Nitrogen forms more oxides than any other element. The percents by mass of N in three different nitrogen oxides are (I) 46.69%; (II) 36.85%; (III) 25.94%. (a) Determine the empirical formula of each compound. (b) How many grams of oxygen per 1.00 g of nitrogen are in each compound?

Nitrogen (N) is a non-metallic element in Periodic Table Group 15 [VA]. It is a translucent, scentless, and tasteless gas that makes up the bulk of the Earth's atmosphere and is present in all living things.

Let us set the base mass to 100 g.

So, the mass of nitrogen is 46.69 g and the mass of oxygen is 53.31 g.

$$\begin{gathered} \mathrm{Moles}\mathrm{of}\mathrm{nitrogen}\mathbf{=}\mathbf{46}\mathbf{.}\mathbf{69}\mathbf{}\mathbf{g}\mathbf{\times }\frac{\mathbf{1}\mathbf{mol}}{\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{33}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{N} \\ \mathrm{Moles}\mathrm{of}\mathrm{oxygen}\mathbf{=}\mathbf{53}\mathbf{.}\mathbf{31}\mathbf{}\mathbf{g}\mathbf{\times }\frac{\mathbf{1}\mathbf{}\mathbf{mol}}{\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{33}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{O} \end{gathered}$$

Then theempirical formulawill be,

role="math" localid="1657089890061" $$\begin{gathered} \mathbf{N}\mathbf{:}\mathbf{O}\mathbf{:}\mathbf{:}\mathbf{3}\mathbf{.}\mathbf{33}\mathbf{:}\mathbf{3}\mathbf{.}\mathbf{33}\Rightarrow \mathbf{1}\mathbf{.}\mathbf{1} \\ \mathbf{Empirical}\mathbf{}\mathbf{formula}\mathbf{:}\mathbf{NO} \end{gathered}$$

To find out how manygrams of oxygenthere is for every 1.00 gram of nitrogenin NO, use the ratio.

role="math" localid="1657090108588" $$\begin{gathered} \frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{oxygen}}{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{nitrogen}}\mathbf{=}\frac{\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}}{\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{143}\mathbf{}\mathbf{g} \end{gathered}$$

Here,the mass of nitrogen is 36.85 g and the mass of oxygen is 63.15 g.

To get the empirical formula,divide the subscripts by the lowest subscript. In addition, the subscript should be lowered to the fewest whole integers that preserve the atom ratio.

${\mathit{N}}_{\frac{2.63}{2.63}}{\mathit{O}}_{\frac{3.94}{2.63}}\to \mathit{N}{\mathit{O}}_{1.5}={\mathit{N}}_2{\mathit{O}}_3$

To find out how many grams of oxygen there are for every 1.00 gram of nitrogen in N₂O₃, use the ratio.

$$\begin{gathered} \frac{\mathbf{3}\mathbf{}\mathbf{(}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{oxygen}\mathbf{)}}{\mathbf{2}\mathbf{}\mathbf{(}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{nitrogen}\mathbf{)}}\mathbf{=}\frac{\mathbf{3}\mathbf{(}\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}\mathbf{)}}{\mathbf{2}\mathbf{(}\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}\mathbf{)}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{714}\mathbf{g} \end{gathered}$$

Here,the mass of nitrogen is 25.94g and the mass of oxygen is 74.06 g.

$$\begin{gathered} \mathrm{Moles}\mathrm{of}\mathrm{nitrogen}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{94}\mathbf{}\mathbf{g}\mathbf{\times }\frac{\mathbf{1}\mathbf{mol}}{\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{83}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{N} \\ \mathrm{Moles}\mathrm{of}\mathrm{oxygen}\mathbf{=}\mathbf{74}\mathbf{.}\mathbf{06}\mathbf{}\mathbf{g}\mathbf{\times }\frac{\mathbf{1}\mathbf{mol}}{\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{63}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{O} \end{gathered}$$

To get the empirical formula,divide the subscripts by the lowest subscript. In addition, the subscript should be lowered to the fewest whole integers that preserve the atom ratio.

${\mathbf{N}}_{\frac{\mathbf{1}\mathbf{.}\mathbf{83}}{\mathbf{1}\mathbf{.}\mathbf{83}}}{\mathbf{O}}_{\frac{\mathbf{4}\mathbf{.}\mathbf{63}}{\mathbf{1}\mathbf{.}\mathbf{83}}}\mathbf{\to }{\mathbf{NO}}_{\mathbf{2}\mathbf{.}\mathbf{5}}\mathbf{=}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}}$

To find out how many grams of oxygen there are every 1.00 grams of nitrogen in N₂O₅, use the ratio.

$$\begin{gathered} \frac{\mathbf{5}\mathbf{(}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{oxygen}\mathbf{)}}{\mathbf{2}\mathbf{(}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{nitrogen}\mathbf{)}}\mathbf{=}\frac{\mathbf{5}\mathbf{(}\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}\mathbf{)}}{\mathbf{2}\mathbf{(}\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}\mathbf{)}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{587}\mathbf{g} \end{gathered}$$