Chapter 3, Problem 36E

How many atoms of nitrogen are present in 5.00 g ofeach of the following?
a. glycine, ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\text{N}$    c. calcium nitrate
b. magnesium nitride    d. dinitrogen tetroxide

Interpretation Introduction

Interpretation: The number of atoms of nitrogen in $5.00\text{g}$ of glycine $\left({\text{C}}_2{\text{H}}_5{\text{O}}_2\text{N}\right)$ is to be calculated.

Concept introduction: In a particular mass corresponding to the substance, the entities specifically as its atoms or as its molecules can be evaluated via mole concept.These respective entities are computed using a specific constant designated as Avogadro’s number and is widely utilized in stoichiometric calculations. Numerical value relative to this constant is $6.023\times {10}^{23}{\text{mol}}^{-1}$ .

Answer to Problem 36E

The number of atoms of nitrogen in $5.00\text{g}$ of glycine $\left({\text{C}}_2{\text{H}}_5{\text{O}}_2\text{N}\right)$ is $4.011\times {10}^{22}\text{atoms}$ .

Explanation of Solution

The molar mass value of glycine $\left({\text{C}}_2{\text{H}}_5{\text{O}}_2\text{N}\right)$ is $75.07\text{g/mol}$ .

The moles of glycine $\left({\text{C}}_2{\text{H}}_5{\text{O}}_2\text{N}\right)$ can be calculated as follows.

  $\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values in above formula.

  $\begin{array}{c}\text{Moles}=\frac{5.00\text{g}}{75.07\text{g/mol}}\\ =0.0666\text{mol}\end{array}$

The glycine molecule ${\text{C}}_2{\text{H}}_5{\text{O}}_2\text{N}$ contain $1$ mole of $\text{N}$ atom. It means that the moles of $\text{N}$ will be same as that of glycine that is $0.0666\text{mol}$ .

The number of nitrogen atoms can be calculated as follows.

  $\text{Number}\text{of}\text{atoms}=\text{Moles}\times \text{Avogadro's}\text{number}$

The value of Avogadro’s number is $6.023\times {10}^{23}{\text{mol}}^{-1}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Number}\text{of}\text{atoms}=0.0666\text{mol}\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ =4.011\times {10}^{22}\text{atoms}\end{array}$

Therefore, the number of nitrogen atoms in glycine are $4.011\times {10}^{22}\text{atoms}$ .

Interpretation Introduction

Interpretation: The number of atoms of nitrogen in $5.00\text{g}$ of magnesium nitride is to be calculated.

Concept introduction:In a particular mass corresponding to the substance, the entities specifically as its atoms or as its molecules can be evaluated via mole concept. These respective entities are computed using a specific constant designated as Avogadro’s number and is widely utilized in stoichiometric calculations. Numerical value relative to this constant is $6.023\times {10}^{23}{\text{mol}}^{-1}$ .

Answer to Problem 36E

The number of atoms of nitrogen in $5.00\text{g}$ of magnesium nitride is $5.96\times {10}^{22}\text{atoms}$ .

Explanation of Solution

The chemical formula corresponding to magnesium nitride is ${\text{Mg}}_3{\text{N}}_2$ .

The molar mass value of ${\text{Mg}}_3{\text{N}}_2$ is $100.9494\text{g/mol}$ .

The moles of ${\text{Mg}}_3{\text{N}}_2$ can be calculated as follows.

  $\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values in above formula.

  $\begin{array}{c}\text{Moles}=\frac{5.00\text{g}}{100.9494\text{g/mol}}\\ =0.04953\text{mol}\end{array}$

The ${\text{Mg}}_3{\text{N}}_2$ molecule contains $2$ moles of $\text{N}$ atoms. The number of moles of nitrogen can be calculated as follows.

  $\begin{array}{c}\text{Moles}\text{of}\text{N}=\text{Moles}\text{of}{\text{Mg}}_3{\text{N}}_2\times \frac{2\text{mol}\text{N}}{1\text{mol}{\text{Mg}}_3{\text{N}}_2}\\ =\left(0.04953\times 2\right)\text{mol}\\ =\text{0}\text{.09906}\text{mol}\end{array}$

The number of nitrogen atoms can be calculated as follows.

  $\text{Number}\text{of}\text{atoms}=\text{Moles}\times \text{Avogadro's}\text{number}$

The value of Avogadro’s number is $6.023\times {10}^{23}{\text{mol}}^{-1}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Number}\text{of}\text{atoms}=0.09906\text{mol}\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ =5.96\times {10}^{22}\text{atoms}\end{array}$

Therefore, the number of nitrogen atoms in $5.00\text{g}$ magnesium nitride are $5.96\times {10}^{22}\text{atoms}$ .

