Respuesta :
Answer: The empirical formula of one that contains 30.45% nitrogen is [tex]NO_{2}[/tex].
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.
[tex]Moles = \frac{mass}{molarmass}\\= \frac{30.45 g}{14.0067 g/mol}\\= 2.17 mol[/tex]
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.
[tex]Moles = \frac{mass}{molarmass}\\= \frac{69.55 g}{15.9994 g/mol}\\= 4.34 mol[/tex]
The ratio of both the atoms is as follows.
[tex]\frac{4.34}{2.17} = 2[/tex]
This means that gas has 2 moles of oxygen to 1 mole of nitrogen. Hence, the formula of oxide is [tex]NO_{2}[/tex].
Thus, we can conclude that the empirical formula of one that contains 30.45% nitrogen is [tex]NO_{2}[/tex].
