Chapter 3, Problem 3.58QP

Calculate the percentage composition for each of the following compounds (three significant figures).

1. a NO₂
2. b H₂N₂
3. c KClO₄
4. d Mg(NO₃)₂

Interpretation Introduction

Interpretation:

Percentage composition of each elements that are present in ${\text{NO}}_{\text{2}}$ should be calculated.

Concept introduction:

Mass percentage:

The percent ration of mass of analyte that is present in a given sample with total mass of sample to give a mass percent of analyte present in a given sample.

$\text{Mass}\text{precent}\text{\%}\text{=}\frac{\text{Mass}\text{of}\text{analyte}}{\text{Mass}\text{of}\text{sample}}\times 100$

Answer to Problem 3.58QP

The percentage composition of $\text{N}$ in ${\text{NO}}_{\text{2}}$ is $\text{30}\text{.4}\text{\%}$

The percentage composition of $\text{O}$ in ${\text{NO}}_{\text{2}}$ is $\text{69}\text{.6}\text{\%}$

Explanation of Solution

To calculate the percentage composition of $\text{N}$ in ${\text{NO}}_{\text{2}}$

Given,

Molar mass of $\text{N}$ is $\text{14}\text{.01 g}$

Molar mass of ${\text{NO}}_{\text{2}}$ is $\text{46}\text{.01 g}$

$\begin{array}{l}\text{\%N}\text{in}{\text{NO}}_{\text{2}}\text{=}\frac{\text{Mass}\text{of}\text{N}}{\text{Mass}\text{of}{\text{NO}}_{\text{2}}}\text{×100}\\ \text{=}\frac{\text{14}\text{.01}\text{g}\text{N}}{46.01\text{g}{\text{NO}}_{\text{2}}}\text{×100}\\ \text{=}30.4\text{\%}\end{array}$

Molar masses of $\text{N}$ and ${\text{NO}}_{\text{2}}$ are plugged in above equation to give percentage composition of $\text{N}$ present in ${\text{NO}}_{\text{2}}$.

To calculate the percentage composition of $\text{O}$ in ${\text{NO}}_{\text{2}}$

Subtract the percentage composition of $\text{N}$ from 100 to gives percentage composition of $\text{O}$ in ${\text{NO}}_{\text{2}}$.

$\begin{array}{l}\text{=}\text{100-30}\text{.4}\\ \text{=}69.6\text{\%O}\end{array}$

Hence,

Mass percentage of $\text{N}$ in ${\text{NO}}_{\text{2}}$ is $\text{30}\text{.4}\text{\%}$

Mass percentage of $\text{O}$ in ${\text{NO}}_{\text{2}}$ is $\text{69}\text{.6}\text{\%}$

Conclusion

Percentage composition of each elements that are present in ${\text{NO}}_{\text{2}}$ was calculated.

Interpretation Introduction

Interpretation:

Percentage composition of each elements that are present in ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$ should be calculated.

Concept introduction:

Mass percentage:

The percent ration of mass of analyte that is present in a given sample with total mass of sample to give a mass percent of analyte present in a given sample.

$\text{Mass}\text{precent}\text{\%}\text{=}\frac{\text{Mass}\text{of}\text{analyte}}{\text{Mass}\text{of}\text{sample}}\times 100$

Answer to Problem 3.58QP

The percentage composition of $\text{H}$ in ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{6}\text{.71\%}$

The percentage composition of $\text{N}$ in ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{93}\text{.3}\text{\%}$

Explanation of Solution

To calculate the percentage composition of $\text{H}$ in ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$

Given,

Molar mass of $\text{H}$ is $\text{1}\text{.008 g}$

Molar mass of ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{30}\text{.04 g}$

$\begin{array}{l}\text{\%H}\text{in}{\text{N}}_{\text{2}}{\text{H}}_{\text{2}}\text{=}\frac{\text{Mass}\text{of}\text{H}}{\text{Mass}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{2}}}\text{×100}\\ \text{=}\frac{\text{2×1}\text{.008}\text{g}\text{H}}{\text{30}\text{.04g}{\text{N}}_{\text{2}}{\text{H}}_{\text{2}}}\text{×100}\\ \text{=}\text{6}\text{.71}\text{\%}\end{array}$

Molar masses of $\text{H}$ and ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$ are plugged in above equation to give percentage composition of $\text{H}$ present in ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$.

To calculate the percentage composition of $\text{N}$ in ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$

Subtract the percentage composition of $\text{H}$ from 100 to gives percentage composition of $\text{N}$ in ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$.

$\begin{array}{l}\text{=}\text{100-6}\text{.71}\\ \text{=}93.3\text{\%O}\end{array}$

Hence,

The percentage composition of $\text{H}$ in ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{6}\text{.71\%}$

The percentage composition of $\text{N}$ in ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{93}\text{.3}\text{\%}$

Conclusion

Percentage composition of each elements that are present in ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$ was calculated.

Interpretation Introduction

Interpretation:

Percentage composition of each elements that are present in ${\text{KClO}}_4$ should be calculated.

Concept introduction:

Mass percentage:

The percent ration of mass of analyte that is present in a given sample with total mass of sample to give a mass percent of analyte present in a given sample.

