# An organic compound contains 49.30% carbon,6.84% hydrogen and its vapour density is 73.Molecular formula of compound is

$\begin{array}{1 1}(a)\;C_6H_{10}O_4&(b)\;C_3H_{10}O_2\\(c)\;C_6H_9O&(d)\;C_4H_{10}O_2\end{array}$

Empirical formula =$C_3H_5O_2$
$n=\large\frac{Mol.mass}{Emp.mass}$
$\;\;=\large\frac{2\times 73}{73}$
$\;\;=2$
Molecular formula $=2\times C_3H_5O_2$
$\qquad\qquad\qquad\;\;=C_6H_{10}O_4$
Hence (a) is the correct answer.