QJEEM25Ch1433: JEE Main 2025 Chemistry - Class 11 JEEM25Ch1433 Solution: Quantitative analysis of an organic compound (X) shows…

To determine the empirical formula mass of compound (X), we calculate the complete elemental composition, including Oxygen, which is often omitted in organic analysis reports if it makes up the remainder of the mass.

In organic compounds, if the sum of given percentages is less than 100%, the difference is attributed to Oxygen:

$$\text{Total given \% (C + Cl + H)}=14.5\mathrm{\%}+64.46\mathrm{\%}+1.8\mathrm{\%}=80.76\mathrm{\%}$$

$$\mathrm{\%}\text{Oxygen (O)}=100\mathrm{\%}-80.76\mathrm{\%}=19.24\mathrm{\%}$$

Divide each percentage by the respective atomic mass to find the molar ratio:

• C: $14.5 / 12 \approx 1.2$
• Cl: $64.46 / 35.5 \approx 1.8$
• H: $1.8 / 1 = 1.8$
• O: $19.24 / 16 \approx 1.2$

Dividing by the smallest value ( $1.2$ ) gives the ratio 1 : 1.5 : 1.5 : 1. Multiplying by 2 to reach whole numbers results in C₂H₃Cl₃O₂.

$$\begin{array}{rl}\text{Mass}& =(2\times 12)+(3\times 1)+(3\times 35.5)+(2\times 16)\\ & =24+3+106.5+32=165.5\\ \text{Mass}& =1655\times {10}^{-1}\end{array}$$