00 - 62.50 = 37.50 cc. of N/10 HCl for NaHCOj -*NaCl, H2O, CO2. The amoimt of bicarbonate is equivalent to the amount of sodium carbonate originally present. 1 cc. of N/10 HCl = .^^^L = 0.010600 g. of sodium carbonate, as only one acid hydrogen is present in the acid carbonate. Then 37.50X0^01060 ^ ^^^ ^ ^^^^^^ ^^^^^ U.5UUU The same amount of hydrochloric acid is used to convert the bicarbonate into sodium chloride, carbon dioxide and water as was required to convert the sodium carbonate into bicarbonate. Therefore, 62.50 - 37.50 = 25.00 cc. of N/10 HCl, used in reacting with the sodium hydroxide. 222 CHEMICAL CALCULATIONS 1 cc. of N/10 HCl = in^n^^in = 0-004001 g. sodium hy- droxide. Then 25.00 X 0.004001 ^, , __ „ ., „ ,, „„ Q^gQQQ X 100 = 20.01% NaOH. Algebraically, according to the equations given, letting X = cc. of N/10 acid combining with NaOH, y = cc. of N/10 acid required to change NaaCOa to NaHCOa and NaHCOg to NaCl, CO2 and H2O. Then X = 125.00 - 100.00 = 25.00 cc. N/10 HCl, y = 100.00 - 62.50 = 37.50 cc. N/10 HCl. The calculation is completed as given above. Volumetric Titration of Oleum. — When an oleum contains free sulphurous anhydride, an interesting and important case of indirect volumetric analysis results. Such an oleum contains sulphuric acid, sulphurous an- hydride (SO2) and sulphuric anhydride (SOa).^ A weighed sample is dissolved in water and titrated with a standard alkali solution, when all the constituents are acted upon. Thus, for sulphuric acid: H2SO4 + 2 NaOH = Na^SOi + 2 H2O. The sulphuric anhydride dissolves in water to form sul- phuric acid and is titrated: SO3 -t- H2O = H2SO4, H2SO4 + 2 NaOH .= Na2S04 + 2 H2O. Likewise, the sulphurous anhydride dissolves: SO2 -f- H2O = H2SO3. ' There may be other impurities, such as soUd particles, etc., but for these calculations, only the three constituents enumerated will be considered as being present. The method is easily extended to cover impurities. VOLUMETRIC ANALYSIS 223 When phenol-phthalein is used as an indicator, the reac- tion with the alkali is H2SO3 + 2 NaOH = NaaSOa + 2 H2O, while with sodium hydroxide, using methyl orange as an indicator, it is H2SO3 + NaOH = NaHSOa + H2O. The sulphurous anhydride is estimated in a separate sample by dissolving the oleum in water and titrating against an iodine solution: H2SO3 + H2O + I2 = H2SO4 + 2 HI. An example of an oleum analysis follows: 5.0000 g. of an olernn are dissolved in water and the volume made up to 500 cc. Of this solution, 100 cc. (= 1.0000 g. of sample) are taken and titrated with N/10 iodine solution, 7.80 cc. being required. A similar portion is taken and titrated with N/5 sodium hydroxide, using phenol- phthalein as an indicator, 122.81 cc. being required. To calculate the composition of the oleum: I:»^;M^X100 = 2.50%SO., 122.81 N/5 sol. = 2 X 122.81 = 245.62 cc. N/10 sol., 245.62 - 7.80 = 237.82 cc. N/10 NaOH required for the titration of the sulphuric acid and sulphur trioxide.^ ' The 7.80 cc. are subtracted, this being the number of cubic centimeters of N/10 solution of sodium hydroxide used in neutral- izing the sulphurous acid. If methyl orange had been used, 253.42 cc. of N/10 sodium hydroxide would have been necessary for -the total acidity titration, but 15.6 co. of N/10 solution would have to be deducted from the 253.42 cc. N/10 sodium hydroxide required for the total acidity, leaving 237.82 cc. necessary for the sulphuric acid and the sulphuric anhydride, as before. 224 CHEMICAL CALCULATIONS 237.82 X 0.0040035 ^ , „. or; oi o/ +„+oi cin X 100 = 95.21% total SO3, 95.21 + 2.50 = 97.71% SO3 + SO2, 100.00 - 97.71 = 2.29% H2O. Calculating this percentage of water into equivalent sul- phuric acid: ^^ X 2.29 = 12.47% H2SO4. Then 100.00 - (12.47 + 2.50) = 85.03% free SO3. To summarize, the oleum is composed of Per cent H2SO4 12.47 SO3 85.03 SO2 2.50 To calculate algebraically, let X = per cent of H2SO4, y = per cent of SO3, z = per cent of SO2, A = percentage total acidity as H2SO4, not including that due to SO2, /.t.*r!||l-- 1^ = 1.22505. Then x + y + z = 100 x'+ y =100-2: X = 100 - (z + y). From the conditions of the problem: x+ y = 100-z x+fy = A ^ A + z- 100 ^ A + z- 100 ^ / - 1 ~ 0.22505 = 4.4436 {A + z- 100). VOLUMETRIC ANALYSIS 225 Solving the given problem by this method: a = 2.50 (as before), ^ _ 237.82 X 0.0049045 ,,,„„ iirr.o/ LOOOO ^ ^ ^ 116.64%, y = 4.4436 (116.64 + 2.50 - 100) = 4.4436 X 19.14 = 85.05%, a; = 100 - (2.50 + 85.05) = 12.45%, which results are the same as were obtained by the method given before.! Adjustment of the Strength of Solutions. — Having a solution containing 47% HNO3, and a weaker solution of 23% HNO3, let it be required to prepare, by mixing these two solutions, 200 kg. of a solution which wiU contain 39% HNO3. To find the weights of the two solutions, which, when mixed, will give the desired quantity of acid of the specified strength, let y = the weight of the weaker acid (23% HNO3) to be taken; then 200 - y = the weight of the stronger acid (47% HNO3) to be taken. Algebraically: 0.23 y + 0.47 (200 - y) = 0.39 (200), y = 66.67 kg. 23% HNO3, 200 - 66.67 = 133.33 kg. 47% HNO3. Using this method, a general formula may be obtained. Let A = strength of stronger solution, B = strength of weaker solution, D = strength of desired solution, X = amount of stronger solution to be taken, y = amount of weaker solution to be taken, 3 = amount of solution desired. * An oleum analyzed by these methods must necessarily foot up to 100% as the free SO3 is calculated by difference. 226 CHEMICAL CALCULATIONS Then By + A{z-y)= Dz, By + Az — Ay = Dz, yiB-A)=z(D-A), _ D-A _ A-D y~ B-A^~ A-B^' x + y = z, X — z — y. Again, suppose the problem is to alter the strength of a definite quantity of acid. Given, an acid of 47% HNO3, let it be required to dilute 200 kg. of this acid to 39% HNO3, using 23%HN03. What weight of 23% HNO3 must be added to 200 kg. to make the mixture 39% HNO3? Using the same symbols as before: 0.47 (200) + 0.23 y = 0.39 (200 + y), y = 100 kg. 23% HNO3. In general terms Ax + By = D{x + y), y{B-D) = x{D- A), _ D-A A -D y- B-D^~ D-B^- The following formulas, derived by the processes given above, are very convenient for adjusting the strength of solutions. In these, let A = actual concentration of the solution that is to be corrected, D = desired concentration, B = concentration of diluting solution, C = concentration of strengthening solution, X = amount of stronger solution to be added, taken or prepared, VUL, UMiSTBIC ANALYSIS 227 y = amount of weaker solution or water to be added or taken, z = amount of solution desired or given. Formulas (a) are to be used when a definite amount of the solution is to be prepared. Formulas (b) are to be used when a definite amount of solution is to be corrected, i.e., strengthened or weakened. I. To dilute a solution with water: ' {a) -^z = x, y = z-x. Q>) — Q— 3: = y. II. To dilute a strong liquid with a weaker liquid: (a) ^ _^ z = 2/, x = z-y. ' The Giles flask is a convenient, instrument for preparing a solution of definite strength. The neck of the flask is blown out into a bulb. The lower mark below the bulb is the mark for the indicated capacity of the flask, while above the bulb is another mark indicating a capacity between it and the lower mark of one- tenth the indicated capacity of the flask. To make up a standard solution, 11/10 plus a small amount required for the volume indicated are weighed out and introduced into the flask. Water is filled to the upper mark and a sample pipetted out of the flask, such sample not exceeding the capacity of the bulb between the upper and lower marks. This is titrated and the solution drawn off to the lower mark (the mark of the indicated capacity). As the solution was purposely made a trifle strong, water must be added to adjust to the strength desired. The amount of liquid being known, the amount of water to add is calculated by (6). 228 CHEMICAL CALCULATIONS III. To strengthen a weak solution with a stronger solution: D- -A C - -A D- -A (^) c^rp2/ = ^-' PROBLEMS 390. How many grams per liter must a solution of potassium dichromate contain so that each cubic centimeter shall show 0.005000 g. of iron according to the reaction : 6 FeCl2+K2Cr207+l4 HC1=6 FeCl3+2 KCl+2 CrCla+T HjO? Ans. 4.3905 g. 391. How much potassium permanganate per liter must a solution contain to be of such strength that each cubic centi- meter is equivalent to 0.001000 g. of manganese? The reaction is 2 KMn04 + 3 MnSOi + 7 H^O = 2 KHSO4 + H2SO4 + 5 H^MnOa. Ans. 1.9180 g. 392. It is desired to make up a solution of potassium dichromate of such strength that one cubic centimeter will indi- cate 0.001000 g. of iron, (a) How many grams of the potas- sium dichromate must be contained in a liter? (6) How many 1 These formulas are universally accurate when the weights of solution are considered. If the solutions are dilute, the specific gravities may be considered, with but Uttle error, as being the same as water. In this case, the formulas may be used for volumes. When this assumption is not permissible, the weights may be cal- culated, and knowing the specific gravities of the components, the volumes requisite calculated from the formula, volume = sp. gr. On mixing such concentrated solutions, to use these formulas it must be assumed that the volumes are additive, i.e., no change of volume takes place. VULiUMJUTJilC AJNALiYiSlS 229 grams of potassium dichromate, that each cubic centimeter shall show 0.001000 g. Fe203? The reaction is 6 FeClj+KzCraOr + 14 HCl = 6 FeCU + 2 KC1 + 2 OrCla + 7 H2O. Ans. (o) 0.8781 g. (6) 0.6141 g. 393. How many grams of potassium permanganate per liter must a solution contain, each cubic centimeter of which is to be equivalent to 0.001000 g. of iron according to the equation: 2 KMnOi + 10 FeSOi + 8 H2SO4 = K2SO4 + 2 MnSOi + 5Fe2(S04)3 + 8H20. Ans. 0.5660 g. 394. Calcium is separated as the oxalate which is treated with sulphuric acid and the oxalic acid liberated is titrated with potassium permanganate. The reactions are CaCl2 + (NH4)2C204 = CaC204 + 2 NH4CI, CaC204 + H2SO4 = CaS04 + H2C2O4, 5 H2C2O4 + 2 KMn04 + 3 H2SO4 = K2SO4 + 2 MnSOi + IOCO2+8H2O. How many grams per liter of potassium permanganate are re- quired that one cubic centimeter shall show (a) 0.001000 g. of calcium? (6) 0.001000 g. of calcium oxide? (c) 0.01000 g. of CaSOi • 2 H2O? Ans. (a) 1.5776 g. (6) 1.1300 g. (c) 3.6715 g. 395. What weight of silver nitrate must be taken, dissolved and made up to a liter that each cubic centimeter shall be equiva- lent to a milligram of sodium chloride according to the reaction: NaCI -1- AgNOs = NaNOs + AgCl. Ans. 2.9060 g. 396. Arsenious oxide acts as a reducing agent: AS2O3 + 20 = AS2O5. How much arsenious oxide must be weighed out, dissolved and made up to a liter that each cubic centimeter shall take up 0.001000 g. of oxygen or its equivalent? Ans. 6.1850 g. 230 CHEMICAL CALCULATIONS 397. A solution contains 67.0000 g. of sodium oxalate in a liter. If it reacts as follows: NasCjOi + H2SO4 = NaaSOi + H2C2O4, H2C2O4 = H2O + CO2 + CO, CO + = CO2, how much oxygen is taken up per cubic centimeter? Ans. 0.0080 g. 398. A solution contains 10.0000 g. of silver nitrate made up to a liter. What is the content of silver per cubic centimeter? Ans. 0.006350 g. 399. 