Q. The elemental composition of a compound is $54.2\%$C, $9.2\%$H and $36.6\%$O. If the molar mass of the compound is $132\ g\ mol^{-1}$, the molecular formula of the compound is : [Given : The relative atomic mass of C : H : O = 12 : 1 : 16]
A) $C_4H_9O_3$
B) $C_6H_{12}O_6$
C) $C_4H_8O_2$
D) $C_6H_{12}O_3$
Explanation
[Solution by Dr. Abubacker] Moles: C $=\frac{54.2}{12}=4.52$, H $=9.2$, O $=\frac{36.6}{16}=2.29$. Ratio $\approx2:4:1$, so the empirical formula is $C_2H_4O$ (mass $44$). $\frac{132}{44}=3$, so the molecular formula is $C_6H_{12}O_3$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.