JEE Main · 2022 · Shift-IImediumMOLE-178

Compound A: 8.7\% H, 74\% C, 17.3\% N; molar mass =162\,g/mol. Molecular formula is: (C=12, H=1, N=14)

Some Basic Concepts (Mole Concept) · Class 11 · JEE Main Previous Year Question

Question

Compound A: 8.7%8.7\% H, 74%74\% C, 17.3%17.3\% N; molar mass =162 g/mol=162\,\text{g/mol}. Molecular formula is: (C=12, H=1, N=14)

Options
  1. a

    \ceC4H6N2\ce{C4H6N2}

  2. b

    \ceC2H3N\ce{C2H3N}

  3. c

    \ceC5H7N\ce{C5H7N}

  4. d

    \ceC10H14N2\ce{C10H14N2}

    ✓
Correct Answerd

\ceC10H14N2\ce{C10H14N2}

Detailed Solution

🧠 Empirical formula first, then scale to molar mass Convert the percentages to moles, find the simplest ratio, get the empirical formula. Then check how many empirical units fit inside the given molar mass of 162 g/mol162\,\text{g/mol}.

🗺️ Step 1 — moles per 100 g100\,\text{g}

  • C: 74/12≈6.1774 / 12 \approx 6.17
  • H: 8.7/1=8.78.7 / 1 = 8.7
  • N: 17.3/14≈1.2417.3 / 14 \approx 1.24

Divide by the smallest (1.241.24):

  • C: 55, H: 77, N: 11.

Empirical formula =\ceC5H7N= \ce{C5H7N}, mass =60+7+14=81 g/mol= 60 + 7 + 14 = 81\,\text{g/mol}.

🗺️ Step 2 — scale to the real molecule n=16281=2.n = \frac{162}{81} = 2.

So the molecular formula is (\ceC5H7N)2=\ceC10H14N2(\ce{C5H7N})_2 = \ce{C10H14N2}.

⚡ Quick scan 162162 is exactly twice 8181. The only option with double the empirical atoms is (d).

Answer: (d)\boxed{\text{Answer: (d)}}

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