Chapter 7, Problem 33E

Interpretation Introduction

(a)

Interpretation:

The number of molecules in $\text{0}\text{.0023}\text{g}$ iodine 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 33E

The number of molecules in $\text{0}\text{.0023}\text{g}$ iodine is $\text{5}{\text{.5×10}}^{\text{18}}\text{molecules}$.

Explanation of Solution

The molecular formula for iodine molecule is ${\text{I}}_{\text{2}}$.

The molar mass of iodine is $126.9\text{g}{\text{mol}}^{-1}$.

The molar mass of ${\text{I}}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \text{2}\times \text{126}\text{.9}\\ & =& 253.8\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the molar mass of ${\text{I}}_{\text{2}}$ is $253.8\text{g}{\text{mol}}^{-1}$.

Thus, one mole of iodine is $\text{253}\text{.8}\text{g}$.

The number of moles in $\text{0}\text{.0023}\text{g}$ of iodine is given below.

$\begin{array}{rll}\text{Moles}\text{of}\text{iodine}& =& \frac{\text{1}\text{mole}\text{of}\text{iodine}}{\text{253}\text{.8}\text{g}\text{of}\text{iodine}}\times \text{weight}\text{of}\text{iodine}\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}\text{iodine}}{\text{253}\text{.8}\text{g}\text{of}\text{iodine}}\times \text{0}\text{.0023}\text{g}\text{of}\text{iodine}\\ & =& 9.062\times {10}^{-6}\text{moles}\text{of}\text{iodine}\end{array}$

The number of molecules in iodine is calculated below.

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

Substitute the number of moles of iodine in the above equation.

$\begin{array}{rll}\text{Molecules}\text{of}\text{iodine}& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}\text{of}\text{iodine}\times \\ 9.062\times {10}^{-6}\text{moles}\text{of}\text{iodine}\end{array}}{\text{1}\text{mole}\text{of}\text{iodine}}\right)\\ & =& 5.5{\text{×10}}^{\text{18}}\text{molecules}\text{of}\text{iodine}\end{array}$

Conclusion

The number of molecules in $\text{0}\text{.0023}\text{g}$ iodine is $\text{5}{\text{.5×10}}^{\text{18}}\text{molecules}$.

Interpretation Introduction

(b)

Interpretation:

The number of molecules in $\text{114}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}$ 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 33E

The number of molecules in $\text{114}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}$ are $1.1\times {\text{10}}^{\text{24}}\text{molecules}$.

Explanation of Solution

The molecular formula is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}$.

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

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

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

The molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times \text{12}\text{.01}\right)+\left(6\times 1.008\right)+\left(2\times 16.00\right)\\ & =& 62.068\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}$ is $62.068\text{g}{\text{mol}}^{-1}$.

Thus one mole of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}$ is $\text{62}\text{.068}\text{g}$.

The number of moles in $\text{114}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}$ is given below.

$\begin{array}{rll}\text{Moles}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}& =& \frac{\text{1}\text{mole}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}}{\text{62}\text{.068}\text{g}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}}\times \text{weight}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}}{\text{62}\text{.068}\text{g}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}}\times \text{114}\text{g}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\\ & =& 1.84\text{moles}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\end{array}$

The number of molecules in ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}$ is calculated below.

$\text{Molecules}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}=\left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{moleculesof}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\text{×}\\ \text{moles}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\end{array}}{\text{1}\text{mole}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}}\right)$

Substitute the number of moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}$ in above equation.

$\begin{array}{rll}\text{Molecules}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\times \\ 1.84\text{moles}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\end{array}}{\text{1}\text{mole}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}}\right)\\ & =& 1.1\times {\text{10}}^{\text{24}}\text{molecules}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\end{array}$

Conclusion

The number of molecules in $\text{114}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}$ is $1.1\times {\text{10}}^{\text{24}}\text{molecules}$.

Interpretation Introduction

(c)

Interpretation:

The number of formula units in $\text{9}\text{.81}\text{g}$ chromium $\left(\text{III}\right)$ 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 33E

The number of formula units in $\text{9}\text{.81}\text{g}$ chromium $\left(\text{III}\right)$ sulfate is $1.51\times {\text{10}}^{\text{22}}\text{formula units}$.

Explanation of Solution

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

The molar mass of chromium is $52.00\text{g}{\text{mol}}^{-1}$.

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

The molar mass of sulfur is $\text{32}\text{.06}\text{g}{\text{mol}}^{-1}$.

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

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times 52.00\right)+\left(3\times 32.06\right)+\left(12\times 16.00\right)\\ & =& 392.18\text{g}{\text{mol}}^{-1}\end{array}$

The formula mass of $\text{9}\text{.81}\text{g}$ chromium $\left(\text{III}\right)$ sulfate, ${\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is $392.18\text{g}{\text{mol}}^{-1}$.

Thus, one mole of ${\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is $392.18\text{g}$.

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

$\begin{array}{rll}\text{Moles}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}& =& \frac{\text{1}\text{mole}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}{392.18\text{g}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}\times \text{weight}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}{392.18\text{g}\text{g}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}\times \text{9}\text{.81}\text{g}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\\ & =& 0.025\text{moles}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\end{array}$

The number of formula units in ${\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is calculated below.

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

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

$\begin{array}{rll}\text{Formula units}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\times \\ 0.025\text{moles}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\end{array}}{\text{1}\text{mole}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}}\right)\\ & =& 1.51\times {\text{10}}^{\text{22}}\text{formula units}\text{of}{\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\end{array}$

Conclusion

The number of formula units in $\text{9}\text{.81}\text{g}$ chromium $\left(\text{III}\right)$ sulfate is $1.51\times {\text{10}}^{\text{22}}\text{formula units}$.