of Hydrocarbons and their Mixtures. 127 



mixtures form the best example of this. They all obey the 

 law pretty closely except marsh-gas itself, which has been 

 considered. Supposing, for the sake of an example, that we 

 admit the " theoretical value " 12*4 C.P. for marsh-gas, the 

 illuminating ratio may be obtained at once by taking A?/ and 

 Ax for the two substances. 



A*/=65-12-4 = 52'6 



Illuminating ratio = A?//A<r = 210*4. 



Then this ratio is the same or nearly so, from whatever 

 pair of the mixtures it is derived. 



I will now employ these principles to calculate the illu- 

 minating-power of the Gas Light Coke gas from its analysis, 

 as given in Chem. Journal, 1884, p. 193, as far as our present 

 materials permit. 



I take from the analysis : — 



Percentage of heavy hydrocarbons 4*41 



a i f 2-98 



Average value < tt -on 



H 47-99 

 CH 4 37-64 

 The H is of course H 2 . 

 From this I calculate, 



For heavy hydrocarbons x= "690 



For entire gas *198 



The disilluminate point is determined as } 

 before described by the two mixtures of >*132 

 disilluminate gas with ethylene ) 



For Ethylene C. P. = 65 



For Benzene 255 



These are the data. 



First form the equation of the line joining the ethylene 

 and benzene points. 



#" = 1 y" = 255 Benzene, 



#' = •5 z/ f = 65 Ethylene. 

 The equation is, 



y-65 = 0--5) x _. 5 . 



In this put # = '69, and we get for the C.P. of the heavy 

 hydrocarbons y= 137. Combining this with the disilluminate 



