| Whence M = | 56100 | (2). |
| K |
Substituting this value of M in (1) we have
| F = | A × 56100 | = | 100A | (3). |
| 561 × K | 561 × K |
It is evident that it is not necessary to calculate the acid value (A) of the sample and the saponification value (K) of the free fat acids, the ratio A/K alone being required. It will be sufficient therefore to substitute for A and K the number of cubic centimeters of alkali solutions required for one gram of the fat and one gram of the fat acids, respectively. If a and b represent these numbers the formula may be written
| F = | 100a | (4); |
| b |
| and N = 100 - F = | 100a | (5). |
| b |
To simplify the determinations, it may be assumed that the free fat acids have the same molecular weight as those still in combination with the glycerol in any given sample. On this assumption, the process may be carried on by determining the acid value A and the saponification value K for the total fat acids. The mean molecular weight M, the percentage of free fat acids F, and the proportion of neutral fat N, may then be calculated from the formulas (2), (3), (4), and (5).
Further, let G = the quantity of glycerol and L that of fat acids obtainable from one gram of neutral fat, that is, ¹/₁₀₀ of H the percentage of total fat acids.
The molecular weight of the neutral fat in each case is 3M + 38. Therefore, 3M + 38 parts of neutral fat yield 3M parts of fat acids and ninety-two parts of glycerol (C₃H₈O₃ = 92).
| Then L = | H | = | 3M | (6); |
| 100 | 3M + 38 |