10 mL of a gaseous hydrocarbon is completely combusted with 60 mL of oxygen by passing electric spark. The volume of the gaseous mixture thus yielded on cooling becomes 46 mL, iff the vapour density of the compound is 15, what will be its molecular formula? [All volumes were measured under the same conditions of temperature and pressure.]

Explanation

Let, molecular formula of the hydrocarbon: `C_(x)H_(y)`
Molecular mass: `2xx15=30" "therefore 12x+y=30`
`underset(10mL)(C_(x)H_(y))+underset(10(x+(y)/(4))mL)((x+(y)/(4))O_(2)) to underset (10xmL)(xCO_(2))+(y)/(2)H_(2)O`
Ignoring the volume of water, construction in the volume of the mixture`=[10+10(x+(y)/(4))-10x]mL`. as given in the question,
`10+10(x+(y)/(4))-10x=[(60+10)-45]mL`
or, `10+(10)/(4)y=25`
`therefore y=6`, Again 12x+y=30 or, 12x+6=30
`thereforex=2`
`therefore `Formula of the hydrocarbon: `C_(2)H_(6)`.