Chapter 3, Problem 3.80QP

Interpretation Introduction

Interpretation: The nitrogen oxides that have more than $50\%$ oxygen by mass are to be predicted. The nitrogen oxides, if any, with same empirical formula are to be stated.

Concept introduction: Molar mass of a compound is sum of the atomic mass of each elements present in the compound.

To determine: The nitrogen that have more than $50\%$ oxygen by mass and nitrogen oxides, if any, with same empirical formula.

Answer to Problem 3.80QP

Solution

The compounds that have the same empirical formula are shown in Figure 1. The nitrogen oxides that have more than $50\%$ oxygen by mass are $\text{NO}$, ${\text{N}}_{\text{2}}{\text{O}}_3$ , ${\text{N}}_{\text{2}}{\text{O}}_4$, ${\text{NO}}_2$ and ${\text{N}}_{\text{2}}{\text{O}}_2$. The compounds that have the same empirical formula are $\text{NO}$, ${\text{N}}_{\text{2}}{\text{O}}_2$ and ${\text{NO}}_2$ and ${\text{N}}_{\text{2}}{\text{O}}_4$.

Explanation of Solution

Explanation

Molecular mass of ${\text{N}}_{\text{2}}\text{O}$ is calculated by the formula,

$\text{Molecular mass of} {\text{N}}_{\text{2}}\text{O}=\left(\begin{array}{c}\left(\text{Number}\text{of}\text{N}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{N}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)$

Number of nitrogen atoms is $2$.

Number of oxygen atoms is $1$.

Molar mass of nitrogen is $\text{14}\text{.00}\text{g/mol}$.

Molar mass of oxygen is $\text{16}\text{.00}\text{g/mol}$.

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

$\begin{array}{c}\text{Molecular mass of} {\text{N}}_{\text{2}}\text{O}=\left(2\times \text{14}\text{.00}\text{g/mol}\right)+\left(1\times 16.00\right)\\ =44\text{g/mol}\end{array}$

The percent composition of oxygen (O) in ${\text{N}}_{\text{2}}\text{O}$ is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of N}}_{\text{2}}\text{O}}\times 100$

Substitute the value of mass of O atoms and mass of ${\text{N}}_{\text{2}}\text{O}$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{16.00g}{44\text{g}}\times 100\\ =36.36\text{\% O}\end{array}$

Therefore, the percent composition of oxygen in ${\text{N}}_{\text{2}}\text{O}$ is $36.36\text{\% O}$.

Molecular mass of $\text{NO}$ is calculated by the formula,

$\text{Molecular mass of} \text{NO}=\left(\begin{array}{c}\left(\text{Number}\text{of}\text{N}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{N}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)$

Number of nitrogen atoms is $1$.

Number of oxygen atoms is $1$.

Molar mass of nitrogen is $\text{14}\text{.00}\text{g/mol}$.

Molar mass of oxygen is $\text{16}\text{.00}\text{g/mol}$.

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

$\begin{array}{c}\text{Molecular mass of} {\text{N}}_{\text{2}}\text{O}=\left(1\times \text{14}\text{.00}\text{g/mol}\right)+\left(1\times 16.00\right)\\ =30\text{g/mol}\end{array}$

The percent composition of oxygen (O) in $\text{NO}$ is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{\text{Mass of NO}}\times 100$

Substitute the value of mass of O atoms and mass of $\text{NO}$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{16g}{30\text{g}}\times 100\\ =53.33\text{\% O}\end{array}$

Therefore, the percent composition of oxygen in $\text{NO}$ is $53.33\text{\% O}$.

Molecular mass of ${\text{N}}_{\text{2}}{\text{O}}_3$ is calculated by the formula,

$\begin{array}{c}\text{Molecular mass of} {\text{N}}_{\text{2}}{\text{O}}_3=\left(\text{Number}\text{of}\text{N}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{N}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}$

Number of nitrogen atoms is $2$.

Number of oxygen atoms is $3$.

Molar mass of nitrogen is $\text{14}\text{.00}\text{g/mol}$.

Molar mass of oxygen is $\text{16}\text{.00}\text{g/mol}$.

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

$\begin{array}{c}\text{Molecular mass of} {\text{N}}_{\text{2}}{\text{O}}_3=\left(2\times \text{14}\text{.00}\text{g/mol}\right)+\left(3\times 16.00\right)\\ =76\text{g/mol}\end{array}$

The percent composition of oxygen (O) in ${\text{N}}_{\text{2}}{\text{O}}_3$ is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of N}}_{\text{2}}{\text{O}}_3}\times 100$

Mass of $\text{3}\text{O}$ atoms is $3\times 16\text{g/mol}=\text{48}\text{g/mol}$.

Substitute the value of mass of O atoms and mass of ${\text{N}}_{\text{2}}{\text{O}}_3$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{48g}{76\text{g}}\times 100\\ =63.15\text{\% O}\end{array}$

Therefore, the percent composition of oxygen in ${\text{N}}_{\text{2}}{\text{O}}_3$ is $63.15\text{\% O}$.

Molecular mass of ${\text{N}}_{\text{2}}{\text{O}}_2$ is calculated by the formula,

$\begin{array}{c}\text{Molecular mass of} {\text{N}}_{\text{2}}{\text{O}}_2=\left(\text{Number}\text{of}\text{N}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{N}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}$

Number of nitrogen atoms is $2$.

Number of oxygen atoms is $2$.

Molar mass of nitrogen is $\text{14}\text{.00}\text{g/mol}$.

Molar mass of oxygen is $\text{16}\text{.00}\text{g/mol}$.

