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).

3.  It is sometimes necessary to calculate the value of solutions according to the principles just explained, when several successive reactions are involved.  Such problems may be solved by a series of proportions, but it is usually possible to eliminate the common factors and solve but a single one.  For example, the amount of MnO_{2} in a sample of the mineral pyrolusite may be determined by dissolving the mineral in hydrochloric acid, absorbing the evolved chlorine in a solution of potassium iodide, and measuring the liberated iodine by titration with a standard solution of sodium thiosulphate.  The reactions involved are: 

MnO_{2} + 4HCl —­> MnCl_{2} + 2H_{2}O + Cl_{2}
Cl_{2} + 2KI —­> I_{2} + 2KCl
I_{2} + 2Na_{2}S_{2}O_{3} —­> 2NaI + Na_{2}S_{4}O_{6}

Assuming that the weight of thiosulphate corresponding to the volume of sodium thiosulphate solution used is known, what is the corresponding weight of manganese dioxide?  From the reactions given above, the following proportions may be stated: 

2Na_{2}S_{2}O_{3}:I_{2} = 316.4:253.9,

I_{2}:Cl_{2} = 253.9:71,

Cl_{2}:MnO_{2} = 71:86.9.

After canceling the common factors, there remains 2Na_{2}S_{2}O_{3}:MnO_{2} = 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