= 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