Chapter 2, Problem 2.138P

Interpretation Introduction

Interpretation:

The mass percentages of $\text{NO}$ in the molecules of nitroglycerin and isoamyl nitrate are to be determined.

Concept introduction:

The molecular mass of a compound is defined as the sum of the atomic masses of atoms of all the elements present in it.

The general formula to calculate the molecular mass of a compound is,

$\text{Molecular mass of the compound}=\left[\begin{array}{l}\Sigma \left(\text{number of atoms of the element}\right)\\ \left(\text{atomic mass of the element}\right)\end{array}\right]$ (1)

The mass percent of an element in a compound is defined as the total concentration of the element in the compound.

The general formula to calculate the total mass of a molecule in a compound is,

$\text{Total mass of a molecule}=\left[\left(\text{total number of molecules}\right)\left(\text{molecular mass }\right)\right]$ (2)

The general formula to calculate the mass percent of a molecule in a compound is,

$\text{Mass percent}=\left(\frac{\text{total mass of the molecule in the compound}}{\text{molecular mass of the compound}}\right)\left(100\right)$ (3)

Answer to Problem 2.138P

The mass percentages of $\text{NO}$ in the molecules of nitroglycerin and isoamyl nitrate are $39.64$ and $22.54$ respectively.

Explanation of Solution

The formula to calculate the molecular mass of $\text{NO}$ is,

$\text{Molecular mass}=\left[\begin{array}{l}\left(1\right)\left(\text{atomic mass of nitrogen}\right)+\\ \left(1\right)\left(\text{atomic mass of oxygen}\right)\end{array}\right]$ (4)

Substitute $14.01\text{ amu}$ for the atomic mass of nitrogen and $16\text{ amu}$ for the atomic mass of oxygen in equation (4).

$\begin{array}{l}\text{Molecular mass}=\left[\left(1\right)\left(14.01\text{ amu}\right)+\left(1\right)\left(16\text{ amu}\right)\right]\\ =14.01\text{ amu}+16\text{ amu}\\ =30.01\text{ amu}\end{array}$

The molecular formula for nitroglycerin is ${\text{C}}_3{\text{H}}_5{\text{N}}_3{\text{O}}_9$.

The formula to calculate the molecular mass of ${\text{C}}_3{\text{H}}_5{\text{N}}_3{\text{O}}_9$ is,

$\text{Molecular mass}=\left[\begin{array}{l}\left(3\right)\left(\text{atomic mass of carbon}\right)+\\ \left(5\right)\left(\text{atomic mass of hydrogen}\right)+\\ \left(3\right)\left(\text{atomic mass of nitrogen}\right)+\\ \left(9\right)\left(\text{atomic mass of oxygen}\right)\end{array}\right]$ (5)

Substitute $12.01\text{ amu}$ for the atomic mass of carbon, $1.008\text{ amu}$ for the atomic mass of hydrogen, $14.01\text{ amu}$ for the atomic mass of nitrogen and $16\text{ amu}$ for the atomic mass of oxygen in equation (5).

$\begin{array}{c}\text{Molecular mass}=\left[\left(3\right)\left(12.01\text{ amu}\right)+\left(5\right)\left(1.008\text{ amu}\right)+\left(3\right)\left(14.01\text{ amu}\right)+\left(9\right)\left(16\text{ amu}\right)\right]\\ =36.03\text{ amu}+5.04\text{ amu}+42.03\text{ amu}+144\text{ amu}\\ =227.1\text{ amu}\end{array}$

The total number of $\text{NO}$ molecules present in one molecule ${\text{C}}_3{\text{H}}_5{\text{N}}_3{\text{O}}_9$ is 3.

The total mass of 3 molecules of $\text{NO}$ in a molecule of ${\text{C}}_3{\text{H}}_5{\text{N}}_3{\text{O}}_9$ is calculated as follows:

$\text{Total mass}=\left[\left(\text{total number of molecules}\right)\left(\text{molecular mass }\right)\right]$ (6)

Substitute 3 for the total number of molecules and $30.01\text{ amu}$ for the molecular mass in equation (6).

