A Textbook of Assaying | Page 6

Cornelius Berenger
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 is necessary to carry the reduction to ozs. and dwts. to the ton; and since gold and silver are sold by troy weight, whilst the ton is avoirdupois, it is of importance to remember that the ounces in the two systems are not the same. A ton contains 15,680,000 grains, which equal 653,333.3 dwts. or 32,666.6 ozs. (troy). The following rules are useful:--
To get ozs. (troy) per ton, multiply parts per 100,000 by 0.327; To get dwts. per ton, multiply parts per 100,000 by 6.53; To get grains per ton, multiply parts per 100,000 by 156.8.
Where liquids are being assayed, cubic centimetres are held to be equivalent to grams, and the usual method of statement is, "so many parts by weight in so many by measure." Where the statement is made as grams per litre or grains per gallon, there can be no doubt as to what is meant; and even if it be expressed in parts per 100,000, parts
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