Chapter ST, Problem 5PQ

Interpretation Introduction

Interpretation:

The compound that has highest percentage of nitrogen has to be identified.

Concept Introduction:

Mass percent is mass of solute divided by mass of the whole compound multiplied by 100. The formula for mass percent of an element in a compound is as follows:

  $\left(\begin{array}{l}\text{mass}\text{percent}\text{of}\text{element}\\ \text{in}\text{compound}\end{array}\right)=\frac{\text{mass of}\text{element}\text{in}\text{1mole}\text{compound}}{\text{mass}\text{of}1\text{mole}\text{compound}}\times 100$

The formula for mass percent of solution is as follows:

  $\text{mass}\text{percent}\text{of}\text{solution}=\frac{\text{mass of}\text{solute}}{\text{mass}\text{of}\text{solution}}\times 100$

Answer to Problem 5PQ

Option (C) is the correct one.

Explanation of Solution

Reason for incorrect option:

The formulafor mass percent of an element in a compound is as follows:

  $\left(\begin{array}{l}\text{mass}\text{percent}\text{of}\text{element}\\ \text{in}\text{compound}\end{array}\right)=\frac{\text{mass of}\text{element}\text{in}\text{1}\text{mole}\text{compound}}{\text{mass}\text{of}1\text{mole}\text{compound}}\times 100$        (1)

For ${\text{C}}_6{\text{H}}_3{\text{N}}_3{\text{O}}_7$,

The moles of $\text{N}$ in ${\text{C}}_6{\text{H}}_3{\text{N}}_3{\text{O}}_7$ is 3 and the mass of $\text{N}$ in 1 mole is $14\text{g}$. So the mass of  3 moles of $\text{N}$ is calculated as follows:

  $\begin{array}{c}\text{mass}\text{of}\text{3}\text{moles of N}=\left(\text{mass}\text{of}\text{1mole of N}\right)\left(\text{total}\text{number}\text{of}\text{moles}\text{of N}\right)\\ =\left(\text{14}\text{g}\right)\left(3\text{mol}\right)\\ =42\text{g/mol}\end{array}$

Substitute $42\text{g/mol}$ for mass of element in 1 mole compound and $229\text{g/mol}$ for mass of 1 mole compound in equation(1).

  $\begin{array}{c}\left(\begin{array}{l}\text{mass}\text{percent}\text{of}\text{element}\\ \text{in}\text{compound}\end{array}\right)=\frac{42\text{g/mol}}{229\text{g/mol}}\times 100\\ =18.3406\%\end{array}$

For ${\text{CH}}_4{\text{N}}_2\text{O}$,

The moles of $\text{N}$ in ${\text{CH}}_4{\text{N}}_2\text{O}$ is 2 and the mass of $\text{N}$ in 1 mole is $14\text{g}$. So the mass 2 moles of $\text{N}$ is calculated as follows:

  $\begin{array}{c}\text{mass}\text{of}\text{2}\text{moles of N}=\left(\text{mass}\text{of}\text{1mole of N}\right)\left(\text{total}\text{number}\text{of}\text{moles}\text{of N}\right)\\ =\left(\text{14}\text{g}\right)\left(2\text{mol}\right)\\ =28\text{g/mol}\end{array}$

Substitute $28\text{g/mol}$ for mass of element in 1 mole compound and $60.1\text{g/mol}$ for mass of 1 mole compound in equation (1).

  $\begin{array}{c}\left(\begin{array}{l}\text{mass}\text{percent}\text{of}\text{element}\\ \text{in}\text{compound}\end{array}\right)=\frac{28\text{g/mol}}{60.1\text{g/mol}}\times 100\\ =46.5890\%\end{array}$

For ${\text{LiNH}}_2$,

The moles of $\text{N}$ in ${\text{LiNH}}_2$ is 1 and the mass of $\text{N}$ in 1 mole is $14\text{g}$.

Substitute $14\text{g/mol}$ for mass of element in 1 mole compound and $23.0\text{g/mol}$ for mass of 1 mole compound in equation (1).

  $\begin{array}{c}\left(\begin{array}{l}\text{mass}\text{percent}\text{of}\text{element}\\ \text{in}\text{compound}\end{array}\right)=\frac{14.0\text{g/mol}}{23.0\text{g/mol}}\times 100\\ =60.8695\%\end{array}$

For $\text{Pb}{\left({\text{N}}_3\right)}_2$,

The moles of $\text{N}$ in $\text{Pb}{\left({\text{N}}_3\right)}_2$ is 6 and the mass of $\text{N}$ in 1 mole is $14\text{g}$. So the mass of 6 moles of $\text{N}$ is calculated as follows:

  $\begin{array}{c}\text{mass}\text{of}\text{6}\text{moles of N}=\left(\text{mass}\text{of}\text{1mole of N}\right)\left(\text{total}\text{number}\text{of}\text{moles}\text{of N}\right)\\ =\left(\text{14}\text{g}\right)\left(6\text{mol}\right)\\ =84\text{g/mol}\end{array}$

Substitute $84\text{g/mol}$ for mass of element in 1 mole compound and $291.0\text{g/mol}$ for mass of 1 mole compound in equation (1).

  $\begin{array}{c}\left(\begin{array}{l}\text{mass}\text{percent}\text{of}\text{element}\\ \text{in}\text{compound}\end{array}\right)=\frac{84\text{g/mol}}{291\text{g/mol}}\times 100\\ =28.8659\%\end{array}$

The compound that has highest percentage of nitrogen is ${\text{LiNH}}_2$. So the correct option is (C).

Reason for incorrect option:

Percentage of $\text{Pb}{\left({\text{N}}_3\right)}_2$ is $28.8659\%$, ${\text{CH}}_4{\text{N}}_2\text{O}$ is $46.5890\%$, ${\text{LiNH}}_2$ is $60.8695\%$ and ${\text{C}}_6{\text{H}}_3{\text{N}}_3{\text{O}}_7$ is $18.3406\%$.

Since compound that has highest percentage of nitrogen is ${\text{LiNH}}_2$. Hence option (A), (B) and (D) are incorrect.

Conclusion

Compound that has highest percentage of nitrogen is ${\text{LiNH}}_2$. So the correct option is (C).