PART IV
STOICHIOMETRY
The problems with which the analytical chemist has to deal are not, as a matter of actual fact, difficult either to solve or to understand. That they appear difficult to many students is due to the fact that, instead of understanding the principles which underlie each of the small number of types into which these problems may be grouped, each problem is approached as an individual puzzle, unrelated to others already solved or explained. This attitude of mind should be carefully avoided.
It is obvious that ability to make the calculations necessary for the interpretation of analytical data is no less important than the manipulative skill required to obtain them, and that a moderate time spent in the careful study of the solutions of the typical problems which follow may save much later embarrassment.
1. It is often necessary to calculate what is known as a "chemical factor," or its equivalent logarithmic value called a "log factor," for the conversion of the weight of a given chemical substance into an equivalent weight of another substance. This is, in reality, a very simple problem in proportion, making use of the atomic or molecular weights of the substances in question which are chemically equivalent to each other. One of the simplest cases of this sort is the following: What is the factor for the conversion of a given weight of barium sulphate (BaSO_{4}) into an equivalent weight of sulphur (S)? The molecular weight of BaSO_{4} is 233.5. There is one atom of S in the molecule and the atomic weight of S is 32.1. The chemical factor is, therefore, 32.1/233.5, or 0.1375 and the weight of S corresponding to a given weight of BaSO_{4} is found by multiplying the weight of BaSO_{4} by this factor. If the problem takes the form, "What is the factor for the conversion of a given weight of ferric oxide (Fe_{2}O_{3}) into ferrous oxide (FeO), or of a given weight of mangano-manganic oxide (Mn_{3}O_{4}) into manganese (Mn)?" the principle involved is the same, but it must then be noted that, in the first instance, each molecule of Fe_{2}O_{3} will be equivalent to two molecules of FeO, and in the second instance that each molecule of Mn_{3}O_{4} is equivalent to three atoms of Mn. The respective factors then become
(2FeO/Fe_{2}O_{3}) or (143.6/159.6) and (3Mn/Mn_{3}O_{4}) or (164.7/228.7).
It is obvious that the arithmetical processes involved in this type of problem are extremely simple. It is only necessary to observe carefully the chemical equivalents. It is plainly incorrect to express the ratio of ferrous to ferric oxide as (FeO/Fe_{2}O_{3}), since each molecule of the ferric oxide will yield two molecules of the ferrous oxide. Mistakes of this sort are easily made and constitute one of the most frequent sources of error.
2. A type of problem which is slightly more complicated in appearance, but exactly comparable in principle, is the following: "What is the factor for the conversion of a given weight of ferrous sulphate (FeSO_{4}), used as a reducing agent against potassium permanganate, into the equivalent weight of sodium oxalate (Na_{2}C_{2}O_{4})?" To determine the chemical equivalents in such an instance it is necessary to inspect the chemical reactions involved. These are:
10FeSO_{4} + 2KMnO_{4} + 8H_{2}SO_{4} —> 5Fe_{2}(SO_{4}){3} + K{2}SO_{4} + 2MnSO_{4} + 8H_{2}O,
5Na_{2}C_{2}O_{4} + 2KMnO_{4} + 8H_{2}SO_{4} —> 5Na_{2}SO_{4} + 10CO_{2} + K_{2}SO_{4} + 2MnSO_{4} + 8H_{2}O.
It is evident that 10FeSO_{4} in the one case, and 5Na_{2}C_{2}O_{4} in the other, each react with 2KMnO_{4}. These molecular quantities are therefore equivalent, and the factor becomes (10FeSO_{4}/5Na_{2}C_{2}O_{4}) or (2FeSO_{4}/Na_{2}C_{2}O_{4}) or (303.8/134).
Again, let it be assumed that it is desired to determine the factor required for the conversion of a given weight of potassium permanganate (KMnO_{4}) into an equivalent weight of potassium bichromate (K_{2}Cr_{2}O_{7}), each acting as an oxidizing agent against ferrous sulphate. The reactions involved are:
10FeSO_{4} + 2KMnO_{4} + 8H_{2}SO_{4} —> 5Fe_{2}(SO_{4}){3} + K{2}SO_{4} + 2MnSO_{4} + 8H_{2}O,
6FeSO_{4} + K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4} —> 3Fe_{2}(SO_{3}){3} + K{2}SO_{4} + Cr_{2}(SO_{4}){3} + 7H{2}O.
