$${C_x}{H_y} + \left( {x + {y \over 4}} \right){O_2} \to xC{O_2} + {y \over 2}{H_2}O$$

From the reaction,

Produced CO₂ = x mol

and produced H₂O = $${y \over 2}$$ mol

Given produced CO₂ = 330 g

$$\therefore$$ moles of CO₂ $$ = {{330} \over {44}} = {{30} \over 4} = x$$

Also given produced H₂O = 270 gm

$$\therefore$$ Moles of H₂O $$ = {{270} \over {18}} = 15 = {y \over 2}$$.

$$\Rightarrow$$ y = 30

$$\therefore$$ $$x:y = {{30} \over 4}:30 = 1:4$$

Formula of the compound = $${(C{H_4})_n}$$

$$\therefore$$ Weight of C in $${(C{H_4})_n}$$ = 12 n

Weight of H in $${(C{H_4})_n}$$ = 4 n

$$\therefore$$ Weight ratio of C and H

= 12 n : 4 n

= 3 : 1

$$\therefore$$ % of C = $${3 \over 4}$$ $$\times$$ 100 = 75

and % of H = $${1 \over 4}$$ $$\times$$ 100 = 25