Chapter 7, Problem 34E

Interpretation Introduction

(a)

Interpretation:

The number of atoms in $\text{7}\text{.70}\text{g}$ iodine atoms is to be calculated.

Concept introduction:

Mole is a basic unit used in the International system of units (SI). It is abbreviated as $\text{mol}$. It can be defined as $6.022\times {10}^{23}$ units of some quantity which can be atoms, molecules or ions. The quantity $6.022\times {10}^{23}$ is known as the Avogadro’s number.

Answer to Problem 34E

The number of atoms in $\text{7}\text{.70}\text{g}$ iodine is $\text{3}\text{.65}\times {\text{10}}^{\text{22}}\text{atoms}$.

Explanation of Solution

The molar mass of iodine atom is $\text{126}\text{.9}\text{g}{\text{mol}}^{-1}$.

Therefore, one mole of iodine atom is $\text{126}\text{.9}\text{g}$.

The number of moles in $\text{7}\text{.70}\text{g}$ of iodine is given as shown below.

$\begin{array}{rll}\text{Moles}\text{of}\text{iodine}\text{atom}& =& \frac{\text{1}\text{mole}\text{of}\text{iodine}\text{atom}}{\text{126}\text{.9}\text{g}\text{of}\text{iodine}\text{atom}}\times \text{weight}\text{of}\text{iodine}\text{atom}\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}\text{iodine}\text{atom}}{\text{126}\text{.9}\text{g}\text{of}\text{iodine}\text{atom}}\times \text{7}\text{.70}\text{g}\text{of}\text{iodine}\text{atom}\\ & =& 6.06\times {10}^{-2}\text{moles}\text{of}\text{iodine}\text{atom}\end{array}$

The number of atoms in iodine atom is calculated as shown below.

$\text{Atoms}\text{of}\text{iodine}\text{atom}=\left(\frac{\begin{array}{l}\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{atoms}\text{of}\text{iodine}\text{atom}\times \\ \text{moles}\text{of}\text{iodine}\text{atom}\end{array}}{\text{1}\text{mole}\text{of}\text{iodine}\text{atom}}\right)$

Substitute the number of atoms of iodine atom in above equation as shown below.

$\begin{array}{rll}\text{Atoms}\text{of}\text{iodine}\text{atom}& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{atoms}\text{of}\text{iodine}\text{atom}\times \\ 6.06\times {10}^{-2}\text{moles}\text{of}\text{iodine}\text{atoms}\end{array}}{\text{1}\text{mole}\text{of}\text{iodine}\text{atom}}\right)\\ & =& 3.65\times {\text{10}}^{\text{22}}\text{atoms}\text{of}\text{iodine}\text{atom}\end{array}$

Conclusion

The number of atoms in $\text{7}\text{.70}\text{g}$ iodine is calculated as $3.65\times {\text{10}}^{\text{22}}\text{atoms}$.

Interpretation Introduction

(b)

Interpretation:

The number of molecules in $0.447\text{g}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}$ is to be calculated.

Concept introduction:

Mole is a basic unit used in the International system of units (SI). It is abbreviated as $\text{mol}$. It can be defined as $6.022\times {10}^{23}$ units of some quantity which can be atoms, molecules or ions. The quantity $6.022\times {10}^{23}$ is known as the Avogadro’s number.

Answer to Problem 34E

The number of molecules in $0.447\text{g}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}$ is $2.1\times {\text{10}}^{\text{21}}\text{molecules}$.

Explanation of Solution

The molecular formula is ${\text{C}}_{\text{9}}{\text{H}}_{\text{20}}$.

The molar mass of carbon is $12.01\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $\text{1}\text{.008}\text{g}{\text{mol}}^{-1}$.

The molar mass of ${\text{C}}_{\text{9}}{\text{H}}_{\text{20}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(9\times \text{12}\text{.01}\right)+\left(20\times 1.008\right)\\ & =& 128.25\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the molar mass of ${\text{C}}_{\text{9}}{\text{H}}_{\text{20}}$ is $128.25\text{g}{\text{mol}}^{-1}$.

This means one mole of ${\text{C}}_{\text{9}}{\text{H}}_{\text{20}}$ is $\text{128}\text{.25}\text{g}$.

The number of moles in $0.447\text{g}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}$ is given as shown below.

