Chapter 5, Problem 5.77QP

Interpretation Introduction

Interpretation:

The percent composition by mass of ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ has to be calculated.

Concept introduction:

Percent composition by mass of each element present in a compound is known as the compound’s percent composition by mass.

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{Where}\text{n}\text{is}\text{the}\text{number}\text{of}\text{atoms}\text{of}\text{the}\text{element}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$

 To determine: The percent composition by mass of ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$

Answer to Problem 5.77QP

The percent composition by mass is $\text{\%}\text{Ca}\text{=}\text{39}\text{.8936}\text{\%}\text{;}\text{\%}\text{P}\text{=}\text{18}\text{.4988}\text{\%}\text{;}\text{\%}\text{O}\text{=}\text{41}\text{.4070}\text{\%}\text{;}\text{\%}\text{O}\text{=}\text{0}\text{.2007}\text{\%}$

Explanation of Solution

Percent composition by mass:

Percent composition by mass of each element present in a compound is known as the compound’s percent composition by mass. The formula is expressed mathematically as follows:

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{Where}\text{n}\text{is}\text{the}\text{number}\text{of}\text{atoms}\text{of}\text{the}\text{element}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$ By using the above formula percent composition by mass is calculated as follows:

$\begin{array}{l}\text{Formula}\text{mass}\text{of}\text{\%}{\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)\text{=}\text{5}\left(\text{mass}\text{of}\text{Ca}\text{in}\text{amu}\right)\text{+}\text{3}\left(\left(\text{mass}\text{of}\text{P}\text{in}\text{amu}\right)\text{+}\text{4}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\right)\\ \text{+}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\text{+}\text{mass}\text{of}\text{H}\text{in}\text{amu}\right)\\ \text{=}\text{5}\left(\text{40}\text{.078}\text{amu}\right)\text{+}\text{3}\left(\left(\text{30}\text{.9738}\text{amu}\right)\text{+}\text{4}\left(\text{15}\text{.9994}\text{amu}\right)\right)\\ \text{+}\left(\text{15}\text{.9994}\text{amu}\text{+}\text{1}\text{.0079}\text{amu}\right)\\ \text{=}\text{200}\text{.39}\text{amu}\text{+}\text{3}\left(\text{30}\text{.9738}\text{+}\text{63}\text{.9976}\right)\text{amu}\text{+}\text{17}\text{.0073}\text{amu}\\ \text{=}\left(\text{200}\text{.39}\text{+}\text{3}\left(\text{94}\text{.9714}\right)\text{+}\text{17}\text{.0073}\right)\text{amu}\\ \text{=}\left(\text{200}\text{.39}\text{+}\text{284}\text{.9142}\text{+}\text{17}\text{.0073}\right)\text{amu}\\ \text{=}\text{502}\text{.3115}\text{amu}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{Ca}\text{=}\frac{\text{5}\text{×atomic}\text{mass}\text{Ca}\text{in}\text{}\text{amu}}{\text{formula}\text{mass}\text{of}{\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)\text{in}\text{amu}}\text{ ×}\text{100\%}\\ \text{=}\frac{\text{5}\text{×40}\text{.078}\text{amu}}{\text{502}\text{.3115}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{200}\text{.39}\text{amu}}{\text{502}\text{.3115}\text{amu}}\text{×}\text{100}\text{\%}\\ \text{=}\text{39}\text{.8936}\text{\%}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{P}\text{=}\frac{\text{3}\text{×atomic}\text{mass}\text{P}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)\text{in}\text{amu}}\text{ ×}\text{100\%}\\ \text{=}\frac{\text{3×30}\text{.9738}\text{amu}}{\text{502}\text{.3115}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{92}\text{.9214}\text{amu}}{\text{502}\text{.3115}\text{amu}}\text{×}\text{100}\text{\%}\\ \text{=}\text{18}\text{.4988}\text{\%}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{O}\text{=}\frac{\text{13}\text{×atomic}\text{mass}\text{O}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{13×15}\text{.9994}\text{amu}}{\text{502}\text{.3115}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{207}\text{.9922}\text{amu}}{\text{502}\text{.3115}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{41}\text{.4070}\text{\%}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{H}\text{=}\frac{\text{1}\text{×atomic}\text{mass}\text{H}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{1}\text{.0079}\text{amu}}{\text{502}\text{.3115}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{0}\text{.2007}\text{\%}\end{array}$

The percent composition by mass of $\text{\%}\text{Ca}\text{=}\text{39}\text{.8936}\text{\%}\text{;}\text{\%}\text{P}\text{=}\text{18}\text{.4988}\text{\%}\text{;}\text{\%}\text{O}\text{=}\text{41}\text{.4070}\text{\%}\text{;}\text{\%}\text{O}\text{=}\text{0}\text{.2007}\text{\%}$

Conclusion

The percent composition by mass of ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ was calculated.

$\text{percent}\text{mass}\text{of}\text{\%}\text{Ca}\text{=}\text{39}\text{.8936}\text{\%}$

$\text{percent}\text{mass}\text{of}\text{\%}\text{P}\text{=}\text{18}\text{.4988}\text{\%}$

$\text{percent}\text{mass}\text{of}\text{\%}\text{O}\text{=}\text{41}\text{.4070}\text{\%}$

$\text{percent}\text{mass}\text{of}\text{\%}\text{H}\text{=}\text{0}\text{.2007}\text{\%}$