Question

Butyric acid contains only $C,H$ and $O$. A 4.24 mg sample of butyric acid is completely burned. It gives 8.45 mg of carbon dioxide $\left(C{O}_2\right)$ and 3.46 mg of water. What is the mass percentage of each element in butyric acid?The molecular mass of butyric acid was determined by experiment to be88. What is the molecular formula?

• A. Molecular formula = $C_4{H}_8{O}_2$
• B. Mass percentage of $C$ = 44.35 %, $H$ = 19.06 %, $O$ = 36.59 %
• C. Molecular formula = $C_3{H}_{8}O$
• D. Mass percentage of $C$ = 54.35 %, $H$ = 9.06 %, $O$ = 36.59 %

Answer 1

Answer: (A), (D)

The correct options are
A Molecular formula = $C_4{H}_8{O}_2$
D Mass percentage of $C$ = 54.35 %, $H$ = 9.06 %, $O$ = 36.59 %

(i) $\%C=\frac{12\times 8.45}{44\times 4.24}\times 100=54.35\%$ $\%H=\frac{2\times 3.46}{18\times 4.24}\times 100=9.06\%$ $\%O=[100-54.35-9.06]=36.59\%$

(ii)

$\%$	$\%/at.wt.=Factor$	Factor/Lowest number
$C=54.2$	$\frac{54.2}{12}=4.52$	$\frac{4.52}{2.29}=2$
$H=9.2$	$\frac{9.2}{1}=9.2$	$\frac{9.2}{2.29}=4$
$O=36.6$	$\frac{36.6}{16}=2.29$	$\frac{2.29}{2.29}=1$

Empirical formula of butyric acid is $C_2{H}_{4}O$

(iii) Empirical formula mass $=44$
Molecular mass $=88$
$\frac{\text{Molecular mass}}{\text{Empirical mass}}=\frac{88}{44}=2$
$\therefore$ Moleformula $=$ Empirical formula $\times n=C_4{H}_8{O}_2$