Chapter 7, Problem 31E

Interpretation Introduction

(a)

Interpretation:

The number of formula units in $\text{29}\text{.6}\text{g}$ of lithium nitrate is to be calculated.

Concept introduction:

A mole is a basic unit that is 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 called the Avogadro’s number.

Answer to Problem 31E

The number of formula units in $\text{29}\text{.6}\text{g}$ of lithium nitrate is $\text{2}{\text{.58×10}}^{\text{23}}\text{formula}\text{units}$.

Explanation of Solution

The molecular formula for lithium nitrate is ${\text{LiNO}}_{\text{3}}$.

The molar mass of nitrogen is $14.01\text{g}{\text{mol}}^{-1}$.

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

The molar mass of lithium is $\text{6}\text{.94}\text{g}{\text{mol}}^{-1}$.

The molar mass of ${\text{LiNO}}_{\text{3}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 6.94+14.01+\left(3\times 16.00\right)\\ & =& 68.95\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the molar mass of ${\text{LiNO}}_{\text{3}}$ is $68.95\text{g}{\text{mol}}^{-1}$.

Thus, one mole of ${\text{LiNO}}_{\text{3}}$ is $\text{68}\text{.95}\text{g}$.

The number of moles in $\text{29}\text{.6}\text{g}$ of ${\text{LiNO}}_{\text{3}}$ is given below.

$\begin{array}{rll}\text{Moles}\text{of}{\text{LiNO}}_{\text{3}}& =& \frac{\text{1}\text{mole}\text{of}{\text{LiNO}}_{\text{3}}}{\text{68}\text{.95}\text{g}\text{of}{\text{LiNO}}_{\text{3}}}\times \text{weight}\text{of}{\text{LiNO}}_{\text{3}}\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}{\text{LiNO}}_{\text{3}}}{\text{68}\text{.95}\text{g}\text{of}{\text{LiNO}}_{\text{3}}}\times \text{29}\text{.6}\text{g}\text{of}{\text{LiNO}}_{\text{3}}\\ & =& 0.429\text{moles}\text{of}{\text{LiNO}}_{\text{3}}\end{array}$

The number of formula units in ${\text{LiNO}}_{\text{3}}$ is calculated as,

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

Substitute the number of moles of ${\text{LiNO}}_{\text{3}}$ in above equation,

$\begin{array}{rll}\text{Formula units}\text{of}{\text{LiNO}}_{\text{3}}& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}{\text{LiNO}}_{\text{3}}\\ \times \text{0}\text{.429}\text{moles}\text{of}{\text{LiNO}}_{\text{3}}\end{array}}{\text{1}\text{mole}\text{of}{\text{LiNO}}_{\text{3}}}\right)\\ & =& 2.58\times {\text{10}}^{\text{23}}\text{formula units}\text{of}{\text{LiNO}}_{\text{3}}\end{array}$

Conclusion

The number of formula units in $\text{29}\text{.6}\text{g}$ of lithium nitrate is $\text{2}{\text{.58×10}}^{\text{23}}\text{formula}\text{units}$.

Interpretation Introduction

(b)

Interpretation:

The number of formula units in $\text{0}\text{.151}\text{g}$ of lithium sulfide is to be calculated.

Concept introduction:

A mole is a basic unit that is 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 called the Avogadro’s number.

Answer to Problem 31E

The number of formula units in $\text{0}\text{.151}\text{g}$ of lithium sulfide is $1.98\times {\text{10}}^{\text{21}}\text{formula}\text{units}$.

Explanation of Solution

The molecular formula for lithium sulfide is ${\text{Li}}_2\text{S}$.

The molar mass of lithium is $\text{6}\text{.94}\text{g}{\text{mol}}^{-1}$.

The molar mass of sulfur is $32.06\text{g}{\text{mol}}^{-1}$

The molar mass of ${\text{Li}}_2\text{S}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times 6.94\right)+32.06\\ & =& 45.94\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the molar mass of ${\text{Li}}_2\text{S}$ is $45.94\text{g}{\text{mol}}^{-1}$.

Thus one mole of ${\text{Li}}_2\text{S}$ is $\text{45}\text{.94}\text{g}$.

The number of moles in $\text{0}\text{.151}\text{g}$ of ${\text{Li}}_2\text{S}$ is given below.