An inspection of these equations shows that 2KMO_{4} react with 10FeSO_{4}, while K_{2}Cr_{2}O_{7} reacts with 6FeSO_{4}. These are not equivalent, but if the first equation is multiplied by 3 and the second by 5 the number of molecules of FeSO_{4} is then the same in both, and the number of molecules of KMnO_{4} and K_{2}Cr_{2}O_{7} reacting with these 30 molecules become 6 and 5 respectively. These are obviously chemically equivalent and the desired factor is expressed by the fraction (6KMnO_{4}/5K_{2}Cr_{2}O_{7}) or (948.0/1471.0).
3. It is sometimes necessary to calculate the value of solutions according to the principles just explained, when several successive reactions are involved. Such problems may be solved by a series of proportions, but it is usually possible to eliminate the common factors and solve but a single one. For example, the amount of MnO_{2} in a sample of the mineral pyrolusite may be determined by dissolving the mineral in hydrochloric acid, absorbing the evolved chlorine in a solution of potassium iodide, and measuring the liberated iodine by titration with a standard solution of sodium thiosulphate. The reactions involved are:
MnO_{2} + 4HCl —> MnCl_{2} + 2H_{2}O + Cl_{2}
Cl_{2} + 2KI —> I_{2} + 2KCl
I_{2} + 2Na_{2}S_{2}O_{3} —> 2NaI + Na_{2}S_{4}O_{6}
Assuming that the weight of thiosulphate corresponding to the volume of sodium thiosulphate solution used is known, what is the corresponding weight of manganese dioxide? From the reactions given above, the following proportions may be stated:
2Na_{2}S_{2}O_{3}:I_{2} = 316.4:253.9,
I_{2}:Cl_{2} = 253.9:71,
Cl_{2}:MnO_{2} = 71:86.9.
After canceling the common factors, there remains 2Na_{2}S_{2}O_{3}:MnO_{2} = 316.4:86.9, and the factor for the conversion of thiosulphate into an equivalent of manganese dioxide is 86.9/316.4.
4. To calculate the volume of a reagent required for a specific operation, it is necessary to know the exact reaction which is to be brought about, and, as with the calculation of factors, to keep in mind the molecular relations between the reagent and the substance reacted upon. For example, to estimate the weight of barium chloride necessary to precipitate the sulphur from 0.1 gram of pure pyrite (FeS_{2}), the proportion should read
488. 120.0 2(BaCl_{2}.2H_{2}O):FeS_{2} = x:0.1,
where !x! represents the weight of the chloride required. Each of the two atoms of sulphur will form upon oxidation a molecule of sulphuric acid or a sulphate, which, in turn, will require a molecule of the barium chloride for precipitation. To determine the quantity of the barium chloride required, it is necessary to include in its molecular weight the water of crystallization, since this is inseparable from the chloride when it is weighed. This applies equally to other similar instances.
If the strength of an acid is expressed in percentage by weight, due regard must be paid to its specific gravity. For example, hydrochloric acid (sp. gr. 1.12) contains 23.8 per cent HCl !by weight!; that is, 0.2666 gram HCl in each cubic centimeter.
5. It is sometimes desirable to avoid the manipulation required for the separation of the constituents of a mixture of substances by making what is called an "indirect analysis." For example, in the analysis of silicate rocks, the sodium and potassium present may be obtained in the form of their chlorides and weighed together. If the weight of such a mixture is known, and also the percentage of chlorine present, it is possible to calculate the amount of each chloride in the mixture. Let it be assumed that the weight of the mixed chlorides is 0.15 gram, and that it contains 53 per cent of chlorine.
The simplest solution of such a problem is reached through algebraic methods. The weight of chlorine is evidently 0.15 x 0.53, or 0.0795 gram. Let x represent the weight of sodium chloride present and y that of potassium chloride. The molecular weight of NaCl is 58.5 and that of KCl is 74.6. The atomic weight of chlorine is 35.5. Then
x + y = 0.15 (35.5/58.5)x + (35.5/74.6)y = 0.00795
Solving these equations for x shows the weight of NaCl to be 0.0625 gram. The weight of KCl is found by subtracting this from 0.15.
The above is one of the most common types of indirect analyses. Others are more complex but they can be reduced to algebraic expressions and solved by their aid. It should, however, be noted that the results obtained by these indirect methods cannot be depended upon for high accuracy, since slight errors in the determination of the common constituent, as chlorine in the above mixture, will cause considerable variations in the values found for the components. They should not be employed when direct methods are applicable, if accuracy is essential.