JEE Main 2025 Apr 8 Shift 2, Chemistry Q13: Stoichiometry

JEE Main2025Apr 8, Shift 2Chemistry

On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g H $_{2}$ O and 0.307 g CO $_{2}$ . The percentages of hydrogen and oxygen in the given organic compound respectively are:

A
53.41, 39.6
B
6.72, 53.41
C
7.55, 43.85
D
6.72, 39.87

Solution

In combustion analysis, every carbon atom of the sample ends up in CO $_{2}$ and every hydrogen atom ends up in H $_{2}$ O.

Mass of hydrogen $= \frac{2}{18} \times 0.127 = 0.0141$ g, so $% H = \frac{0.0141}{0.210} \times 100 \approx 6.72%$ .

Mass of carbon $= \frac{12}{44} \times 0.307 = 0.0837$ g.

Mass of oxygen $=$ total sample mass $-$ (mass of C $+$ mass of H) $= 0.210 - (0.0837 + 0.0141) = 0.1122$ g.

$% O = \frac{0.1122}{0.210} \times 100 \approx 53.41%$ .

Thus $% H = 6.72$ and $% O = 53.41$ , matching option (2).

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JEE Main 2025 Apr 8 Shift 2, Chemistry Q13: Stoichiometry

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