$\begin{array}{rll}\text{Moles}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}& =& \frac{\text{1}\text{mole}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}}{\text{128}\text{.25}\text{g}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}}\times \text{weight}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}}{\text{128}\text{.25}\text{g}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}}\times \text{0}\text{.447}\text{g}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}\\ & =& 3.48\times {10}^{-3}\text{moles}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}\end{array}$

The number of molecules in ${\text{C}}_{\text{9}}{\text{H}}_{\text{20}}$ is calculated as shown below.

$\text{Molecules}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}=\left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}\text{of}\text{128}\text{.25}\text{g}\times \\ \text{moles}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}\end{array}}{\text{1}\text{mole}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}}\right)$

Substitute the number of moles of ${\text{C}}_{\text{9}}{\text{H}}_{\text{20}}$ in above equation as shown below.

$\begin{array}{rll}\text{Molecules}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}\times \\ 3.48\times {10}^{-3}\text{moles}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}\end{array}}{\text{1}\text{mole}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}}\right)\\ & =& 2.1\times {\text{10}}^{\text{21}}\text{molecules}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}\end{array}$

Conclusion

The number of molecules in $0.447\text{g}{\text{C}}_{\text{9}}{\text{H}}_{\text{20}}$ is calculated as $2.1\times {\text{10}}^{\text{21}}\text{molecules}$.

Interpretation Introduction

(c)

Interpretation:

The number of formula units in $\text{72}\text{.6}\text{g}$ manganese $\left(\text{II}\right)$ carbonate is to be calculated.

Concept introduction:

Mole is a basic unit used in the International system of units (SI). It is abbreviated as $\text{mol}$. It can be defined as $6.022\times {10}^{23}$ units of some quantity which can be atoms, molecules or ions. The quantity $6.022\times {10}^{23}$ is known as the Avogadro’s number.

Answer to Problem 34E

The number of formula units in $\text{72}\text{.6}\text{g}$ manganese $\left(\text{II}\right)$ carbonate is $\text{3}\text{.80}\times {\text{10}}^{\text{23}}\text{formula}\text{units}$.

Explanation of Solution

The molecular formula for manganese $\left(\text{II}\right)$ carbonate is $\text{Mn}\left({\text{CO}}_{\text{3}}\right)$.

The molar mass of manganese is $54.94\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of carbon is $12.01\text{g}{\text{mol}}^{-1}$.

The molar mass of $\text{Mn}\left({\text{CO}}_{\text{3}}\right)$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 54.94+12.01+\left(3\times 16.00\right)\\ & =& \text{114}\text{.95}\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the molar mass of $\text{Mn}\left({\text{CO}}_{\text{3}}\right)$ is $\text{114}\text{.95}\text{g}{\text{mol}}^{-1}$.

This means one mole of $\text{Mn}\left({\text{CO}}_{\text{3}}\right)$ is $\text{114}\text{.95}\text{g}$.

The number of moles in $\text{Mn}\left({\text{CO}}_{\text{3}}\right)$ is given as shown below.

$\begin{array}{rll}\text{Moles}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)& =& \frac{\text{1}\text{mole}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)}{\text{114}\text{.95}\text{g}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)}\times \text{weight}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)}{\text{114}\text{.95}\text{g}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)}\times \text{72}\text{.6}\text{g}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)\\ & =& 0.632\text{moles}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)\end{array}$

The number of formula units in $\text{Mn}\left({\text{CO}}_{\text{3}}\right)$ is calculated as shown below.

$\text{Formula units}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)=\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)}{\text{1}\text{mole}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)}\times \text{moles}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)$ Substitute the number of moles of $\text{Mn}\left({\text{CO}}_{\text{3}}\right)$ in above equation,

$\begin{array}{rll}\text{Formula units}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)\times \\ \text{0}\text{.632}\text{moles}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)\end{array}}{\text{1}\text{mole}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)}\right)\\ & =& 3.80\times {\text{10}}^{\text{23}}\text{formula units}\text{of}\text{Mn}\left({\text{CO}}_{\text{3}}\right)\end{array}$

Conclusion

The number of formula units in $\text{72}\text{.6}\text{g}$ manganese $\left(\text{II}\right)$ carboante is calculated as $3.80\times {\text{10}}^{\text{23}}\text{formula}\text{units}$.