$\begin{array}{rll}\text{Moles}\text{of}{\text{Li}}_2\text{S}& =& \frac{\text{1}\text{mole}\text{of}{\text{Li}}_2\text{S}}{\text{45}\text{.94}\text{g}\text{of}{\text{Li}}_2\text{S}}\times \text{weight}\text{of}{\text{Li}}_2\text{S}\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}{\text{Li}}_2\text{S}}{\text{45}\text{.94}\text{g}\text{of}{\text{Li}}_2\text{S}}\times \text{0}\text{.151}\text{g}\text{of}{\text{Li}}_2\text{S}\\ & =& 0.00328\text{moles}\text{of}{\text{Li}}_2\text{S}\end{array}$

The number of formula units in ${\text{Li}}_2\text{S}$ is calculated as,

$\text{Formula units}\text{of}{\text{Li}}_2\text{S}=\left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}{\text{Li}}_2\text{S}\times \\ \text{moles}\text{of}{\text{Li}}_2\text{S}\end{array}}{\text{1}\text{mole}\text{of}{\text{Li}}_2\text{S}}\right)$

Substitute the number of moles of ${\text{Li}}_2\text{S}$ in above equation,

$\begin{array}{rll}\text{Formula units}\text{of}{\text{Li}}_2\text{S}& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}{\text{Li}}_2\text{S}\times \\ 0.00328\text{moles}\text{of}{\text{LiNO}}_{\text{3}}\end{array}}{\text{1}\text{mole}\text{of}{\text{Li}}_2\text{S}}\right)\\ & =& 1.98\times {\text{10}}^{\text{21}}\text{formula units}\text{of}{\text{Li}}_2\text{S}\end{array}$

Conclusion

The number of formula units in $\text{0}\text{.151}\text{g}$ of lithium sulfide is $1.98\times {\text{10}}^{\text{21}}\text{formula}\text{units}$.

Interpretation Introduction

(c)

Interpretation:

The number of formula units in $\text{457}\text{g}$ of iron $\left(\text{III}\right)$ sulfate is to be calculated.

Concept introduction:

A mole is a basic unit that is 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 called the Avogadro’s number.

Answer to Problem 31E

The number of formula units in $\text{457}\text{g}$ of iron $\left(\text{III}\right)$ sulphate is $\text{6}{\text{.88×10}}^{\text{23}}\text{formula}\text{units}$.

Explanation of Solution

The molecular formula for iron $\left(\text{III}\right)$ sulfate is ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$.

The molar mass of iron is $\text{55}\text{.85}\text{g}{\text{mol}}^{-1}$.

The molar mass of sulfur is $32.06\text{g}{\text{mol}}^{-1}$

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

The molar mass of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times \text{55}\text{.85}\right)+\left(3\times 32.06\right)+\left(12\times 16\right)\\ & =& 399.88\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the molar mass of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is $399.88\text{g}{\text{mol}}^{-1}$.

Thus one mole of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is $399.88\text{g}{\text{mol}}^{-1}$.

The number of moles in $\text{457}\text{g}$ of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is given below.

$\begin{array}{rll}\text{Moles}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}& =& \left(\frac{\text{1}\text{mole}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\times \text{weight}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\text{in}\text{g}}{\text{399}\text{.88g}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}\right)\\ & =& \left(\frac{\text{1}\text{mole}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\text{×457}\text{g}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}{\text{399}\text{.88}\text{g}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}\right)\\ & =& 1.1428\text{moles}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\end{array}$

The number of formula units in ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is calculated as,

$\text{Formula units}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}=\left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\times \\ \text{moles}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\end{array}}{\text{1}\text{mole}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}\right)$

Substitute the number of moles of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ in above equation,

$\begin{array}{rll}\text{Formula units}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\text{×}\\ \text{1}\text{.1428}\text{moles}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\end{array}}{\text{1}\text{mole}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}\right)\\ & =& 6.88{\text{×10}}^{\text{23}}\text{formula units}\text{of}{\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\end{array}$

Conclusion

The number of formula units in $\text{457}\text{g}$ of iron $\left(\text{III}\right)$ sulphate is $6.88{\text{×10}}^{\text{23}}\text{formula}\text{units}$.