Chapter 2, Problem 2.134P

(a)

Interpretation Introduction

Interpretation:

The different molecular masses that are possible for the compound ${\text{N}}_2\text{O}$ are to be determined.

Concept introduction:

Isotopes are atoms with the same atomic number but a different mass number. The difference in the mass numbers of isotopes arises due to the difference in the number of neutrons in them. The mass number of an atom is the sum of the protons and neutrons in them. The number of protons of isotopes is the same but the neutron number varies. Due to the same number of protons in the isotopes, they occupy the same position in the periodic table.

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

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

Answer to Problem 2.134P

The different molecular masses in $\text{amu}$ possible for the compound ${\text{N}}_2\text{O}$ are 48, 46, 44, 47 and 45.

Explanation of Solution

The isotopes of nitrogen and oxygen are ${}^{14}\text{N}$, ${}^{15}\text{N}$ and ${}^{16}\text{O}$, ${}^{18}\text{O}$.

The different molecules for the compound ${\text{N}}_2\text{O}$ possible with different isotopic compositions are ${}^{15}{\text{N}}_2{}^{18}\text{O}$, ${}^{15}{\text{N}}_2{}^{16}\text{O}$, ${}^{14}{\text{N}}_2{}^{18}\text{O}$, ${}^{14}{\text{N}}_2{}^{16}\text{O}$, ${}^{15}{\text{N}}^{14}{\text{N}}^{18}\text{O}$ and ${}^{15}{\text{N}}^{14}{\text{N}}^{16}\text{O}$.

The formula to calculate the molecular mass of ${}^{15}{\text{N}}_2{}^{18}\text{O}$ is,

$\text{Molecular mass}=\left(2\right)\left({\text{atomic mass of }}^{15}\text{N}\right)+\left(1\right)\left({\text{atomic mass of }}^{18}\text{O}\right)$ (2)

Substitute $15\text{ amu}$ for the atomic mass of ${}^{15}\text{N}$ and $18\text{ amu}$ for the atomic mass of ${}^{18}\text{O}$ in equation (2).

$\begin{array}{c}\text{Molecular mass}=\left(2\right)\left(15\text{ amu}\right)+\left(1\right)\left(18\text{ amu}\right)\\ =30\text{ amu}+18\text{ amu}\\ \text{=4}8\text{ amu}\end{array}$

The formula to calculate the molecular mass of ${}^{15}{\text{N}}_2{}^{16}\text{O}$ is,

$\text{Molecular mass}=\left(2\right)\left({\text{atomic mass of }}^{15}\text{N}\right)+\left(1\right)\left({\text{atomic mass of }}^{16}\text{O}\right)$ (3)

Substitute $15\text{ amu}$ for the atomic mass of ${}^{15}\text{N}$ and $16\text{ amu}$ for the atomic mass of ${}^{16}\text{O}$ in equation (3).

$\begin{array}{c}\text{Molecular mass}=\left(2\right)\left(15\text{ amu}\right)+\left(1\right)\left(16\text{ amu}\right)\\ =30\text{ amu}+16\text{ amu}\\ \text{=4}6\text{ amu}\end{array}$

The formula to calculate the molecular mass of ${}^{14}{\text{N}}_2{}^{18}\text{O}$ is,

$\text{Molecular mass}=\left(2\right)\left({\text{atomic mass of }}^{14}\text{N}\right)+\left(1\right)\left({\text{atomic mass of }}^{18}\text{O}\right)$ (4)

Substitute $14\text{ amu}$ for the atomic mass of ${}^{14}\text{N}$ and $18\text{ amu}$ for the atomic mass of ${}^{18}\text{O}$ in equation (4).

$\begin{array}{c}\text{Molecular mass}=\left(2\right)\left(14\text{ amu}\right)+\left(1\right)\left(18\text{ amu}\right)\\ =28\text{ amu}+18\text{ amu}\\ \text{=4}6\text{ amu}\end{array}$

The formula to calculate the molecular mass of ${}^{14}{\text{N}}_2{}^{16}\text{O}$ is,

$\text{Molecular mass}=\left(2\right)\left({\text{atomic mass of }}^{14}\text{N}\right)+\left(1\right)\left({\text{atomic mass of }}^{16}\text{O}\right)$ (5)

Substitute $14\text{ amu}$ for the atomic mass of ${}^{14}\text{N}$ and $16\text{ amu}$ for the atomic mass of ${}^{16}\text{O}$ in equation (5).

