An organic compound having $C,H$ and $O$ has $13.13\%H,52.14\%C$. Its molar mass is $46.068$ g. What are its empirical and molecular formulae?

Percentage composition $=\frac{\mathrm{mass} \mathrm{of} \mathrm{element}}{\mathrm{total} \mathrm{mass} \mathrm{of} \mathrm{compound}}\times 100$

mass of carbon $=\frac{52.14\times 46.068}{100}\approx 24 g$

$=2$ atoms of $C$.

max of Hydrogen $=\frac{13.13\times 46.068}{100}=6 g$

$=6$ atoms of ' $H$ "'

mass of oxygen $=\frac{[100-(13.13+52.14\left)\right]\times 46.068}{100}$

$\approx 16=1$ atom of ' $O$ '.

So, molecular formula $=C_2{H}_{6}O$

Since it cannot be further simplified, the empirical Formula is also the same.

Hence, option (C) is correct.