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]

Moles of C = \( \frac{78}{12} = 6.5 \), moles of H = \( \frac{22}{1} = 22 \). Molar ratio C:H = \( 6.5 : 22 \approx 1 : 3.38 \)... Wait, \( 78\% \) C and \( 22\% \) H gives C:H = \( \frac{78}{12} : \frac{22}{1} = 6.5 : 22 \approx 1 : 3.38 \). Let's check \( \mathrm{CH}_{3} \): C:H ratio is 1:3, which fits empirical formula \( \mathrm{CH}_{3} \).