Question

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

• A. CH₄
• B. CH
• C. CH₂
• D. CH₃

Answer 1

The Correct Option is D

To determine the empirical formula of the organic compound, we need to calculate the ratio of moles of carbon (C) and hydrogen (H) in the compound.

1. Given that the compound contains 78% carbon by weight, the remaining percentage must be hydrogen:
   • Percentage of hydrogen = 100% - 78% = 22%
2. Calculate the moles of each element per 100 grams of the compound:
   • Moles of carbon, \(n_C = \frac{78}{12} = 6.5\)
   • Moles of hydrogen, \(n_H = \frac{22}{1} = 22\)
3. Find the simplest whole number ratio by dividing each by the smallest number of moles calculated:
   • Ratio of moles of carbon to hydrogen:
     • Carbon: \(\frac{6.5}{6.5} = 1\)
     • Hydrogen: \(\frac{22}{6.5} \approx 3.38\)
4. Since 3.38 is approximately equal to 3, adjust it to the nearest whole number:
   • The empirical formula is thus CH₃.

Hence, the correct option for the empirical formula of the compound is CH₃.