30.20 cc. of a solution of sulphuric acid neutralizes 0.1708 g. of pure anhydrous sodium carbonate. What is the amoimt of sulphuric acid in the solution per cubic centimeter? Ans. 0.0052335 g. 400. 0.5000 g. of pure arsenious oxide are dissolved: 100.00 cc. of a solution of iodine are necessary to oxidize the arsenious oxide : AS2O3 + 2 12 + 2 H2O = AS2O5 + 4 HI. What is the strength of the iodine solution per cubic centimeter? Ans. 0.012826 g. 401. 0.1062 g. of iron 99.8% pure are converted into ferrous sulphate and require 19.00 cc. of potassium permanganate to oxidize to ferric sulphate: 2 KMn04 + 10 FeS04 + 8 H2SO4 = K2SO4 + 2 MnSOi + 5Fe2(S04)3 + 8H20. What is the content of potassium permanganate per cubic centi- meter in this solution? Ans. 0.003157 g. 402. 0.2680 g. of pure sodium oxalate are weighed out and dissolved in water. 40.00 cc. of potassium permanganate solution are required to react with this amount of sodium oxa- late: 2 KMn04 -I- 5 NajCzOi -|- 8 H2SO4 = K2SO4 + 2 MnSOi + 5 Na2S04 -i- 10 CO2 -I- 8 HjO. VOLUMETRIC ANALYSIS 231 What is the content of potassium permanganate per cubic centimeter? Ans. 0.0031606 g. 403. 50.00 cc. of a solution of hydrochloric acid are treated with silver nitrate, yielding 0.7167 g. of silver chloride. What is the weight of hydrochloric acid per cubic centimeter? Ans. 0.003647 g. 404. A solution of sulphuric acid is standardized by precipi- tation as barium sulphate. 30.00 cc. gave 0.3500 g. of barium sulphate. What is the strength of the sulphuric acid solution per cubic centimeter? Ans. 0.004903 g. 405. 45.00 cc. of a solution of barium hydrate when precipi- tated with sulphuric acid yield 0.6842 g. of barium sulphate, (a) What is the content of barium hydroxide per cubic centi- meter? (6) To how many grams of hydrochloric acid per cubic centimeter is this solution equivalent? Ans. (a) 0.011164 g. (6) 0.004751 g. 406. 25.00 cc. of a solution of sulphuric acid yield 0.5852 g. of barium sulphate. What is the content of sulphuric acid per cubic centimeter of this solution? Ans. 0.0098355 g. 407. What is the strength of a solution of barium hydroxide if 30.00 cc. require for neutralization 59.80 cc. of a solution of sulphuric acid known to contain 0.004900 g. per cubic centi- meter? Ans. 0.01707 g. 408. It is found that 32.50 cc. of a solution of sodium hydrox- ide exactly neutrahze 30.00 cc. of a solution of sulphuric acid known to contain 0.04902 g. per cubic centimeter. What is the strength of the sodiiun hydroxide solution per cubic centimeter? Ans. 0.03691 g. 409. A solution of sodium hydroxide is equivalent to 0.004875 g. of sulphuric acid per cubic centimeter, (a) What is its equiv- alent in nitric acid per cubic centimeter? (6) Hydrochloric acid? Ans. (a) 0.006264 g. (6) 0.003625 g. 232 CHEMICAL CALCULATIONS 410. One cubic centimeter of a solution of potassium perman- ganate is equivalent to 0.005000 g. of iron. What is its equiv- alent of KH3(C204)2 • 2 H2O, the reaction being 10 KHaCCjOOa + 8 KMn04 + 17 H2SO4 = 9 K2SO4 + 8 MnS04 -I- 40 CO2 + 32 H2O? Ans. 0.005689 g. 411. One cubic centimeter of a solution of potassium perman- ganate is equivalent to 0.005585 g. of iron. To how much potassium ferrocyanide, K4Fe(CN)6 • 3 H2O, per cubic centi- meter, is this solution equivalent, the reaction being 10 K4Fe(CN)6 -I- 2 KMn04 -|- 8 H2SO4 = 10 KjFe(CN)6 -I- 6 K2SO4 + 2 MnS04 + 8 H2O? Ans. 0.042236 g. 412. A solution of potassium dichromate is equivalent to 0.005000 g. of iron per cubic centimeter : KzCrjOy-l-e reCl2+ 14 HCl = 2 KC-|-2 CrCl3-l-6 FeCls-f-T H2O; if the solution is used to determine antimony: K2Cr207-|-3 SbCl3+14 HC1= 3 SbCl5+2 CrCl3+2 KCl+7 H2O, what is the equivalent of antimony per cubic centimeter of this solution? Ans. 0.005381 g. 413. A permanganate solution is equivalent to 0.005000 g. of iron per cubic centimeter according to the reaction: 2 KMn04 + 10 FeS04 -|- 8 H2SO4 = K2SO4 + 2 MnS04 -f5Fe2(S04)3 + 8H20; if it is used to determine nitrous acid according to the equation: 2 KMn04 -I- 5 HNO2 -|- 3 H2SO4 = K2SO4 + 2 MnS04 4-5HN03 + 3H20: (a) What is the value of the permanganate solution per cubic centimeter in terms of nitrous acid? (6) In terms of nitrous anhydride, (N2O3)? Ans. (a) 0.002105 g. (6) 0.001702 g. 414. How many grams per cubic centimeter are contained in an N/2 solution of (a) potassium carbonate and (6) potassium VOLUMETRIC ANALYSIS 233 bicarbonate, when the solutions are used to neutralize acids? (The reaction results in decomposition to carbon dioxide.) Ans. (o) 0.03455 g. (6) 0.05054 g. 415. How many grams per cubic centimeter are contained in a 0.1 N solution of acetic acid? Ans. 0.006003 g. 416. What is the normal quantity of (o) barium sulphate? (6) Bariiun chloride? (c) Silver chloride? Ans. (a) 116.72 g. (6) 104.15 g. (c) 143.34 g. 417. How many grams per cubic centimeter in an N/10 solu- tion of (a) ferrous sulphate (FeS04 • 7 H2O), and (6) ferrous ammonium sulphate (FeS04 • (NH4)2S04 • 6 H2O), when used with an oxidizing agent? Ans. (a) 0.027802 g. (6) 0.039216 g. 418. Potassium tetroxalate, KH3(C204)2 • 2 H2O, as an acid reacts: 2 KH3(C204)2 + 6 NaOH = K2C2O4 + 3 Na2C204 + 6 H2O, and as a reducing agent: KH3(C204)2 + 2 Mn02 + 3 H2SO4 = 2 MnSOi + KHSO4 + 4CO2 + 4H2O. (a) How many grams of the crystallized salt are contained in a liter when used as a standard N/10 acid solution? (6) As a reducing agent? Ans. (a) 8.472 g. (6) 6.354 g. 419. A solution of sodium hydroxide containing 0.02003 g. per cubic centimeter is to be used in titrating sulphuric acid. What weight of the acid should be taken so that the reading of the burette in cubic centimeters shall equal the percentage of sulphuric acid in the sample? Ans. 2.4553 g. 420. A solution of potassium dichromate is equivalent to 0.00559 g. of iron per cubic centimeter. How many grams of 234 CHEMICAL CALCULATIONS iron ore must be taken so that each cubic centimeter used in the determination shall indicate (a) 0.1% Fe? (6) 0.5% Fe? Ans. (a) 5.59 g. (6) 1.118 g. 421. When phenol-phthalein is used as an indicator, potas- sium dichromate may be titrated with an acid : H2SO4 + KzCrjO, = KHCrzOj + KHSO4. How many grams of potassium dichromate must be contained in a Uter to make a normal solution? Ans. 147.10 g. 422. Potassium ferrocyanide, K4Fe(CN)6 • 3 H2O, acts as a reducing agent : 10 K4Fe(CN)e + 2 KMn04 + 8 H2SO4 = 10 KsFeCCN), + 6 K2SO4 + 2 MnS04 + 8 H2O. How much potassium ferrocyanide is contained in a cubic centi- meter of the normal solution? Ans. .42235 g. 423. How much arsenic is shown by 12.00 cc. of N/20 iodine solution, the reaction being AS2O3 + 4 1 + 2 H2O = AS2O5 + 4 HI. Ans. 0.022488 g. 424. To what weight of calcium oxide does 30.00 cc. of 0.1 N potassium permanganate correspond when the calcium has been separated as the oxalate, acidified and the oxalic acid titrated by potassium permanganate? Ans. 0.084105 g. 425. (a) How many grams of silver nitrate in a cubic centi- meter of N/5 solution? (6) To how many grams of sodium chloride are 20.00 cc. of this solution equivalent? Ans. (a) 0.033978 g. (6) 0.2338 g. 426. According to the reaction: NH4CNS + AgNOa = NH4NO3 + AgCNS. (o) What is the weight of ammonium sulpho cyanate in a cubic centimeter of a 0.1 N solution? (6) How many grams of silver are shown by 15.00 cc. of 0.1 N ammonium sulpho cyanate solution? ^ Ans. (a) 0.007612 g. (b) 0.1618 g. VOLUMETRIC ANALYSIS 235 427. What is the value in terms of manganese dioxide, of a cubic centimeter of an N/10 solution of ferrous sulphate acting as a reducing agent, the reaction being 2 FeSOi + Mn02 + 2 H2SO4 = MnSOi + FeaCSO^s + 2 H2O. Ans. 0.004347 g. 428. Phosphoric acid may be determined by adding an excess of silver nitrate and sodium acetate, when the following reac- tion takes place: H3PO4 + 3 AgNOa + 3 CHaCOONa, = AgaPOi + 3 NaNOa + 3 CH3COOH, the acetic acid being titrated with barium hydroxide, Ba(0H)2 + 2 CH3COOH = Ba(CH3COO)2 + 2 H2O. If 41.20 cc. of 0.1 N barium hydroxide are required to neutraUze the acetic acid, how much phosphoric acid is present? Ans. 0.13467 g. 429. Antimony may be determined by digestion with potas- sium iodide and hydrochloric acid, the iodine distUled over and collected in potassium iodide and titrated with sodixun thiosul- phate. The reactions are KI + HCl = KCl -I- HI, 4 HI + SbjOs = SbaOs + 2 H2O -1- 4 1, 414-4 Na2S203 = 4 Nal + 2 NaaSiOe. If 30.25 cc. of thiosulphate are required to determine the iodine liberated, each cubic centimeter of which indicates 0.01282 g. of iodine, how much antimony is present? Ans. 0.18365 g. 430. A solution of sulphuric acid contains 0.004912 g. of acid per cubic centimeter. What is the factor to 0.1 N? Ans. 1.0014. 