A 10.000g sample of a compound of nitrogen and oxygen contains 3.686 g of nitrogen. What is the empirical formula? I'm unsure if it is N2O.

Jun 14, 2017

$\text{Empirical formula} = {N}_{2} {O}_{3}$

Explanation:

The $\text{empirical formula}$ is the simplest whole number ratio that represents constituent elements in a species....

And given the composition, we simply take the molar quantities of each element ......

$\text{Moles of nitrogen}$ $=$ $\frac{3.686 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1} = 0.2630 \cdot m o l$

$\text{Moles of oxygen}$ $=$ $\frac{10.00 \cdot g - 3.686 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 0.3947 \cdot m o l$

And we divide thru by the smallest molar quantity to get......

$N {O}_{1.50}$, but since we require whole numbers we get......

${N}_{2} {O}_{3}$.

This nitrogen oxide is probably not one we would commonly encounter.....$O = N - \stackrel{+}{N} \left(= O\right) {O}^{-}$ is a possibility, and also $\text{trinitramide}$ ${N}_{4} {O}_{6}$, a potential rocket fuel. Neither is something you would want to play with..........

These are unstable nitrogen oxides that would only be found in specialist labs........and certainly not in the quantity quoted in the question.