Chapter 3, Problem 3.57QP

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

1. a NO
2. b SO₃
3. c NaH₂PO₃
4. d Pb(NO₃)₂

Interpretation Introduction

Interpretation:

Percentage composition of each elements that are present in $\text{NO}$ 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 3.57QP

The percentage composition of $\text{N}$ in $\text{NO}$ is $\text{46}\text{.7}\text{\%}$

The percentage composition of $\text{O}$ in $\text{NO}$ is $\text{53}\text{.31}\text{\%}$

Explanation of Solution

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

Given,

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

Molar mass of $\text{NO}$ is $\text{30}\text{.01 g}$

$\begin{array}{l}\text{\%N}\text{in}\text{NO}\text{=}\frac{\text{Mass}\text{of}\text{N}}{\text{Mass}\text{of}\text{NO}}\text{×100}\\ \text{=}\frac{\text{14}\text{.01}\text{g}\text{N}}{\text{30}\text{.01}\text{g}\text{NO}}\text{×100}\\ \text{=}\text{46}\text{.7}\text{\%}\end{array}$

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

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

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

$\begin{array}{l}\text{=}\text{100-46}\text{.7}\\ \text{=}\text{53}\text{.31}\text{\%O}\end{array}$

Hence,

Mass percentage of $\text{N}$ in $\text{NO}$ is $\text{46}\text{.7}\text{\%}$

Mass percentage of $\text{O}$ in $\text{NO}$ is $\text{53}\text{.31}\text{\%}$

Conclusion

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

Interpretation Introduction

Interpretation:

Percentage composition of each elements that are present in ${\text{SO}}_{\text{3}}$ 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 3.57QP

The percentage composition of $\text{S}$ in ${\text{SO}}_{\text{3}}$ is $\text{40}\text{.0\%}$

The percentage composition of $\text{O}$ in ${\text{SO}}_{\text{3}}$ is $\text{60}\text{.0}\text{\%}$

Explanation of Solution

To calculate the percentage composition of $\text{S}$ in ${\text{SO}}_{\text{3}}$

Given,

Molar mass of $\text{S}$ is $\text{32}\text{.06 g}$

Molar mass of ${\text{SO}}_{\text{3}}$ is $\text{80}\text{.06 g}$

$\begin{array}{l}\text{\%N}\text{in}{\text{SO}}_{\text{3}}\text{=}\frac{\text{Mass}\text{of}\text{S}}{\text{Mass}\text{of}{\text{SO}}_{\text{3}}}\text{×100}\\ \text{=}\frac{\text{32}\text{.06}\text{g}\text{S}}{\text{80}\text{.06 g}{\text{SO}}_{\text{3}}}\text{×100}\\ \text{=}\text{40}\text{.0}\text{\%}\end{array}$

Molar masses of $\text{S}$ and ${\text{SO}}_{\text{3}}$ are plugged in above equation to give percentage composition of $\text{S}$ present in ${\text{SO}}_{\text{3}}$.

To calculate the percentage composition of $\text{O}$ in ${\text{SO}}_{\text{3}}$

Subtract the percentage composition of $\text{S}$ from 100 to gives percentage composition of $\text{O}$ in ${\text{SO}}_{\text{3}}$.

$\begin{array}{l}\text{=}\text{100-40}\text{.0}\\ \text{=}60.0\text{\%O}\end{array}$

Hence,

The percentage composition of $\text{S}$ in ${\text{SO}}_{\text{3}}$ is $\text{40}\text{.0\%}$

The percentage composition of $\text{O}$ in ${\text{SO}}_{\text{3}}$ is $\text{60}\text{.0}\text{\%}$

Conclusion

Percentage composition of each elements that are present in ${\text{SO}}_{\text{3}}$ was calculated.

Interpretation Introduction

Interpretation:

Percentage composition of each elements that are present in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ 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 3.57QP

The percentage composition of $\text{Na}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ is $\text{22}\text{.1}\text{\%}$

The percentage composition of $\text{H}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ is $\text{1}\text{.939}\text{\%}$

The percentage composition of $\text{P}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ is $\text{29}\text{.784}\text{\%}$

The percentage composition of $\text{O}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ is $\text{46}\text{.2}\text{\%}$

Explanation of Solution

To calculate the percentage composition of $\text{Na}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$

Molar mass of $\text{Na}$ is $\text{22}\text{.99 g}$

Molar mass of ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ is $\text{103}\text{.98 g}$

$\begin{array}{l}\text{\%Na}\text{in}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{3}}\text{=}\frac{\text{Mass}\text{of}\text{Na}}{\text{Mass}\text{of}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{3}}}\text{×100}\\ \text{=}\frac{\text{22}\text{.99}\text{g}\text{Na}}{\text{103}\text{.98 g}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{3}}}\text{×100}\\ \text{=}\text{22}\text{.1}\text{\%}\end{array}$

Molar masses of $\text{Na}$ and ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ are plugged in above equation to give percentage composition of $\text{Na}$ present in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$.

