Chemistry · Practical Organic Chemistry

JEE Main 2025 — 7 April, Morning Shift — Question 20

An organic compound weighing 500 mg , produced 220 mg of $\mathrm{CO}_{2}$ , on complete combustion. The percentage composition of carbon in the compound is $\_\_\_\_\_$ %. (nearest integer) (Given molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ of $\mathrm{C}: 12, \mathrm{O}: 16$ )

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

$\%$ of $C=\frac{\frac{12}{44} \times 220}{500} \times 100=12$

Answer key and solution verified before publishing.

Practise Practical Organic Chemistry

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Chemistry
Chapter: Practical Organic Chemistry
Topic: Quantitative organic analysis