431. A solution of sulphuric acid is standardized against a standard solution of approximately 0.5 N sodium hydroxide, the factor of which is 0.9992 to 0.5 N. 45.00 cc. of the sodium hydroxide solution are required to neutralize 44.70 cc. of the sulphuric acid solution. What is the factor of the sulphuric acid solution to 0.5 N? Ans. 1.006. 236 CHEMICAL CALCULATIONS 432. A solution of sodium hydroxide is to be standardized. 2.4530 g. of pure anhydrous sodium carbonate are weighed out and it requires 45.70 cc. of a solution of sulphuric acid for neu- tralization. 45.15 cc. of this solution of sulphuric acid require 44.90 cc. of the solution of sodium hydroxide. What is the normahty of this solution of sodium hydroxide? Ans. 1.018 N. 433. A solution of potassium permanganate is found to be equivalent to 0.005600 g. of iron per cubic centimeter, (a) What is its normality? (6) If 20.20 cc. of this solution are used in a determination, to how many cubic centimeters of 0.1 N solution is this amount equivalent? Ans. (a) 0.1003 N. (6) 20.21 cc. 434. 45.00 cc. of hydrochloric acid (factor to 0.1 N = 0.9987) require 87.25 cc. of a solution of barium hydroxide for neutral- ization, (a) What is the normality of the barium hydroxide? (6) What is the factor to 0.05 N? (c) How many grams of barium hydroxide are present per cubic centimeter? Ans. (a) 0.05151 N. (6) 1.030. (c) 0.004414 g. 435. The factor of a solution of sodium chloride is 1.007 to 0.1 N. 40.00 cc. of this solution are equivalent to 39.90 cc. of a solution of silver nitrate according to the reaction: AgNOs -t- NaCl = AgCl -|- NaNOa. (as) What is the factor of the solution of silver nitrate to 0.1 N? (6) How many grams per cubic centimeter does the silver ni- trate solution contain? Ans. (a) 1.0095. (6) 0.01715 g. 436. 0.2118 g. of arsenious oxide are weighed out to stand- ardize solutions of ioduie and thiosulphate, the reactions being AS2O3 -t- 4 1 -1- 2 H2O = AS2O5 + 4 HI, 21-1-2 NazSsOs = 2 Nal -f NazSiOe. It requires 42.40 cc. of iodine to react with the arsenious oxide, (o) What is the normahty of the iodine solution? (6) If 43.60 VOLUMETRIC ANALYSIS 237 cc. of the thiosulphate solution are equivalent to 43.00 cc. of the iodine solution, what is the normaUty of the thiosulphate solution? Am. (a) 0.1009 N. (6) a09961 N. 437. A solution of sulphuric acid being precipitated with an excess of barium chloride gives 0.05841 g. of barium. sulphate per cubic centimeter, (a) What is the normality of this solu- tion? (6) What is the factor to half normal? (c) To how many cubic centimeters of half normal acid are 18.20 cc. of this solution equivalent? (d) If 2.5000 g. of sodium carbonate require 45.00 cc. of this solution of sulphuric acid for neutral- ization, what is the per cent of sodium carbonate in the sample? Ans. (a) 0.5004 N. (b) 1.001. (c) 18.22. id) 47.73%. 438. How many grams of sodium thiosulphate, NaaSzOs^ 5 H2O must be present in a liter of solution to make it N/20, the reaction being 2 NajSzOs -h I2 = 2 Nal -I- NazSA. Ans. 12.411 g. 439. Oxalic acid may be used as an acid and as a reducing agent. When used as an acid, the reaction may be symboUzed: H2C2O4 -I- 2 NaOH = Na2C204 + 2 H2O. As a reducing agent, the reaction may be typified: H2C204-|-0 = H20 + 2C02. How many grams of the crystallized oxalic acid, correspond- ing to the formula H2C2O4 • 2 H2O, must be taken per liter to prepare (o) a liter of a 0.1 N solution in which the oxalic acid acts as an acid? (6) To prepare a 0.5 N solution, the acid acting as a reducing substance? Ans. (a) 6.3025 g. (6) 31.^125 g. 440. Potassium tetroxalate, KHC2O4 • H2C2O4 • 2 H2O, crys- tallizes well and may be used to prepare standard solutions. 238 CHEMICAL CALCULATIONS (a) How many grams of this salt should be taken to make a liter of normal solution, the salt acting as an acid according to the reaction: KHC2O4 • H2C2O4 + 3 NaOH = KNaCzOi + Na2C204 + 3 H2O. (6) If used as a reducing solution, how many grams should be taken per liter to make a normal solution, the reaction being symbolized : KHC2O4 + 20 = KOH + H2O + 4 CO2. Ans. (a) 84.72 g. (6) 63.5325 g. 441. Crude cream of tartar, KHC4H4O6, is to be titrated with 0.4876 N NaOH solution. What weight of this substance must be taken for titration so that the burette reading shall Ladicate the percentage of KHC4H4O6 directly? KHC4H4O6 + NaOH = KNaCiHiOe + H2O. Ans. 9.1737 g. 442. A sample of oxalic acid, H2C2O4 ■ 2 H2O, is to be titrated with 0.1095 N potassium permanganate: 2 KMn04 + 5 H2C2O4 + 3 H2SO4 = K2SO4 + 2 MnSOi + IOGO2 + 8H2O. What weight of the sample must be taken so that the biu"ette reading multiplied by two shall give the percentage of C2H2O4 • 2 H2O? Ans. 0.3451 g. 443. Manganese is titrated with potassium permanganate in neutral solution according to the foUowing reaction: 2 KMnOi + 3 MnS04 + 2 H2O = K2SO4 + 5 MnOz + 5 H2SO4. (a) How many grams of potassium permanganate are contained in a normal solution according to this reaction? (6) How many grams of manganese are equivalent to a cubic centimeter of a tenth normal solution of potassiiun permanganate made up on this basis? (c) If the potassium permanganate is standardized for use in an acid solution, to how many grams of manganese per cubic centimeter is such a solution equivalent? (In acid VOLUMETRIC ANALYSIS 239 solution, two molecules of potassium pennanganate liberate five available oxygen atoms.) Ans. (a) 52.6767 g. (6) 0.0027465 g. (c) 0.001648 g. 444. A silver nitrate solution is 0.1032 N. What weight of sample of a chloride must be taken so that each cubic centimeter of silver nitrate solution used in titration shall indicate one per cent of chlorine? Ans. 0.3659 g. 445. A sample containing iron is to be titrated with 0.2016 N potassium permanganate solution, (a) What weight of material should be weighed out for titration so that each cubic centi- meter used shall show one per cent of metallic iron? (6) So that the burette reading multipUed by two shall give the per- centage of iron as ferric oxide? Ans. (a) 1.1258 g. (6) 0.8048 g. 446. How many cubic centimeters of a 1.093 N solution are equivalent to 42.82 cc. of a 1.047 N solution? Ans. 41.02 cc. 447. How many cubic centimeters of a 0.02534 N solution are equivalent to 13.47 cc. of a 0.1073 N solution? Ans. 57.04 cc. 448. How many cubic centimeters of a solution whose factor to N/2 is 1.003 are equivalent to 19.42 cc. of a solution whose factor to N/5 is 0.9942? Ans. 7.70 cc. 449. How many cubic centimeters of a solution, the factor of which to N/10 is 0.9982, are equivalent to 15.94 cc. of a solu- tion whose factor to N/2 is 0.9743? Ans. 79.63 cc. 450. (a) 30.00 cc. of a 0.5018 N solution of sulphuric acid neutrahze 59.02 cc. of a solution of sodium hydroxide. What is the normaUty of the sodium hydroxide solution? Ans. 0.2551 N. 451. (6) 50.00 cc. of a solution of sulphuric acid yield 2.9250 g. of barium sulphate. 30.00 cc. of this solution neutralize 61.20 cc. of a solution of sodium hydroxide. What is the nor- mahty of the sodium hydroxide solution? Ans. 0.2457 N. 240 CHEMICAL CALCULATIONS 452. Hydrogen peroxide is titrated with potassium perman- ganate according to the reaction: 2 KMnOi + 3 H2SO4 + 5 HA = K2SO4 + 2 MnSOi + 8 H2O + 5 O2. 25.60 cc. of the hydrogen peroxide solution decolorized 45.90 cc. of the potassium permanganate solution, the strength of which is 0.6432 N. How many grams of hydrogen peroxide are present per liter? Ans. 1.9614 g. 453. 3.0153 g. of a solution of hydrogen peroxide require 53.15 cc. of a solution of 0.09920 N potassium permanganate to produce the pink color. The reaction is 2 KMn04 + 3 H2SO4 + 5 H2O2 = K2SO4 + 2 MnS04 + 8H2O + 5O2. What is the percentage of hydrogen peroxide in this sample? Ans. 3.74%. 454. Bleaching powder is to be analyzed for available chlo- rine. 7.0920 g. are rubbed up with water and made up to one liter. Of this suspension, 50.00 cc. are taken, treated with an excess of potassium iodide, and the liberated iodine titrated with N/10 solution of sodium thiosulphate. The reaction is Cl2-|-2KI = 2KCl + l2. If 35.50 cc. of the thiosulphate solution are required, what is the percentage of available chlorine? Ans. 35.50%. 455. A sample of material containing pyrolusite (Mn02) ij treated with hydrochloric acid: Mn02 + 4 HCl = MnClz + CI2 + 2 H2O. The chlorine is led into a solution of potassium iodide libera 1- ing iodine: 2KI-|-Cl2 = 2KCl + l2. The iodine is determined by a solution of sodium thiosulphate: 12 + 2 NajSaOa = 2 Nal + Na2S406. 25.40 cc. of 0.1016 N thiosulphate solution are required. How much pyrolusite is present? Ans. 0.1122 g. VOLUMETRIC ANALYSIS 241 456. (a) What is the equivalent in 0.5 N solution of 35.00 cc. of 0.1052 N solution? (6) How many cubic centimeters of 0.2 N solution are equivalent to 42.00 cc. of a 0.5063 N solution? Ans. (a) 7.364 cc. (6) 106.3 cc. 4B7. Manganese dioxide is reduced with an excess of oxalic acid (50.00 cc. of 0.2016 N): MnOj + H2SO4 + C2H2O4 = MnSOi + 2 CO2 + 2 H2O. The excess of oxalic acid requires 10.15 cc. of 0.2008 N potassium permanganate: 2 KMnOi + 5 C2H2O4 + 3 H2SO4 = K2SO4 + 2 MnSOi + IOCO2 + 8H2O. How much manganese dioxide is present? Ans. 0.3495 g. 458. 5.0000 g. of ammonium sulphate are treated in a flask with 100.00 cc. of sodium hydroxide solution (factor to 1 N = 1.009). After boiling tiU all the ammonia is expelled, the solu- tion is titrated with sulphuric acid, 30.25 cc. (factor to 1 N = 1.002) being required. What is the percentage of ammonia (NH3) hidicated? Ans. 24.04%. 459. One cubic centimeter of a potassium permanganate solution is equivalent to 0.005000 g. of iron. To 40.00 cc. of the permanganate solution an excess of potassium iodide is added which causes a Uberation of iodine according to the reac- tion: 2 KMn04 -I- 10 KI + 16 HCl = 12 KCl -f- 8 H2O -I- 2 MnClj -I- 10 1. The free iodine is titrated with a solution of sodium thiosulphate, 2 Na2S203 -I- 2 1 = 2 Nal -h Na2S406, 35.90 cc. of the thiosulphate being required. What is the normality of the thiosulphate solution? Ans. 0.09977 N. 460. 