JEE Main 2025 Apr 7 Shift 1, Chemistry Q21: Qualitative and Quantitative Analysis

All carbon in the organic sample ends up as CO${}_2$ on complete combustion. Convert CO${}_2$ mass to mass of carbon:

Moles of CO${}_2$ produced $=\frac{220\times 10^{-3}\text{g}}{44\text{g/mol}}=5\times 10^{-3}$ mol

Each mole of CO${}_2$ contains one mole of C, so

Mass of C $=5\times 10^{-3}\times 12=0.060$ g $=60$ mg

Percentage of C in the original sample:

$\%\text{C}=\frac{60}{500}\times 100=12\%$

$\therefore$ Answer = 12.