If molecular formula of a compound is ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$.  Then its Empirical formula is      

Among the given choices, correct choice is D

   $\text{EF=}\frac{\text{MF}}{\text{n}}$, $\text{EF=}\frac{{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}{\text{n}}$, $\text{EF=}\frac{{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}{2}$

Therefore, its Empirical formula is $\frac{{\text{C}}_3{\text{H}}_6{\text{O}}_3}{3}={\text{CH}}_{\text{2}}\text{O}$.