Let mass of methane in original mixture be x g
So total number of moles in original mixture n1 = (12-x)/4+x/16
When 8g O2 is added to it total number of moles become n2= (12-x)/4+x/16+ 8/32
Temperature and volume remaining constant pressure is proportional to number of moles as we know PV=nRT.
So P1V=n1RT...(1)
P2V=n2RT...(2)
Dividing (2) by (1) we get
n2/n1=P2/P1
But by the problem P2/P1=7/6
Hence n2/n1=7/6
=>[ (12-x)/4+x/16+ 8/32]/[(12-x)/4+x/16]=7/6
=> x= 8g
so mass of original mixture is 12g
Hence the amount of Methane in original mixture =8/12*100%=66.67%