2.1200 g. of pure anhydrous sodium carbonate are weighed out and treated with 50.00 cc. of a solution of sulphuric acid: NaaCOs + H2SO4 = NajSOi -f- H2O -|- CO2. 242 CHEMICAL CALCULATIONS The excess of acid is titrated' back with sodium hydroxide solu- tion, 10.05 CO. beiag required: H2SO4 + 2 NaOH = NajSOi + 2 H2O. The solution of sulphuric acid and the sodium hydroxide solu- tion are titrated against each other when it is found that 45.00 cc. of the sulphiu'ic acid solution require 45.10 cc. of sodium hydroxide for neutralization. If the sodium hydroxide solution is used for determining oxaUc acid (C2H2O4 -21120), how many grams of oxalic acid are indicated by one cubic centimeter of the sodium hydroxide solution? H2C2O4 + 2 NaOH = NajCsOi + 2 H2O. Ans. 0.06286 g. 461. Sulphur dioxide is determined by adding an excess of iodine (50.00 cc. of 0.1009 N) : Na2S03 + I2 + H2O = Na.SOi + 2 HI, the excess of iodine being titrated back with sodium thiosulphate (20.15 cc. of 0.1022 N): 2 Na2S203 + I2 = NajSiOe + 2 Nal. How much sulphur dioxide is present? Ans. 0.09566 g. 462. One gram of a sample of nitrate of soda is examined by treating with Devada's alloy (which produces nascent hydrogen when acted upon by sodium hydroxide) which reduces the nitrate to ammonia. The ammonia is collected in 100.00 cc. of 0.1157 N sulphuric acid. 10.15 cc. of 0.1063 N sodium hydrox- ide are necessary to neutralize the excess of sulphuric acid. What is the per cent of sodimn nitrate m the sample? Ans. 89.18%. 463. Potassium persulphate is estimated by potassium per- manganate indirectly. When treated with a ferrous salt in acid solution: K2S2O8 + 2 reS04 = Fe2(S04)3 + K2SO4, VOLUMETRIC ANALYSIS 243 the excess of ferrous sulphate being determined by perman- ganate: 2 KMnOi + 10 FeSOi + 8 H2SO4 = K2SO4 + 2 MnSOi + 5Fe2(S04)3 + 8H20. 30.00 CO. of the ferrous sulphate solution are equivalent to 31.25 CO. of 0.1000 N KMnOi solution. 30.00 cc. of the ferrous sulphate solution together with the sample of potassium per- sulphate require 7.35 cc. of 0.1000 N potassium permanganate solution. How much potassium persulphate is present? Ans. 0.3230 g. 464. Iodine in iodides is determined by adding an excess of silver nitrate: KI + AgNOa = Agl -I- KNO3, and the excess of silver nitrate is found with potassium thio- cyanate: AgNOs + KCNS = KNO3 -f AgCNS. 42.00 cc. of silver nitrate solution are equivalent to 40.15 cc. of thiocyanate solution. 1.00 cc. of the silver nitrate solution contains 0.010820 g. of silver. How much potassium iodide is indicated if 32.00 cc. of the silver nitrate solution were added and 2.80 cc. of the thiocyanate solution were required for the back titration? Ans. 0.4841 g. 465. Iron in the ferric condition is treated with an excess of sodium thiosulphate (50.00 cc.) : 2 NajSzOs + 2 FeCIj -)- 2 HCl = 2 FeClz -|- 4 NaCl -|- H2S4O6. The excess of thiosulphate is found to require 7.85 cc. of iodine solution. The solution of iodine was standardized by titrating it against a standard arsenic solution, each cubic centimeter of which contained 0.005162 g. of arsenious oxide: AS2O3 + 2 12 + 2 H2O = AsaOs + 4 HI, 45.00 cc. of the arsenic solution requiring 45.20 cc. of the iodine solution. The thiosulphate solution was standardized by titrating it against the iodine solution: I2 -1- 2 NaaSaOa = 2 Nal + NazSiOe, 244 CHEMICAL CALCULATIONS 45.00 cc. of the iodine solution requiriag 45.95 cc. of the thio- sulphate solution. What is the weight of iron present? Am. 0.2385 g. 466. A solution of sodium carbonate and bicarbonate is to be examined for the content of these salts. Phenol-phthalein is added and the solution titrated with 0.1000 N hydrochloric acid in the cold, requiring 30.00 cc. Phenol-phthalein reacts neutral when all the carbonate is converted to bicarbonate: NaaCOs + HCl = NaHCOs + NaCl. Methyl orange is added which reacts acid when all the bicarbon- ate is decomposed : NaHCOa + HCI = NaCl -f- CO2 + H2O, 35.00 cc. more being required. What are the amounts of car- bonate and bicarbonate of soda present? Ans. 0.3180 g. NazCOa, 0.0420 g. NaHCOa. 467. A sample of quicklime is to be analyzed for calcium oxide and calcium carbonate. 14.0000 g. are weighed out, slaked with warm water (free from carbon dioxide) and made up to 500 cc. Of this emulsion, 50 cc. are pipetted out and again made up to 500 cc, 50.00 cc. of this latter solution being taken for analysis. 60.00 cc. of N/10 hydrochloric acid solution are added and the whole heated to boiling or till there is no further evolution of carbon dioxide. Ca(0H)2 + 2 HCl = CaCU + 2 H2O, CaCOs + 2 HCl = CaCl2 + H2O + CO2. The excess of acid is found by titrating back with N/10 sodium hydroxide solution, 20 cc. being required, methyl orange being used as an indicator. A second portion of 50 cc. is taken and titrated with N/10 hydrochloric acid solution, using phenol- phthalein as an indicator, 35 cc. being required. Under these conditions only the calcium hydroxide is titrated. Calculate the percentages of calcium oxide and calcium carbonate in the quicklime. Ans. 70.09% CaO, 17.87% CaCOs. VOLtJMETRlC ANALYSIS 246 468. 1.5000 g. of iron are dissolved in hydrochloric acid and treated with 0.7300 g. of a nitrate. 2 MNO3 + 6 FeCl2 + 8 HCl = 2 MCI + 2 NO + 6 FeCIa + 4 H2O. After reaction, the excess of ferrous chloride is determined by titration with N/2 potassium permanganate. 2 KMn04 + 10 FeCls + 16 HCl = 2 KCl + 2 MnCU + 10FeCl3 + 8H2O. An independent determination shows that 1.5000 g. of iron re- quire 53.55 cc. of the N/2 potassium permanganate solution. The excess of ferrous chloride requires 3.55 cc. of the perman- ganate solution, (a) If the nitrate taken is sodium nitrate (NaNOs), calculate the percentage present. (6) The per- centage of N2O5. Ans. (a) 97.05% NaNOa, (6) 61.66% N2O6. 469. Phenol-phthalein is neutral to acid sodium carbonate. By using methyl orange, the acid sodium carbonate may be titrated to completion, i.e., to sodium chloride, carbon dioxide and water. A sample of sodium carbonate containing sodium hydroxide is titrated with hydrochloric acid (factor to 0.1 N = 0.9985) using phenol-phthalein, when 50.00 cc. are required. Methyl orange is added and 20.00 cc. more hydrochloric acid are needed to produce acid reaction with this indicator. What are the amounts of sodium hydroxide and sodium carbonate present? Ans. 0.1198 g. NaOH, 0.2117 g. NaaCOs. 470. Calculate the composition of a mixed acid from the fol- lowing: For total acid: Weight of acid taken 4.8970 g. NaOH to neutralize 81 . 60 cc. For H2SO4 (other acids being driven off by evaporation) : Weight of acid taken 8.5420 g. NaOH to neutralize 80.75 cc. 246 CHEMICAL CALCULATIONS For N2O3: Weight of acid taken 17.000 g. 0.1 N KMn04 required. 10.00 cc. Factor NaOH solution to normal is 1.039. Ans. H2SO4 = 48.17%, HNO3 = 47.03%, N2O3 = 0.11%, H2O = 4.69 %, 100.00%. 471. Calculate the composition of a mixed acid from the following : For total acid : Weight of acid taken 5.9216 g. 1.012 NNaOH to neutralize 99,92 oc. For H2SO4 (other acids being driven off by evaporation) : Weight of acid taken 10 . 4387 g. 1.012 N NaOH to neutralize 104.00 cc. For N2O3: Weight of acid taken 18.000 g. 0. 1 N KMn04 required 9.00 cc. Ans. H2SO4 = 49.45%, HNO3 = 43.92%, N2O3 = 0.10%, H2O = 6.53 %, 100.00%. 472. Let X = percentage of free acetic anhydride (C2H30)20, y = percentage of total acidity calculated to acetic anhydride. Derive the formula X = 6.6643 (y - 84.995). 473. One gram of a sample of acetic anhydride containing acetic acid was treated with 200 cc. of a solution of barium hydroxide. (C2H30)20 + H2O = 2 CH3COOH, Ba(0H)2 + 2 CH3COOH = Ba(C2H302)2+ 2 H2O. vOLXXMETRIC ANALYSIS 247 200.00 cc. of the barium hydroxide solution was found to be equivalent to 190.00 cc. of an N/10 solution of hydrochloric acid. The excess of barium hydroxide was titrated back with N/10 hydrochloric acid, 1.37 cc. being required. What are the per- centages of acetic anhydride and acetic acid present? ^ Ans. 74.99% (CsHsOjO, 25.01% CH3COOH. 474. 0.8585 g. of a sample of acetic anhydride containing acetic acid is treated with 200.00 cc. of a solution of barium hydroxide. This volume of barium hydroxide solution is equiv- alent to 170.00 cc. of N/10 hydrochloric acid. The excess of barium hydroxide is titrated back with N/10 hydrochloric acid, 8.46 cc. being required. Calculate the percentages of acetic anhydride and acetic acid present in this sample. (See the two problems above.) Ans. 73.41% (CaHsOsO, 26.59% CH3COOH. 475. Sodium hydroxide and trisodium phosphate are to bo determined in the presence of each other. Phenol-phthalein reacts neutral to disodium phosphate; therefore, in titrating a mixture of these two salts with sulphuric acid and phenol- phthalein: 2 X NaOH + 2 2/ Na3P04 +{x + y) H2SO4 = 2y Na^HPOi + ix + y) NaaSOi + 2x H2O. 45.00 cc. of 0.5000 N H2SO4 are required for this titration. Methyl orange is now added, which is alkaline toward disodium phosphate, but neutral to monosodium phosphate. The titra- tion is continued with 0.5000 N H2SO4: 2 y Na^HPOi + y H2SO4 = 2y NaH2P04 -1- y Na2S04, 35.00 cc. being required. What are the amounts of trisodium phosphate and sodium hydroxide? Ans. 0.2000 g. NaOH, 2.8707 g. Na3P04. 476. (For the conduct of the sodium salts of phosphoric acid with indicators, see problem 475.) What is the composition of a solution of mixed tri and disodimn phosphate if the phenol- ■ See problem above; also note similar case of oleum, p. 191. 248 CHEMICAL CALCULATIONS phthalein titration required 25.00 cc. of 0.5000 N H2SO4 and the titration with methyl orange requires 35.00 cc. 0.5000 N H2SO4 in addition? Ans. 0.7103 g. NajHPO^, 2.0505 g. Na3P04. 