Question

Determine the molecular formula of an organic compound containing 49.3% carbon 6.84% hydrogen and its vapour density is $73.$

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

Answer 1

Answer: (B)

Element	Atomic mass	Simple ratio
C	12	$\frac{49.3}{12}=4.1$	$\frac{4.1}{2.7}=1.5;1.5\times 2=3$
H	1	$\frac{6.84}{1}=6.84$	$\frac{6.84}{2.7}=2.5\times 2.5$
O	16	$\frac{43.86}{16}=2.7$	$\frac{2.7}{2.7}=1\times 2=2$

Empirical formula = $C_3{H}_5{O}_2$

Empirical Formula wt. $=12\times 3+1\times 5+16\times 2=73$

Molecular wt $=V.D.\times 2=73\times 2=146$

$n=\frac{M.wt}{E.F.wt}=\frac{146}{73}=2$

Molecular formula $=(E.F{)}_n={\left(C_3{H}_5{O}_2\right)}_2=C_6{H}_{10}{O}_4$.