------ 18.531 "

add this to (3):

18.531 grams 73.146 "

------ 91.677 "

and deduct (4) to find the weight of the equal bulk of paraffin.

91.677 grams 84.665 "

------ 7.012 "

divide the weight of the nitre by the weight of the paraffin:

7.012)18.531(2.6427 ------ ......

The sp. g., taking paraffin as the standard instead of water, is 2.6427.

Multiply this by the sp. g. of paraffin, 0.7948, and the result is 2.1004 as the sp. g. of nitre compared with water.

Similarly, a sp. g. compared with water at say 50 C. can be converted into one compared with water at standard temperature, by multiplying by the sp. g. of water at 50 C. The following table gives the sp. g. of water at various temperatures:--

-----------+------++-----------+------++-----------+------- Degrees | || Degrees | || Degrees | Centigrade.|Sp. G.||Centigrade.|Sp. G.||Centigrade.|Sp. G.

-----------+------++-----------+------++-----------+------- 4 |1.0000|| 20 |0.9982|| 40 |0.9923 10 |0.9997|| 25 |0.9971|| 50 |0.9881 15 |0.9991|| 30 |0.9957|| 100 |0.9586 -----------+------++-----------+------++-----------+-------

If, for example, a substance at 50 C. has a sp. g. of 0.9010 as compared with water at 50 C., it will have (compared with water at 4 C.) a sp. g. of 0.90100.9881; or 0.8903. The figures 0.8903 represent the sp. g. of the substance at 50 C. compared with water at 4 C.

Except in comparing the sp. gravities of the same substance at different temperatures, a calculation of this kind serves no useful purpose.

_In taking the specific gravity of a solid not in powder_, a lump of it is freed from loose particles and its exact weight determined. By means of a horse hair with a slip knot it is suspended to the balance, and beneath it is placed, out of contact with the balance pan, a beaker of distilled water. The horse hair must be long enough to keep the mineral well beneath the surface of the water so as to allow the balance to vibrate. Air bubbles are removed by touching with a camel-hair pencil.

Whilst the mineral is suspended in water the weight is again taken. It will weigh less than before, and the difference between the two weighings gives the weight of water (and consequently the volume) displaced by the mineral. The weight in air divided by the difference is the specific gravity. Thus

Weight in air 3.2170 grams Weight in water 2.7050 "

------ Difference 0.5120 gram

3.2170/0.5120 equals 6.28, the sp. g.

The sp. g. of a substance depends mainly on its composition, but is affected by certain conditions. The effect of temperature has been already considered. Air holes and empty s.p.a.ces lessen the specific gravity of otherwise solid bodies; and metals, which after fusion become imperfect solids, have their density increased by hammering or rolling.

But metals when free from pores have their density diminished when rolled, without annealing. The effects of these conditions are slight when compared with those due to the presence of impurities.

For simple substances, or mixtures of only two substances, a determination of sp. g. is a sufficient check on the composition for many practical purposes; and with more complex mixtures, such as slags and some of the products of dressing operations in which the material does not differ much in its nature from time to time, such a determination will yield information of considerable value, and afford a check upon the proper working of a process.

When the mixing of two substances is accompanied by a change in volume, the sp. g. of the mixture can only be learnt by experiment. But when the substances have no such action on each other the resulting sp. g. can be calculated. Some of these calculations have a practical interest as well as an educational value. Students should practise them so as to become familiar with the relations between weight and volume.

_When substances are mixed by volume_, the sp. g. of the mixture is the mean of those of its const.i.tuents, and may be calculated in the usual way for obtaining averages. 1 c.c. of a substance having a sp. g. of 1.4 mixed with 1 c.c. of another having a sp. g. of 1.0 will yield 2 c.c. of a substance having a sp. g. of 1.2. If, however, we write gram instead of c.c. in the above statement, the resulting sp. g. will be 1.16. The simplest plan is to remember that the sp. g. is the weight divided by the volume (sp. g. = w/v) and the sp. g. of a mixture is the sum of the weights divided by the sum of the volumes (sp. g. = (w + w" + w", &c.)/(v + v" + v", &c.)). In the above example the sum of the volumes is 2 c.c.; the weights (got by multiplying each volume by its corresponding sp. g.) are 1.4 gram and 1 gram. The sum of the weights divided by the sum of the volumes is 2.4/2 or 1.2.

