A Textbook of Assaying | Page 6

can be dried, and their moisture
determined by the loss in this way. When, however, the substance
contains another somewhat volatile ingredient, it is exposed over
sulphuric acid in a desiccator for two days (if in vacuo, all the better),
and the loss determined. Moisture in dynamite should be determined in
this way.
When water is simply mechanically mixed with a substance it presents
but little difficulty. The combined water is a different matter. Slaked
lime, even when perfectly dry, contains much water; and if the water of
soda crystals were separated and frozen, it would occupy a volume
equal to that of the original crystals. Perfectly dry substances may
contain much water, and this combined water is retained by different
materials with very unequal vigour. Sodium sulphate and sodium
phosphate crystals lose water even when exposed under ordinary
conditions to dry air. Soda crystals when heated melt, and at a moderate
temperature give off their water with ebullition. The temperature at
which all the water is given up varies with each particular salt; the
actual determination of the water in each case will require somewhat

different treatment. Such determinations, however, are seldom required;
and from a practical point of view this combined water causes no
trouble.
In assaying ores, we term "moisture" all water which is lost by
exposure in a water-oven at 100° C., and the "dry ore" is the ore which
has been dried at this temperature. No advantage, but rather endless
confusion, would be caused by varying the temperature with the object
of estimating the whole of the water which a hydrated salt may contain.
The results of the assay of the other components should be calculated
on the "dry ore." One advantage of this is obvious:--The dry ore has a
constant composition, and the results of all assays of it will be the same,
no matter when made; the moisture, however, may vary from day to
day, and would be influenced by a passing shower of rain. It is well to
limit this variability to the moisture by considering it apart, and thus
avoid having the percentage, say, of copper rising and falling under the
influence of the weather.
In the case of certain salts, however, such as soda crystals and hydrated
sulphate of copper (when these constitute the bulk of the substance to
be assayed), it is as well to perform the assay on the moist, or at any
rate air-dried, substance.[2] It would be equally convenient to calculate
on the substance dried at 100° C.; but in this case it would be well, in
order to avoid a somewhat shallow criticism, to replace the term
"moisture" by the longer but equivalent phrase "water lost at 100° C."
~Calculation and Statement of Results.~--By far the most generally
convenient method of stating the results of an assay is that of the
percentage or parts in a hundred, and to avoid a needlessly troublesome
calculation it is well to take such a quantity of ore for each assay as by
a simple multiplication will yield the percentage. In these calculations
decimals are freely employed, and students should make themselves
familiar with the methods of using them.
Other methods of statement are in use, and have advantages in certain
special cases. With bullion the parts in a thousand are given, and in
those cases in which the percentage is very small, as in water analysis,
it is convenient to report on parts in 100,000, or even on parts per

1,000,000. These are easily got from the corresponding percentages by
shifting the decimal point one, three, or four places to the right. Thus
92.5 per cent. is 925 per thousand; and 0.0036 per cent. is 3.6 per
100,000, or 36 per million.
With ores of tin, silver, and gold, the result is stated as so many cwts.,
lbs., or ozs., in the ton. With dressed tin ores as they are sent to the
smelter, the produce is given in cwts. and quarters to the ton. The
corresponding percentage may be obtained by multiplying by five; or,
inversely, if the percentage is given, the produce may be got by
dividing by five. A produce of 13-1/2 equals a percentage of 13.5×5 =
67.5; and a percentage of 70.0 equals a produce of 70/5 = 14. With tin
ores as raised (in which the percentage is small) the reduction must be
carried to pounds per ton. One per cent. equals 22.4 lbs. to the ton;
consequently, if we multiply the percentage by 22.4, the produce will
be given. Thus, if an ore contains 6.7 per cent. of oxide of tin, the
produce is 6.7×22.4 = 150 lbs. (or 1 cwt., 1 quarter, and 10 lbs.) to the
ton. With gold and silver ores, the proportion of precious metal is small,
and it