Question

An organic compound contains carbon, hydrogen and oxygen. Its chemical analysis gave $C,38.71\%$ and $H,9.67\%$. The empirical formula of compound would be

• A. $C{H}_{3}O$
• B. $C{H}_{2}O$
• C. $CH{O}_2$
• D. $CHO$

Answer 1

The correct option is A $C{H}_{3}O$

Percentage of Carbon=$38.71$
Percentage of Hydrogen=$9.67$
Total percentage $=100\%$
As the compound contains Carbon, hydrogen and oxygen only.
Then percentage of oxygen $=100-38.71-9.67=51.62$
$\therefore C:H:O=\frac{38.71}{12}:\frac{9.67}{1}:\frac{51.62}{16}$
$=3.225:9.67:3.226=1:3:1$
$\therefore$ Empirical formula $=C{H}_{3}O$