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
`therefore %C=(12)/(44)xx(3.08)/(w)xx100=(84)/(w), %H=(2)/(18)xx(0.72)/(w)xx100=(8)/(w)`
Rate of C : H atoms = `(84)/(w)xx(I)/(12):(8)/(w)xx(1)/(1)=7:8` Thus, E.F = `C_(7)H_(8)`.