Question

The elemental analysis of an organic compound containing $C,H\text{and}O$ gives the percentage compositions of $C$ and $H$ to be $39.9\%$ and $6.7\%$ respectively. If the molecular weight of the compound is $180\text{u}$, then the number of carbon atoms present in the molecule is:

• A. $6$
• B. $8$
• C. $4$
• D. $3$

Answer 1

The correct option is A $6$

$$\begin{array}{cccccc}\text{Element}& \text{symbol}& \text{percentage}& \text{At. mass}& \text{Mole of element}& \text{Simplest whole number molar ratio}\\ \text{Carbon}& C& 39.9& 12& 3.325& 1\\ \text{Hydrogen}& H& 6.7& 1& 6.7& 2\\ \text{Oxygen}& O& 53.4& 16& 3.337& 1\end{array}$$
The empirical formula of the compound is $C{H}_{2}O$.
Empirical formula mass $=12+2\times 1+16=30$
$n=\frac{\text{molecular mass}}{\text{Empirical formula mass}}=\frac{180}{30}\Rightarrow n=6$
$\text{Molecular formula}=n\times \text{(empirical formula)}=6\times C{H}_{2}O=C_6{H}_{12}{O}_6$
Thus, the number of carbon atoms present is $6$.