120 g of an organic compound that contains only carbon and hydrogen gives 330 g of CO₂ and 270 g of water on complete combustion. The percentage of carbon and hydrogen, respectively are

<p>$${C_x}{H_y} + \left( {x + {y \over 4}} \right){O_2} \to xC{O_2} + {y \over 2}{H_2}O$$</p> <p>From the reaction,</p> <p>Produced CO₂ = x mol</p> <p>and produced H₂O = ${y \over 2}$ mol</p> <p>Given produced CO₂ = 330 g</p> <p>$\therefore$ moles of CO₂ $= {{330} \over {44}} = {{30} \over 4} = x$</p> <p>Also given produced H₂O = 270 gm</p> <p>$\therefore$ Moles of H₂O $= {{270} \over {18}} = 15 = {y \over 2}$.</p> <p>$\Rightarrow$ y = 30</p> <p>$\therefore$ $x:y = {{30} \over 4}:30 = 1:4$</p> <p>Formula of the compound = ${(C{H_4})_n}$</p> <p>$\therefore$ Weight of C in ${(C{H_4})_n}$ = 12 n</p> <p>Weight of H in ${(C{H_4})_n}$ = 4 n</p> <p>$\therefore$ Weight ratio of C and H</p> <p>= 12 n : 4 n</p> <p>= 3 : 1</p> <p>$\therefore$ % of C = ${3 \over 4}$ $\times$ 100 = 75</p> <p>and % of H = ${1 \over 4}$ $\times$ 100 = 25</p>