Chapter 2, Problem 2.17QP

(a)

Interpretation Introduction

Interpretation: The pair in which the heavier isotope has more abundance is to be identified.

Concept introduction: The natural abundance of an isotope is calculated from the formula,

$\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3$

To determine: If the heavier isotope has more abundance among the given pair of isotopes.

Answer to Problem 2.17QP

Solution

The heavier isotope ${}^{11}\text{B}$ has more abundance.

Explanation of Solution

Explanation

The natural abundance of an isotope is calculated by the formula,

$\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3$

Where,

• $m$ is the mass of an isotope.
• $x$ is natural abundance, expressed in decimals.

The given isotopes are ${}^{10}\text{B}\text{and}{}^{11}\text{B}$ and the average atomic mass is $10.81$ atomic mass unit.

The given isotopes have mass $10\text{and}11$ atomic mass unit respectively.

The natural abundance of ${}^{10}\text{B}$ is assume as $x$ and hence, the natural abundance of ${}^{11}\text{B}$ becomes $\left(100-x\right)$.

Substitute the values in above formula,

$\begin{array}{c}\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3\\ 10.81=10\times \frac{x}{100}+11\times \frac{\text{100}-x}{100}\\ 10.81=\frac{10x+11\left(100-x\right)}{100}\\ 1081=10x+\text{1100}-11x\end{array}$

Simplify the above equation.

$\begin{array}{c}x=1100-1081\\ =19\%\end{array}$

So, the natural abundance of ${}^{11}\text{B}$ becomes,

$100-19=81\%$

Hence, the heavier isotope ${}^{11}\text{B}$ has more abundance.

(b)

Interpretation Introduction

To determine: The heavier isotope that has more abundance for the given pairs of isotopes.

Answer to Problem 2.17QP

Solution

The heavier isotope ${}^7\text{Li}$ has more abundance.

Explanation of Solution

Explanation

The natural abundance of an isotope is calculated by the formula,

$\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3$

Where,

• $m$ is the mass of an isotope.
• $x$ is natural abundance, expressed in decimals.

The given isotopes are ${}^6\text{Li}\text{and}{}^7\text{Li}$ and its average atomic mass is $6.941$ atomic mass unit.

So, the given isotopes have mass $6\text{and}7$ atomic mass unit respectively.

The natural abundance of ${}^6\text{Li}$ is assume as $\text{x}$ and hence, the natural abundance of ${}^7\text{Li}$ becomes $\left(100-x\right)$.

Substitute the values in above formula,

$\begin{array}{c}\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3\\ 6.941=6\times \frac{x}{100}+7\times \frac{\text{100}-x}{100}\\ 6.941=\frac{\text{6}x+7\left(100-x\right)}{100}\\ 694.1=6x+\text{700}-7x\end{array}$

Simplify the above equation.

$\begin{array}{c}x=700-694.1\\ =5.9\%\end{array}$

So, the natural abundance of ${}^7\text{Li}$ becomes,

$100-5.9=94.1\%$

Hence, the heavier isotope ${}^7\text{Li}$ has more abundance.

(c)

Interpretation Introduction

To determine: The heavier isotope that has more abundance for the given pairs of isotopes.

Answer to Problem 2.17QP

Solution

The heavier isotope ${}^{15}\text{N}$ has not more abundance.

Explanation of Solution

Explanation

The natural abundance of an isotope is calculated by the formula,

$\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3$

Where,

• $m$ is the mass of an isotope.
• $x$ is natural abundance, expressed in decimals.

The given isotopes are ${}^{14}\text{N}\text{and}{}^{15}\text{N}$ and its average atomic mass is $14.01$ atomic mass unit.

So, the given isotopes have mass $14\text{and}15$ atomic mass unit respectively.

The natural abundance of ${}^{14}\text{N}$ is assume as $\text{x}$ and hence, the natural abundance of ${}^{15}\text{N}$ becomes $\left(100-x\right)$.

Substitute the values in above formula,

$\begin{array}{c}\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3\\ 14.01=14\times \frac{x}{100}+15\times \frac{\text{100}-x}{100}\\ 14.01=\frac{\text{14}x+15\left(100-x\right)}{100}\\ 1401=14x+\text{1500}-15x\end{array}$

Simplify the above equation

$\begin{array}{c}x=1500-1401\\ =99\%\end{array}$

So, the natural abundance of ${}^{15}\text{N}$ becomes,

$100-99=1\%$

Hence, the heavier isotope ${}^{15}\text{N}$ has not more abundance.

(d)

Interpretation Introduction

To determine: The heavier isotope that has more abundance for the given pairs of isotopes.

Answer to Problem 2.17QP

Solution

The heavier isotope ${}^{22}\text{Ne}$ has not more abundance.

Explanation of Solution

Explanation

The natural abundance of an isotope is calculated by the formula,

$\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3$

Where,

• $m$ is the mass of an isotope.
• $x$ is natural abundance, expressed in decimals.

The given isotopes are ${}^{20}\text{Ne}\text{and}{}^{22}\text{Ne}$ and its average atomic mass is $20.18$ atomic mass unit.

So, the given isotopes have mass $20\text{and}22$ atomic mass unit respectively.

The natural abundance of ${}^{20}\text{Ne}$ is assume as $\text{x}$ and hence, the natural abundance of ${}^{22}\text{Ne}$ becomes $\left(100-x\right)$.

Substitute the values in above formula,

$\begin{array}{c}\text{Average}\text{atomic}\text{mass}=m_1\times x_1+m_2\times x_2+m_3\times x_3\\ 20.18=20\times \frac{x}{100}+22\times \frac{\text{100}-x}{100}\\ 20.18=\frac{\text{20}x+22\left(100-x\right)}{100}\\ 2018=20x+\text{2200}-22x\end{array}$

Simplify the above equation.

$\begin{array}{c}2x=2200-2018\\ 2x=182\\ x=\frac{182}{2}\\ x=91\%\end{array}$

So, the natural abundance of ${}^{22}\text{Ne}$ becomes,

$100-91=9\%$

Hence, the heavier isotope ${}^{22}\text{Ne}$ has not more abundance.

Conclusion

The pair in which the heavier isotope has more abundance has been rightfully identified.