Molecular formula	Empirical formula
A) ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$	1) ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$
B) ${\text{CH}}_3\text{COOH}$	2) ${\text{CH}}_2$
C) ${\text{C}}_4{\text{H}}_{\text{10}}$	3) ${\text{C}}_1{\text{H}}_2{\text{O}}_1$
D) ${\text{C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$	4) ${\text{C}}_2{\text{H}}_5$

Empirical Formula (E.F): The simplest ratio of the number of atoms of different elements

present in a given compound is known as empirical formula.

Consider ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ molecule, the empirical formula is ${\text{C}}_{\frac{\text{2}}{2}=1}{\text{H}}_{\frac{\text{4}}{2}=2}={\text{CH}}_2$.

For the molecule ${\text{CH}}_3\text{COOH}$, the empirical formula is ${\text{C}}_{\frac{2}{2}=1}{\text{H}}_{\frac{4}{2}=1}{\text{O}}_{\frac{2}{2}=1}{\text{=\hspace{0.17em}CH}}_2\text{O}$.

For the molecule ${\text{C}}_4{\text{H}}_{\text{10}}$, the empirical formula is  ${\text{C}}_{\frac{4}{2}=2}{\text{H}}_{\frac{10}{2}=5}={\text{C}}_{\text{2}}{\text{H}}_5$.

For the molecule, ${\text{C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ the empirical formula is ${\text{C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$.

Note: There is no common factor for the numbers given in above molecule.