$\left(\mathrm{Organic} \mathrm{Compound} \mathrm{containg} \mathrm{C},\mathrm{H},\mathrm{O}\right)\stackrel{\mathrm{dissociati}on}{⟶}\underset{7.5\mathrm{gram}}{\mathrm{H}}+\underset{60\mathrm{gram}}{\mathrm{O}}+\underset{48.5\mathrm{gram}}{\mathrm{C}}$$\%$ of $\mathrm{H}$ in the given organic compound $=\frac{7.5}{116}\times 100=6.465\%$

(A) $\%$ of $\mathrm{H}=\frac{4}{60}\times 100=6.6\%$

(B) $\%$ of $\mathrm{H}=\frac{2}{30}\times 100=6.6$%

(c) $\%$ of $\mathrm{H}=\frac{6}{62}\times 100=7.677$%

(D) $\%$ of $\mathrm{H}=\frac{4}{44}\times 100=9.09$%

The percentage of Hydrogen in the given organic compound is equal to $6.465\%$ which is approximately equal to the percentage of hydrogen in ${\mathrm{CH}}_3\mathrm{COOH} \mathrm{and} \mathrm{HCHO}$.