In a Liebig's method of estimation of carbon and hydrogen , 0.56 g of an organic compound on ignitio

Question

In a Liebig's method of estimation of carbon and hydrogen , 0.56 g of an organic compound on ignition gave 1.76 g of $\mathrm{CO}_{2}$ and 0.72 g of water. The organic compound is\

Professor PC Thomas Classes · Answer

Percentage of carbon $$ =\frac{12 	imes 1.76 	imes 100}{44 	imes 0.56}=85.7 \% $$ Percentage of hydrogen $$ =\frac{2 	imes 0.72 	imes 100}{18 	imes 0.56}=14.35 \% $$ $\mathrm{C}: \mathrm{H}=\frac{85.7}{12}: \frac{14.3}{1}$\ $=7.14: 14.3=1: 2$ (mole ratio)\ Empirical formula $=\mathrm{C}_{1} \mathrm{H}_{2}$\ Molecular formula $=\left(\mathrm{CH}_{2}ight)_{\mathrm{n}}=\mathrm{CnH}_{2} \mathrm{n}$\