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 ;…

$$\begin{array}{|c|c|c|c|c|}\hline \text{Element} & \begin{array}{c} \text{Mass} \\ \text{percentage} \%\end{array} & \begin{array}{c} \text{No. of} \\ \text{moles} \end{array} & \begin{array}{c} \text{No. of moles} / \\ \text{Smallest number} \end{array} & \begin{array}{c} \text{Simplest whole} \\ \text{number} \end{array} \\\hline A & 32 \% & \frac{32}{64}=\frac{1}{2} & \frac{1}{2} \times 2 & =1 \\\hline B & 20 \% & \frac{20}{40}=\frac{1}{2} & \frac{1}{2} \times 2 & =1 \\\hline C & 48 \% & \frac{48}{32}=\frac{3}{2} & \frac{3}{2} \times 2 & =3 \\\hline\end{array}$$ElementABC​Masspercentage%​32%20%48%​No. ofmoles​6432​=21​4020​=21​3248​=23​​No. of moles/Smallest number​21​×221​×223​×2​Simplest wholenumber​=1=1=3​​ So, empirical formula of $X=\begin{array}{clll}A & : & B & \text { C } \\ 1 & : & 1 & :\end{array}$X=A1​::​B1​ C :​ $\therefore$∴ The correct empirical formula of compound X is $\mathrm{ABC}_3$ABC3​