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