Question

A compound composed of 2.1% H, 29.8% N, and 68.1% O has a molar mass of approximately 50 g/mol. What is the formal charge of the nitrogen?

Answer 1

Nitrogen has a ZERO formal charge in nitrous acid, `HNO_2`

Explanation:

We need (i) to obtain the emprical, and (ii) the molecular formula of the nitrogen species:

If we assume 100 g of stuff, there are,

`H:` `(2.1*g)/(1.008*g*mol^-1` `=` `2.08" moles of hydrogen"`

`N:` `(29.8*g)/(14.01*g*mol^-1` `=` `2.13" moles of nitrogen"`

`O:` `(68.1*g)/(16.0*g*mol^-1` `=` `4.26" moles of oxygen"`

We divide thru by the lowest molar quantity to give `HNO_2` as the empirical formula, the simplest whole number ratio that defines constituent elements in a species.

But we know that the chemical formula is ALWAYS a mulitple of the empirical formula:

And thus, `50.0*g*mol^-1` `~=` `(1.008+14.01+2xx16.0)*g*mol^-1xxn`

In nitrous acid, the charge on `N` is formally zero; it has the 5 valence electrons, with which the 2 inner core electrons, balance the nuclear charge of `+7`. In nitric acid, `HNO_3`, nitrogen would be quaternized and carry a formal `+1` charge.