Chapter 3.1, Problem 3.5BFP

(a)

Interpretation Introduction

Interpretation:

The formula units present in $75.5\text{ g}$ of calcium phosphate is to be calculated.

Concept introduction:

One mole is defined as the amount of substance that contains the same number of entities such as molecules, ions, atoms as the number of atoms in $12\text{ g}$ of $\text{C-12}$. This number is called Avogadro’s number. The value of Avogadro’s number is $\text{6}\text{.022}\times {\text{10}}^{23}\left(\text{entitites/mol}\right)$.

Molar mass is defined as the mass of $1\text{ mol}$ of a chemical substance in grams. The molar mass of a compound is calculated by the addition of the molar mass of each element multiplied by its number of atoms present in the chemical formula.

A formula unit is used for the ionic compound to represent their empirical formula. The steps to determine the formula unit of an ionic compound from the given mass are as follows:

Step 1: Divide the given mass of the ionic compound with the molar mass to calculate the moles. The expression to calculate the moles of an ionic compound when the mass is given is as follows:

$\text{Amount}\left(\text{mol}\right)=\left(\text{Mass }\left(\text{g}\right)\right)\left(\frac{1\text{ mol}}{\text{No}\text{. of grams}}\right)$

Step 2: Multiply the calculated moles with the Avogadro’s number to determine the formula units of an ionic compound. The expression to determine the formula unit is as follows:

$\text{Number of formula units}=\left(\text{Amount }\left(\text{mol}\right)\right)\left(\frac{6.022\times {10}^{23}\text{ formula units}}{1\text{ mol}}\right)$

Answer to Problem 3.5BFP

In $75.5\text{ g}$ of calcium phosphate, $1.47\times {10}^{23}\text{ formula units}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ are present.

Explanation of Solution

The molecular formula of calcium phosphate is ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$. The expression to calculate the molar mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is as follows:

${\text{Molar mass of Ca}}_3{\left({\text{PO}}_4\right)}_2=\left(3\right)\left(\text{M of Ca}\right)+\left(2\right)\left(\text{M of P}\right)+\left(8\right)\left(\text{M of O}\right)$

Substitute $40.08\text{ g/mol}$ for $\text{M}$ of $\text{Ca}$ , $30.97\text{ g/mol}$ for $\text{M}$ of $\text{P}$ , and $16.00\text{ g/mol}$ for $\text{M}$ of $\text{O}$  in the above equation as follows:

$\begin{array}{c}{\text{Molar mass of Ca}}_3{\left({\text{PO}}_4\right)}_2=\left(3\right)\left(40.08\text{ g/mol}\right)+\left(2\right)\left(30.97\text{ g/mol}\right)+\left(8\right)\left(16.00\text{ g/mol}\right)\\ =120.24\text{ g/mol}+\text{61}\text{.94 g/mol}+\text{128}\text{.00 g/mol}\\ =\text{310}\text{.18 g/mol}\end{array}$

Divide $75.5\text{ g}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ with the molar mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ to calculate its moles as follows:

$\begin{array}{c}{\text{Amount of Ca}}_3{\left({\text{PO}}_4\right)}_2\left(\text{mol}\right)=\left(75.5{\text{ g Ca}}_3{\left({\text{PO}}_4\right)}_2\right)\left(\frac{1{\text{ mol Ca}}_3{\left({\text{PO}}_4\right)}_2}{\text{310}{\text{.18 g Ca}}_3{\left({\text{PO}}_4\right)}_2}\right)\\ =0.243407{\text{ mol Ca}}_3{\left({\text{PO}}_4\right)}_2\end{array}$

One mole of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ contains $\text{6}\text{.022}\times {\text{10}}^{23}\text{ formula units}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$. Multiply $0.243407{\text{ mol Ca}}_3{\left({\text{PO}}_4\right)}_2$ with the Avogadro’s number to calculate the number of formula units of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ as follows:

$\begin{array}{c}{\text{Number of formula units Ca}}_3{\left({\text{PO}}_4\right)}_2=\left(0.24341{\text{ mol Ca}}_3{\left({\text{PO}}_4\right)}_2\right)\left(\frac{\begin{array}{l}6.022\times {10}^{23}\text{ formula units}\\ {\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\end{array}}{1{\text{ mol Ca}}_3{\left({\text{PO}}_4\right)}_2}\right)\\ =1.4658\times {10}^{23}{\text{ formula units Ca}}_3{\left({\text{PO}}_4\right)}_2\\ \approx 1.47\times {10}^{23}{\text{ formula units Ca}}_3{\left({\text{PO}}_4\right)}_2\end{array}$

Conclusion

In $75.5\text{ g}$ of calcium phosphate, $1.47\times {10}^{23}\text{ formula units}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ are present.

(b)

Interpretation Introduction

Interpretation:

The number of phosphate $\left({\text{PO}}_4^{3-}\right)$ ions in $75.5\text{ g}$ of calcium phosphate is to be calculated.

Concept introduction:

A formula unit is used for the ionic compound to represent their empirical formula. The molecular formula of a compound tells the number of atoms/ions of each element present in the compound.

Answer to Problem 3.5BFP

In $1.47\times {10}^{23}\text{ formula units}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$, $2.94\times {10}^{23}{\text{ PO}}_4^{3-}\text{ ions}$ are present.

Explanation of Solution

From the molecular formula of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$, it is concluded that 2phosphate $\left({\text{PO}}_4^{3-}\right)$ ions are present in one formula unit of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$. Calculate the number of ${\text{PO}}_4^{3-}$ ions in $1.47\times {10}^{23}\text{ formula units}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ as follows:

$\begin{array}{c}{\text{Number of PO}}_4^{3-}\text{ ions}=\left(1.47\times {10}^{23}{\text{ formula units Ca}}_3{\left({\text{PO}}_4\right)}_2\right)\left(\frac{2{\text{ PO}}_4^{3-}\text{ ions}}{1{\text{ formula unit Ca}}_3{\left({\text{PO}}_4\right)}_2}\right)\\ =2.94\times {10}^{23}{\text{ PO}}_4^{3-}\text{ ions}\end{array}$

Conclusion

In $1.47\times {10}^{23}\text{ formula units}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$, $2.94\times {10}^{23}{\text{ PO}}_4^{3-}\text{ ions}$ are present.