JEE Main 2024 — 5 April, Shift 1 — Question 73

Given: $\%C = 42.1\%$, $\%H = 6.4\%$, $\%O = 100\% - 42.1\% - 6.4\% = 51.5\%$ Molecular weight $= 342\, \mathrm{g/mol}$

Step 1: Calculate moles of each element in 100 g compound $n_C = \frac{42.1}{12} = 3.5083$ mol, $n_H = \frac{6.4}{1} = 6.4$ mol, $n_O = \frac{51.5}{16} = 3.21875$ mol

Step 2: Find mole ratio relative to smallest number $C: \frac{3.5083}{3.21875} = 1.089$, $H: \frac{6.4}{3.21875} = 1.988$, $O: \frac{3.21875}{3.21875} = 1$

Step 3: Approximate to whole numbers by multiplying by 11 $C: 1.089 \times 11 \approx 12$, $H: 1.988 \times 11 \approx 22$, $O: 1 \times 11 = 11$

Molecular formula approximately: $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$

Step 4: Check molar mass $12 \times 12 = 144$, $22 \times 1 = 22$, $11 \times 16 = 176$ $144 + 22 + 176 = 342\, \mathrm{g/mol}$

Hence, molecular formula is $\mathrm{C_{12}H_{22}O_{11}} \text{ (Option\;D).}$