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 find the empirical formula of the compound, we need to determine the simplest whole number ratio of carbon to hydrogen in the compound.

1. Determine the Mass of Carbon and Hydrogen:
   The compound contains 78% carbon by weight. Therefore, the percentage of hydrogen in the compound is: 
   \(100\% - 78\% = 22\%\)
2. Calculate Moles of Each Element:
   For carbon, with an atomic weight of 12:  
   \(\text{Moles of C} = \frac{78}{12} = 6.5\) 
   For hydrogen, with an atomic weight of 1: 
   \(\text{Moles of H} = \frac{22}{1} = 22\)
3. Determine the Simplest Whole Number Ratio:
   To find the simplest ratio, divide each by the smallest number of moles: 
   \(\frac{6.5}{6.5} = 1 \quad \text{(for C)}\)
   \(\frac{22}{6.5} \approx 3.38\)
   Round \(3.38\) to the nearest whole number, which is 3.
4. Formulate the Empirical Formula:
   The empirical formula is determined as \(\text{CH}_3\).

Conclusion: The most simplified ratio of carbon to hydrogen in the compound is 1:3. Therefore, the empirical formula of the compound is \(\text{CH}_3\).