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NEET CHEMISTRYEasy

An organic compound contains 80 % (by wt.) carbon and the remaining percentage of hydrogen. The empirical formula of this compound is: [Atomic wt. of C is 12, H is 1]

A

CH₃

B

CH₄

C

CH

D

CH₂

Step-by-Step Solution

Determine the percentage of Hydrogen: $\% \text{ Hydrogen} = 100\% - \% \text{ Carbon} = 100 - 80 = 20\%$
Calculate Moles of each element (assuming 100 g sample):

Moles of Carbon ( $C$ ) = $\frac{\text{Mass}}{\text{Atomic Mass}} = \frac{80}{12} \approx 6.67 \text{ mol}$
Moles of Hydrogen ( $H$ ) = $\frac{20}{1} = 20 \text{ mol}$

Determine the simplest molar ratio: Divide both values by the smallest number of moles ( $6.67$ ):

Ratio for $C = \frac{6.67}{6.67} = 1$
Ratio for $H = \frac{20}{6.67} \approx 3$

Conclusion: The simplest whole number ratio of $C:H$ is $1:3$ . Therefore, the empirical formula is $CH_3$ .

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