Chapter 4, Problem 56QP

How many atoms (or ions) of each element are in 140.0 g of the following substances?

$\begin{array}{l}\left(\text{a}\right){\text{ BaSO}}_{\text{4}}\\ \left(\text{b}\right){\text{ Mg}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\\ \left(\text{c}\right){\text{ O}}_{\text{2}}\\ \left(\text{d}\right)\text{ KBr}\end{array}$

Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in $140.0\text{ g}$ of ${\text{BaSO}}_4$ is to be determined.

Explanation of Solution

The number of formula units present is determined as follows.

$\text{Number of formula units}=\frac{m}{MM}\times {\text{N}}_{\text{A}}$ ...(1)

Here, $m$ is the given mass, $MM$ is the molar mass, $n$ is the number of moles, and ${\text{N}}_{\text{A}}$ is Avogadro’s number with a value of $6.022\times {10}^{23}{\text{mol}}^{-1}$ .

The number of atoms is determined as follows.

$\text{Number of atoms/ions}=\frac{m}{MM}\times {\text{N}}_{\text{A}}\times \text{Present ions}$ ...(2)

By combining equations (1) and (2), the number of atoms can be determined as follows.

$\text{Number of atoms/ions}=\text{Number of formula units}\times \text{Present ions}$ ...(3)

The molar mass of one mole of ${\text{BaSO}}_4$ is $233.38\text{ g/mol}$ and its weight is $140.0\text{ g}$ . Substitute these values in equation (1) to determine the number of formula units present in $140.0\text{ g}$ of ${\text{BaSO}}_4$ as follows.

$\begin{array}{c}\text{Formula unit}{\text{of BaSO}}_4=\frac{140.0\cancel{\text{g}}}{233.38\cancel{\text{g/mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =3.613\times {10}^{22}{\text{ Formula unit BaSO}}_4\end{array}$

In the ${\text{BaSO}}_4$ molecule, there is one barium ion $\left({\text{Ba}}^{2+}\right)$ and one sulfate ion $\left({\text{SO}}_4^{2-}\right)$ . By using equation (3), the number of barium ions $\left({\text{Ba}}^{2+}\right)$ in $140.0\text{ g}$ of ${\text{BaSO}}_4$ is determined as follows.

$\begin{array}{c}{\text{Number of Ba}}^{2+}\text{ions}=3.613\times {10}^{23}\times {\text{1 Ba}}^{2+}\\ =3.613\times {10}^{23}{\text{ Ba}}^{2+}\text{ ions}\end{array}$

Similarly, the number of sulfate ions is calculated as follows.

$\begin{array}{c}{\text{Number of SO}}_4^{2-}\text{ ions}=3.613\times {10}^{23}\times {\text{1 SO}}_4^{2-}\\ =3.613\times {10}^{23}{\text{ SO}}_4^{2-}\text{ ions}\end{array}$

Therefore, there are $3.613\times {10}^{22}{\text{ Ba}}^{2+}\text{ ions}$ and $5.14\times {10}^{23}{\text{ SO}}_4^{2-}\text{ ions}$ present in $140.0\text{ g}$ of ${\text{BaSO}}_4$ .

The total number of barium and sulfate ions in ${\text{BaSO}}_4$ is calculated as follows.

$\begin{array}{c}{\text{Total ions of BaSO}}_4{}_2=\left[3.613\times {10}^{23}{\text{ Ba}}^{2+}\text{ ions}+3.613\times {10}^{23}{\text{ SO}}_4^{2-}\text{ ions}\right]\\ =7.225\times {10}^{23}\text{ ions KBr}\end{array}$

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140.0\text{ g}$ of ${\text{Mg}}_3{\left({\text{PO}}_4\right)}_2$ is to be determined.

Explanation of Solution

The molar mass of one mole of ${\text{Mg}}_3{\left({\text{PO}}_4\right)}_2$ is $233.38\text{ g/mol}$ and its weight is $140.0\text{ g}$ . Substitute these values in equation (1) to determine the number of formula units present in $140.0\text{ g}$ of ${\text{Mg}}_3{\left({\text{PO}}_4\right)}_2$ as follows.

$\begin{array}{c}\text{Formula unit}{\text{of Mg}}_3{\left({\text{PO}}_4\right)}_2=\frac{140.0\cancel{\text{g}}}{265.85\cancel{\text{g/mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =3.2\times {10}^{23}{\text{ Formula unit Mg}}_3{\left({\text{PO}}_4\right)}_2\end{array}$

In the ${\text{Mg}}_3{\left({\text{PO}}_4\right)}_2$ molecule, there are three magnesium ions $\left({\text{3 Mg}}^{2+}\right)$ and two phosphate ions $\left({\text{2 PO}}_{\text{4}}^{\text{3-}}\right)$ . By using equation (3), the number of magnesium ions $\left({\text{3 Mg}}^{2+}\right)$ in $140.0\text{ g}$ of ${\text{Mg}}_3{\left({\text{PO}}_4\right)}_2$ is determined as follows.

