If the vapour of ether is assumed, then _f_ = 10.64, and we have (_p_ - _f_) (30 - 10.64) 19.36 ----------- = ----------- = ----- = _p_ 30 30 .645, ... or 64½ per cent dry air.[28] 35½ per cent ethereal vapour. ----- 100 [Footnote 28: The aqueous vapour in this case maybe considered as insignificant.] 4. Suppose we find by trial the weight of 100 cubic inches of common air saturated with vapour at 60°, the barometer standing at 30 inches to be 30.5 grains, and the weight of hydrogen gas in like circumstances to be 2.118 grains; query the weights of 100 cubic inches of each gas free from vapour, and their specific gravities, the temperature and pressure being as above? If 30.5 ∶ 2.118 ∷ 1 ∶ .0694 = sp. gr. of vapourized hydrogen, that of vapourized air being 1. Subtracting .42 grs. (weight of vapour per table) from 30.5 grs., leaves 30.08 grains; and subtracting .65 parts of an inch from 30 inches, leaves 29.35 inches. Hence 100 cubic inches of dry air at the pressure of 29.35 inches, weigh 30.08 grains; and we have 29.35 ∶ 30 ∷ 30.08 ∶ 30.746 grains, the weight of 100 inches of dry air. Again, subtracting .42 grs. from 2.118, leaves 1.698 grains = weight of 100 cubic inches of hydrogen of 60° and sustaining the pressure of 29.35 inches; whence if 29.35 ∶ 30 ∷ 1.698 ∶ 1.736 grains, weight of 100 inches of dry hydrogen; and 30.746 ∶ 1.736 ∷ 1 ∶ .05645 = sp. gr. of dry hydrogen, that of dry air being unity. Or the results may be exhibited as under: Weight of 100 cubic inches. Sp. Gravities. Vap. air 30.5 grains 1 14.4 Vap. hydrogen 2.118 ---- .0694 1 Dry air 30.746 grains 1 17.7 Dry hydrogen 1.736 ---- .05645 1 FORMULÆ FOR DETERMINING THE PROPORTIONS OF COMBUSTIBLE GASES IN MIXTURES. It frequently happens, especially in the decomposition of vegetable substances by heat, that the product consists of several combustible gases in mixture, and it is desirable to determine the proportions of each of those which collectively constitute the mixture. The following forms will be found useful for this purpose. 1. _Carbonic oxide and hydrogen._ Let _x_ = the volume of carbonic oxide, _y_ = that of hydrogen, _w_ = that of mixture, and _a_ = that of carbonic acid, produced by exploding the mixed gases with oxygen over mercury. Then the carbonic oxide, or _x_ = _a_, and the hydrogen, or _y_ = _w_ - _a_. 2. _Sulphuretted hydrogen and hydrogen._ Let _x_ = the volume of sulphuretted hydrogen, _y_ = that of hydrogen, _w_ = that of the mixture, and _g_ = the oxygen spent in the combustion of _w_. Then because _x_ + _y_ = _w_, and 1½_x_ + ½_y_ = _g_; we have _x_ = _g_ - ½_w_, and _y_ = 1½_w_ - _g_.