Chapter 5, Problem 72E

Dimethylnitrosamine, (CH₃)₂N₂O , is a carcinogenic (cancer-causing) substance that may be formed in foods, beverages, or gastric juices from the reaction of nitrite ion (used as a food preservative) with other substances.

a. What is the molar mass of dimethylnitrosamine?

b. How many moles of (CH3h N20 molecules are present in 250 mg dimethylnitrosamine?

c. What is the mass of 0.050 mole of dimethylnitrosamine?

d. How many atoms of hydrogen are in 1.0 mole of dimethylnitrosamine?

e. What is the mass of 1.0 × 10⁶ molecules of dimethylnitrosamine?

f. What is the mass in grams of one molecule of dimethylnitrosamine?

Interpretation Introduction

Interpretation: The answers are to be given for the given options.

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.

Explanation of Solution

To determine: The molar mass of Dimethylnitrosamine $\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\right]$.

The atomic weight of carbon $\left(\text{C}\right)$ is $\text{12}\text{.01}\text{g}/\text{mol}$.

The atomic weight of oxygen $\left(\text{O}\right)$ is $\text{16}\text{.00}\text{g}/\text{mol}$.

The atomic weight of hydrogen $\left(\text{H}\right)$ is $\text{1}\text{.008}\text{g}/\text{mol}$.

The atomic weight of nitrogen $\left(\text{N}\right)$ is $\text{14}\text{.01}\text{g}/\text{mol}$.

In a compound ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$, two carbon, one oxygen, two nitrogen and six hydrogen atoms are present. Hence, molar mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is calculated as,

$\left[\begin{array}{l}\text{6}\left(\text{1}\text{.008}\text{g}/\text{mol}\right)+\text{2}\left(\text{12}\text{.01}\text{g}/\text{mol}\right)\\ +\left(\text{16}\text{.00}\text{g}/\text{mol}\right)+\text{2}\left(\text{14}\text{.01}\text{g/mol}\right)\end{array}\right]=\underset{\_}{74.09g/mol}$

The molar mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\underset{\_}{74.09g/mol}$.

Conclusion

Molar mass is defined as mass of $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ atoms, molecules of that substance.

Interpretation Introduction

Interpretation: The answers are to be given for the given options.

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 moles in $\text{250}\text{mg}$ of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$.

Explanation of Solution

Given

The mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{3}}{\text{O}}_{\text{2}}$ is $\text{250}\text{mg}$.

The molar mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\text{74}\text{.09}\text{g/mol}$.

The conversion of milligram $\left(\text{mg}\right)$ into gram $\left(\text{g}\right)$ is done as,

$\text{1}\text{mg}={\text{10}}^{-\text{3}}\text{g}$

Hence, the conversion of $\text{250}\text{mg}$ into gram is,

$\begin{array}{c}\text{250}\text{mg}=\left(\text{250}\times {\text{10}}^{-\text{3}}\right)\text{g}\\ =\text{0}\text{.25}\text{g}\end{array}$

Formula

The number of moles in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is calculated as,

$\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\frac{\text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}}{\text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}}$

Substitute the values of mass and molar mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ in above equation,

$\begin{array}{c}\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\frac{\text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}}{\text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}}\\ =\frac{\text{0}\text{.25}\text{g}}{\text{74}\text{.09}\text{g/mol}}\\ =\underset{\_}{3.37\times {10}^{-3}mol}\end{array}$

The number of moles in $\text{250}\text{mg}$ of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\underset{\_}{3.37\times 10^{-3}mol}$.

Conclusion

Number of moles is calculated by dividing the given mass with the molar mass of that compound.

Interpretation Introduction

Interpretation: The answers are to be given for the given options.

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 mass of $\text{0}\text{.050}\text{mol}$ ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$.

Explanation of Solution

Given

The moles of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\text{0}\text{.050}\text{mol}$.

The molar mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\text{74}\text{.09}\text{g/mol}$.

Formula

The mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is calculated as,

$\begin{array}{l}\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\frac{\text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}}{\text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}}\\ \text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\times \text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\end{array}$

Substitute the values of moles and molar mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ in above equation.

$\begin{array}{c}\text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\times \text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\\ =\text{74}\text{.09}\text{g/mol}\times \text{0}\text{.050}\text{mol}\\ =\underset{\_}{3.7g}\end{array}$

The mass of $\text{0}\text{.050}\text{mol}$ ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\underset{\_}{3.7g}$.