Interpretation Introduction

Interpretation: The number of atoms of nitrogen in $5.00\text{g}$ of calcium nitrate is to be calculated.

Concept introduction:In a particular mass corresponding to the substance, the entities specifically as its atoms or as its molecules can be evaluated via mole concept. These respective entities are computed using a specific constant designated as Avogadro’s number and is widely utilized in stoichiometric calculations. Numerical value relative to this constant is $6.023\times {10}^{23}{\text{mol}}^{-1}$ .

Answer to Problem 36E

The number of atoms of nitrogen in $5.00\text{g}$ of calcium nitrate is $3.67\times {10}^{22}\text{atoms}$ .

Explanation of Solution

The chemical formula corresponding to calcium nitrate is $\text{Ca}{\left({\text{NO}}_3\right)}_2$ .

The molar mass value of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is $164.088\text{g/mol}$ .

The moles of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ can be calculated as follows.

  $\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values in above formula.

  $\begin{array}{c}\text{Moles}=\frac{5.00\text{g}}{164.088\text{g/mol}}\\ =0.03047\text{mol}\end{array}$

The $\text{Ca}{\left({\text{NO}}_3\right)}_2$ molecule contains $2$ moles of $\text{N}$ atoms. The number of moles of nitrogen can be calculated as follows.

  $\begin{array}{c}\text{Moles}\text{of}\text{N}=\text{Moles}\text{of}\text{Ca}{\left({\text{NO}}_3\right)}_2\times \frac{2\text{mol}\text{N}}{1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}\\ =\left(0.03047\times 2\right)\text{mol}\\ =\text{0}\text{.06094}\text{mol}\end{array}$

The number of nitrogen atoms can be calculated as follows.

  $\text{Number}\text{of}\text{atoms}=\text{Moles}\times \text{Avogadro's}\text{number}$

The value of Avogadro’s number is $6.023\times {10}^{23}{\text{mol}}^{-1}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Number}\text{of}\text{atoms}=0.06094\text{mol}\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ =3.67\times {10}^{22}\text{atoms}\end{array}$

Therefore, the number of nitrogen atoms in $5.00\text{g}$ calcium nitrate are $3.67\times {10}^{22}\text{atoms}$ .

Interpretation Introduction

Interpretation: The number of atoms of nitrogen in $5.00\text{g}$ of dinitrogen tetroxide is to be calculated.

Concept introduction:In a particular mass corresponding to the substance, the entities specifically as its atoms or as its molecules can be evaluated via mole concept. These respective entities are computed using a specific constant designated as Avogadro’s number and is widely utilized in stoichiometric calculations. Numerical value relative to this constant is $6.023\times {10}^{23}{\text{mol}}^{-1}$ .

Answer to Problem 36E

The number of atoms of nitrogen in $5.00\text{g}$ of dinitrogen tetroxide is $6.54\times {10}^{22}\text{atoms}$ .

Explanation of Solution

The chemical formula corresponding to dinitrogen tetroxide is ${\text{N}}_2{\text{O}}_4$ .

The molar mass value of ${\text{N}}_2{\text{O}}_4$ is $92.011\text{g/mol}$ .

The moles of ${\text{N}}_2{\text{O}}_4$ can be calculated as follows.

  $\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values in above formula.

  $\begin{array}{c}\text{Moles}=\frac{5.00\text{g}}{92.011\text{g/mol}}\\ =0.05434\text{mol}\end{array}$

The ${\text{N}}_2{\text{O}}_4$ molecule contains $2$ moles of $\text{N}$ atoms. The number of moles of nitrogen can be calculated as follows.

  $\begin{array}{c}\text{Moles}\text{of}\text{N}=\text{Moles}\text{of}{\text{N}}_2{\text{O}}_4\times \frac{2\text{mol}\text{N}}{1\text{mol}{\text{N}}_2{\text{O}}_4}\\ =\left(0.05434\times 2\right)\text{mol}\\ =\text{0}\text{.10868}\text{mol}\end{array}$

The number of nitrogen atoms can be calculated as follows.

  $\text{Number}\text{of}\text{atoms}=\text{Moles}\times \text{Avogadro's}\text{number}$

The value of Avogadro’s number is $6.023\times {10}^{23}{\text{mol}}^{-1}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Number}\text{of}\text{atoms}=0.10868\text{mol}\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ =6.54\times {10}^{22}\text{atoms}\end{array}$

Therefore, the number of nitrogen atoms in $5.00\text{g}$ dinitrogen tetroxide are $6.54\times {10}^{22}\text{atoms}$ .