$\text{Mass}\text{precent}\text{\%}\text{=}\frac{\text{Mass}\text{of}\text{analyte}}{\text{Mass}\text{of}\text{sample}}\times 100$

Answer to Problem 3.58QP

The percentage composition of $\text{K}$ in ${\text{KClO}}_4$ is $\text{28}\text{.22}\text{\%}$

The percentage composition of $\text{Cl}$ in ${\text{KClO}}_4$ is $\text{25}\text{.6}\text{\%}$

The percentage composition of $\text{O}$ in ${\text{KClO}}_4$ is $\text{46}\text{.2}\text{\%}$

Explanation of Solution

To calculate the percentage composition of $\text{K}$ in ${\text{KClO}}_4$

Molar mass of $\text{K}$ is $\text{39}\text{.10 g}$

Molar mass of ${\text{KClO}}_4$ is $\text{138}\text{.55 g}$

$\begin{array}{l}\text{\%K}\text{in}{\text{KClO}}_{\text{4}}\text{=}\frac{\text{Mass}\text{of}\text{K}}{\text{Mass}\text{of}{\text{KClO}}_{\text{4}}}\text{×100}\\ \text{=}\frac{\text{39}\text{.10 g}\text{K}}{\text{138}\text{.55 g}{\text{KClO}}_{\text{4}}}\text{×100}\\ \text{=}\text{28}\text{.2}\text{\%}\end{array}$

Molar masses of $\text{K}$ and ${\text{KClO}}_4$ are plugged in above equation to give percentage composition of $\text{K}$ present in ${\text{KClO}}_4$.

To calculate the percentage composition of $\text{Cl}$ in ${\text{KClO}}_4$

Molar mass of $\text{Cl}$ is $\text{35}\text{.45 g}$

$\begin{array}{l}\text{\%Cl}\text{in}{\text{KClO}}_{\text{4}}\text{=}\frac{\text{Mass}\text{of}\text{Cl}}{\text{Mass}\text{of}{\text{KClO}}_{\text{4}}}\text{×100}\\ \text{=}\frac{\text{35}\text{.45}\text{g}\text{Cl}}{\text{138}\text{.55 g}{\text{KClO}}_{\text{4}}}\text{×100}\\ \text{=}\text{25}\text{.6\%}\end{array}$

Molar masses of $\text{Cl}$ and ${\text{KClO}}_4$ are plugged in above equation to give percentage composition of $\text{Cl}$ present in ${\text{KClO}}_4$.

To calculate the percentage composition of $\text{O}$ in ${\text{KClO}}_4$

Subtract the percentage compositions of $\text{K and }\text{Cl}$ from 100 to gives percentage composition of $\text{O}$ in ${\text{KClO}}_4$.

$\begin{array}{l}\text{=}\text{100-(28}\text{.22+25}\text{.586)\%}\\ \text{=}\text{46}\text{.2}\text{\%}\text{O}\end{array}$

Hence,

The percentage composition of $\text{K}$ in ${\text{KClO}}_4$ is $\text{28}\text{.22}\text{\%}$

The percentage composition of $\text{Cl}$ in ${\text{KClO}}_4$ is $\text{25}\text{.6}\text{\%}$

The percentage composition of $\text{O}$ in ${\text{KClO}}_4$ is $\text{46}\text{.2}\text{\%}$

Conclusion

Percentage composition of each elements that are present in ${\text{KClO}}_4$ was calculated.

Interpretation Introduction

Interpretation:

Percentage composition of each elements that are present in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ should be calculated.

Concept introduction:

Mass percentage:

The percent ration of mass of analyte that is present in a given sample with total mass of sample to give a mass percent of analyte present in a given sample.

$\text{Mass}\text{precent}\text{\%}\text{=}\frac{\text{Mass}\text{of}\text{analyte}}{\text{Mass}\text{of}\text{sample}}\times 100$

Answer to Problem 3.58QP

The percentage composition of $\text{Mg}$ in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{16}\text{.4}\text{\%}$

The percentage composition of $\text{N}$ in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{18}\text{.9}\text{\%}$

The percentage composition of $\text{O}$ in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $64.7\text{\%}$

Explanation of Solution

To calculate the percentage composition of $\text{Mg}$ in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Mass of $\text{Mg}$ is $\text{24}\text{.31 g}$

Mass of ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{148}\text{.32 g}$

$\begin{array}{l}\text{\%}\text{Mg}\text{in}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{=}\frac{\text{Mass}\text{of}\text{Mg}}{\text{Mass}\text{of}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{×100}\\ \text{=}\frac{\text{24}\text{.31}\text{g}\text{Mg}}{\text{148}\text{.32g}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{×100}\\ \text{=}\text{16}\text{.4}\text{\%}\end{array}$

Molar masses of $\text{Mg}$ and ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ are plugged in above equation to give percentage composition of $\text{Mg}$ present in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$.

To calculate the percentage composition of $\text{N}$ in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Molar mass of $\text{N}$ is $\text{14}\text{.01 g}$

$\begin{array}{l}\text{\%N}\text{in}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{=}\frac{\text{Mass}\text{of}\text{N}}{\text{Mass}\text{of}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{×100}\\ \text{=}\frac{\text{2 x 14}\text{.01 g N}}{\text{148}\text{.32 g}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{×100}\\ \text{=}\text{18}\text{.9}\text{\%}\end{array}$

Molar masses of $\text{N}$ and ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ are plugged in above equation to give percentage composition of $\text{N}$ present in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$.

To calculate the percentage composition of $\text{O}$ in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Subtract the percentage compositions of $\text{Mg and}\text{N}$ from 100 to gives percentage composition of $\text{O}$ in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$.

$\begin{array}{l}\text{=}\text{100-(16}\text{.4}\text{+18}\text{.9)\%}\\ \text{=}64.7\text{\%}\text{O}\end{array}$

Hence,

The percentage composition of $\text{Mg}$ in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{16}\text{.4}\text{\%}$

The percentage composition of $\text{N}$ in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{18}\text{.9}\text{\%}$

The percentage composition of $\text{O}$ in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $64.7\text{\%}$

Conclusion

Percentage composition of each elements that are present in ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ was calculated.