Chapter 3, Problem 62E

(a)

Interpretation Introduction

Interpretation: The mass of compound is given. By using the mass, the number of phosphorous atoms present in 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.

To determine: The number of phosphorous $\left(\text{P}\right)$ atoms 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 $\text{219}\text{.866}\text{g/mol}$.

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.

$\begin{array}{c}\text{Moles}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}=\frac{\text{Mass}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}{\text{Molar}\text{mass}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}\\ =\frac{\text{1}\text{.00}\text{g}}{\text{219}\text{.866}\text{g/mol}}\\ =\text{0}\text{.0045}\text{mol}\end{array}$

Since, one mole of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is equal to four moles of phosphorous, Therefore, $\text{0}\text{.0045}\text{mol}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is equal to,

$\text{4}\times \text{0}\text{.0045}\text{mol}=\text{0}\text{.018}\text{mol}$ Of phosphorous $\left(\text{P}\right)$

The number of phosphorous atoms in ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is calculated using the formula,

$\text{Number}\text{of}\text{P}\text{atom}=\text{Moles}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{atoms}$

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

$\begin{array}{c}\text{Number}\text{of}\text{P}\text{atoms}=\text{Moles}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\text{0}\text{.018}\text{mol}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\underset{\_}{1.08\times 10^{22}atoms}\end{array}$

Conclusion

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

(b)

Interpretation Introduction

Interpretation: The mass of compound is given. By using the mass, the number of phosphorous atoms present in 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.

To determine: The number of phosphorous $\left(\text{P}\right)$ atoms in $\text{1}\text{.00}\text{g}$ of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$.

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 ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{310}\text{.172}\text{g/mol}$.

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.

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

Since, one mole of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is equal to two moles of phosphorous, Therefore, $\text{0}\text{.0032}\text{mol}$ of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is equal to,

$\text{2}\times \text{0}\text{.0032}\text{mol}=\text{0}\text{.0064}\text{mol}$ of phosphorous $\left(\text{P}\right)$

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

$\text{Number}\text{of}\text{P}\text{atoms}=\text{Moles}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}$

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

$\begin{array}{c}\text{Number}\text{of}\text{P}\text{atom}=\text{Moles}\text{of}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{atoms}\\ =\text{0}\text{.0064}\text{mol}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\underset{\_}{3.85\times 10^{21}atoms}\end{array}$

Conclusion

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

(c)

Interpretation Introduction

Interpretation: The mass of compound is given. By using the mass, the number of phosphorous atoms present in 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.

To determine: The number of phosphorous $\left(\text{P}\right)$ atoms 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 ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is $\text{141}\text{.955}\text{g/mol}$.

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.

$\begin{array}{c}\text{Moles}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}=\frac{\text{Mass}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}}{\text{Molar}\text{mass}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}}\\ =\frac{\text{1}\text{.00}\text{g}}{\text{141}\text{.955}\text{g/mol}}\\ =\text{0}\text{.0070}\text{mol}\end{array}$

Since, one mole of ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is equal to one mole of phosphorous, Therefore, $\text{0}\text{.0070}\text{mol}$ of ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is equal to $\text{0}\text{.0070}\text{mol}$ of phosphorous $\left(\text{P}\right)$.

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

$\text{Number}\text{of}\text{P}\text{atoms}=\text{Moles}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}$

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

$\begin{array}{c}\text{Number}\text{of}\text{P}\text{atoms}=\text{Moles}\text{of}{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\text{0}\text{.0070}\text{mol}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\underset{\_}{4.215\times 10^{21}atoms}\end{array}$

Conclusion

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