in the process under consideration are those of potassium bichromate
and ferrous sulphate. The reaction between them, in the presence of an
excess of sulphuric acid, may be expressed as follows:

6FeSO_{4} + K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4} --> 3Fe_{2}(SO_{4})_{3} +
K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + 7H_{2}O.

If the compounds of iron and chromium, with which alone we are now
concerned, be written in such a way as to show the oxides of these
elements in each, they would appear as follows: On the left-hand side
of the equation 6(FeO.SO_{3}) and K_{2}O.2CrO_{3}; on the right-hand
side, 3(Fe_{2}O_{3}.3SO_{3}) and Cr_{2}O_{3}.3SO_{3}. A careful
inspection shows that there are three less oxygen atoms associated
with chromium atoms on the right-hand side of the equation than on the
left-hand, but there are three more oxygen atoms associated with iron
atoms on the right than on the left. In other words, a molecule of
potassium bichromate has given up three atoms of oxygen for oxidation
purposes; i.e., a molecular weight in grams of the bichromate (294.2)
will furnish 3 X 16 or 48 grams of oxygen for oxidation purposes.
As this 48 grams is six times 8 grams, the basis of the system, the
normal solution of potassium bichromate should contain per liter one
sixth of 294.2 grams or 49.03 grams.

A further inspection of the dissected compounds above shows that six
molecules of FeO.SO_{3} were required to react with the three atoms of
oxygen from the bichromate. From the two equations

3H_{2} + 3O --> 3H_{2}O
6(FeO.SO_{3}) + 3O --> 3(Fe_{2}O_{3}.3SO_{3})