Chapter 3, Problem 60E

What number of molecules (or formula units) are present in 1.00 g of each of the compounds in Exercise 52?

Interpretation Introduction

Interpretation: The mass of compound is given. By using the mass, the number of molecules present of each of the compound given in exercise 52 is to be determined.

Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.

Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

The amount of substance containing $\text{12}\text{g}$ of pure carbon is called a mole. One mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ molecules. The number of molecules in one mole is also called Avogadro’s number.

Hence,

$\begin{array}{l}\left(\text{6}\text{.022}\times {\text{10}}^{\text{23}} \text{atoms}\right)\left(\frac{\text{12}\text{u}}{\text{1}\text{atom}}\right)=\text{12}\text{g}\\ \Rightarrow \text{1}\text{u}=\frac{\text{1}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}} }\text{g}\end{array}$

To determine: The number of molecules in $\text{1}\text{.00}\text{g}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$.

Explanation of Solution

Given

The mass of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is $\text{1}\text{.00}\text{g}$.

The molar mass of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is,

$\left(\text{4}\times \text{30}\text{.973}+6\times 15.999\right)\text{g}/\text{mol}=\text{219}\text{.886}\text{g}/\text{mol}$

Formula

The number of moles in ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is calculated as,

$\text{Moles}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}=\frac{\text{Mass}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}{\text{Molar}\text{mass}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}$

Substitute the values of mass and molar mass of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ in above equation.

$\begin{array}{c}\text{Moles}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}=\frac{\text{Mass}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}{\text{Molar}\text{mass}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}\\ =\frac{\text{1}\text{.00}\text{g}}{\text{219}\text{.886}\text{g}/\text{mol}}\\ =\text{0}\text{.0045}\text{mol}\end{array}$

The number of molecules in $\text{1}\text{.00}\text{g}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is calculated using the formula,

$\text{Molecules}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}=\text{Moles}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}\right)$

Substitute the value of number of moles in ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ in above equation.

$\begin{array}{c}\text{Molecules}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}=\text{Moles}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}\right)\\ =\text{0}\text{.0045}\text{mol}\times \left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}\right)\\ =\underset{\_}{2.709\times 10^{21}molecules}\end{array}$

Conclusion

The number of molecules is calculated by multiplying the number of moles with Avogadro’s number.

Interpretation Introduction

Interpretation: The mass of compound is given. By using the mass, the number of molecules present of each of the compound given in exercise 52 is to be determined.

Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.

Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

The amount of substance containing $\text{12}\text{g}$ of pure carbon is called a mole. One mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ molecules. The number of molecules in one mole is also called Avogadro’s number.

Hence,

$\begin{array}{l}\left(\text{6}\text{.022}\times {\text{10}}^{\text{23}} \text{atoms}\right)\left(\frac{\text{12}\text{u}}{\text{1}\text{atom}}\right)=\text{12}\text{g}\\ \Rightarrow \text{1}\text{u}=\frac{\text{1}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}} }\text{g}\end{array}$

To determine: The number of molecules in $\text{1}\text{.00}\text{g}$ of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_2$.

Explanation of Solution

Given

The mass of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{1}\text{.00}\text{g}$.

The molar mass of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_2$ is,

$\left(3\times 40.078+2\times 30.973+8\times 15.999\right)\text{g/mol}=310.172\text{g/mol}$

Formula

The number of moles in ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is calculated as,

$\text{Moles}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}=\frac{\text{Mass}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}{\text{Molar}\text{mass}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}$

Substitute the values of mass and molar mass of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ in above equation.

$\begin{array}{c}\text{Moles}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}=\frac{\text{Mass}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}{\text{Molar}\text{mass}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}\\ =\frac{\text{1}\text{.00}\text{g}}{\text{310}\text{.172}\text{g}/\text{mol}}\\ =\text{0}\text{.0032}\text{mol}\end{array}$

The number of molecules in of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is calculated using the formula,

$\text{Molecules}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}=\text{Moles}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\times \left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}\right)$

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

$\begin{array}{c}\text{Molecules}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}=\text{Moles}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\times \left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}\right)\\ =\text{0}\text{.0032}\text{mol}\times \left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}\right)\\ =\underset{\_}{1.927\times 10^{21}molecules}\end{array}$

Conclusion

The number of molecules is calculated by multiplying the number of moles with Avogadro’s number.

Interpretation Introduction

Interpretation: The mass of compound is given. By using the mass, the number of molecules present of each of the compound given in exercise 52 is to be determined.

Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.

Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

The amount of substance containing $\text{12}\text{g}$ of pure carbon is called a mole. One mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ molecules. The number of molecules in one mole is also called Avogadro’s number.

Hence,

$\begin{array}{l}\left(\text{6}\text{.022}\times {\text{10}}^{\text{23}} \text{atoms}\right)\left(\frac{\text{12}\text{u}}{\text{1}\text{atom}}\right)=\text{12}\text{g}\\ \Rightarrow \text{1}\text{u}=\frac{\text{1}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}} }\text{g}\end{array}$

To determine: The number of molecules in $\text{1}\text{.00}\text{g}$ of ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$.

Explanation of Solution

Given

The mass of ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is $\text{1}\text{.00}\text{g}$.

The molar mass of ${\text{Na}}_2{\text{HPO}}_4$ is,

$\left(\text{2}\times \text{22}\text{.989}+\text{1}\text{.0079}+\text{30}\text{.973}+4\times 15.999\right)\text{g}/\text{mol}=\text{141}\text{.995}\text{g}/\text{mol}$

Formula

The number of moles in ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is calculated as,

$\text{Moles}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}=\frac{\text{Mass}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}}{\text{Molar}\text{mass}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}}$

Substitute the values of mass and molar mass of ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ in above equation.

$\begin{array}{c}\text{Moles}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}=\frac{\text{Mass}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}}{\text{Molar}\text{mass}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}}\\ =\frac{\text{1}\text{.00}\text{g}}{\text{141}\text{.995}\text{g}/\text{mol}}\\ =\text{0}\text{.0070}\text{mol}\end{array}$

The number of molecules in ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is calculated using the formula,

$\text{Molecules}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}=\text{Moles}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\times \left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}}\right)$

Substitute the value of number of moles in ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ in above equation.

$\begin{array}{c}\text{Molecules}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}=\text{Moles}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\times \left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}}\right)\\ =\text{0}\text{.0070}\text{mol}\times \left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}}\right)\\ =\underset{\_}{4.215\times 10^{21}molecules}\end{array}$

Conclusion

The number of molecules is calculated by multiplying the number of moles with Avogadro’s number.