Chemistry - Mole Concept Question with Solution | TestHub

$$\begin{gathered} {\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}{\mathrm{O}}_{\mathrm{z}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}-\frac{\mathrm{z}}{2}\right){\mathrm{O}}_2\to {\mathrm{xCO}}_2+\frac{\mathrm{y}}{2}{\mathrm{H}}_2\mathrm{O} \\ \begin{array}{ccc}0.3 \mathrm{g}& 0.2 \mathrm{g}& 0.1 \mathrm{g}\end{array} \end{gathered}$$

$\frac{{}^{\mathrm{n}}{\mathrm{CO}}_2}{{}^{\mathrm{n}}{\mathrm{H}}_2\mathrm{O}}=\frac{\mathrm{x}}{{\displaystyle \frac{\mathrm{y}}{2}}}=\frac{{\displaystyle \frac{0.2}{44}}}{{\displaystyle \frac{0.1}{18}}}=\frac{9}{11}$

$\mathrm{x}=\frac{9\mathrm{y}}{22}$

Now, $\frac{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}{\mathrm{O}}_2}{{}^{\mathrm{n}}{\mathrm{CO}}_2}=\frac{1}{\mathrm{x}}$

$\Rightarrow \frac{0.3}{12\mathrm{x}+\mathrm{y}+16\mathrm{z}}\times \frac{44}{0.2}=\frac{1}{\mathrm{x}}$

$$\begin{gathered} 66\mathrm{x} = 12 \mathrm{x} + \mathrm{y} + 16 \mathrm{z} \\ 54\mathrm{x} = \mathrm{y} + 16 \mathrm{z} \end{gathered}$$

$\frac{54\times 9y}{22} – y = 16 z$

$z = \frac{29y}{22}$

$$\begin{gathered} \mathrm{CxHyOz} = \\ {\mathrm{C}}_{\frac{9\mathrm{y}}{22}}{\mathrm{H}}_{\mathrm{y}}{\mathrm{O}}_{\frac{29\mathrm{y}}{22}} \\ = {\mathrm{C}}_9{\mathrm{H}}_{22}{\mathrm{O}}_{29} \end{gathered}$$

$\% \mathrm{of} \mathrm{C} = \frac{12\times 9}{\left(12 \times 9 +22 +29\times 16\right)}\times 100 = 18.18\%$