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?

Aug 15, 2016

Nitrogen has a ZERO formal charge in nitrous acid, $H N {O}_{2}$

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 $=$ $2.08 \text{ moles of hydrogen}$

$N :$ (29.8*g)/(14.01*g*mol^-1 $=$ $2.13 \text{ moles of nitrogen}$

$O :$ (68.1*g)/(16.0*g*mol^-1 $=$ $4.26 \text{ moles of oxygen}$

We divide thru by the lowest molar quantity to give $H N {O}_{2}$ 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, $50.0 \cdot g \cdot m o {l}^{-} 1$ $\cong$ $\left(1.008 + 14.01 + 2 \times 16.0\right) \cdot g \cdot m o {l}^{-} 1 \times n$

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, $H N {O}_{3}$, nitrogen would be quaternized and carry a formal $+ 1$ charge.