Question

A certain alkaloid has $70.8\%C,6.2\%H$ and $4.1\%$ of N and rest O. The empirical formula is

• A. $C_{20}{H}_{21}N{O}_4$
• B. $C_{20}{H}_{20}N{O}_4$
• C. $C_{21}{H}_{20}N{O}_3$
• D. $C_{20}{H}_{10}N{O}_3$

Answer 1

The correct option is A $C_{20}{H}_{21}N{O}_4$

$\%$ of O = $100-(\%C+\%H+\%N)=100-(70.8+6.2+4.1)=18.9$
$\begin{array}{ccccc}\text{Element}& \%& \%\text{comp/at.mass}& \text{Simplest ratio}& \text{Empirical formula}\\ \text{C}& 70.8& \frac{70.8}{12}=5.9& \frac{5.9}{0.2928}=20.15& 20\\ \text{H}& 6.2& \frac{6.2}{1}=6.2& \frac{6.2}{00.2928}=21.17& 21\\ \text{N}& 4.1& \frac{4.1}{14}=0.2928& \frac{0.2928}{0.2928}=1& 1\\ \text{O}& 18.9& \frac{18.9}{16}=1.181& \frac{1.181}{0.2928}=4.033& 4\end{array}$
Therefore empirical formula is : $C_{20}{H}_{21}N{O}_4$