0.262 g of a substance gave, on combustion, 0.361g of ${\mathrm{CO}}_2$ and 0.147 g of ${\mathrm{H}}_2\mathrm{O}$. What is the empirical formula of the substance

$\mathrm{\%}\text{ of carbon }=\frac{12}{44}\times \frac{wt. of C{O}_2}{wt. of organic compound}\times 100=\frac{12}{44}\times \frac{0.361}{0.262}\times 100=37.57$

$\% of hydrogen$

$=\frac{2}{18}\times \frac{\text{ wt. of }{\mathrm{H}}_2\mathrm{O}}{\text{ wt .of organic compound }}\times 100=\frac{2}{18}\times \frac{0.147}{0.262}\times 100=6.23$

$\mathrm{\%}\text{ of oxygen }=100-(37.57+6.23)=56.2C=\frac{37.57}{12}=3.13;H=\frac{6.23}{1}=6.23$

$$\begin{gathered} O=\frac{56.2}{16}=3.51, \\ \text{ Empirical formula }={\mathrm{CH}}_2\mathrm{O} \end{gathered}$$