(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_{ 4} in the other, each react with 2KMnO_{ 4}. These molecular quantities are therefore equivalent, and the factor becomes (10FeSO_{ 4}/5Na_{ 2}C_{ 2}O_{ 4}) or (2FeSO_{ 4}/Na_{ 2}C_{ 2}O_{ 4}) or (303.8/134).
Again, let it be assumed that it is desired to determine the factor required for the conversion of a given weight of potassium permanganate (KMnO_{ 4}) into an equivalent weight of potassium bichromate (K_{ 2}Cr_{ 2}O_{ 7}), each acting as an oxidizing agent against ferrous sulphate. The reactions involved are:
10FeSO_{ 4} + 2KMnO_{ 4} + 8H_{ 2}SO_{ 4} --> 5Fe_{ 2}(SO_{ 4})_{ 3} + K_{ 2}SO_{ 4} + 2MnSO_{ 4} + 8H_{ 2}O,
6FeSO_{ 4} + K_{ 2}Cr_{ 2}O_{ 7} + 7H_{ 2}SO_{ 4} --> 3Fe_{ 2}(SO_{ 3})_{ 3} + K_{ 2}SO_{ 4} + Cr_{ 2}(SO_{ 4})_{ 3} + 7H_{ 2}O.
An inspection of these equations shows that 2KMO_{ 4} react with 10FeSO_{ 4}, while K_{ 2}Cr_{ 2}O_{ 7} reacts with 6FeSO_{ 4}. These are not equivalent, but if the first equation is multiplied by 3 and the second by 5 the number of molecules of FeSO_{ 4} is then the same in both, and the number of molecules of KMnO_{ 4} and K_{ 2}Cr_{ 2}O_{ 7} reacting with these 30 molecules become 6 and 5 respectively. These are obviously chemically equivalent and the desired factor is expressed by the fraction (6KMnO_{ 4}/5K_{ 2}Cr_{ 2}O_{ 7}) or (948.0/1471.0).