Chapter 5, Problem 61E

What number of atoms of nitrogen are present in 1.00 g of each of the compounds in Exercise 51?

Interpretation Introduction

Interpretation: The mass of compound is given. By using the mass, the number of nitrogen atoms present in each of the compound given in exercise 51 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 nitrogen $\left(\text{N}\right)$ atoms in $\text{1}\text{.00}\text{g}$ of ${\text{NH}}_{\text{3}}$.

Explanation of Solution

Given

The mass of ${\text{NH}}_{\text{3}}$ is $\text{1}\text{.00}\text{g}$.

The molar mass of ${\text{NH}}_{\text{3}}$ is $\text{17}\text{.034}\text{g/mol}$.

Formula

The number of moles of ${\text{NH}}_{\text{3}}$ is calculated as,

$\text{Moles}\text{of}{\text{NH}}_{\text{3}}=\frac{\text{Mass}\text{of}{\text{NH}}_{\text{3}}}{\text{Molar}\text{mass}\text{of}{\text{NH}}_{\text{3}}}$

Substitute the values of mass and molar mass of ${\text{NH}}_{\text{3}}$ in above equation.

$\begin{array}{c}\text{Moles}\text{of}{\text{NH}}_{\text{3}}=\frac{\text{Mass}\text{of}{\text{NH}}_{\text{3}}}{\text{Molar}\text{mass}\text{of}{\text{NH}}_{\text{3}}}\\ =\frac{\text{1}\text{.00}\text{g}}{\text{17}\text{.034}\text{g/mol}}\\ =\text{0}\text{.0587}\text{mol}\end{array}$

Since, one mole of ${\text{NH}}_{\text{3}}$ is equal to one mole of nitrogen, Therefore, $\text{0}\text{.0587}\text{mol}$ of ${\text{NH}}_{\text{3}}$ is equal to $\text{0}\text{.0587}\text{mol}$ of nitrogen $\left(\text{N}\right)$.

The number of nitrogen atoms in ${\text{NH}}_{\text{3}}$ is calculated using the formula,

$\text{Number}\text{of}\text{N}\text{atoms}=\text{Moles}\text{of}{\text{NH}}_{\text{3}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}$

Substitute the value of number of moles of ${\text{NH}}_{\text{3}}$ in above equation.

$\begin{array}{c}\text{Number}\text{of}\text{N}\text{atoms}=\text{Moles}\text{of}{\text{NH}}_{\text{3}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\text{0}\text{.0587}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\underset{\_}{3.535\times 10^{22}atoms}\end{array}$

Conclusion

The number of atoms 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 nitrogen atoms present in each of the compound given in exercise 51 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 nitrogen $\left(\text{N}\right)$ atoms in $\text{1}\text{.00}\text{g}$ of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$.

Explanation of Solution

Given

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

The molar mass of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is $\text{32}\text{.052}\text{g/mol}$.

Formula

The number of moles of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is calculated as,

$\text{Moles}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}=\frac{\text{Mass}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}{\text{Molar}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}$

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

$\begin{array}{c}\text{Moles}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}=\frac{\text{Mass}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}{\text{Molar}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\\ =\frac{\text{1}\text{.00}\text{g}}{\text{32}\text{.052}\text{g/mol}}\\ =\text{0}\text{.0312}\text{mol}\end{array}$

Since, one mole of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is equal to two moles of nitrogen, Therefore, $\text{0}\text{.0311}\text{mol}$ of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is equal to,

$\text{2}\times \text{0}\text{.0311}\text{mol}=\text{0}\text{.0623}\text{mol}$ of nitrogen $\left(\text{N}\right)$

The number of nitrogen atoms in ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is calculated using the formula,

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

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

$\begin{array}{c}\text{Number}\text{of}\text{N}\text{atoms}=\text{Moles}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\text{0}\text{.0623}\text{mol}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\underset{\_}{3.758\times 10^{22}atoms}\end{array}$

Conclusion

The number of atoms 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 nitrogen atoms present in each of the compound given in exercise 51 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 nitrogen $\left(\text{N}\right)$ atoms in $\text{1}\text{.00}\text{g}$ of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

Explanation of Solution

Given

The mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is $\text{1}\text{.00}\text{g}$.

The molar mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is $\text{250}\text{.28}\text{g/mol}$.

Formula

The number of moles in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is calculated as,

$\text{Moles}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}=\frac{\text{Mass}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{Molar}\text{mass}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}$

Substitute the values of mass and molar mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ in above equation.

$\begin{array}{c}\text{Moles}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}=\frac{\text{Mass}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{Molar}\text{mass}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}\\ =\frac{\text{1}\text{.00}\text{g}}{\text{250}\text{.28}\text{g/mol}}\\ =\text{3}\text{.99}\times {\text{10}}^{-\text{3}}\text{mol}\end{array}$

Since, one mole of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is equal to two moles of nitrogen, Therefore, $\text{3}\text{.99}\times {\text{10}}^{-\text{3}}\text{mol}$ of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is equal to,

$\text{2}\times \text{3}\text{.99}\times {\text{10}}^{-\text{3}}\text{mol}=\text{7}\text{.98}\times {\text{10}}^{-\text{3}}\text{mol}$ of nitrogen $\left(\text{N}\right)$

The number of nitrogen atoms in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is calculated using the formula,

$\text{Number}\text{of}\text{N}\text{atoms}=\text{Moles}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}$

Substitute the value of number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in above equation.

$\begin{array}{c}\text{Number}\text{of}\text{N}\text{atoms}=\text{Moles}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\text{7}\text{.98}\times {\text{10}}^{-\text{3}}\text{mol}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\underset{\_}{4.81\times 10^{21}atoms}\end{array}$

Conclusion

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