Q24 / 75 · JEE Main 2025 (28 Jan Shift 1) · Chemistry
Quantitative analysis of an organic compound (X) shows following % composition.
C : $14.5 \%$
Cl : 64.46%
H: 1.8 %
(Empirical formula mass of the compound $(\mathrm{X})$ is __________ $\times 10^{-1}$
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ of $\mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16, \mathrm{Cl}: 35.5$)
Correct answer: 1655
Solution
$\begin{array}{ll:cccl} & \mathrm{C} & \mathrm{Cl} & \mathrm{H} & \mathrm{O} & \\ \text { %mass } & 14.5 & 64.46 & 1.8 & 19.24\end{array}$
Molar ratio $\frac{14.5}{12} \quad \frac{64.46}{35.5} \quad \frac{1.8}{1} \quad \frac{19.24}{16}$
$\begin{aligned}
&\begin{array}{ccccc}
& 1.2 & 1.8 & 1.8 & 1.2 \\
\text { Minimum } & 2 & 3 & 3 & 2
\end{array}\\
&\text { integral ratio }
\end{aligned}$
$\begin{aligned} & \text { Empiricial formula }=\mathrm{C}_2 \mathrm{H}_3 \mathrm{Cl}_3 \mathrm{O}_2 \\ & \text { Mass }=165.5 \\ & \text { Mass }=1655 \times 10^{-1}\end{aligned}$
Hence, the correct answer is 1655.