JEE Main 2025 Apr 2 Shift 1, Chemistry Q17: Stoichiometry

JEE Main2025Apr 2, Shift 1Chemistry

On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of $C O_{2}$ and 0.567 g of $H_{2} O$ . The empirical formula mass of compound (X) is _________ g.

(Given molar mass in g $m o l^{- 1}$ C : 12, H : 1, O : 16)

A
30
B
45
C
60
D
15

Solution

Mass of carbon: all C in 1 g of X ends up in $C O_{2}$ .

$n_{C} = \frac{1.46}{44} = 0.0332 mol \Rightarrow m_{C} = 0.0332 \times 12 = 0.398 g$

Mass of hydrogen: each $H_{2} O$ holds 2 H.

$n_{H} = 2 \times \frac{0.567}{18} = 0.063 mol \Rightarrow m_{H} = 0.063 \times 1 = 0.063 g$

Mass of oxygen by difference:

$m_{O} = 1.000 - 0.398 - 0.063 = 0.539 \approx 0.541 g$

$n_{O} = \frac{0.541}{16} = 0.0338 mol$

Mole ratio $C : H : O \approx 0.033 : 0.063 : 0.033 \approx 1 : 2 : 1$ .

Empirical formula: $C H_{2} O$ , with empirical formula mass $= 12 + 2 + 16 = 30$ g.

$∴$ Option (1) is correct.

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JEE Main 2025 Apr 2 Shift 1, Chemistry Q17: Stoichiometry

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