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

Options

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

Answer

C

Explanation

To get the empricial formula, we need to take the simplest whole number ratio of the moles of `C` to `H`. Let `C_(X) H_(Y)` be the molecular formula of the hydrocarbon. Then the balanced chemical equation for its combustion will be
`C_(X) H_(Y) + (X + (Y)/(4)) O_(2) rarr X CO_(2) + (Y)/(2) H_(2) O`
`n_(CO_(2)) = ("Mass"_(CO_(2)))/("Molar mass"_(CO_(2))) = (3.08g)/(44g mol^(-1)) = 0.07 mol`
`n_(H_(2)O) = ("Mass"_(H_(2)O))/("Molar mass"_(H_(2) O)) = (0.72g)/(18g mol^(-1)) = 0.04 mol`
Combining this information with that of the balanced equation, we have
`(n_(CO_(2)))/(n_(H_(2)O)) = (X)/(Y//2) = (0.07)/(0.04)`
`implies (2X)/(Y) = (7)/(4) implies (X)/(Y) = (7)/(8)`