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

Options

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

Answer

A

Explanation

Let the mass of gaseous hydrocarbon taken = w gm
`:. %C = (12)/(44) xx (3.08)/(w) xx 100 = (84)/(w)`
`%H = (2)/(18) xx (0.72)/(w) xx 100 = (8)/(w)`
Ratio of C : H atoms `= (84)/(w) xx(1)/(12) : (8)/(w) xx (1)/(1) = 7 : 8`
Thus, E.F. = `C_(7)H_(8)`.