weight of metal used; and calculations from the mean standard will be
incorrect.

By subtracting the lowest standardising from the highest, a third result
is got free from any error common to the other two; thus:--

0.4 gram = 48.6 c.c. "cyanide."
0.1 " = 12.9 " "
--- ----
0.3 " = 35.7 " "

And the standard calculated from this corrected result is 0.8404.
Further, if 0.3 gram requires 35.7 c.c., then 0.1 gram should require
11.9 c.c., or 1.0 c.c. less than that actually found.

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 |