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_(6)`

Answer

A

Explanation

Moles of `CO_(2)` produced `=(3.08)/44 = 0.07`
Moles of `H_(2)O` produced `=0.72/18 = 0.04`
Combustion of hydrocarbons `C_(x)H_(y)` may be represented as:
`C_(x)H_(y) + (x+y/4)O_(2) to xCO_(2) + y/2H_(2)O`
`x =0.07, y/2=0.04`
`therefore y=0.08`
`x/y = 0.07/0.08= 7/8`
`therefore` Empirical formula of hydrocarbons `C_(7)H_(6)`.