JEE Main · 2024 · Shift-ImediumMOLE-157

An organic compound has 42.1\% carbon, 6.4\% hydrogen and remainder is oxygen. If its molecular weight is 342, then its…

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

Question

An organic compound has 42.1%42.1\% carbon, 6.4%6.4\% hydrogen and remainder is oxygen. If its molecular weight is 342342, then its molecular formula is:

Options
  1. a

    \ceC11H18O12\ce{C11H18O12}

  2. b

    \ceC12H20O12\ce{C12H20O12}

  3. c

    \ceC12H22O11\ce{C12H22O11}

  4. d

    \ceC14H20O10\ce{C14H20O10}

Correct Answerc

\ceC12H22O11\ce{C12H22O11}

Detailed Solution

🧠 Use the molecular weight to get atom counts directly Total mass per molecule =342= 342. Take 42.1%42.1\% as carbon, 6.4%6.4\% as hydrogen, the rest as oxygen, and divide by atomic masses.

🗺️ Three counts C atoms: 0.421×342/12144/12=120.421 \times 342 / 12 \approx 144 / 12 = 12. H atoms: 0.064×342/122/1=220.064 \times 342 / 1 \approx 22 / 1 = 22. O atoms: (34214422)/16=176/16=11(342 - 144 - 22) / 16 = 176 / 16 = 11.

So the formula is \ceC12H22O11\ce{C12H22O11} — which is sucrose.

Speed scan M=342M = 342 is the textbook fingerprint of disaccharides (sucrose, lactose, maltose). All three share \ceC12H22O11\ce{C12H22O11}. Recognise it on sight.

Answer: (c)\boxed{\text{Answer: (c)}}

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