A compound $X$ contains $32 \%$ of $A, 20 \%$ of $B$ and remaining percentage of $C$. Then, the empirical…
A compound $X$ contains $32 \%$ of $A, 20 \%$ of $B$ and remaining percentage of $C$. Then, the empirical formula of X is :
(Given atomic masses of $A=64 ; B=40 ; C=32 u$)
$\mathrm{ABC}_3$
$\mathrm{AB}_2 \mathrm{C}_2$
$\mathrm{ABC}_4$
$\mathrm{A}_2 \mathrm{BC}_2$
Solution
$\begin{array}{|c|c|c|c|c|}\hline \text{Element} & \begin{array}{c} \text{Mass} \\ \text{percentage} \%\end{array} & \begin{array}{c} \text{No. of} \\ \text{moles} \end{array} & \begin{array}{c} \text{No. of moles} / \\ \text{Smallest number} \end{array} & \begin{array}{c} \text{Simplest whole} \\ \text{number} \end{array} \\\hline A & 32 \% & \frac{32}{64}=\frac{1}{2} & \frac{1}{2} \times 2 & =1 \\\hline B & 20 \% & \frac{20}{40}=\frac{1}{2} & \frac{1}{2} \times 2 & =1 \\\hline C & 48 \% & \frac{48}{32}=\frac{3}{2} & \frac{3}{2} \times 2 & =3 \\\hline\end{array}$
So, empirical formula of
$X=\begin{array}{clll}A & : & B & \text { C } \\ 1 & : & 1 & : 3\end{array}$
$\therefore$ The correct empirical formula of compound X is $\mathrm{ABC}_3$