Chemistry · Practical Organic Chemistry

JEE Main 2026 — 24 January, Evening Shift — Question 67

0.25 g of an organic compound "A" containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of $\mathrm{CaCl}_{2}$ tube and potash tube at the end of the experiment. The amount was found to be 0.15 g and 0.1837 g , respectively. The percentage of oxygen in compound A is $\_\_\_\_$ $\%$ . (Nearest integer) (Given : molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{O}: 16$ )

Answer: 73

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}} \mathrm{O}_{\mathrm{z}}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$ $\begin{array}{llll}\mathrm{t}=0 & 0.25 \mathrm{gm} & & \mathrm{t}_{\mathrm{eq}} & - & 0.18 \mathrm{gm} & 0.15 \mathrm{gm}\end{array}$ Mass of ' C ' $=\frac{0.18}{44} \times 12=0.049 \simeq 0.05 \mathrm{gm}$ Mass of ' H ' $=\frac{0.15}{18} \times 2=0.016 \simeq 0.017 \mathrm{gm}$ . Mass of ' O ' $=0.25-0.05-0.017=0.1833 \mathrm{gm}$ Mass % of ' O ' $=\frac{0.1833}{0.25} \times 100=73.32 \%$ .

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Chemistry
Chapter: Practical Organic Chemistry
Topic: Quantitative organic analysis