214 Mr. Sylvester on a Method of expresshig [Sept. 



Atoms. Sp. Gr. 



Carbon c .... 



Sulphur s .... 



Phosphorus f .... 



Steam. oh . ... oh 



Nitric oxide ao . . . . ao 



Nitrous gas ao" .... (ao a -) % 



Muriatic acid Ch .... (C//)*" 



Carbonic oxide co . . . . co 



Carbonic acid co- . . . . co" 



Sulphuret hydrogen .... sh . . . . sh 



Phosphuretted hydrogen FA .... F/t 



Ammonia ah- .... (ah 3 ) r 



defiant gas ch .... (cA) J 



Carburetted hydrogen . . ch- .... chr 



If the ratio by weight of two gases, which act upon each other, 

 be known, the ratio of their volumes may be known by multi- 

 plying the ratio of their weights by the inverted ratio of their 

 specific gravities. 



In forming water, the ratio bv weight will be -, and the ratio 



oftheir specific gravities will be -, and inverted is-. Hence -x 

 5= i-, which is one volume of oxygen to two volumes of hydrogen. 







It will be seen that in reducing the above ratios, it is only neces- 

 sary to take away the letters, and leave the exponents ; as in the 



above r, X -^ = 4U^X 4- = 4- as before. 



A 1 o a ' 



a 



c« 



The ratio by weight of oxygen to carbonic oxide will be 

 , or o to co. The inverted ratio of their specific gravities will be 



-. Then - x- ai x is one volume of oxygen to two of 

 carbonic oxide. The ratio of oxygen to defiant gas is - . Then — 



(ch)' 2 



x —j- = -f x f = three volumes oxygen to one olefiant gas. 



Oxygen to carburetted hydrogen = — -. Then — x — = 



% x -|- = -S-, which is two volumes oxygen to one of carburetted 

 hydrogen. In order to see what quantity of oxygen will be 

 required for the saturation of an inflammable gas, write down 

 the gas first, as in olefiant gas, ch ; then it will be seen that 

 c the carbon will want two atoms of oxygen, and h the hydrogea 



