QJEEM22Ch3a00: JEE Main 2022 Chemistry - Class 11 JEEM22Ch3a00 Solution: 120 g of an organic compound that contains only carbon and…

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

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