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 (X) is × 10 -1 (Given molar mass in g mol -1 of C : 12, H : 1 , O : 16, Cl : 35.5 )

Chemistry · Practical Organic Chemistry

JEE Main 2025 — 28 January, Morning Shift — Question 46

Quantitative analysis of an organic compound (X) shows following % composition.

C : 14.5%

$\mathrm{Cl}: 64.46 \%$

H: 1.8%

(Empirical formula mass of the compound (X) is___ $\times 10^{-1}$

(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ of $\mathrm{C}: 12, \mathrm{H}: 1$ , $\mathrm{O}: 16, \mathrm{Cl}: 35.5$ )

Numerical answer — enter this value.

$\mathrm{C}: \mathrm{Cl}: \mathrm{H}$ : O $\begin{array}{lllll}\text { \%mass } & 14.5 & 64.46 & 1.8 & 19.24\end{array}$

Molar ratio $\frac{14.5}{12} \quad \frac{64.46}{35.5} \quad \frac{1.8}{1} \quad \frac{19.24}{16}$

$\quad\quad\quad\quad\quad 1.2\quad\quad1.8\quad\quad 1.8\quad\quad1.2$

Minimum $\quad\quad\quad 2 \quad\quad 3 \quad\quad 3 \quad\quad 2$

integral ratio

Empiricial formula $=\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}$

Mass $=165.5$

Mass $=1655 \times 10^{-1}$

Answer key and solution verified before publishing.

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