477. Oleum often contains sulphur dioxide. When the sample is dissolved in water and titrated with standard sodium hydroxide solution, using phenol-phthalein, the sulphur dioxide combines with the water, forming sulphurous acid, which takes up sodium hydroxide: H2SO3 + 2 NaOH = NajSOs + 2 H2O. What is the analysis of an oleum from the following data: For total acid as SO3: Weight of oleiun taken 3 . 0570 g. 1 N NaOH required 74. 30 cc. For sulphur dioxide: Weight of oleum taken 7.0510 g. 0.1 N I solution required 46. 80 cc. The reaction for the sulphurous acid and iodine is H2SO3 + I2 + H2O = H2SO4 + 2 HI. Ans. Free SO3 = 80.31%, H2SO4 = 17.56%, SO2 = 2.13 %, 100.00%. 478. What is the composition of an oleum titrated with phenol-phthalein as an indicator, giving the following: For total acid: Weight of acid taken 5.0000 g. 1.112 N NaOH to neutralize 9^.95 cc. For sulphur dioxide: Weight of acid taken 5.0000 g. 0.1 N I solution required 39 . 00 cc. Ans. Free SO3 = 34.17%, H2SO4 = 63.33%, SO2 = 2.50%. VOLUMETRIC ANALYSIS 249 479. A solution shows 0.049205 g. of sulphuric acid per cubic centimeter. How many cubic centimeters of water must be added to a kilogram of this solution to make it 1.000 N? Ans. 3.20 cc. 480. How much 0.2019 N sodium hydroxide and water must be taken to make 5 liters of 0.1000 N sodium hydroxide?' Ans. 2391.2 cc. NaOH, 2608.8 cc. H2O. 481. 50.00 cc. of a solution (factor to N/10 = 1.005) corre- spond to 48.90 cc. of another solution. How many cubic centi- meters of water per liter must be added to this second solution to make it N/10? Ans. 28.00 cc. 482. How many grams each of a 0.5012 N solution must be mixed with a 0.1078 N solution to make one kilogram of a 0.2000 N solution? Ans. 765.6 g. 0.1078 N, 234.4 g. 0.5012 N. 483. How many pounds of 80.00% acetic acid must be added to 92.60% acetic acid to make 600 lbs. of 90.00% acetic acid? Ans. 123.8 lbs. 80.00%, 476.2 lbs. 92.60%. 484. How many pounds of water and how many pounds of 60.00% sulphuric acid must be mixed to prepare 400 lbs. of a 34.20% sulphuric acid? Ans. 288 lbs. 60% H2SO4, 172 lbs. water. 485. How many pounds of water must be added to 800 pounds of 73.00% sulphuric acid to make the whole 70.00% sulphuric acid? Ans. 34.29 lbs. water. 486. How much water must be added to 1000 cc. of a 0.1128 N solution to make it 0.1000 N? (When the densities of the solu- tions mixed, and the density of the resultant solution are very nearly the same, as in this case, volumes may be substituted in the formulas without sensible error.) Ans. 128.00 cc. water. 1 Consider the densities of the two Kquids and the resultant to be the same in this, and in other similar problems, unless otherwise mentioned. 250 CHEMICAL CALCULATIONS 487. How many pounds of 62.18% sulphuric acid and of 98.00% sulphuric acid must be taken to make 1000 lbs. of 93.00% acid? Ans. 139.6 lbs. 62.18% H2SO4, 860.4 lbs. 98.00% H2SO4. 488. How many pounds of 62.18% sulphuric acid must be added to 1000 lbs. of 98.00% sulphuric acid to dilute the whole . to 93.00% acid? Ans. 162.2 lbs. 62.18% H2SO4. 489. How much 0.1012 N solution must be added to 1000 cc. of a 0.5009 N solution to make it 0.2000 N? Ans. 3045.5 cc. 0.1012 N. 490. How many pounds of 80.00% acetic acid and 60.00% acetic acid should be mixed to make 500 lbs. of 65.00% acetic acid?i Ans. 125 lbs. 80%, CH3COOH, ' 375 lbs. 60% CH3COOH. 491. How many cubic centimeters of 0.0957 N and 0.1120 N solution must be taken to make 1000 cc. of 0.1000 N solution? (Consider the 0.0957 N solution as being strengthened by the 0.1120 N solution.) Ans. 263.8 cc. 0.1120 N, 736.2 cc. 0.0957 N. 492. How many pounds of 80.00% sulphuric acid must be added to 980 lbs. of 35.00% sulphuric acid to strengthen the whole to 40.00% acid? Ans. 122.5 lbs. 80% H2SO4. 493. How many pounds of a 20.00% hydrochloric acid must be added to 800 lbs. of a 43.00% hydrochloric acid to convert this weight into a 30.00% hydrochloric acid? Ans. 1040 lbs. 494. How many pounds each of a 30.00% oleum and a 98.00% sulphuric acid must be mixed to prepare 100 lbs. of 100.00% sulphuric acid? (Calculate the percentages of SO3 present in each.) Ans. 22.82 lbs. 30.00% oleum, 77.18 lbs. 98.00% H2SO4. ' In a problem of this kind in which a definite amount is called for, (a) of formula II of page 227 serves as weU as (a) of for- mula III. In one case the formula gives the amount of diluting solution to add; in the other, the amount of strengthening solution required. VOLUMETKIC ANALiSIS 251 495. A solution of potassium permanganate is equivalent to 0.01060 g. of Na2C204 per cubic centimeter. A weaker solu- tion is equivalent to 0.002120 g. of Na2C204 per cubic centi- meter. How many cubic centimeters of the stronger solution must be mixed with how many cubic centimeters of the weaker solution to prepare three liters of a solution equivalent to 0.002792 g. of iron per cubic centimeter? Ans. 435 cc. stronger, 2566 cc. weaker. 496. A solution of potassium permanganate is equivalent to 0.005600 g. of iron per cubic centimeter. What volume of this solution and what volume of water must be mixed to prepare three liters of a solution of which each cubic centimeter shall be equivalent to 0.005000 g. of iron? Ans. 2678 cc. sol., 322 cc. H2O. 497. A solution of sodium hydroxide is equivalent to 0.001000 g. of sulphuric acid per cubic centimeter. How much water must be added to one liter of this solution to make it equivalent to 0.001000 g. of nitric acid per cubic centimeter? Ans. 285.7 cc. CHAPTER X USE OF SPECIFIC GRAAHTY TABLES AND ACID CALCULATIONS Calculation of Data in Specific Gravity Tables. — Large shipments of acid, particularly sulphuric acid, are usually billed and paid for on the basis of 66° B6., 60° B6., etc. It is, therefore, necessary to calculate the actual strength of the acid to its equivalent in 66° B6., or what- ever the strength basis of the acid is billed and paid for. The table of Ferguson and Talbot,^ for sulphuric acid will be used as a model, the same serving as an example of the use of such data. The methods of calculation of the data will be explained, the same being appUcable to other solutions, the tables of which have not been so fully arranged.^ Specific gravity tables do not represent individual determinations for the densities listed,* but are con- structed from curves and methods of interpolation al- ready considered.^ Twaddell's Hydrometers ^ consist of a series of hydrom- eters so arranged that the graduations permit of easy reading over a wide range. They are graduated only for ' See Appendix, also Chem. Aim., pp. 388-393. 2 See Chem. Ann., pp. 392-456. ' The Baum6 is given in Ferguson and Talbot's table to two significant figures; it is accurate to two places of decimals. * Pp. 32-41. ' See table of Ferguson and Talbot, Chem. Ann., pp. 388-392, ooltuun 3. 252 USE OF SPECIFIC GRAVITY TABLES 253 liquids heavier than water. The unit of the Twaddell scale is 0.005 specific gravity; hence if X is the reading Twaddell, Sp.gr. = 1 + 0.005 Z. The weight of a cubic foot of water at 60° F. is 62.37 pounds. The weight of a cubic foot of a liquid is its specific gravity multiplied by 62.37.^ The acid content corresponding -to 66° B6. (Oil of Vitriol, O. V.) has been carefully ascertained and found to be 93.19% H2SO4. A sample of sulphuric acid of 65.75° Be. containing 91.80% HaSO^^ is equivalent to ^X 100 = 98.51% O.V.,' and as a cubic foot of liquid of 65.75° B6., which is equiva- lent to 1.8297 sp.gr., weighs 114.12 lbs. (62.37 X 1.8297),^ the number of pounds of Oil of Vitriol equivalent to one cubic foot of this acid is ^^ X 114.12 = 112.42 lbs." The equivalent in 60° Be. acid (77.67% H2SO4) of an acid of 64° BL (85.66% H2SO4) is • ^-^ X 100 = 110.29%," ? See table of Ferguson and Talbot, Chem. Ann., pp. 388-392, column 3. 2 Ibid., p. 392, column 4. ' Ibid., p. 392, column 6. * Ibid., p. 392, column 5. 6 Ibid., Pv392, column 7. 6 Ibid., p. 393, column 3. 254 CHEMICAL CALCULATIONS and as 60° B6. corresponds to 1.7059 sp. gr., the number of pounds of 60° Be. equivalent to a cubic foot of 64° B6. sulphuric acid is 85.66 77.67 X 1.7059 X 62.37 = 123.14 Ibs.i Use of Specific Gravity Tables. — As an example illustrating the use to which specific gravity tables may be put, suppose it is required to calculate the number of pounds of 50° B6. sulphuric acid in a shipment, the fol- lowing data being given: 42 inches of sulphuric acid are drawn out of a tank at a temperature of 101° F. Suppose that it has been deter- mined by calculating the capacity of the tank from the inside measurements; that 1 inch in the depth of the tank corresponds to 50.00 cubic feet. A sample of the acid taken from the tank and brought into the labo- ratory showed 56.88° B6. at 92° F. Correction must be made for temperature in order to reduce to 60° F., the temperature for which the tables are constructed: 92° - 60° = 32° difference.^ From the table under the heading "Allowance for Temper- ature,'" it is seen that the allowance for 60° B6. acid is 0.026° B6. for each degree Fahrenheit, and that the correc- tion for 50° Be. acid is 0.028° Be. As the acid in ques- tion is about midway between these pointH) the allowance for each degree Fahrenheit is very nearly 0.027° B6. The correction for temperature is 32 X 0.027 = 0.86° B6., ' See table of Ferguson and Talbot, Chem. Ann., p. 393, column 4; also from the product of the corresponding figure in column 5, p. 392, and the fraction given. " Ibid., p. 388, note to table. ' Ibid., p. 391. USE OF SPECIFIC GRAVITY TABLES 255 and as the standard temperature, 60° F., is lower than 92° F., the temperature at which the Baume of the sample was taken, the acid becoming denser as the temperature is lowered, this amount must be added. The Baum6 of the acid at 60° F. is, then, 56.88 + 0.86 = 57.74° B6. at 60° F. The Bamn6 of the acid at 101° F., the temperature at which the acid was drawn off, is calculated: 101 - 60 = 41° F. difference, 41 X 0.027 = 1.11° B6. correction; and as the density of the acid is lowered as the temper- ature is raised, 57.74 - 1.11 = 56.63 B^. at 101° F. The easiest way to get the specific gravity corresponding to this degree Baume is by interpolating the given data: 57° Be. = 1.6477 sp. gr.i 56° Be. = 1.6292 sp. gr. diff. = 0.0185 sp. gr., 56.63 - 56.00 = 0.63° B6. diff., 0.0185 X 0.63 = 0.0117 sp. gr., 1.6292 + 0.0117 = 1.6409 sp. gr. corresponding to 56.63° B^. Then, as 42 inches were drawn from the tank, the poimds drawn off are 42 X 50.00 X 62.37 X 1.6409 = 214,920 lbs. As the acid is sold on the basis of 50° B6., the pounds of 50° B6. corresponding to 57.74° Be. acid are easily foimd by interpolating from the table: 1 See table of Ferguson and Talbot, Chem. Ann., p. 390, column 2. 