Question

An organic substance containing carbon, hydrogen and oxygen gave the percentage composition as $C=40.687$ %$,H=5.085$% and the remaining oxygen. The vapour density of the compound is 59. Calculate the molecular formula of the compound.

• A. $C_4{H}_{6}O$
• B. $C_4{H}_6{O}_4$
• C. $C_2{H}_6{O}_4$
• D. $C_4{H}_{2}6O_4$

Answer 1

Answer: (B)

$C_4{H}_6{O}_4$

Consider $\to 100g$ of compound

	Masses	Molar mass	Moles	Ratios	Ratio
$C$	$40.3687$	$12$	$3.3916$	$3.4/3.4=1$	$2$
$H$	$5.085$	$1$	$5.085$	$5.1/3.4=1.5$	$3$
$O$	$54.228$	$16$	$3.392$	$3.4/3.4=1$	$2$

Emperical formula $=C_2{H}_2{O}_2$

Emperical mass $=2\times 12+3\times 1+2\times 16$

$=24+3+32$

$=59$

Molecular mass $=2\times$vapor density $=2\times 59=118$

$n=\frac{M.M}{E.M}=\frac{118}{59}=2$

Molecular formula $=n\times E.F=2\times C_2{H}_3{O}_2$

$=C_4{H}_6{O}_4$