Explanation

(c)
[table] Element | % | Relative no. of atom | Simplest ratio ; C
H
O | 49.3
6.84
43.86 | $\frac{49.3}{12} = 4.1$
$\frac{6.84}{1} = 6.84$
$\frac{43.86}{16} = 2.74$ | $\frac{4.1}{2.74} = 1.5 \times 2 = 3$
$\frac{6.84}{2.74} = 2.5 \times 2 = 5$
$\frac{2.74}{2.74} = 1 \times 2 = 2$ [/table]
The empirical formula is $C_{3}H_{5}O_{2}$
Empirical formula weight
$= (3 \times 12) + (5 \times 1) + (2 \times 16)$
$= 36 + 5 + 32$
$= 73$
Molecular wt. of the compound
$= 2 \times VD$
$= 2 \times 73$
=$146$
$n = \frac{mol.\ \ wt.}{empirical\ formula\ wt.}$
$= \frac{146}{73}$
$= 2$
Molecular formula=empirical formula$\times 2$
$= (C_{3}H_{5}O_{2}) \times 2$
$= C_{6}H_{10}O_{4}$