| Cupric oxide (CuO) | 10.58 |
| Ferrous oxide (FeO) | 15.69 |
| Zinc oxide (ZnO) | 0.35 |
| Sulphuric oxide (SO2) | 28.82 |
| Water (H2O) | 44.71 |
| —————— | |
| 100.15 |
First find the molecular weights of CuO, FeO, &c., and divide the corresponding percentages by these figures. Thus, CuO = 63.3+16 = 79.3 and 10.58 divided by 79.3 gives 0.1334. Similarly FeO = 56+16 = 72 and 15.69 divided by 72 gives 0.2179. Treated in the same way the oxide of zinc, sulphuric oxide and water give as results 0.0043, 0.3602 and 2.484.
Classify the results as follows:—
| Bases. | Acids. | Water. |
| CuO 0.1334 | SO3 0.3602 | H2O 2.484 |
| FeO 0.2179 | ||
| ZnO 0.0043 | ||
| —————————— | —————————— | —————————— |
| RO 0.3556 | RO3 0.3602 | R2O 2.484 |
The figures 0.3556, 0.3602 and 2.484 should be then divided by the lowest of them—i.e., 0.3556; or where, as in this case, two of the figures are very near each other the mean of these may be taken—i.e., 0.3579. Whichever is taken the figures got will be approximately 1, 1 and 7. The formula is then RO.SO3.7H2O in which R is nearly 2/5ths copper, 3/5ths iron and a little zinc.
This formula requires the following percentage composition, which for the sake of comparison is placed side by side with the actual results.
| Calculated. | Found. | |
| Cupric oxide | 11.29 | 10.58 |
| Ferrous oxide | 15.37 | 15.69 |
| Zinc oxide | nil | 0.35 |
| Sulphuric oxide | 28.47 | 28.82 |
| Water | 44.84 | 44.71 |
| ——— | ——— | |
| 99.97 | 100.15 |
Trimming the results of an analysis to make them fit in more closely with the calculations from the formula would be foolish as well as dishonest. There can be no doubt that the actual analytical results represent the composition of the specimen much more closely than the formula does; although perhaps other specimens of the same mineral would yield results which would group themselves better around the calculated results than around those of the first specimen analysed. It must be remembered that substances are rarely found pure either in nature or in the arts; so that in most cases the formula only gives an approximation to the truth. In the case of hydrated salts there is generally a difficulty in getting the salt with exactly the right proportion of water.
PRACTICAL EXERCISES.
The following calculations may be made:—