Chapter 4, Problem 55QP

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

$\begin{array}{l}\left(\text{a}\right){\text{ H}}_{\text{2}}\\ \left(\text{b}\right)\text{ Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\\ \left(\text{c}\right){\text{ N}}_{\text{2}}{\text{O}}_{\text{2}}\\ \left(\text{d}\right){\text{ K}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in $140\text{ g}$ ${\text{H}}_2$ is to be determined.

Explanation of Solution

For the number of formula units:

$m\left(\text{g}\right)\stackrel{MM}{\to }n\stackrel{{\text{N}}_{\text{A}}}{\to }\text{Number of formula units}$

Here, $m$ is the given mass, $MM$ is the molar mass and $n$ is the number of moles.

For the number of atoms or ions:

$\text{Number of formula units}\stackrel{\text{formula ratio}}{\to }\text{Atoms or ions}$

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, ${\text{N}}_{\text{A}}$ is Avogadro’s number with a value of $6.022\times {10}^{23}{\text{mol}}^{-1}$ .

The number of atoms is determined as follows:

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

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

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

The molar mass of one mole of ${\text{H}}_2$ is $2.016\text{g}$ . The weight of ${\text{H}}_2$ is $140\text{ g}$ . In the ${\text{H}}_2$ molecule, there are two hydrogen atoms. Substitute these values in equation (2) as follows.

$\begin{array}{c}\text{Number of H atoms}=\frac{140\cancel{\text{g}}}{\text{2}\text{.016}\cancel{\text{g/mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\times 2\text{ H atoms}\\ =8.364\times {10}^{25}\text{ H atoms}\end{array}$

Therefore, there are $8.364\times {10}^{25}$ H atoms present in $140\text{ g}$ of ${\text{H}}_2$ .

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140\text{ g}$ $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is to be determined.

Explanation of Solution

The molar mass of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is $164.10\text{g/mol}$ . The weight of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is $140\text{ g}$ . Substitute these values in equation (1) to determine the number of formula units present in $140\text{ g}$ of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ as follows.

$\begin{array}{c}\text{Formula units}\text{Ca}{\left({\text{NO}}_3\right)}_2=\frac{140\cancel{\text{g}}}{164.1\text{g/mol}\cancel{\text{g}}\text{/}\cancel{\text{mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =5.14\times {10}^{23}\text{ formula units}\text{Ca}{\left({\text{NO}}_3\right)}_2\end{array}$

There is one ${\text{Ca}}^{2+}$ in $\text{Ca}{\left({\text{NO}}_3\right)}_2$ . By using equation (3), the number of ${\text{Ca}}^{2+}$ ions in $140\text{ g}$ $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is determined as follows.

$\begin{array}{c}\text{Number of atoms}=5.14\times {10}^{23}\times {\text{1 Ca}}^{2+}\\ =5.14\times {10}^{23}{\text{ Ca}}^{2+}\end{array}$

There are two ${\text{NO}}_3^{-}$ ions in $\text{Ca}{\left({\text{NO}}_3\right)}_2$ . By using equation (3), the number of ${\text{NO}}_3^{-}$ in $140\text{ g}$ $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is determined as follows.

$\begin{array}{c}\text{Number of atoms}=5.14\times {10}^{23}\times {\text{2 NO}}_3^{-}\\ =1.028\times {10}^{24}{\text{ NO}}_3^{-}\end{array}$

Therefore, there are $5.14\times {10}^{23}{\text{ Ca}}^{2+}$ and $1.028\times {10}^{24}{\text{ NO}}_3^{-}$ present in $140\text{ g}$ of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ .

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140\text{ g}$ ${\text{N}}_2{\text{O}}_2$ is to be determined.

Explanation of Solution

The molar mass of ${\text{N}}_2{\text{O}}_2$ is $60.02\text{g/mol}$ . The weight of ${\text{N}}_2{\text{O}}_2$ is $140\text{ g}$ . Substitute these values in equation (1) to determine the number of formula units present in $140\text{ g}$ of ${\text{N}}_2{\text{O}}_2$ as follows.

$\begin{array}{c}\text{Formula units}{\text{N}}_2{\text{O}}_2=\frac{140\cancel{\text{g}}}{60.02\text{g/mol}\cancel{\text{g}}\text{/}\cancel{\text{mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =1.405\times {10}^{24}\text{ formula units}{\text{N}}_2{\text{O}}_2\end{array}$

There are two N in ${\text{N}}_2{\text{O}}_2$ . By using equation (3), the number of N atoms in $140\text{ g}$ ${\text{N}}_2{\text{O}}_2$ is determined as follows.

$\begin{array}{c}\text{Number of atoms}=1.405\times {10}^{24}\times \text{1 N}\\ =2.8\times {10}^{24}\text{ N}\text{atoms}\end{array}$

There are two O ions in ${\text{N}}_2{\text{O}}_2$ . By using equation (3), the number of O in $140\text{ g}$ ${\text{N}}_2{\text{O}}_2$ is determined as follows.

$\begin{array}{c}\text{Number of atoms}=1.405\times {10}^{24}\times \text{2 O}\\ =2.8\times {10}^{24}\text{ O}\text{atoms}\end{array}$

Therefore, there are $2.8\times {10}^{24}\text{ N}$ and $2.8\times {10}^{24}\text{ O}$ atoms present in $140\text{ g}$ of ${\text{N}}_2{\text{O}}_2$ .

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140\text{ g}$ ${\text{K}}_2{\text{SO}}_4$ is to be determined.

Explanation of Solution

The molar mass of ${\text{K}}_2{\text{SO}}_4$ is $174.26\text{g/mol}$ . The weight of ${\text{K}}_2{\text{SO}}_4$ is $140\text{ g}$ . Substitute these values in equation (1) to determine the number of formula units present in $140\text{ g}$ of ${\text{K}}_2{\text{SO}}_4$ as follows.

$\begin{array}{c}\text{Formula units}{\text{K}}_2{\text{SO}}_4=\frac{140\cancel{\text{g}}}{174.26\text{g/mol}\cancel{\text{g}}\text{/}\cancel{\text{mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =4.838\times {10}^{23}\text{ formula units}{\text{K}}_2{\text{SO}}_4\end{array}$

There are 2 ${\text{K}}^{+}$ in ${\text{K}}_2{\text{SO}}_4$ . By using equation (3), the number of ${\text{K}}^{+}$ ions in $140\text{ g}$ ${\text{K}}_2{\text{SO}}_4$ is determined as follows.

$\begin{array}{c}\text{Number of atoms}=4.838\times {10}^{23}\times {\text{2 K}}^{+}\\ =9.676\times {10}^{23}{\text{ K}}^{+}\text{ions}\end{array}$

There is one ${\text{SO}}_2{}^{4-}$ in ${\text{K}}_2{\text{SO}}_4$ . By using equation (3), the number of ${\text{SO}}_2{}^{4-}$ in $140\text{ g}$ ${\text{K}}_2{\text{SO}}_4$ is determined as follows.

$\begin{array}{c}\text{Number of atoms}=4.838\times {10}^{23}\times {\text{1 SO}}_2{}^{4-}\\ =4.838\times {10}^{23}{\text{ SO}}_2{}^{4-}\end{array}$

Therefore, there are $9.676\times {10}^{23}{\text{ K}}^{+}$ and $4.838\times {10}^{23}{\text{ SO}}_2{}^{4-}$ present in $140\text{ g}$ of ${\text{K}}_2{\text{SO}}_4$ .