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.