We may therefore use the following rules for working processes which do not yield proportional results. Make a series of two or three titrations, using very different quantities of metal in each. Subtract the lowest of these from the highest, and calculate the standard with the remainder. Calculate the volume required by this standard in any case, and find the excess or deficit, as the case may be. If an excess, subtract it from the result of each titration; if a deficit, add it; and use the standard in the usual way. The following table shows an actual example:-- +--------------+---------------+-----------+ | Chalk taken. | Gas obtained. | Standard. | +--------------+---------------+-----------+ | 0.0873 gram | 17.8 c.c. | 0.4904 | | 0.1305 " | 27.3 " | 0.4780 | | 0.1690 " | 35.8 " | 0.4721 | | 0.1905 " | 40.4 " | 0.4715 | | 0.2460 " | 52.5 " | 0.4686 | | 0.3000 " | 64.0 " | 0.4687 | +--------------+---------------+-----------+ It will be seen that the standard decreases as the quantity of chalk increases; this points to a deficiency in the quantity of gas evolved. Then 0.3000 = 64.0 c.c. 0.0873 = 17.8 " ------ = ---- 0.2127 = 46.2 " and 0.2127×100/46.2 = 0.4604. Then, multiplying the weight of chalk taken by 100, and dividing by 0.4604, we get the calculated results of the following table:-- +--------------+------------+-----------------+-------------+ | Chalk taken. | Gas found. | Gas calculated. | Difference. | +--------------+------------+-----------------+-------------+ | 0.0873 gram | 17.8 c.c. | 18.9 c.c. | -1.1 c.c. | | 0.1305 " | 27.3 " | 28.3 " | -1.0 " | | 0.1690 " | 35.8 " | 36.7 " | -0.9 " | | 0.1905 " | 40.4 " | 41.4 " | -1.0 " | | 0.2460 " | 52.5 " | 53.4 " | -0.9 " | | 0.3000 " | 64.0 " | 65.1 " | -1.1 " | +--------------+------------+-----------------+-------------+ By adding 1 c.c. to the quantity of gas obtained, and taking 0.4604 as the standard, the calculated results will agree with those found with a variation of 0.1 c.c. When a large number of assays of the same kind are being made, this method of calculation is convenient; when, however, only one or two determinations are in question, it is easier to make a couple of standardisings, taking quantities as nearly as possible the same as those present in the assays.