256 CHEMICAL CALCULATIONS 58° B^. = 119.59% 50° B6. acid^ 57° bI = 117.00% 50° Be. acid diff. = 2.59% 50° Be. acid, 2.59 X 0.74 = 1.92, 117 + 1.92 = 118.92% 50° Be. acid corresponding to 57.74° B^. acid, 214,920 X 1.1892 = 255,827 lbs. 50° B6. acid. Calculation of Mixed Acid. — " Mixed Acid " is a commercial term, generally meaning a mixture of nitric and sulphuric acids, and is extensively used in manu- facturing processes. On account of tiie relative costs of concentrated and dilute nitric acids, the water called for is added in the form of dilute nitric acid, using a minimum of concentrated nitric acid and a maximum of dilute nitric acid. Water, as such, is seldom added. For example, let it be required to calculate the quantities of acids necessary to make a mixture ("mix") of 60,000 lbs. of a mixed acid to consist of Per cent H2SO4 (add as 98% H2SO4) 46.00 HNO3 (add as 61.4% and as 95.5% HNO3) 49 . 00 H2O 5.00 100.00 60,000 X 0.46 = 27,600 lbs. H2SO4 called for 60,000 X 0.49 = 29,400 lbs. HNO3 called for 60,000 X 0.05 = 3,000 lbs. H2O called for 60,000 lbs. ^^^ = 28,163 lbs. 98% H2SO4 to take, 60,000 - 28,163 = 31,837 lbs. still to add. * See table of Ferguson and Talbot, Chem. Ann., p. 391, column 5. USE OF SPECIFIC GRAVITY TABLES 257 Let X = number of pounds of 95.5% HNO3 to add; then 31,837 — X will represent the number of pounds of 61.4% HNO3 to add. By algebra: 0.955 X + 0.614 (31,837 - a;) = 29,400. Solving: X = 28,891 lbs. 95.5% HNO3 to take, 31,837 - 28,891 = 2946 lbs. 61.4% HNO3 to take. So, to make the mix, take H2SO4 28,163 lbs. 98% HNO3 , 28,891 lbs. 95.5% HNO3 2,946 lbs. 61.4% 60,000 lbs. This same result might have been reached by means of the formulas given to adjust the strengths of liquids.' 29,400 lbs. of 100% nitric acid are called for; the weight of material still to be added, after the 98% sulphuric acid is added, is 31,837 lbs. as before. This makes 90 4.nn ^^ X 100 = 92.35% HNO3 to be added. ol,oo7 To make 31,837 lbs. of an acid of this concentration from 95.5% and 61.4% nitric acid, using III (a) of page 227, n^'S ~ «!'! X 31,837 = 28,896 lbs. 95.5% HNO3 to take, as before. 31,837 - 28,896 = 2941 lbs. 61.4% HNO3 to take, as before. Strengthening a Mixed Acid by Means of Oleum. — An example involving the use of olpum will next be considered. Let it be required to make 61,320 lbs. of a mixed acid of the composition: • Pp. 227-228, II (a), or III (a), according to whether the acid is regarded as being diluted or strengthened. 258 CHEMICAL CALCULATIONS Per cent HNO3 (add as 94.5% HNO3) 56.00 H2SO4 (add as 98.56% H2SO4 and as 20.00% oleum, a minimum of which is to be taken) 41 . 00 H2O 3.00 100.00 The tank in which the acid is to be mixed already contains 2604 lbs. of the remains of a previous mix of the com- position: Per cent HNO3 52.00 • H2SO4 42.50 H2O 5.50 100.00 Consequently, HjSO, HNO, H2O Called for Lba. 25,141 1,107 24,034 Lbs. 34,339 1,354 32,985 Lbs. 1840 In tank 143 To be added 1697 If the attempt were made to calculate the weights of acid to add by the previous method, it would be seen that the method would not work, as too much water would be added with the sulphuric acid and, hence, a nitric acid stronger than 94.5% HNO3 would have to be used to complete the mix; hence, oleum will have to be employed. Thus: 24,034 0.9856 ^ 24,385 lbs. 98.56% H2SO4, 24,385 - 24,034 = 351 lbs. H2O added with the 98.56% H2SO4, 1697 - 351 = 1346 lbs. H2O remaining. USE OF SPECIFIC GRAVITY TABLES 269 Adding this water with the nitric acid would call for a stronger nitric acid than the 94.5% HNO3, as is seen from the following: 32,985 + 1346 = 34,331 lbs. HNO3 and H2O still to add, 32 985 g^^ X 100 = 96.08% HNO3 required to complete the mix. Going back to the original figures after this preliminary- calculation which has shown the necessity of using oleum; first calculating the weight of nitric acid to be added: ^g = 34,905 lbs. 94.5% HNO3 to add, 34,905-32,985 = 19201bs. H2O added with the94. 5%HN03. But the mix only calls for 1697 lbs. of water, hence 1920 — 1697 = 223 lbs. of water will be added in excess. This water must be taken up with oleum. Now, to the acid already in the tank the following quantities of acid must be added: H2SO4 24,034 lbs. 100% HaSd HNOs 32,985 lbs. 100% HNO3 H2O 1,697 lbs. 100% H2O Total. ^ 68,716 In adding 34,905 lbs. of 94.5% HNO3 there remain only 58,716 - 34,905 = 23,811 lbs. of acid (H2SO4) to add. To adjust the proportions and not add more acid than called for is done by adding oleum which takes up water from the nitric acid. The percentage strength of the sulphmic acid requisite is m^ X 100 = 100.93% H2SO4. 260 CHEMICAL CALCULATIONS The percentage of SO3 in 100.93% H2SO4 is in 98.56% H2SO4 the percentage of SO3 is in 20.00% oleum the percentage of SO3 is ' Per cent SO3 = 0.8163 (100 - 20) + 20 = 85.31% SO3; then, to make 23,811 lbs. of a 100.93% H2SO4, from 20.00% oleum and 98.56% H2SO4, require:^ llf^ I 80I5 X 23,811 = 9506 lbs. 20% oleum, 23,811 - 9506 = 14,305 lbs. 98.56% H2SO4. So, to make the mix, add to the acid already in the tank: HNO3 34,905 lbs. 94.50% H2SO4 14,305 lbs- 98.56% Oleum 9,506 lbs. 20.00% It is frequently desired to prepare a "mix" from a mixed acid already on hand by adding to it the requisite amounts of sulphuric acid and nitric acid to bring it up to the desired concentration. Thus, it may be required to fortify a " spent " mixed acid, or it may be that after add- ing the calculated amounts of ingredients to make a batch of mixed acid that the mixed acid resulting does not ana- lyze up to specifications. It must then be adjusted by a further addition of the deficient constituent. Thus, sup- pose a mixed acid of the following composition is desired : Per cent HjSOi 46.00 HNOa 49.00 H2O 5.00 1 See p. 191. 2 See p. 228. USE OF SPECIFIC GRAVITY TABLES 261 and there is on hand a supply of mixed acid of the com- position : Per cent H2SO4 45.10 HNO3 48.00 H2O 6.90 How many pounds of 98% sulphuric acid and 96% nitric acid should be added to the mixed acid for each 1000 lbs. of the corrected acid produced? An algebraic method of calculation is as follows : Let X = weight of mixed acid to take, y = weight of 98%o H2SO4 to take, z = weight of 96% HNO3 to take. The total sulphuric acid in the corrected mix must be (0.451) X = weight of H2SO4 added with the mixed acid, (0.98 ) y = weight of H2SO4 added as 98%, H2SO4. The total nitric acid in the corrected mix must be (0.48) X = weight of HNO3 added with the mixed acid, (0.96) z = weight of HNO3 added as 96% HNO3. The total water in the corrected mix must be (0.069) X = weight of H2O added with the mixed acid, (1.00 - 0.98) y = (0.02) y = weight of H2O added with the 98%, H2S4O, (1.00 - 0.96) z = (0.04) z = weight of H2O added with the 96% HNO3 acid. In each 1000 lbs. of the corrected mix there must be 1000 X 0.46 = 460 lbs. H2SO4, 1000 X 0.49 = 490 lbs. HNO3, 1000 X 0.05 = 50 lbs. H2O. 262 CHEMICAL CALCULATIONS Hence, the following equations result: (1) (0.451) X + (0.98) y = 460. (2) (0.48) X + (0.96) z = 490. (3) (0.069) a; + (0.02) y + (0.04) 2 = 50. Expressing (1) and (2) in terms of x by solving for y and z in these equations: _ 460- (0.451) a: ^ 460 _ (0.451) x ^^ ^ ~ 0.98 0.98 0.98 (5) z = = 469.388 - (0.460204) x. 490 - (0.48) x _ 490 (0.48) x 0.96 0.96 0.96 = 510.417 - (0.500000) x. Substituting in (3) gives (0.069) X + 0.02 [469.388 - (0.460204) x] + 0.04 [510.417 - (0.500000) x\ = 50, (0.069) X + 9.38776 - (0.00920408) x + 20.41668 - (0.02) x = 50, (0.03979512) x = 20.19556, 20.19556 X = = 507.49 lbs. mixed acid to 0.03979512 take per 1000 lbs. Substituting this value of x in (4) and (5) gives y = 469.388 - (0.460204) (507.49) = 235.84 lbs. 98% H2SO4 to take, z = 510.417 - (0.500000) (507:49) = 256.67 lbs. 96% HNO3 to take. The ratios of these values may be used either to pre- pare a definite amount of mixed acid or to correct a defi- nite amount of " spent " acid. Knowing the ratios per 1000 lbs. the quantities requisite for any weight of acid are readily calculated. After making a calculation, it is USE OF SPECIFIC GRAVITY TABLES 263 often a good plan to check up the results. This insures the accuracy of the calculation. Thus in the above cal- culation: H,S0, HNO2 HiO Mixed acid Lbs. 228.88 231.12 Lba. 243.60 Lba. 35.01 98% H2SO4 4.72 96% HNOs 246.40 490.00 10.27 460.00 50.00 PROBLEMS 498. What is the Twaddell reading corresponding to (a) 1.6117 sp. gr.? (6) To 66.00° B6.1 Ans. (a) 122.2 Tw., (6) 167.1 Tw. 499. What is the reading of 141.2 Tw. in (o) specific gravity and (6) in Baum6? Ans. (a) 1.7060 sp. gr., (6) 60.00° B6. BOO. (a) What is the percentage 0. V. (Oil of Vitriol, 93.19% H2SO4) .equivalent to 62.18% H2SO4? (6) What is the per- centage of 50° B6. sulphuric acid (62.18% H2SO4) equivalent toO.V.? Ans. (a) 66.72%, (6) 149.87%. 501. (o) What is the equivalent in Oil of Vitriol (93.19% H2SO4) of 600 lbs. of a sulphuric acid of 89.55% H2SO4? (6) In 50° B6. sulphuric acid (62.18% H2SO4)? Ans. (a) 576.6 lbs., (6) 864.12 lbs. 602. Find the percentage of 100% H2SO4 equivalent to (o) a 20.00% oleum? (6) To a 30.00% oleum? Ans. (a) 104.5%, (6) 106.75%. 603. What is the percentage of Oil of Vitriol equivalent to 25.00% oleum? (&) What is the percentage of 98.00% sul- phuric acid equivalent to a 35.00% oleum? Am. (a) 113.34%, (6) 110.08%. 