Explanation

(a)
Atom At mass $\mathbf{(a)}$ $\mathbf{\%\ (b)}$ $\frac{\mathbf{b}}{\mathbf{a}}$ Ratio
C 12 49.3 $\frac{49.3}{12} = 4.10$ 2
H 1 6.84 $\frac{6.84}{1} = 6.84$ 3
O 16 43.86 $\frac{43.86}{16} = 2.74$ 1
Hence, empirical formula $= (C_{2}H_{3}O)$
Molecular mass $= 2 \times VD = 2 \times 73$
$= 146$
$n = \frac{molar\ mass}{empirical\ formula\ mass} = \frac{146}{43} \approx 3$
So, formula$= \left( C_{2}H_{3}O \right)_{3} \approx C_{6}H_{9}O_{3}$