Question

A sample of gas contains `0.7*g` of nitrogen, and `1.6*g` of oxygen. What are (i) its `"empirical formula"`, and (ii) its `"molecular formula"` if its molecular mass is `92*g*mol^-1`?

Answer 1

The stuff is likely `N_2O_4` `"..dinitrogen tetroxide.."`

Explanation:

We gots....`(0.7*g)/(14.01*g*mol^-1)=0.0500*mol` with respect to nitrogen...

And....`(1.6*g)/(15.999*g*mol^-1)=0.100*mol` with respect to oxygen...

And CLEARLY, we gots an empirical formula, the simplest whole number ratio defining constituent atoms in a species, of `NO_2`.

But the molecular formula is a whole number multiple of the empirical formula....

And thus `92*g*mol^-1-=nxx(14.01+2xx15.999)*g*mol^-1`

i.e. `n=(92*g*mol^-1)/(46.0*g*mol^-1)`..so `n=2`...and we gots a molecular formula of `N_2O_4`, `"dinitrogen tetroxide"`. The gas has formal charge separation in its Lewis structure...

`(""^(-)O)O=stackrel(+)N-stackrel(+)N=O(O^-)`, and this is the dimer of the NEUTRAL `NO_2` radical....`O=stackrel(*)N^(+)-O^(-)`...