Question

An organic compound containing carbon, hydrogen and oxygen contains $52.2\%$ carbon and $13.04\%$ hydrogen. Vapour density of the compound is $23.$ Its molecular formula will be

• A. $C_2{H}_{6}O$
• B. $C_3{H}_{6}O$
• C. $C_4{H}_{8}O$
• D. $C_5{H}_{10}O$

Answer 1

Answer: (A)

Element	C	H	O
Percentage	52.2	13.04	$100-(52.2+13.04)$ $=34.76$
Atomic ratio	$\frac{52.2}{12}=4.35$	$\frac{13.4}{1}=13.04$	$\frac{34.76}{16}=2.17$
Simplest ratio	$\frac{4.35}{2.17}=2$	$\frac{13.04}{2.17}=6$	$\frac{2.17}{2.17}=1$
Simple whole no. ratio	2	6	1

Hence, empirical formula is $C_2{H}_{6}O$.

Empirical formula mass $=2\times 12+6\times 1+16=46$

Molecular mass $=2\times 23=46$

$n=\frac{46}{46}=1$

Molecular formula $=(\text{ Empirical formula }{)}_n={\left(C_2{H}_{6}O\right)}_1=C_2{H}_{6}O$