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 organic compound, we start by understanding that empirical formulas represent the simplest whole-number ratio of atoms in a compound. Given: 

• The compound contains \(78\%\) carbon by weight.
• The remaining percentage is hydrogen, which is \(22\%\) (since \(100\% - 78\% = 22\%\)).
• The atomic weights are: \(C = 12\) and \(H = 1\).

Next, calculate the moles of each element in 100 grams of the compound:

For Carbon: \(\text{Mass of C} = 78 \, \text{g} \, \Rightarrow \, \text{Moles of C} = \frac{78}{12} = 6.5\)

For Hydrogen: \(\text{Mass of H} = 22 \, \text{g} \, \Rightarrow \, \text{Moles of H} = \frac{22}{1} = 22\)

Next, determine the simplest whole-number ratio of carbon to hydrogen by dividing both by the smallest number of moles:

Moles of C: \(\frac{6.5}{6.5} = 1\)

Moles of H: \(\frac{22}{6.5} \approx 3.38\), which is approximately 3 when rounded to the nearest whole number.

The simplest whole-number ratio of carbon to hydrogen is \(1:3\). Therefore, the empirical formula of the compound is \(CH_3\).

Therefore, the correct option is: \(CH_3\).