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

$\begin{array}{c}\text{Molecular mass of} {\text{N}}_{\text{2}}{\text{O}}_2=\left(2\times \text{14}\text{.00}\text{g/mol}\right)+\left(2\times 16.00\right)\\ =60\text{g/mol}\end{array}$

The percent composition of oxygen (O) in ${\text{N}}_{\text{2}}{\text{O}}_2$ is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of N}}_{\text{2}}{\text{O}}_2}\times 100$

Mass of $2\text{O}$ atoms is $2\times 16\text{g/mol}=32\text{g/mol}$.

Substitute the value of mass of O atoms and mass of ${\text{N}}_{\text{2}}{\text{O}}_2$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{32g}{60\text{g}}\times 100\\ =53.33\text{\% O}\end{array}$

Therefore, the percent composition of oxygen in ${\text{N}}_{\text{2}}{\text{O}}_2$ is $53.33\text{\% O}$.

Molecular mass of ${\text{NO}}_2$ is calculated by the formula,

$\text{Molecular mass of} {\text{NO}}_2=\left(\begin{array}{c}\left(\text{Number}\text{of}\text{N}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{N}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)$

Number of nitrogen atoms is $1$.

Number of oxygen atoms is $2$.

Molar mass of nitrogen is $\text{14}\text{.00}\text{g/mol}$.

Molar mass of oxygen is $\text{16}\text{.00}\text{g/mol}$.

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

$\begin{array}{c}\text{Molecular mass of} {\text{NO}}_2=\left(1\times \text{14}\text{.00}\text{g/mol}\right)+\left(2\times 16.00\right)\\ =46\text{g/mol}\end{array}$

The percent composition of oxygen (O) in ${\text{NO}}_2$ is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of N}}_{\text{2}}{\text{O}}_2}\times 100$

Mass of $2\text{O}$ atoms is $2\times 16\text{g/mol}=32\text{g/mol}$.

Substitute the value of mass of O atoms and mass of ${\text{NO}}_2$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{32g}{46\text{g}}\times 100\\ =69.56\text{\% O}\end{array}$

Therefore, the percent composition of oxygen in ${\text{NO}}_2$ is $69.56\text{\% O}$.

Molecular mass of ${\text{N}}_{\text{2}}{\text{O}}_4$ is calculated by the formula,

$\text{Molecular mass of}{\text{N}}_{\text{2}}{\text{O}}_4=\left(\begin{array}{c}\left(\text{Number}\text{of}\text{N}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{N}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)$

Number of nitrogen atoms is $2$.

Number of oxygen atoms is $4$.

Molar mass of nitrogen is $\text{14}\text{.00}\text{g/mol}$.

Molar mass of oxygen is $\text{16}\text{.00}\text{g/mol}$.

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

$\begin{array}{c}\text{Molecular mass of} {\text{N}}_{\text{2}}{\text{O}}_4=\left(2\times \text{14}\text{.00}\text{g/mol}\right)+\left(4\times 16.00\right)\\ =92\text{g/mol}\end{array}$

The percent composition of oxygen (O) in ${\text{N}}_{\text{2}}{\text{O}}_4$ is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of N}}_{\text{2}}{\text{O}}_4}\times 100$

Mass of $4\text{O}$ atoms is $4\times 16\text{g/mol}=64\text{g/mol}$.

Substitute the value of mass of O atoms and mass of ${\text{N}}_{\text{2}}{\text{O}}_4$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{64}\text{g}}{\text{92}\text{g}}\times 100\\ =69.56\text{\% O}\end{array}$

Therefore, the percent composition of oxygen in ${\text{N}}_{\text{2}}{\text{O}}_4$ is $69.56\text{\% O}$.

The nitrogen oxides that have more than $50\%$ oxygen by mass are $\text{NO}$, ${\text{N}}_{\text{2}}{\text{O}}_3$ , ${\text{N}}_{\text{2}}{\text{O}}_4$, ${\text{NO}}_2$ and ${\text{N}}_{\text{2}}{\text{O}}_2$.

The ratio of $\text{N:O}$ in $\text{NO}$ is $1:1$. Thus the empirical formula of $\text{NO}$ is $\text{NO}$. The ratio of $\text{N:O}$ in ${\text{N}}_{\text{2}}{\text{O}}_2$ is $1:1$. Thus the empirical formula of ${\text{N}}_{\text{2}}{\text{O}}_2$ is $\text{NO}$. The compounds that have the same empirical formula are $\text{NO}$ and ${\text{N}}_{\text{2}}{\text{O}}_2$.

The ratio of $\text{N:O}$ in ${\text{NO}}_2$ is $1:2$. Thus the empirical formula of ${\text{NO}}_2$ is ${\text{NO}}_2$. The ratio of $\text{N:O}$ in ${\text{N}}_{\text{2}}{\text{O}}_4$ is $1:2$. Thus the empirical formula of ${\text{N}}_{\text{2}}{\text{O}}_4$ is ${\text{NO}}_2$. The compounds that have the same empirical formula are ${\text{NO}}_2$ and ${\text{N}}_{\text{2}}{\text{O}}_4$.

Conclusion

The nitrogen oxides that have more than $50\%$ oxygen by mass are $\text{NO}$, ${\text{N}}_{\text{2}}{\text{O}}_3$ , ${\text{N}}_{\text{2}}{\text{O}}_4$, ${\text{NO}}_2$ and ${\text{N}}_{\text{2}}{\text{O}}_2$. The compounds that have the same empirical formula are $\text{NO}$, ${\text{N}}_{\text{2}}{\text{O}}_2$ and ${\text{NO}}_2$ and ${\text{N}}_{\text{2}}{\text{O}}_4$