The sp. g. of a mixture of 10 c.c. of a substance having a sp. g. of 1.2, with 15 c.c. of another having a sp. g. of 1.5 may be thus found:--

sp. g. = (12+22.5)/(10+15) = 1.38

multiply each volume by its sp. g. to get its weight:

101.2 = 12 151.5 = 22.5

add these together (12+22.5 = 34.5) and divide by the sum of the volumes (10+15 = 25):

25)34.5(1.38 25 -- 95, &c.

The sp. g. will be 1.38, provided the mixture is not accompanied by any change of volume.

The same formula will serve when the proportion of the ingredients is given by weight. A mixture of 4 parts by weight of galena (sp. g. 7.5) with 5 parts of blende (sp. g. 4) will have a sp. g. of 5.06:

sp. g. = (4+5)/(0.53+1.25) = 9/1.78 = 5.06

It is necessary in this case to calculate the volumes of the galena and of the blende, which is done by dividing the weights by the sp.

gravities: thus, 4 divided by 7.5 gives 0.53 and 5 divided by 4 gives 1.25.

The converse problem is a little more difficult. Given the sp. g. of a mixture and of each of the two ingredients, the percentage by weight of the heavier ingredient may be ascertained by the following rule, which is best expressed as a formula. There are three sp. gravities given; if the highest be written H, the lowest L and that of the mixture M, then:

Percentage of heavier mineral = (100H(M-L))/(M(H-L))

Suppose a sample of tailings has a sp. g. of 3.0, and is made up of quartz (sp. g. 2.6) and pyrites (sp. g. 5.1): then the percentage of pyrites is 27:

(1005.1(3-2.6))/(3(5.1-2.6)) = (5100.4)/(32.5) = 204/7.5 = 27.2

The same problem could be solved with the help of a little algebra by the rule already given, as thus: the sp. g. of a mixture equals the sum of the _weights_ of the const.i.tuents divided by the sum of the _volumes_. Then 100 grams of the tailings with _x_ per cent. of pyrites contain 100-_x_ per cent. of quartz. The sum of the weights is 100. The volume of the pyrites is _x_/5.1 and of the quartz (100-_x_)/2.6.

Then we have by the rule

3 = 100/((_x_/5.1)+(100-_x_)/2.6) 3 = 1326/(510-2.5_x_) 204 = 7.5_x_ and _x_ = 27.2

If the percentage (P) and sp. g. (H) of one const.i.tuent and the sp. g.

(M) of the mixture are known, the sp. g. of the other const.i.tuent may be calculated by the following formula, in which _x_ is the required sp.

g.:

_x_ = ((100-P)MH)/((100H)-(PM))

For example, "tailings" (sp. g. 3.0) containing 27.2 per cent. of pyrites (sp. g. 5.1) will contain (100-27.2), 72.8 per cent. of earthy matter having a mean sp. g. of _x_:

_x_ = ((100-27.2)35.1)/((1005.1)-(27.23)) = 1113.84/428.4 = 2.6

The differences in sp. g. corresponding to differences in strength have been carefully determined and tabulated in the case of the stronger acids and of many other liquids. Such tables are given at the end of this book.

_To Calculate the Weight of a Measured Volume of Mineral or Rock._--Multiply the cubic feet by 62.4 and then multiply by the sp. g.

of the stuff, the answer gives the weight in pounds. For example, 100 cubic feet of quartz weighs 10062.42.6 = 16,224 lbs. The weight of any ma.s.s of mineral of known extent and sp. g. is ascertained in this way.

The following table gives the specific gravities of some of the commoner minerals.

Barytes 4.5 Blende 4.0 Calcite 2.6 Ca.s.siterite 6.9 Chalybite 3.8 Copper pyrites 4.2 Fluor 3.1 Galena 7.5 Haemat.i.te 5.0 Mispickel 6.2 Pyrites 5.0 Quartz 2.6

FOOTNOTES: