0.262g of a substance gave, on combustion, 0.361g of CO₂ and 0.147g of H₂O. What is the empirical formula of the substance

% of carbon 
$\begin{array}{l}=\frac{12}{44}\times \frac{\mathrm{Wt}.\mathrm{of} {\mathrm{CO}}_2}{ \mathrm{Wt}.\mathrm{of} \mathrm{organic} \mathrm{compound} }\times 100\\ =\frac{12}{44}\times \frac{0.361}{0.262}\times 100=37.57\end{array}$
% of hydrogen
$\begin{array}{l}=\frac{2}{18}\times \frac{ \mathrm{wt}.\mathrm{of} {\mathrm{H}}_2\mathrm{O}}{ \mathrm{wt}.\mathrm{of} \mathrm{organic} \mathrm{compound} }\times 100\\ =\frac{2}{18}\times \frac{0.147}{0.262}\times 100=6.23\end{array}$
% of oxygen = 100 – (37.57+6.23) = 56.2
$\begin{array}{l}\mathrm{C}=\frac{37.57}{12}=3.13;\mathrm{H}=\frac{6.23}{1}=6.23\\ \mathrm{O}=\frac{56.2}{16}=3.51\end{array}$
Empirical formula = CH₂O