# What is the empirical formula of a compound with the molecular formula N_2O_4?

The empirical formula is the SIMPLEST whole number ratio that defines constituent atoms in a species. Does ${N}_{2} {O}_{4}$ satisfy that definition?
Of course, the empirical formula of ${N}_{2} {O}_{4}$ is $N {O}_{2}$. Why?
The equilibrium ${N}_{2} {O}_{4} \left(g\right) r i g h t \le f t h a r p \infty n s 2 N {O}_{2} \left(g\right)$ has been well-studied. Dimerization of $N {O}_{2}$ to give ${N}_{2} {O}_{4}$ can be rationalized on the basis that the Lewis structure of $N {O}_{2}$ (with 17 valence electrons!) places a single electron on the cationic nitrogen centre: ""^(-)O-^*N^(+)=O.