Chapter 8, Problem 120AP

Interpretation Introduction

Interpretation:

Among the given compounds or molecules, the species that possesses $46.7\text{\%}$ percent by mass of nitrogen atom is to be identified.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$.

Answer to Problem 120AP

The correct option is (d).

Explanation of Solution

Reason for correct option:

(d) The given molecule is $\text{NO}$.

The mass of $1\text{mole}$ of $\text{N}$ present in $\text{NO}$ is $14\text{g/mol}\times \text{1}\text{mol}=14\text{g}$.

The mass of $1\text{mole}$ of $\text{O}$ present in $\text{NO}$ is $15.99\text{g/mol}\times 1\text{mol}=15.99\text{g}$.

Thus, the total mass of $1\text{mole}$

$\text{NO}$ is,

$\text{Total}\text{mass}\text{of}\text{NO}=14+15.99=29.99\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of nitrogen and mass of $\text{NO}$ in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of N}=\frac{\text{14}\text{g}}{29.99\text{g}}\times 100\text{\%}\\ =46.7\text{\%}\end{array}$

Thus, the percent by mass of nitrogen in $\text{NO}$ is $46.7\text{\%}$.

Conclusion

Reasons for incorrect options:

(a) The given molecule is ${\text{N}}_2{\text{O}}_5$.

The mass of $2\text{mole}$ of $\text{N}$ present in ${\text{N}}_2{\text{O}}_5$ is $14\text{g/mol}\times \text{2}\text{mol}=28\text{g}$.

The mass of $5\text{mole}$ of $\text{O}$ present in ${\text{N}}_2{\text{O}}_5$ is $15.99\text{g/mol}\times 5\text{mol}=79.95\text{g}$.

Thus, the total mass of $1\text{mole}$

${\text{N}}_2{\text{O}}_5$ is,

$\text{Total}\text{mass}\text{of}{\text{N}}_2{\text{O}}_5=28+79.95=107.95\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of nitrogen and mass of ${\text{N}}_2{\text{O}}_5$ in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of N}=\frac{\text{28}\text{g}}{107.95\text{g}}\times 100\text{\%}\\ =25.94\text{\%}\end{array}$

Thus, the percent by mass of nitrogen in ${\text{N}}_2{\text{O}}_5$ is $25.94\text{\%}$.

(b) The given molecule is ${\text{N}}_2\text{O}$.

The mass of $2\text{mole}$ of $\text{N}$ present in ${\text{N}}_2\text{O}$ is $14\text{g/mol}\times \text{2}\text{mol}=28\text{g}$.

The mass of $1\text{mole}$ of $\text{O}$ present in ${\text{N}}_2\text{O}$ is $15.99\text{g/mol}\times 1\text{mol}=15.99\text{g}$.

Thus, the total mass of $1\text{mole}$

${\text{N}}_2\text{O}$ is,

$\text{Total}\text{mass}\text{of}{\text{N}}_2\text{O}=28+15.99=43.99\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of nitrogen and mass of ${\text{N}}_2\text{O}$ in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of N}=\frac{\text{28}\text{g}}{43.99\text{g}}\times 100\text{\%}\\ =63.7\text{\%}\end{array}$

Thus, the percent by mass of nitrogen in ${\text{N}}_2\text{O}$ is $63.7\text{\%}$.

(c) The given molecule is ${\text{NO}}_2$.

The mass of $1\text{mole}$ of $\text{N}$ present in ${\text{NO}}_2$ is $14\text{g/mol}\times \text{1}\text{mol}=14\text{g}$.

The mass of $2\text{mole}$ of $\text{O}$ present in ${\text{NO}}_2$ is $15.99\text{g/mol}\times 2\text{mol}=31.98\text{g}$.

Thus, the total mass of $1\text{mole}$

${\text{NO}}_2$ is,

$\text{Total}\text{mass}\text{of}{\text{NO}}_2=14+31.98=45.98\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of nitrogen and mass of ${\text{NO}}_2$ in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of N}=\frac{\text{14}\text{g}}{45.98\text{g}}\times 100\text{\%}\\ =30.45\text{\%}\end{array}$

Thus, the percent by mass of nitrogen in ${\text{NO}}_2$ is $30.45\text{\%}$.

(e) The given molecule is ${\text{NO}}_3$.

The mass of $1\text{mole}$ of $\text{N}$ present in ${\text{NO}}_3$ is $14\text{g/mol}\times \text{1}\text{mol}=14\text{g}$.

The mass of $1\text{mole}$ of $\text{O}$ present in ${\text{NO}}_3$ is $15.99\text{g/mol}\times 3\text{mol}=47.97\text{g}$.

Thus, the total mass of $1\text{mole}$

${\text{NO}}_3$ is,

$\text{Total}\text{mass}\text{of}{\text{NO}}_3=14+47.97=61.97\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of nitrogen and mass of ${\text{NO}}_3$ in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of N}=\frac{\text{14}\text{g}}{61.97\text{g}}\times 100\text{\%}\\ =22.6\text{\%}\end{array}$

Thus, the percent by mass of nitrogen in ${\text{NO}}_3$ is $22.6\text{\%}$.