Q133CP - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

Each compound's empirical formula and grams of oxygen per 1.00 g of nitrogen are as follows:

i) $\mathbf{N}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{143}\mathbf{ }\mathbf{g}$

ii) ${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{714}\mathbf{ }\mathbf{g}$

iii) ${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{857}\mathbf{ }\mathbf{g}$.

Step 1: About Nitrogen

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.

Step 2: Finding the empirical formula of each compound and grams of oxygen per 1.00 g of nitrogen with 46.69%

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 the empirical formula will be,

$$\begin{gathered} \mathbf{N}\mathbf{:}\mathbf{O}\mathbf{:}\mathbf{:}\mathbf{3}\mathbf{.}\mathbf{33}\mathbf{:}\mathbf{3}\mathbf{.}\mathbf{33}\Rightarrow \mathbf{1}\mathbf{.}\mathbf{1} \\ \mathbf{Empirical}\mathbf{ }\mathbf{formula}\mathbf{:}\mathbf{NO} \end{gathered}$$

To find out how many grams of oxygen there is for every 1.00 gram of nitrogen in NO, use the ratio.

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

Step 3: Finding the empirical formula of each compound and grams of oxygen per 1.00 g of nitrogen with 36.85%

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

$$\begin{gathered} \mathrm{Moles} \mathrm{of} \mathrm{nitrogen}\mathbf{=}\mathbf{36}\mathbf{.}\mathbf{85}\mathbf{ }\mathbf{g}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mol}}{\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{g}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{63}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{N} \\ \mathrm{Moles} \mathrm{of} \mathrm{oxygen}\mathbf{=}\mathbf{63}\mathbf{.}\mathbf{15}\mathbf{ }\mathbf{g}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mol}}{\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{g}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{94}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{O} \end{gathered}$$

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.

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

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

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

Step 4: Finding the empirical formula of each compound and grams of oxygen per 1.00 g of nitrogen for 25.94%

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.

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