Chapter F, Problem E.21E

(a)

Interpretation Introduction

Interpretation:

Amount in moles, number of molecules and formula units present in $\text{10}\text{.0}\text{g}$ of alumina has to be calculated.

Explanation of Solution

Number of moles of alumina:

Formula of alumina is given as ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$.  The mass of alumina is $\text{10}\text{.0}\text{g}$.  Molar mass of alumina is $101.96\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of alumina is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}})}=\frac{10.0\text{g}}{101.96\text{g}\cdot {\text{mol}}^{-1}}\\ =0.098\text{mol}\end{array}$

Thus the number of moles of alumina in $\text{10}\text{.0}\text{g}$ is $\text{0}\text{.098}\text{mol}$.

Number of molecules in alumina:

One mole of any substance contains $6.0221\times {10}^{23}\text{molecules}$.  Therefore, the number of molecules present in $\text{0}\text{.098}\text{mol}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{alumina molecules}=0.098\times \text{6}\text{.0221}\times {\text{10}}^{\text{23}}\text{molecules}\\ \text{=5}\text{.90}\times {\text{10}}^{\text{22}}\text{molecules}\end{array}$

Thus the number of molecules of alumina in $\text{10}\text{.0}\text{g}$ is $\text{5}\text{.90}\times {\text{10}}^{\text{22}}\text{molecules}$.

Number of formula units:

As this is given species is a compound, the number of formula units will be equal to the number of molecules.  Therefore, the number of formula units in $\text{10}\text{.0}\text{g}$ of alumina is $5.90\times {10}^{22}$.

(b)

Interpretation Introduction

Interpretation:

Amount in moles, number of molecules and formula units present in $\text{25}\text{.92}\text{mg}$ of hydrogen fluoride has to be calculated.

Explanation of Solution

Number of moles of hydrogen fluoride:

Formula of hydrogen fluoride is given as $\text{HF}$.  The mass of hydrogen fluoride is $\text{25}\text{.92}\text{mg}$.  Molar mass of hydrogen fluoride is $20.01\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of hydrogen fluoride is calculated as shown below;

    $\begin{array}{c}{n}_{(\text{HF})}=\frac{25.92\times {10}^{-3}\text{g}}{20.01\text{g}\cdot {\text{mol}}^{-1}}\\ =1.30\times {10}^{-3}\text{mol}\end{array}$

Thus the number of moles of hydrogen fluoride in $\text{25}\text{.92}\text{mg}$ is $1.30\times {10}^{-3}\text{mol}$.

Number of molecules in hydrogen fluoride:

One mole of any substance contains $6.0221\times {10}^{23}\text{molecules}$.  Therefore, the number of molecules present in $1.30\times {10}^{-3}\text{mol}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{HF molecules}=1.30\times {10}^{-3}\text{mol}\times \text{6}\text{.0221}\times {\text{10}}^{\text{23}}\text{molecules}\\ \text{=7}\text{.83}\times {\text{10}}^{\text{20}}\text{molecules}\end{array}$

Thus the number of molecules of hydrogen fluoride in $\text{25}\text{.92}\text{mg}$ is $\text{7}\text{.83}\times {\text{10}}^{\text{20}}\text{molecules}$.

Number of formula units:

As this is given species is a compound, the number of formula units will be equal to the number of molecules.  Therefore, the number of formula units in $\text{25}\text{.92}\text{mg}$ of hydrogen fluoride is $\text{7}\text{.83}\times {\text{10}}^{\text{20}}$.

(c)

Interpretation Introduction

Interpretation:

Amount in moles, number of molecules and formula units present in $\text{1}\text{.55}\text{mg}$ of hydrogen peroxide has to be calculated.

Explanation of Solution

Number of moles of hydrogen peroxide:

Formula of hydrogen peroxide is given as ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$.  The mass of hydrogen peroxide is $\text{1}\text{.55}\text{mg}$.  Molar mass of hydrogen peroxide is $34.01\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of hydrogen peroxide is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{H}}_{\text{2}}{\text{O}}_{\text{2}})}=\frac{\text{1}\text{.55}\times {10}^{-3}\text{g}}{34.01\text{g}\cdot {\text{mol}}^{-1}}\\ =0.046\times {10}^{-3}\text{mol}\end{array}$

Thus the number of moles of hydrogen peroxide in $\text{1}\text{.55}\text{mg}$ is $0.046\times {10}^{-3}\text{mol}$.

