# An Introductory Course of Quantitative Chemical Analysis eBook

This eBook from the Gutenberg Project consists of approximately 220 pages of information about An Introductory Course of Quantitative Chemical Analysis.

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.