Question

There is a series of nitrogen oxides with the general formula N?O?. What is the empirical formula of one that contains 30.45% nitrogen?
Question 13 options:

N2O


NO


NO2­


N2O3


N2O5

Answer 1

Answer: The empirical formula of one that contains 30.45% nitrogen is $NO_{2}$.

Explanation:

Given: Mass of nitrogen = 30.45 g

Let us assume that the mass of given oxide is 100 grams.

As the atomic mass of nitrogen is 14.0067 g. So, moles of nitrogen will be calculated as follows.

$Moles = \frac{mass}{molarmass}\\= \frac{30.45 g}{14.0067 g/mol}\\= 2.17 mol$

Also, mass of oxygen = (100 - 30.45) g = 69.55 g

Atomic mass of oxygen is 15.9994 g/mol. So, moles of oxygen will be as follows.

$Moles = \frac{mass}{molarmass}\\= \frac{69.55 g}{15.9994 g/mol}\\= 4.34 mol$

The ratio of both the atoms is as follows.

$\frac{4.34}{2.17} = 2$

This means that gas has 2 moles of oxygen to 1 mole of nitrogen. Hence, the formula of oxide is $NO_{2}$.

Thus, we can conclude that the empirical formula of one that contains 30.45% nitrogen is $NO_{2}$.