A gaseous hydrocarbon gives upon combustion 0.72g water and 3.08g `CO_(2)`. The empirical formula of the hydrocarbon-

Options

A. `C_(2)H_(4)`
B. `C_(3)H_(4)`
C. `C_(6)H_(5)`
D. `C_(7)H_(8)`

Answer

D

Explanation

Moles of water pdouced`=(0.72)/(18)=0.04`
Moles of `CO_(2)` produced`=(3.08)/(44)=0.07`
Equation for combustion of an unknown hydrocarbon,
`C_(x)H_(y)` is- `C_(x)Y_(h)+(x+(y)/(4))O_(2) to xCO_(2)+(y)/(2)H_(2)O`
`therefore x=0.07 and (y)/(2)=0.04" "therefore y=0.08`
`(x)/(y)=0.07//0.08=(7)/(8),` The empirical formula is `C_(7)H_(8)`.