$\begin{array}{c}\text{Total mass}=\left(\text{3}\right)\left(30.01\text{ amu}\right)\\ =90.03\text{ amu}\end{array}$

The formula to calculate the mass percent of $\text{NO}$ in a molecule of ${\text{C}}_3{\text{H}}_5{\text{N}}_3{\text{O}}_9$ is,

$\text{Mass percent}=\left(\frac{{\text{total mass of the NO in C}}_3{\text{H}}_5{\text{N}}_3{\text{O}}_9}{{\text{molecular mass of C}}_3{\text{H}}_5{\text{N}}_3{\text{O}}_9}\right)\left(100\right)$ (7)

Substitute $90.03\text{ amu}$ for the total mass of $\text{NO}$ in ${\text{C}}_3{\text{H}}_5{\text{N}}_3{\text{O}}_9$ and $227.1\text{ amu}$ for the molecular mass of ${\text{C}}_3{\text{H}}_5{\text{N}}_3{\text{O}}_9$ in equation (7).

$\begin{array}{c}\text{Mass percent}=\left(\frac{90.03\text{ amu}}{227.1\text{ amu}}\right)\left(100\right)\\ =\left(0.3964\right)\left(100\right)\\ =39.64\end{array}$

The molecular formula for isoamyl nitrate is ${\left({\text{CH}}_3\right)}_2{\text{CHCH}}_2{\text{CH}}_2{\text{ONO}}_2$.

The formula to calculate the molecular mass of ${\left({\text{CH}}_3\right)}_2{\text{CHCH}}_2{\text{CH}}_2{\text{ONO}}_2$ is,

$\text{Molecular mass}=\left[\begin{array}{l}\left(5\right)\left(\text{atomic mass of carbon}\right)+\\ \left(11\right)\left(\text{atomic mass of hydrogen}\right)+\\ \left(1\right)\left(\text{atomic mass of nitrogen}\right)+\\ \left(3\right)\left(\text{atomic mass of oxygen}\right)\end{array}\right]$ (8)

Substitute $12.01\text{ amu}$ for the atomic mass of carbon, $1.008\text{ amu}$ for the atomic mass of hydrogen, $14.01\text{ amu}$ for the atomic mass of nitrogen and $16\text{ amu}$ for the atomic mass of oxygen in equation (5).

$\begin{array}{c}\text{Molecular mass}=\left[\left(5\right)\left(12.01\text{ amu}\right)+\left(11\right)\left(1.008\text{ amu}\right)+\left(1\right)\left(14.01\text{ amu}\right)+\left(3\right)\left(16\text{ amu}\right)\right]\\ =60.05\text{ amu}+11.088\text{ amu}+14.01\text{ amu}+48\text{ amu}\\ =133.15\text{ amu}\end{array}$

The total number of $\text{NO}$ molecules present in one molecule ${\left({\text{CH}}_3\right)}_2{\text{CHCH}}_2{\text{CH}}_2{\text{ONO}}_2$ is 1.

The total mass of 1 molecule of $\text{NO}$ in a molecule of ${\text{C}}_3{\text{H}}_5{\text{N}}_3{\text{O}}_9$ is calculated as follows:

$\text{Total mass}=\left[\left(\text{total number of molecules}\right)\left(\text{molecular mass }\right)\right]$ (9)

Substitute 1 for the total number of molecules and $30.01\text{ amu}$ for the molecular mass in equation (9).

$\begin{array}{c}\text{Total mass}=\left(\text{1}\right)\left(30.01\text{ amu}\right)\\ =30.01\text{ amu}\end{array}$

The formula to calculate the mass percent of $\text{NO}$ in a molecule of ${\left({\text{CH}}_3\right)}_2{\text{CHCH}}_2{\text{CH}}_2{\text{ONO}}_2$ is,

$\text{Mass percent}=\left(\frac{\text{total mass of the NO in }{\left({\text{CH}}_3\right)}_2{\text{CHCH}}_2{\text{CH}}_2{\text{ONO}}_2}{\text{molecular mass of }{\left({\text{CH}}_3\right)}_2{\text{CHCH}}_2{\text{CH}}_2{\text{ONO}}_2}\right)\left(100\right)$ (10)

Substitute $30.01\text{ amu}$$90.03\text{ amu}$ for the total mass of $\text{NO}$ in ${\left({\text{CH}}_3\right)}_2{\text{CHCH}}_2{\text{CH}}_2{\text{ONO}}_2$ and $133.15\text{ amu}$ for the molecular mass of ${\left({\text{CH}}_3\right)}_2{\text{CHCH}}_2{\text{CH}}_2{\text{ONO}}_2$ in equation (10).

$\begin{array}{c}\text{Mass percent}=\left(\frac{30.01\text{ amu}}{133.15\text{ amu}}\right)\left(100\right)\\ =\left(0.2254\right)\left(100\right)\\ =22.54\end{array}$

Conclusion

The mass percentages of $\text{NO}$ in the molecules of nitroglycerin and isoamyl nitrate are $39.64$ and $22.54$ respectively.