Question

An organic compound contains $49.3\%$ carbon, $6.84\%$ hydrogen and its vapour density is $73$. Molecular formula of the compound is

• A. $C_3{H}_5{O}_2$
• B. $C_4{H}_{10}{O}_2$
• C. $C_6{H}_{10}{O}_4$
• D. $C_3{H}_{10}{O}_2$

Answer 1

Answer: (C)

Element	$\%$	Relative number of atoms	Simplest ratio
C	49.3	$\frac{49.3}{12}=4.10$	$\frac{4.10}{2.74}=1.5\times 2=3$
H	6.84	$\frac{6.84}{1}=6.84$	$\frac{6.84}{2.74}=2.5\times 2=5$
O	43.86	$\frac{43.86}{16}=2.74$	$\frac{2.74}{2.74}=1\times 2=2$

$\therefore$ The empirical formula is $C_3{H}_5{O}_2$
Empirical formula weight
$=12\times 3+1\times 5+16\times 2$
$=73$
Molecular weight of the compound
$=2\times$ vapour density
$=2\times 73=146$
. $n=\frac{\text{ molecular weight }}{\text{ empirical formula weight }}$
$=\frac{146}{73}=2$
Molecular formula $=$ empirical formula $\times 2$
$=2\left(C_3{H}_5{O}_2\right)=C_6{H}_{10}{O}_4$