Chapter 2, Problem 2.24QP

(a)

Interpretation Introduction

To determine: The number of neutrons present in each of the given isotopes.

Interpretation: The number of neutrons present in each of the given isotopes is to be calculated.

Concept introduction: Natural abundance is the proportion of a particular isotope relative to all the isotopes of that element in a natural sample.

Answer to Problem 2.24QP

Number of neutrons in ${}^{\text{84}}\text{Sr}$ is $46$.

Number of neutrons in ${}^{\text{86}}\text{Sr}$ is $48$.

Number of neutrons in ${}^{\text{87}}\text{Sr}$ is $49$.

Number of neutrons in ${}^{\text{88}}\text{Sr}$ is $50$.

Explanation of Solution

Strontium has four isotopes: ${}^{\text{84}}\text{Sr}$, ${}^{\text{86}}\text{Sr}$, ${}^{\text{87}}\text{Sr}$ and ${}^{\text{88}}\text{Sr}$.

Number of neutrons in ${}^{\text{84}}\text{Sr}$ is calculated by the formula,

$\text{Neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}$

Mass number of ${}^{\text{84}}\text{Sr}$ is $84$.

Atomic number of strontium is $38$.

Substitute the value of mass number and atomic number in the above equation.

$\begin{array}{c}\text{Neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}\\ =\text{84}-38\\ =46\end{array}$

Number of neutrons in ${}^{\text{86}}\text{Sr}$ is calculated by the formula,

$\text{Neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}$

Mass number of ${}^{\text{86}}\text{Sr}$ is $86$.

Atomic number of strontium is $38$.

Substitute the value of mass number and atomic number in the above equation.

$\begin{array}{c}\text{Neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}\\ =\text{86}-38\\ =48\end{array}$

Number of neutrons in ${}^{\text{87}}\text{Sr}$ is calculated by the formula,

$\text{Neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}$

Mass number of ${}^{\text{87}}\text{Sr}$ is $87$.

Atomic number of strontium is $38$.

Substitute the value of mass number and atomic number in the above equation.

$\begin{array}{c}\text{Neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}\\ =\text{87}-38\\ =49\end{array}$

Number of neutrons in ${}^{\text{88}}\text{Sr}$ is calculated by the formula,

$\text{Neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}$

Mass number of ${}^{\text{88}}\text{Sr}$ is $88$.

Atomic number of strontium is $38$.

Substitute the value of mass number and atomic number in the above equation.

$\begin{array}{c}\text{Neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}\\ =\text{88}-38\\ =50\end{array}$

(b)

Interpretation Introduction

To determine: The average atomic mass of strontium and the comparison of it to the value in the periodic table on the inside front cover.

Interpretation: The average atomic mass of strontium is to be calculated and the comparison of it to the value in the periodic table on the inside front cover is to be done.

Concept introduction: The average atomic mass of an element is the weighted average mass of all the isotopes of that element. The average atomic mass of an element ( $m$) is calculated by the formula,

$m=a_1{m}_1+a_2{m}_2$

Answer to Problem 2.24QP

The average atomic mass of strontium is $\underset{\_}{87.6161amu}$ and it is approximately equal to the value in the periodic table on the inside front cover.

Explanation of Solution

The average atomic mass of strontium ( $m$) is calculated by the formula,

$m=a_1{m}_1+a_2{m}_2+a_3{m}_3+a_4{m}_4$ (1)

Where,

• $m_1$ is the atomic mass of ${}^{84}\text{Sr}$.
• $m_2$ is the atomic mass of ${}^{86}\text{Sr}$.
• $m_3$ is the atomic mass of ${}^{87}\text{Sr}$.
• $m_4$ is the atomic mass of ${}^{88}\text{Sr}$.
• $a_1$ is the natural abundance of ${}^{84}\text{Sr}$.
• $a_2$ is the natural abundance of ${}^{86}\text{Sr}$.
• $a_3$ is the natural abundance of ${}^{87}\text{Sr}$.
• $a_4$ is the natural abundance of ${}^{88}\text{Sr}$.

The natural abundance of ${}^{84}\text{Sr}$ is,

$\begin{array}{c}0.56\text{\%}{}^{84}\text{Sr}=\frac{0.56}{100}\\ =0.0056\end{array}$

The natural abundance of ${}^{86}\text{Sr}$ is,

$\begin{array}{c}9.86\text{\%}{}^{86}\text{Sr}=\frac{9.86}{100}\\ =0.0986\end{array}$

The natural abundance of ${}^{87}\text{Sr}$ is,

$\begin{array}{c}7.00\text{\%}{}^{87}\text{Sr}=\frac{7.00}{100}\\ =0.07\end{array}$

The natural abundance of ${}^{88}\text{Sr}$ is,

$\begin{array}{c}82.58\text{\%}{}^{88}\text{Sr}=\frac{82.58}{100}\\ =0.8258\end{array}$

The atomic mass of ${}^{84}\text{Sr}$ is $\text{83}\text{.9134}\text{amu}$, the atomic mass of ${}^{86}\text{Sr}$ is $85.9094\text{amu}$, the atomic mass of ${}^{87}\text{Sr}$ is $\text{86}\text{.9089}\text{amu}$ and the atomic mass of ${}^{88}\text{Sr}$ is $\text{87}\text{.9056}\text{amu}$.

Substitute the value of $m_1$, $m_2$, $m_3$, $m_4$, $a_1$, $a_2$, $a_3$ and $a_2$ in equation (1).

$\begin{array}{c}m=a_1{m}_1+a_2{m}_2+a_3{m}_3+a_4{m}_4\\ =\left(\begin{array}{c}\left(0.0056\times \text{83}\text{.9134}\text{amu}\right)+\left(0.0986\times 85.9094\text{amu}\right)+\\ \left(0.07\times \text{86}\text{.9089}\text{amu}\right)+\left(0.8258\times \text{87}\text{.9056}\text{amu}\right)\end{array}\right)\\ =0.4699\text{amu}+8.4706\text{amu}+6.0836\text{amu}+72.592\text{amu}\\ =\underset{\_}{87.6161amu}\end{array}$

Therefore, the average atomic mass of strontium is $\underset{\_}{87.6161amu}$ and it is approximately equal to the value in the periodic table on the inside front cover.

Conclusion

1. a. Number of neutrons in ${}^{\text{84}}\text{Sr}$, ${}^{\text{86}}\text{Sr}$, ${}^{\text{87}}\text{Sr}$ and ${}^{\text{88}}\text{Sr}$ are $46$, $48$, $49$ and $50$ respectively.
2. b. The average atomic mass of strontium is $\underset{\_}{87.6161amu}$.