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]

$\begin{array}{l}\mathrm{C}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} :\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{H}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}:\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{O}\\ \frac{54.2}{12}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} :\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}9.2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} :\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{36.6}{16}\\ 4.516\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} :\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}9.2\hspace{0.25em} \hspace{0.25em}:\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2.287\\ \frac{4.516}{2.287}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{9.2}{2.287}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{2.287}{2.287}\\ 1.97\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4.02\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} :\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}$
C₂H₄O $\Rightarrow$Empirical formula 
E.F. mass = 24 + 4 + 16 = 44 and molar mass = 132 
Hence molecular formula = (C₂H₄O)₃ 
                                              = C₆H₁₂O₃ 
Correct Option (3)