264 CHEMICAL CALCULATIONS 504. Knowing that 60° B6. sulphuric acid contains 77.67% H2SO4 and that 50° B6. sulphuric acid contains 62.18% H2SO4, what is the number of pounds of 50° B6. sulphuric acid equiv- alent to a cubic foot of 60° B6. sulphuric acid? Ans. 132.91 lbs. 605. 50° B^. sulphuric acid contains 62.18% H2SO4, and 52° B6. sulphuric acid contains 65.13% H2SO4. (a) To how many pounds of 50° B6. sulphuric acid are 350 cubic feet of 52° B6. sulphuric acid equivalent? (&) If 60° B6. sulphuric acid contains 77.67% H2SO4, to how many pounds of 60° B6. sulphiu-ic acid are 530 cubic feet of 52° B6. acid equivalent? Ans. (a) 35,647.5 lbs., (&) 43,216.2 lbs. 506. Calculate the equivalent weight in terms of 60° Bd. sulphuric acid equivalent to 2310 cubic feet, measured at 102 F., a sample of which showed 59.66° B6. at 80° F. Ans. 243,150 lbs. 507. Calculate the weight of 50° B6. sulphuric acid equivalent to a shipment of 2160.61 cubic feet, measured at 120° F., a sample of which showed 56.14° B6. at 80° F.' Ans. 252,410 lbs. 508. How many pounds of 60° B^. sulphuric acid are equiv- alent to a shipment of 2507 cubic feet measured at 92° F., a sample of which showed 65.52° B6. at 77° F.? Ans. 282,614 lbs. 509. It is required to make 37,000 lbs. of a mixed acid of the composition: Per cent' H3SO4 41.00 HNO3 52.00 H2O 7.00 there remaining in the mixing tank 6720 lbs. of an acid from a former mix of the composition: 1 In commercial transactions, calculations are often carried to a degree of accuracy unwarranted by the accuracy of the readings. USE OP SPECIFIC GRAVITY TABLES 265 Per cent H2SO4 42.00 HNO3 52.54 H2O 5.46 How many pounds of 98.00% sulphuric acid, and 94.70% and 61.40% of nitric acid must be added to the acid already in the tank to make this mix? Ans. 12,599.6 lbs. 98.00% H2SO4, 14,574.3 lbs. 94.70% HNO3, 3,106.1 lbs. 61.40% HNO3. 510. If nitric acid to the extent of one per cent is lost in mixing the acids by reason of heat generated or other reasons, calculate the quantities called for in Problem 509 so that after mixing the acid will have the desired composition. Ans. 12,600 lbs. 98.00% H2SO4, 15,689 lbs. 94.70% HNO3, 1,991 lbs. 61.40% HNO3. 611. Calculate the amounts of acid required to make 34,000 lbs. of a mixed acid of the composition: Per cent H2SO4 65.90 HNO3 . . '. 18.10 H2O 16.00 there remaining in the tank 3780 lbs. of an acid from a former mix, the composition of which is Per cent H2SO4 42.00 HNO3 52.00 H2O 6.00 it being also desired to use in making this mix, in order to work it off, 7000 lbs. of an acid of Per cent H2SO4 64.00 HNO3 28.00 H2O 8.00 The sulphuric acid is to be added as 93.20% H2SO4, and the nitric acid as 52.30% HNO3; water, if any be required, is to be added 266 CHEMICAL CALCULATIONS as such. How many pounds of these substances must be taken, in addition to the acid abeady in the tank and the 7000 lbs. of the acid which it is desired to work off? Am. 17,531 lbs. 93.2% H2SO4, 4,260 lbs. 52.3% HNO3, 1,429 lbs. water. 612. Calculate how much 93.20% H2SO4, the acid to be worked off, and water must be added in problem (511) that a maximum amount of the acid to be worked off may be used, adding no 52.30% HNO3. Ans. 14,957 lbs. acid to be worked off, 12,067 lbs. 93.2% H2SO4, 3,196 lbs. water. 613. How many pounds of 98.00% sulphuric acid, and 96.00% and 61.4% nitric acid must be taken to make 60,000 lbs. of a mixed acid of the composition: Per cent H2SO4 46.00 HNO3 48.00 H2O 6.00 Ans. 28,163 lbs. 98:0% H2SO4, 26,711 lbs. 96.0% HNO3, 5,126 lbs. 61.4% HNO3. 514. It is required to make a mixed acid of the composition: Per cent HjSOi 46.00 HNO3 49.00 H2O 5.00 from 96.00% and 61.4% nitric acid and 98.00% sulphuric acid. How many pounds of each must be taken to prepare 60,000 lbs.? Ans. 28,163 lbs. 98.0% H2SO4, 28,474 lbs. 96.0% HNO3, 3,363 lbs. 61.4% HNO3. 616. How many pounds of 95.00% nitric acid and 30.00% oleum must be added for each 1000 lbs. of a mixed acid of the composition: USE OP SPECIFIC GRAVITY TABLES 267 Per cent H2SO1 -r 43.00 HNO3 61.00 H2O 6.00 to convert into a mixed acid of the composition: Per cent H2SO4 42.00 HNO3 53.00 H2O 6.00 Ans. 137.07 lbs. 95% HNO3, 71.38 lbs. 30% oleum. 616. It is required to make 61,320 lbs. of a mixed acid of the composition: Per cent HNO3 56.00 H2SO1 41.00 H2O 3.00 there remaining in the tank, from a former mix, 2604 lbs. of an acid of the composition: Per cent HNO3 52.00 HzSOi 42.50 H2O 5.50 How many pounds of a 20.00% oleum, 98.56% sulphuric acid and 94.50 nitric acid must be added to the acid already in the tank? Ans. 9,678 lbs. 20.00% oleum, 14,133 lbs. 98.56% H2SO1, 34,905 lbs. 94.50% HNO3. 388 APPENDIX XXXV. — SUL By W. C. Ferguson Degrees Baume. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Specific Gravity 00° Degrees Twaddell. 1.0000 1.0069 1.0140 1.0211 1.0284 1.0357 1.0432 1.0507 1.0584 1.0662 1.0741 1.0821 1.0902 1.0985 1.1069 1.1154 1.1240 1.1328 1.1417 1.1508 1 . 1600 1.1694 1.1789 1.1885 1.1983 0.0 1.4 2.8 4.2 5,7 7.1 8.6 10.1 11.7 13.2 14.8 16.4 18.0 19.7 21.4 23.1 24.8 26.6 28.3 30.2 32.0 33.9 35.8 37.7 39.7 Per cent H2SO1. 0,00 1.02 2.08 3,13 4.21 5.28 6.37 7.45 8.55 9.66 10.77 11.89 13.01 14.13 15.25 16.38 17.53 18.71 19.89 21.07 22.25 23,43 24,61 25.81 27.03 Weight of I Cu. Ft. in Lbs. Av. 62.37 62.80 63,24 63.69 64.14 64.60 65.06 65.53 66.01 66.50 66.99 67.49 68.00 68.51 69.04 69.57 70.10 70.65 71.21 71.78 72.35 72.94 73.53 74.13 74.74 Per cent O.V.* 0.00 1.09 2.23 3.36 4.52 5.67 6.84 7.99 9.17 10.37 11.56 12.76 13.96 15.16 16.36 17.58 18.81 20.08 21.34 22.61 23.87 25.14 26.41 27.69 29.00 Pounds O. V. in I Cubic Foot. 0.00 0.68 1.41 2.14 2.90 3.66 4.45 5.24 6.06 6.89 7.74 8.61 9.49 10.39 11.30 12.23 13.19 14.18 15.20 16.23 17.27 18.34 19.42 20.53 21.68 Sp. Gr. determinations were made at 60° F., compared with water at 60° F. From the Sp. Grs., the corresponding degrees Baum6 were cal- culated by the followmg formula: BaumI = 145 — 145/Sp. Gr. Baum6 Hydrometers for use with this table must be graduated by the above formula, which formula should always be printed on the scale. * 66° Baum^ = Sp. Gr. 1.8354 = Oil of Vitriol (0. V.). 1 cu. ft. water at 60° F. weighs 62.37 lbs. av. Atomic weights from F. W. Clarke's table of 1901. O = 16. H2SO4 = 100 per cent. %-aSOi %0.V. %60° O. V. = 93.19 = 100.00 = 119.98 60° = 77.67 = 83.35 = 100.00 60° = 62.18 = 66.72 = 80.06 (268) A±'±'J!iiNL»lJi. 389 PHURIC ACID AND H. P. Talbot Degrees Baume. * Freez- ing (Melting) Point. F. APPROXIMATE BOILING 50° B, 295° F. 60° " 386° " 61° " 400° " POINTS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 32.0 31.2 30.5 29.8 28.9 28.1 27.2 26.3 25.1 24.0 22.8 21.5 20.0 18.3 16.6 14.7 12.6 10.2 7.7 4.8 + 1.6 - 1.8 - 6.0 -11 -16 62° " 415° " 63° " 432° " 64° " 451° " 65° " 485° " 66° " 538° " FIXED POINTS Specific Gravity. Per Cent Specific Gravity. Per Cent HjSO,. 1.0000 1.0048 1.0347 1.0649 1.0992 1.1353 1.1736 1.2105 1.2513 1.2951 1.3441 1.3947 1.4307 1.4667 1.4822 .00 .71 5.14 9.48 14.22 19.04 23.94 28.55 33.49 38.64 44.15 49.52 53.17 56.68 58.14 1.5281 1.5440 1.5748 1.6272 1.6679 1.7044 1.7258 1.7472 1.7700 1.7959 1.8117 1.8194 1.8275 1.8354 62.34 63.79 66.51 71.00 74.46 77.54 79.40 81.32 83.47 86.36 88.53 89.75 91.32 93.19 Acids stronger than 66° B6. shoxild have their percentage com- positions determined by chemical analysis. * Calculated from Pickering's results, Jour, of Lon. Ch. Soc, vol. 57, p. 363. „ ^ Authorities — W. C. Ferguson; H. P. Talbot. This table has been approved and adopted as a standard by the Manufacturing Chemists' Association of the United States. W. H. Bower, Henry Howard, Jas. L. Morgan, Arthur Wyman, a. g. rosengarten, Executive Committee. New York, June 23, 1904. (269) 390 APPENDIX Specific Weight of Pounds Degrees Gravity Degrees Per Cent I Cu. Ft. Per Cent O. V. in Baume. 6o° TwaddeU. HjSOj. in Lbs. O. V. I Cubic At. Foot. 25 1.2083 41.7 28.28 75.36 30.34 22.87 26 1.2185 43.7 29.53 76.00 31.69 24.08 27 1.2288 45.8 30.79 76.64 33,04 25.32 28 1.2393 47.9 32.05 77.30 34.39 26.58 29 1.2500 50.0 33.33 77.96 35.76 27.88 30 1.2609 52.2 34.63 78.64 37.16 29.22 31 1.2719 54.4 35.93 79,33 38.55 30.68 32 1.2832 56.6 37.26 80.03 39.98 32,00 33 1.2946 58.9 38.58 80.74 41.40 33.42 34 1.3063 61.3 39.92 81.47 42.83 34.90 35 1.3182 63.6 41.27 82.22 44.28 36.41 36 1.3303 66.1 42.63 82.97 45.74 37.95 37 1.3426 68.5 43.99 83.74 47.20 39.53 38 1.3551 71.0 45.35 84.52 48.66 41.13 39 1.3679 73.6 46.72 85.32 50.13 42.77 40 1.3810 76.2 48.10 86.13 51.61 44.45 41 1.3942 78.8 49.47 86.96 53.08 46.16 42 1.4078 81.6 50.87 87.80 54.58 47.92 43 1.4216 84.3 52.26 88.67 56.07 49.72 44 1.4356 87.1 53.66 89.54 57.58 51.56 45 1.4500 90.0 55.07 90.44 59.09 53.44 46 1.4646 92.9 56.48 91.35 60.60 55.36 47 1.4796 95.9 57.90 92.28 62.13 57.33 48 1.4948 99.0 59.32 93.23 63.65 59.34 49 1.5104 102.1 60.75 94.20 65.18 61.40 50 1.5263 105.3 62.18 95.20 66.72 63.52 51 1.5426 108.5 63.66 96.21 68.31 65.72 52 1.5591 111.8 65.13 97.24 69.89 67.96 53 1.5761 115.2 66.63 98.30 71.50 70.28 54 1.5934 118.7 68.13 99.38 73.11 72.66 55 1.6111 122.2 69.65 100.48 74.74 75.10 56 1.6292 125.8 71.17 101.61 76.37 77.60 57 1.6477 129.5 72.75 102.77 78.07 80.23 58 1.6667 133.3 74.36 103.95 79.79 82.95 59 1.6860 137.2 75.99 105.16 81.54 85.75 (270) ) APPENDIX 391 •Freezing Degrees (Melting) Baume. Point. -F. ALLOWANCE ] Atl0°B6. .029°B< FOR TEMPERATURE 25 —23 5.or.00023Sp.Gr. = l°P. 26 — 30 " 20° " .036° " .00034 " = 1° " 27 -39 " 30° " .035° " .00039 " = 1° " 28 -49 " 40° " .031° " .00041 " = 1° " 29 —61 " 50° " .028° " .00045 " = 1° " " 60° " .026° " .00053 " = 1° " 30 —74 " 63° " .026° " .00057 " = 1° " 31 -82 " 66° " .0235° " .00054 " = 1° " 32 -96 33 -97 34 35 36 37 -91 —81 Pounds Pounds -70 -60 Per Cent 60° Baumg. 60° Baum6 in I Cubic Per Cent So° Baum€. 