To calculate the percentage composition of $\text{H}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$

Molar mass of ${\text{H}}_{\text{2}}$ is $\text{2}\text{.016 g}$

$\begin{array}{l}\text{\%H}\text{in}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{3}}\text{=}\frac{\text{Mass}\text{of}\text{H}}{\text{Mass}\text{of}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{3}}}\text{×100}\\ \text{=}\frac{\text{2}\text{.016}\text{g}\text{H}}{\text{103}\text{.98 g}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{3}}}\text{×100}\\ \text{=}\text{1}\text{.94}\text{\%}\end{array}$

Molar masses of $\text{H}$ and ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ are plugged in above equation to give percentage composition of $\text{H}$ present in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$.

To calculate the percentage composition of $\text{P}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$

Molar mass of $\text{P}$ is $\text{30}\text{.97 g}$

$\begin{array}{l}\text{\%P}\text{in}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{3}}\text{=}\frac{\text{Mass}\text{of}\text{P}}{\text{Mass}\text{of}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{3}}}\text{×100}\\ \text{=}\frac{\text{30}\text{.97 g}\text{P}}{\text{103}\text{.98 g}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{3}}}\text{×100}\\ \text{=}\text{29}\text{.8}\text{\%}\end{array}$

Molar masses of $\text{P}$ and ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ are plugged in above equation to give percentage composition of $\text{P}$ present in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$.

To calculate the percentage composition of $\text{O}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$

Subtract the percentage compositions of $\text{Na,}\text{H,}\text{and }\text{P}$ from 100 to gives percentage composition of $\text{O}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$.

$\begin{array}{l}\text{=}\text{100-(22}\text{.110+1}\text{.939+29}\text{.784)\%}\\ \text{=}\text{46}\text{.2}\text{\%}\text{O}\end{array}$

Hence,

The percentage composition of $\text{Na}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ is $\text{22}\text{.1}\text{\%}$

The percentage composition of $\text{H}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ is $\text{1}\text{.939}\text{\%}$

The percentage composition of $\text{P}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ is $\text{29}\text{.784}\text{\%}$

The percentage composition of $\text{O}$ in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ is $\text{46}\text{.2}\text{\%}$

Conclusion

Percentage composition of each elements that are present in ${\text{NaH}}_{\text{3}}{\text{PO}}_{\text{3}}$ was calculated.

Interpretation Introduction

Interpretation:

Percentage composition of each elements that are present in ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ 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 3.57QP

The percentage composition of $\text{Pb}$ in ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{62}\text{.6}\text{\%}$

The percentage composition of $\text{N}$ in ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{8}\text{.46}\text{\%}$

The percentage composition of $\text{O}$ in ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $29.0\text{\%}$

Explanation of Solution

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

Mass of $\text{Pb}$ is $\text{207}\text{.2 g}$

Mass of ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{331}\text{.22 g}$

$\begin{array}{l}\text{\%}\text{Pb}\text{in}{\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{=}\frac{\text{Mass}\text{of}\text{Pb}}{\text{Mass}\text{of}{\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{×100}\\ \text{=}\frac{\text{207}\text{.}\text{2}\text{g}\text{Pb}}{\text{331}\text{.22}\text{g}{\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{×100}\\ \text{=}\text{62}\text{.6}\text{\%}\end{array}$

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

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

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

$\begin{array}{l}\text{\%N}\text{in}{\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{=}\frac{\text{Mass}\text{of}\text{N}}{\text{Mass}\text{of}{\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{×100}\\ \text{=}\frac{\text{2 x 14}\text{.01 g N}}{\text{331}\text{.22 g}{\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{×100}\\ \text{=}\text{8}\text{.46}\text{\%}\end{array}$

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

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

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

$\begin{array}{l}\text{=}\text{100-(62}\text{.6+8}\text{.46)\%}\\ \text{=}29.0\text{\%}\text{O}\end{array}$

Hence,

The percentage composition of $\text{Pb}$ in ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{62}\text{.6}\text{\%}$

The percentage composition of $\text{N}$ in ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{8}\text{.46}\text{\%}$

The percentage composition of $\text{O}$ in ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $29.0\text{\%}$

Conclusion

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