Question

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 $1$]

• A. $CH$
• B. $CH _{2}$
• C. $CH _{3}$
• D. $CH _{4}$

Answer 1

The Correct Option is C

To determine the empirical formula of the given organic compound, we need to find the simplest whole-number ratio of moles of carbon (C) and hydrogen (H) in the compound.

Given:
- The compound contains \(78 \%\) carbon by weight.
- The remaining percentage, therefore, is \((100\% - 78\%) = 22\%\) hydrogen.

Let's calculate the moles of each element in 100 grams of the compound:

1. Weight of Carbon = \(78 \, \text{g}\)
   Moles of Carbon = \(\frac{78 \, \text{g}}{12 \, \text{g/mol}} = 6.5 \, \text{mol}\)
2. Weight of Hydrogen = \(22 \, \text{g}\)
   Moles of Hydrogen = \(\frac{22 \, \text{g}}{1 \, \text{g/mol}} = 22 \, \text{mol}\)

The ratio of moles of carbon to moles of hydrogen is:

\(\frac{6.5}{6.5} = 1\) (for Carbon)
\(\frac{22}{6.5} \approx 3.38\) (for Hydrogen)

By approximating \(3.38\) to the nearest whole number, we get 3. Since approximations in empirical formula directly to the nearest whole numbers, we adjust it to 3. Therefore, the empirical formula is:

\(CH_3\)

Thus, the empirical formula of the compound is \(CH_3\), which is the correct option.

Conclusion: The correct answer is \(CH_3\) because this formula accurately reflects the simplest whole-number ratio of carbon and hydrogen present in the compound given the percentage composition.