50° Baumi in I Cubic 38 -53 Foot. Foot. 39 40 -47 -41 61.93 53.34 77.36 66.63 41 -35 63.69 55.39 79.56 69.19 42 -31 65.50 57.50 81.81 71.83 43 -27 67.28 59.66 84.05 74.53 44 -23 69.09 61.86 86.30 77.27 45 -20 70.90 64.12 88.56 80.10 46 -14 72.72 66.43 90.83 82.98 47 -15 74.55 68.79 93.12 85.93 48 -18 76.37 71.20 95.40 88.94 49 -22 78.22 73.68 97.70 92.03 50 -27 80.06 76.21 100.00 95.20 51 -33 81.96 78.85 102.38 98.50 52 -39 83.86 81.54 104.74 101.85 53 -49 85.79 84.33 107.15 105.33 54 -59 87.72 87.17 109.57 108.89 55 89.67 90.10 112.01 112.55 56 9 91.63 93.11 114.46 116.30 57 1 93.67 96.26 117.00 120.24 58 S 95.74 99.52 119.59 124.31 59 -'7 n 97.84 102.89 122.21 128.52 (271) 392 APPENDIX Specific Weight of Pounds Degrees Gravity Degrees Per Cent I Cu. Ft. Per Cent O.V. in Baumg. 6o° TwaddeU. HsSOi. in Lbs. O.V. I Cubic Av. Foot. 60 1.7059 141.2 77.67 106.40 83.35 88.68 61 1.7262 145.2 79.43 107.66 85.23 91.76 62 1.7470 149.4 81.30 108.96 87.24 95.06 63 1.7683 153.7 83.34 110.29 89.43 98.63 64 1.7901 158.0 85.66 111.65 91.92 102.63 64i 1.7957 159.1 86.33 112.00 92.64 103.75 641 1.8012 160.2 87.04 112.34 93.40 104.93 64f 1.8068 161.4 87.81 112.69 94.23 106.19 65 1.8125 162.5 88.65 113.05 95.13 107.54 65i 1.8182 163.6 89.55 113.40 96.10 108.97 65§ 1.8239 164.8 90,60 113.76 97.22 110.60 65i 1.8297 165.9 91.80 114.12 98.51 112.42 66 1.8354 167.1 93.19 114.47 100.00 114.47 XXXVI. — FUMING SULPHURIC ACID AT 20° Cl. Winkler 100 Parts Contain 100 Parts Contain Total SO3. Specific Grav- ity. Total SOa. Grav- ity. Free SO. * HjSOi. Acid of Free SOb* SSO,. Acid of 66° B. 66° B. 1.835 75.31 92.25 99 1.905 83.57 10.56 89.44 65.68 1.840 77.38 94.79 90.69 1.910 83.73 11.43 88.57 65.25 1.845 79.28 97.11 83.08 1.915 84.08 13.33 86.67 63.84 1.850 80.01 98.01 80.10 1.920 84.56 15.95 84.05 62.10 1.855 80.95 99.16 76.38 1.925 85.06 18.67 81.33 59.90 1.860 81.84 1.54 98.46 72.81 1.930 85.57 21.34 78.66 57.86 1.865 82.12 2.66 97.34 71.71 1.935 86.23 25.65 74.35 55.21 1.870 82.41 4.28 95.76 70.53 1.940 86.78 28.03 71.97 53.00 1.875 82.63 5.44 94.56 69.35 1.945 87.13 29.94 70.06 51.60 1.880 82.81 6.42 93.58 68.92 1.950 87.41 31.46 63.54 50.48 1.885 82.97 7.29 92.71 68.27 1.955 87.65 32.77 67.23 49.52 1.890 83.13 8.16 91.94 67.55 1.960 88.22 35.87 64.13 47.23 1.895 83.43 9.34 90.66 66.81 1.965 88.92 39.68 60.32 44.42 1.900 83.48 10.07 89.93 66.24 1.970 89.83 44.64 55.36 40.78 This column gives the amount of SO3 which may be distilled o£f. (272) APPENDIX 393 Degrees Baumi. * Freezing (Melting) Point. Per Cent 6o° Baum£. Pounds 6o° Baume in Cubic Foot. Per Cent so- Baumg. Pounds So° Banm£ in Cubic Foot. 60 +12.6 100.00 106.40 124.91 132.91 61 27.3 102.27 110.10 127.74 137.52 62 39.1 104.67 114.05 130.75 142.47 63 46.1 107.30 118.34 134.03 147.82 64 46.4 110.29 123.14 137.76 153.81 64i 43.6 111.15 124.49 138.84 155.50 64| 41.1 112.06 125.89 139.98 157.25 64f 37.9 113.05 127.40 141.22 159.14 65 33.1 114.14 129.03 142.57 161.17 65i 24.6 115.30 130.75 144.02 163.32 65i 13.4 116.65 132.70 145.71 165.76 65i -1 118.19 134.88 147.63 168.48 66 -29 119.98 137.34 149.87 171.56 XXXVII. — SULPHURIC ACID 94^100% H2SO4 By H. B. Bishop The acid used in this table was prepared from Baker and Adam- son's c.p. sulphuric acid 95 per cent, which was strengthened to 100 per cent by the addition of fuming sulphuric acid made by dis- tilling fiuning acid (70 per cent free SO3) into a portion of the 95 per cent c.p. acid. The final acid was tested for impurities: residue upon evaporation, chlorine, niter and sulphur dioxide. The only impurity foimd was a trace of sulphur dioxide (0.001 per cent) which was less than the sensitiveness of the determination. The analytical and specific gravity determinations, and the allow- ance for temperature were made in the same manner, and with the same accuracy as in the sulphuric acid table adopted in 1904, the specific gravity 1.8354 and 93.19 per cent H2SO1 being taken as a standard. The actual determinations were made within a few hundredths of a per cent of the points given in the table, the even percentages being calculated by interpolation. Per Cent H2SO,. Sp. Gr. at 60° F. Allowance for Temperature. 66° B6. 93.19 1.8354 At 94% 0.00054 sp. gr. = 1° F. 94.00 1.8381 At 96% 0.00053 sp.gr. = IT. 95.00 1.8407 At 97.5% 0.00052 sp. gr. = 1° F. 96.00 1.8427 At 100% 0.00052 sp. gr. = 1° F. 97.00 1.8437 97.50 1.8439 98.00 1.8437 99.00 1.8424 100.00 1.8391 (273) 384 APPENDIX TENSION OF AQUEOUS VAPOR Temp. Piess., Temp. Press., Temp. Press., "C. mm. °e. mm. °C. mm. 0° 4.60 12° 10.48 24° 22.18 1° 4.92 13° 11.19 25° 23.65 2° 5.29 14° 11.94 26° 24.99 3° 5.68 15° 12.73 27° 26.50 4° 6.09 16° 13.57 28° 28.10 5° 6.53 17° 14.45 29° 29.78 6° 7.00 18° 15.38 30° 31.56 7° 7.49 19° 16.37 31° 33.42 8° S.02 20° 17.41 32° 35.37 9° 8.58 21° 18.50 33° 37.43 10° 9.18 22° 19.66 34° 39.59 11° 9.81 23° 20.88 35° 41.85 Mensuration IT = 3.1416. Length of circle (radius r) = 2 ttt. Area of triangle (base b, altitude a) = -^^ Area of circle (radius r) = ttt^. Surface of sphere (radius r) = 4 irfi. Surface of cylinder (radius r, height K) = 27rr (r + A). Surface of cone (radius r, slant height V) = ?rr (r + 0- 4 Volume of sphere (radius r) = -z wr^. o Volume of cylinder (radius r, height h) = irfih. Gases (Weight at Standabd Conditions, 0° C. and 760 mm.) (Air 1.2926 grams 1 Carbon dioxide 1 . 9768 grams > Hydrogen 0.089873 grams ; Nitrogen 1.2507 grams (Oxygen 1.4292 grams Miscellaneous One pound avoirdupois = 453.59 grams. One meter = 39.37 inches. One gallon = 231 cubic inches. One cubic foot of -water at 60° F. weighs 62.37 pounds. INDEX Analysis: Direct gravimetric, 173. Elimination of constituent, 176. Indirect gravimetric, 177. Volumetric, 202. Approximate numbers: Addition, 28. Definition, 26. Division, 30. Multiplication, 29. Rules, 31. Subtraction, 31. Atomic weights: Calculation, 145. Relation to valence, 142. Table, back of cover. Unit, 91, 141. Barometer: Correction, 84. Baum^: Conversions, 65. Sulphuric acid, 35. Equivalent: Definition, 12. Values of solution, 204. Extrapolation: Method, 40. Factor: Chemical, 9. General, 15. Weights, 176, 205. Formula: From composition, 149. Of compounds, 148. Of minerals, 157. Gas: Analysis, 88. Moist, 85. Thermometer, 81. Volume from weight of sub- stance, 101. Heat: British Thermal Unit, 49. Calorie, 49. Of combustion, 51. Of fusion, 50. Of vaporization, 50. Specific, 50. Hydrometers: Baum6, 64. Constant volume, 63. Constant weight, 62. Interpolation: Method, 37. Law: Avogadro, 90. Boyle, 80. Charles, 80. Charles and Boyle combined, 82. Constancy of composition, 1. Constancy of mass, 1. 275 276 INDEX Law (continued): Dalton, 107. Dulong and Petit, 143. Gay-Lussac, 87. Graham, 94. Liter: Mohr, 106. Standard, 104. Mixed acid: Analysis, 218. Calculations, 256. Strengthening of, 257. Molecular weights: Prom elevation of B.P., 146. From depression of F.P., 146. From vapor density, 91. Of an organic acid, 147. Of an organic base, 147. Normality: Calculation of, 213. Equivalent, 214. Factor, 213. Of acids and bases, 208. Of oxidizing substances, 209. Of salt solutions, 210. Oleum: Calculation of strength, 191. Volumetric titration of, 222. Ratios: Chemical, 2. General, 15. Solutions: Adjustment of strength, 225. Calculation of normality, 213. Calculation with normal, 210. Solutions (ccmlinued) : Equivalent values, 204. Methods of standardizing, 203. Normal, 207. Single factor, 202. Specific gravity: Calculation of tables, 252. Definition, 59. Influence of temperature, 66. Liquids, by pyknometer, 62. Liquids, by sinker, 62. Powders, 61. SoUd, heavier than water, 60. Solid, lighter than water, 61. Use of tables, 254. Temperatvire: Absolute scale, 49. Centigrade scale, 48. Conversions, 51. Fahrenheit scale, 48. Reaumur scale, 49. Standard conditions, 83. Titration: Mixed acid, 218. Oleum, 222. With two indicators, 220. With two solutions, 216. Valence: Calculation of, 144. Definition, 141. Vapor density: Definition, 92. Measurement, 96. Weight: In vacuo, 102. Combining, 141. A SELECTED LIST OF BOOKS ON CHEMISTRY AND CHEMICAL TECHNOLOGY FublisTied by D. VAN NOSTRAND COMPANY 25 Park Place New York American Institute of Chemical Engineers. Transactions. 8vo. cloth. Issued annually. Vol. I., 1908, to Vol. VI., 1913, now ready. each, net, $6.00 Annual Reports on the Progress of Chemistry. Issued annually by the Chemical Society. Svo. cloth. 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Barium .... Bismuth . . . Boron Bromine . . . Cadmium. . Caesium. . . Calcium . . . . Carbon .... Cerium. . . . Chlorine . . . Chromium. . Cobalt Columbiiui. Copper .... Dysprosium. Erbium . . . . Europium . . Fluorine . . . Gadolinium. Gallium . . . . Germanium. Glucinum . . Gold Helium. . . . Holmium. . Hydrogen. . Indium. . . . Iodine Iridium. . . . Iron Krypton Lanthanum. Lead Lithium . . . . Lutecium. . . Magnesium. Manganese. Mercury. . . Symbol. Al Sb A Bi B Br Cd Cs Ca C Ce 01 Cr Co Cb Cu Dy Er Eu F Gd Ga Ge Gl Au He Ho H In I Ir Fe Kr La Pb Li Lu Mg Mn Hg Atomic Weight. 27.1 120.2 39.88 74.96 137.37 208.0 11.0 79.92 112.40' 132.81 40.07 12.00 140.25 35.46 52.0 58.97 93.5 63.57 162.5 167.7 152.0 19.0 157.3 69.9 72.5 9.1 197.2 3.99 163.5 1.008 114.8 126.92 193.1 55.84 82.92 139.0 207.10 6.94 174.0 24.32 54.93 200.6 Name. Molybdenum. . . Neodymium .... Neon Nickel Niton Nitrogen Osmium Oxygen Palladium Phosphorus .... Platinum Potassium Praseodymium. . Radium Rhodium Rubiditun Ruthenium Samarium Scandium Selenium Silicon Silver Sodium Strontium Sulphur Tantalum Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten Uranium Vanadium Xenon Ytterbium (Neoytterbium) Yttrium Zinc Zirconium Symbol. Mo Nd Ne Ni Nt N Os O Pd P Pt K Pr Ra Rh Rb Ru Sa Sc Se Si Ag Na Sr S Ta Te Tb Tl Th Tm Sn Ti W U V Xe Yb Yt Zn Zr Atomic Weight. 96.0 144.3 20.2 58.68 222.4 14.01 190.9 16.00 106.7 31.04 195.2 39.10 140.6 226.4 102.9 85.45 101.7 150.4 44.1 79.2 28.3 107.88 23.00 87.63 32.07 181.5 127.5 159.2 204.0 232.4 168.5 119.0 48.1 184.0 238.5 51.0 130.2 172.0 89.0 65.37 90.6 * Compiled by the International Committee on Atomic Weights consisting of F. W. Clarke, W. Ostwald, T. E. Thorpe, and G. Urbain.