A gaseous hydrocarbon on combustion gives 0 . 72g of water and 3 ....

${\text{C}}_{\text{x}} {\text{H}}_{\text{y}} + \left[\text{x} + \dfrac{\text{y}}{4}\right] {\text{O}}_{2} ⟶ \text{xC} {\text{O}}_{2} + \dfrac{\text{y}}{2} {\text{H}}_{2} \text{O}$

Moles of ${CO}_{2}$ $= \dfrac{3 \text{.} 0 8}{4 4} = 0 \text{.} 0 7 \text{moles} = \text{x}$

Moles of ${H}_{2}O$ $= \dfrac{0 \text{.} 7 2}{1 8} = 0 \text{.} 0 4 \text{moles} = \dfrac{\text{y}}{2}$

$x = 0 . 07g$ atoms of carbon

$y = 0 . 08g$ atoms of hydrogen

$x: y0 . 070 . 0878$

Empirical formula becomes ${C}_{7}{H}_{8}$.