$\begin{array}{c}{\text{Number of Mg}}^{2+}\text{ions}=3.2\times {10}^{23}\times 3{\text{ Mg}}^{2+}\\ =9.6\times {10}^{23}{\text{ Mg}}^{2+}\text{ ions}\end{array}$

Similarly, the number of phosphate ions is calculated as follows.

$\begin{array}{c}{\text{Number of PO}}_{\text{4}}^{\text{3-}}\text{ ions}=3.2\times {10}^{23}\times 2{\text{ PO}}_{\text{4}}^{\text{3-}}\\ =6.4\times {10}^{23}{\text{ PO}}_{\text{4}}^{\text{3-}}\text{ ions}\end{array}$

Therefore, there are $9.6\times {10}^{23}{\text{ Mg}}^{2+}\text{ ions}$ and $6.4\times {10}^{23}{\text{ PO}}_{\text{4}}^{\text{3-}}\text{ ions}$ present in $140.0\text{ g}$ of ${\text{Mg}}_3{\left({\text{PO}}_4\right)}_2$ .

The total number of potassium and bromide ions in ${\text{Mg}}_3{\left({\text{PO}}_4\right)}_2$ is calculated as follows.

$\begin{array}{c}{\text{Total ions of Mg}}_3{\left({\text{PO}}_4\right)}_2=\left[9.6\times {10}^{23}{\text{ Mg}}^{2+}\text{ ions}+6.4\times {10}^{23}{\text{ PO}}_{\text{4}}^{\text{3-}}\text{ ions}\right]\\ =1.6\times {10}^{24}\text{ ions KBr}\end{array}$

Interpretation Introduction

Interpretation:

The number of atoms of each element present in $140.0\text{ g}$ of ${\text{O}}_2$ is to be determined.

Explanation of Solution

The molar mass of two moles of ${\text{O}}_2$ is $32\text{ g/mol}$ and its weight is $140.0\text{ g}$ . In the ${\text{O}}_2$ molecule, there are two oxygen atoms. Substitute these values in equation (2) as follows.

$\begin{array}{c}{\text{Number of O}}_2\text{ atoms}=\frac{140.0\cancel{\text{g}}}{32\cancel{\text{g/mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\times 2{\text{ O}}_2\text{ atoms}\\ =5.269\times {10}^{24}{\text{ O}}_2\text{ atoms}\end{array}$

Therefore, there are $5.269\times {10}^{24}{\text{ O}}_2$ atoms present in $140.0\text{ g}$ of ${\text{O}}_2$ .

Interpretation Introduction

Interpretation:

The number of ions of each element present in $140.0\text{ g}$ of $\text{KBr}$ is to be determined.

Explanation of Solution

The molar mass of one mole of $\text{KBr}$ is $\text{119}\text{.002 g/mol}$ and its weight is $140.0\text{ g}$ . Substitute these values in equation (1) to determine the number of formula units present in $140.0\text{ g}$ of $\text{KBr}$ as follows.

$\begin{array}{c}\text{Formula unit}\text{of KBr}=\frac{140.0\cancel{\text{g}}}{119.002\cancel{\text{g/mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =7.085\times {10}^{23}\text{ Formula unit KBr}\end{array}$

In the $\text{KBr}$ molecule, there is one potassium ion $\left({\text{K}}^{+}\right)$ and one barium ion $\left({\text{Br}}^{-}\right)$ . By using equation (3), the number of potassium ions $\left({\text{K}}^{+}\right)$ in $140.0\text{ g}$ of $\text{KBr}$ is determined as follows.

$\begin{array}{c}{\text{Number of K}}^{+}\text{ions}=7.085\times {10}^{23}\times {\text{1 K}}^{+}\\ =7.085\times {10}^{23}{\text{ K}}^{+}\text{ ions}\end{array}$

Similarly, the number of sulfate ions is calculated as follows.

$\begin{array}{c}{\text{Number of Br}}^{-}\text{ ions}=7.085\times {10}^{23}\times {\text{1 Br}}^{-}\\ =7.085\times {10}^{23}{\text{ Br}}^{-}\text{ ions}\end{array}$

Therefore, there are $5.14\times {10}^{23}{\text{ Ba}}^{2+}\text{ ions}$ and $5.14\times {10}^{23}{\text{ SO}}_4^{2-}\text{ ions}$ present in $140.0\text{ g}$ of ${\text{BaSO}}_4$ .

The total number of potassium and bromide ions in $\text{KBr}$ is calculated as follows.

$\text{Total ions of KBr}=$ $\begin{array}{c}\text{Total ions of KBr}=\left[7.085\times {10}^{23}{\text{ K}}^{+}\text{ ions}+7.085\times {10}^{23}{\text{ Br}}^{-}\text{ ions}\right]\\ =1.417\times {10}^{24}\text{ ions KBr}\end{array}$