$\begin{array}{c}\text{Molecular mass}=\left(2\right)\left(14\text{ amu}\right)+\left(1\right)\left(16\text{ amu}\right)\\ =28\text{ amu}+16\text{ amu}\\ \text{=4}4\text{ amu}\end{array}$

The formula to calculate the molecular mass of ${}^{15}{\text{N}}^{14}{\text{N}}^{18}\text{O}$ is,

$\text{Molecular mass}=\left[\begin{array}{l}\left(1\right)\left({\text{atomic mass of }}^{15}\text{N}\right)+\left(1\right)\left({\text{atomic mass of }}^{14}\text{N}\right)+\\ \left(1\right)\left({\text{atomic mass of }}^{18}\text{O}\right)\end{array}\right]$ (6)

Substitute $14\text{ amu}$ for the atomic mass of ${}^{14}\text{N}$, $15\text{ amu}$ for the atomic mass of ${}^{15}\text{N}$ and $18\text{ amu}$ for the atomic mass of ${}^{18}\text{O}$ in equation (6).

$\begin{array}{c}\text{Molecular mass}=\left[\left(1\right)\left(15\text{ amu}\right)+\left(1\right)\left(14\text{ amu}\right)+\left(1\right)\left(18\text{ amu}\right)\right]\\ =15\text{ amu}+14\text{ amu}+18\text{ amu}\\ =47\text{ amu}\end{array}$

The formula to calculate the molecular mass of ${}^{15}{\text{N}}^{14}{\text{N}}^{16}\text{O}$ is,

$\text{Molecular mass}=\left[\begin{array}{l}\left(1\right)\left({\text{atomic mass of }}^{15}\text{N}\right)+\left(1\right)\left({\text{atomic mass of }}^{14}\text{N}\right)+\\ \left(1\right)\left({\text{atomic mass of }}^{16}\text{O}\right)\end{array}\right]$ (7)

Substitute $14\text{ amu}$ for the atomic mass of ${}^{14}\text{N}$, $15\text{ amu}$ for the atomic mass of ${}^{15}\text{N}$ and $16\text{ amu}$ for the atomic mass of ${}^{16}\text{O}$ in equation (7).

$\begin{array}{c}\text{Molecular mass}=\left[\left(1\right)\left(15\text{ amu}\right)+\left(1\right)\left(14\text{ amu}\right)+\left(1\right)\left(16\text{ amu}\right)\right]\\ =15\text{ amu}+14\text{ amu}+16\text{ amu}\\ =45\text{ amu}\end{array}$

Conclusion

The different molecular masses in $\text{amu}$ possible for the compound ${\text{N}}_2\text{O}$ are 48, 46, 44, 47 and 45.

(b)

Interpretation Introduction

Interpretation:

The least and the most common molecular masses of ${\text{N}}_2\text{O}$ are to be determined.

Concept introduction:

Isotopes are atoms with the same atomic number but a different mass number. The difference in the mass numbers of isotopes arises due to the difference in the number of neutrons in them. The mass number of an atom is the sum of the protons and neutrons in them. The number of protons of isotopes is the same but the neutron number varies. Due to the same number of protons in the isotopes, they occupy the same position in the periodic table.

Answer to Problem 2.134P

The most common molecular mass of ${\text{N}}_2\text{O}$ is $44\text{ amu}$ whereas, the least common molecular mass is $48\text{ amu}$.

Explanation of Solution

The percent abundances of ${}^{14}\text{N}$ and ${}^{16}\text{O}$ are, $99.6$ and $99.8$ respectively.

The molecule that is composed of the isotopes with higher abundances will be the most common and hence, the molecular mass of that molecule will be the most common for the compound ${\text{N}}_2\text{O}$.

Since the isotope ${}^{14}\text{N}$ of nitrogen has more abundance than the ${}^{15}\text{N}$ isotope and the isotope ${}^{16}\text{O}$ of oxygen has more abundance than the ${}^{18}\text{O}$ isotope, therefore the most common molecule is ${}^{14}{\text{N}}_2{}^{16}\text{O}$. The molecular mass of ${}^{14}{\text{N}}_2{}^{16}\text{O}$ is $44\text{ amu}$.

The molecule that is composed of the isotopes with lower abundances will be the most common and hence, the molecular mass of that molecule will be the least common for the compound ${\text{N}}_2\text{O}$. Therefore, the least common molecule is ${}^{15}{\text{N}}_2{}^{18}\text{O}$. The molecular mass of ${}^{15}{\text{N}}_2{}^{18}\text{O}$ is $48\text{ amu}$.

Conclusion

The most common molecular mass of ${\text{N}}_2\text{O}$ is $44\text{ amu}$ whereas, the least common molecular mass is $48\text{ amu}$.