Question

An organic compound contains $86.47$% carbon and $6.33$% hydrogen. The empirical formula of the compound is:

• A. $C_{16}{H}_{14}O$
• B. $C_{10}{H}_{14}O$
• C. $C_6{H}_{12}O$
• D. $C_{14}{H}_{16}O$

Answer 1

The correct option is A $C_{16}{H}_{14}O$

Percentage of $O$ in the compound = $7.2$%
Therefore, ratio of $C$, $H$ and $O$ are
$\frac{C}{12}:\frac{H}{1}:\frac{O}{16}$ $=\frac{86.47}{12}:\frac{6.33}{1}:\frac{7.2}{16}=$$16:14:1$.
Hence, the empirical formula is $C_{16}{H}_{14}O$.