Number of molecules in hydrogen peroxide:

One mole of any substance contains $6.0221\times {10}^{23}\text{molecules}$.  Therefore, the number of molecules present in $0.046\times {10}^{-3}\text{mol}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{ molecules}=0.046\times {10}^{-3}\text{mol}\times \text{6}\text{.0221}\times {\text{10}}^{\text{23}}\text{molecules}\\ \text{=2}\text{.77}\times {\text{10}}^{\text{19}}\text{molecules}\end{array}$

Thus the number of molecules of hydrogen peroxide in $\text{1}\text{.55}\text{mg}$ is $\text{2}\text{.77}\times {\text{10}}^{\text{19}}\text{molecules}$.

Number of formula units:

As this is given species is a compound, the number of formula units will be equal to the number of molecules.  Therefore, the number of formula units in $\text{1}\text{.55}\text{mg}$ of hydrogen peroxide is $\text{2}\text{.77}\times {\text{10}}^{\text{19}}$.

(d)

Interpretation Introduction

Interpretation:

Amount in moles, number of molecules and formula units present in $\text{1}\text{.25}\text{kg}$ of glucose has to be calculated.

Explanation of Solution

Number of moles of glucose:

Formula of glucose is given as ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$.  The mass of glucose is $\text{1}\text{.25}\text{kg}$.  Molar mass of glucose is $180.16\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of glucose is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}})}=\frac{\text{1}\text{.25}\times {10}^3\text{g}}{180.15\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0069\times {10}^3\text{mol}\end{array}$

Thus the number of moles of glucose in $\text{1}\text{.25}\text{kg}$ is $0.0069\times {10}^3\text{mol}$.

Number of molecules in glucose:

One mole of any substance contains $6.0221\times {10}^{23}\text{molecules}$.  Therefore, the number of molecules present in $0.0069\times {10}^3\text{mol}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ molecules}=0.0069\times {10}^3\text{mol}\times \text{6}\text{.0221}\times {\text{10}}^{\text{23}}\text{molecules}\\ \text{=4}\text{.15}\times {\text{10}}^{\text{26}}\text{molecules}\end{array}$

Thus the number of molecules of glucose in $\text{1}\text{.25}\text{kg}$ is $\text{4}\text{.15}\times {\text{10}}^{\text{26}}\text{molecules}$.

Number of formula units:

As this is given species is a compound, the number of formula units will be equal to the number of molecules.  Therefore, the number of formula units in $\text{1}\text{.25}\text{kg}$ of glucose is $\text{4}\text{.15}\times {\text{10}}^{\text{26}}$.

(e)

Interpretation Introduction

Interpretation:

Amount in moles, number of molecules and atoms present in $\text{4}\text{.37}\text{g}$ of nitrogen has to be calculated.

Explanation of Solution

Number of moles of nitrogen:

The mass of nitrogen atom is $\text{4}\text{.37}\text{g}$.  Molar mass of nitrogen is $14.00\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of nitrogen atom is calculated as shown below;

    $\begin{array}{c}{n}_{(\text{N})}=\frac{4.37\text{g}}{14.00\text{g}\cdot {\text{mol}}^{-1}}\\ =0.312\text{mol}\end{array}$

Thus the number of moles of nitrogen atoms in $\text{4}\text{.37}\text{g}$ is $0.312\text{mol}$.

Two atoms of nitrogen combines to give a molecule of nitrogen.  Therefore, the number of moles of nitrogen molecule is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{N}}_{\text{2}})}=\frac{0.312\text{mol}}{2}\\ =0.156\text{mol}\end{array}$

Thus the number of moles of nitrogen molecules in $\text{4}\text{.37}\text{g}$ is $0.156\text{mol}$.

Number of atoms or molecules in nitrogen:

One mole of any substance contains $6.0221\times {10}^{23}\text{atoms}$.  Therefore, the number of atoms present in $0.312\text{mol}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{N}\text{atoms}=0.312\times \text{6}\text{.0221}\times {\text{10}}^{\text{23}}\text{atoms}\\ \text{=1}\text{.88}\times {\text{10}}^{\text{23}}\text{atoms}\end{array}$

Thus the number of atoms of nitrogen in $\text{4}\text{.37}\text{g}$ is $\text{1}\text{.88}\times {\text{10}}^{\text{23}}$.

Two atoms of nitrogen combines to give a molecule of nitrogen.  Therefore, the number of nitrogen molecule is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}{\text{N}}_{\text{2}}\text{molecule}=\frac{\text{1}\text{.88}\times {\text{10}}^{\text{23}}}{2}\\ \text{=}9.4\times {\text{10}}^{\text{22}}\text{molecules}\end{array}$

Thus the number of nitrogen molecules in $\text{4}\text{.37}\text{g}$ is $9.4\times {\text{10}}^{\text{22}}$.