On complete combustion 0 . 30g of an organic compound gave 0 . 20g...

${C}_{x}{H}_{y}{O}_{z} + \left(x + \dfrac{y}{4} - \dfrac{z}{2}\right){O}_{2} \to {xCO}_{2} + \dfrac{y}{2}{H}_{2}O$\begin{cases}0 . 3g & 0 . 2g & 0 . 1g\end{cases}$$</p><p>$\dfrac{{CO}_{2} n}{{H}_{2}O n} = \dfrac{x}{\dfrac{y}{2}} = \dfrac{\dfrac{0 . 2}{44}}{\dfrac{0 . 1}{18}} = \dfrac{9}{11}$</p><p>$x = \dfrac{9y}{22}$</p><p>Now,$\dfrac{{C}_{x}{H}_{y}{O}_{2} n}{{CO}_{2} n} = \dfrac{1}{x}$</p><p>$⇒ \dfrac{0 . 3}{12x + y + 16z} \times \dfrac{44}{0 . 2} = \dfrac{1}{x}$</p><p>$66x = 12x + y + 16z54x = y + 16z$</p><p>$\dfrac{54 \times 9y}{22} – y = 16z$</p><p>$z = \dfrac{29y}{22}$</p><p>$CxHyOz = C\dfrac{9y}{22}{H}_{y}O\dfrac{29y}{22} = {C}_{9}{H}_{22}{O}_{29}$</p><p>$% of C = \dfrac{12 \times 9}{\left(12 \times 9 + 22 + 29 \times 16\right)} \times 100 = 18 . 18 %$