Question

An organic compound containing $C$ and $H$ gave the following analysis: $C=52\%,H=13\%$ and the remaining, oxygen. Its empirical formula would be:

• A. $C{H}_{3}O$
• B. $C_2{H}_{6}O$
• C. $C_2{H}_{4}O$
• D. $C_2{H}_{2}O$

Answer 1

The correct option is B $C_2{H}_{6}O$

Calculation of empirical formula:
% of oxygen will be $[100-(52+13\left)\right]\%=35\%$

$\begin{array}{ccccc}\text{Element}& \text{\% in compound}& \text{Atomic mass}& \text{No. of moles}& \text{Simple ratio}\\ C& 52& 12& \frac{52}{12}=4.33& \frac{4.33}{2.18}=2\\ H& 13& 1& \frac{13}{1}=13& \frac{13}{2.18}=6\\ O& 35& 16& \frac{35}{16}=2.18& \frac{2.18}{2.18}=1\end{array}$

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