Conclusion

Mass of molecule is calculated by multiplying the number of moles and molar mass of that molecule.

Interpretation Introduction

Interpretation: The answers are to be given for the given options.

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 hydrogen $\left(\text{H}\right)$ atoms in $\text{1}\text{.00}\text{mol}$ of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$.

Explanation of Solution

Given

The moles of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\text{1}\text{.00}\text{mol}$.

Since, one mole of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is equal to six moles of hydrogen. Therefore, $\text{1}\text{.00}\text{mol}$ of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is equal to

$\text{6}\times \text{1}\text{.00}\text{mol}=\text{6}\text{.00}\text{mol}$ of hydrogen $\left(\text{H}\right)$.

Formula

The number of hydrogen atoms in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is calculated using the formula,

$\text{Number}\text{of}\text{H}\text{atom}=\text{Moles}\text{of}\text{H}\text{atom}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}$

Substitute the value of number of moles of $\text{H}$ atom in above equation,

$\begin{array}{c}\text{Number}\text{of}\text{H}\text{atom}=\text{Moles}\text{of}\text{H}\text{atom}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\text{6}\text{.00}\text{mol}\times \text{6}\text{.022}\times {\text{10}}^{\text{23}}\\ =\underset{\_}{3.6\times 10^{24}atoms}\end{array}$

The number of hydrogen $\left(\text{H}\right)$ atoms in $\text{1}\text{.00}\text{mol}$ of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\underset{\_}{3.6\times 10^{24}atoms}$.

Conclusion

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

Interpretation Introduction

Interpretation: The answers are to be given for the given options.

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 mass of exactly $\text{1}\text{.0}\times {\text{10}}^{\text{6}}\text{molecules}$ of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$.

Explanation of Solution

Given

The total molecules in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\text{1}\text{.0}\times {\text{10}}^{\text{6}}\text{molecules}$.

The molar mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\text{74}\text{.09}\text{g/mol}$.

Since one mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ molecules. Therefore, number of moles in $\text{1}\text{.0}\times {\text{10}}^{\text{6}}\text{molecules}$ is,

$\frac{\text{1}\text{.0}\times {\text{10}}^{\text{6}}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}}}\text{mol}=\text{1}\text{.6}\times {\text{10}}^{-\text{18}}\text{mol}$

Formula

The mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is calculated as,

$\begin{array}{l}\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\frac{\text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}}{\text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}}\\ \text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\times \text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\end{array}$

Substitute the values of moles and molar mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ in above equation.

$\begin{array}{c}\text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\times \text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\\ =\text{74}\text{.09}\times \text{1}\text{.6}\times {\text{10}}^{-\text{18}}\text{mol}\\ =\underset{\_}{1.2\times 10^{-16}g}\end{array}$

The mass of exactly $\text{1}\text{.0}\times {\text{10}}^{\text{6}}\text{molecules}$ of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\underset{\_}{1.2\times 10^{-16}g}$.

Conclusion

Mass of molecule is calculated by multiplying the number of moles and molar mass of that molecule.

Interpretation Introduction

Interpretation: The answers are to be given for the given options.

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 mass of exactly $\text{1}\text{.0}\text{molecules}$ of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$.

Explanation of Solution

Given

The total molecules in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\text{1}\text{.0}\text{molecules}$.

The molar mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\text{74}\text{.09}\text{g/mol}$.

Since one mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ molecules. Therefore, number of moles in $\text{1}\text{.0}\text{molecules}$ is,

$\frac{\text{1}\text{.0}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}}}\text{mol}=\text{1}\text{.6}\times {\text{10}}^{\text{23}}\text{mol}$

Formula

The mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is calculated as,

$\begin{array}{l}\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\frac{\text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}}{\text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}}\\ \text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\times \text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\end{array}$

Substitute the values of moles and molar mass of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ in above equation.

$\begin{array}{c}\text{Mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}=\text{Moles}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\times \text{Molar}\text{mass}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}\\ =\text{74}\text{.09}\times \text{1}\text{.6}\times {\text{10}}^{\text{23}}\text{mol}\\ =\underset{\_}{1.2\times 10^{-22}g}\end{array}$

The mass of exactly $\text{1}\text{.0}\text{molecules}$ of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{N}}_{\text{2}}\text{O}$ is $\underset{\_}{1.2\times 10^{-22}g}$.

Conclusion

Mass of molecule is calculated by multiplying the number of moles and molar mass of that molecule.