1. CH 2. CH₂ 3. CH₃ 4. CH₄

Answer: 3. CH₃ Key idea: divide percentage by atomic mass, then take the simplest ratio. C: 78/12=6.5 ; H: 22/1=22 . Ratio H : C = 22/6.5≈3.4 . The nearest simple whole-number formula among the options is CH₃ (the data are rounded, so 3.4 is taken as about 3). Empirical formula: CH₃ .

Topic: Empirical and molecular formula calculations· Purification and Characterisation of Organic Compounds

An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]

1. CH

2. \(CH_2\)

3. \(CH_3\)

4. \(CH_4\)

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Answer: 3. \(CH_3\)

Key idea: divide percentage by atomic mass, then take the simplest ratio.

C: \(78/12=6.5\); H: \(22/1=22\).

Ratio H : C = \(22/6.5\approx3.4\). The nearest simple whole-number formula among the options is \(\mathrm{CH_3}\) (the data are rounded, so 3.4 is taken as about 3).

Empirical formula: \(\mathrm{CH_3}\).

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An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]

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