Chapter 2, Problem 2.120AP

(a)

Interpretation Introduction

Interpretation: The masses of the two isotopes of bromine are given. The possible masses of the bromine molecule formed from these isotopes are to be calculated and the natural abundance of each of the molecules of the predicted masses is to be calculated.

Concept introduction: The average mass is calculated by the formula,

$\text{Average}\text{mass}={\displaystyle \sum _{i=1}^n{\left(\text{Natural}\text{abundance}\text{of}\text{isotope}\right)}_i\times {\left(\text{Mass}\text{of}\text{isotope}\right)}_i}$

To determine: The possible masses of the of the individual molecules of ${\text{Br}}_2$.

Answer to Problem 2.120AP

Solution:

The possible masses of ${\text{Br}}_2$ is $157.8366\text{amu}$ $\left({}^{79}\text{Br}{-}^{79}\text{Br}\right)$, $159.8346\text{amu}$ $\left({}^{81}\text{Br}{-}^{79}\text{Br}\right)$ and $161.8326\text{amu}$ $\left({}^{81}\text{Br}{-}^{81}\text{Br}\right)$.

Explanation of Solution

Given

The mass of isotope of bromine, ${}^{79}\text{Br}$ is $78.9183\text{amu}$.

The mass of isotope of bromine, ${}^{81}\text{Br}$ is $80.9163\text{amu}$.

The possibilities of formation of molecules of bromine are,

• ${}^{79}\text{Br}$ and ${}^{79}\text{Br}$
• ${}^{81}\text{Br}$ and ${}^{79}\text{Br}$
• ${}^{81}\text{Br}$ and ${}^{81}\text{Br}$

Therefore, the mass of ${}^{79}\text{Br}{-}^{79}\text{Br}$ is,

$78.9183\text{amu}\times 2=157.8366\text{amu}$

The mass of ${}^{81}\text{Br}{-}^{79}\text{Br}$ is,

$78.9183\text{amu}+80.9163\text{amu}=159.8346\text{amu}$

The mass of ${}^{81}\text{Br}{-}^{81}\text{Br}$ is,

$80.9163\text{amu}\times 2=161.8326\text{amu}$

Therefore, the possible masses of ${\text{Br}}_2$ is $157.8366\text{amu}$, $159.8346\text{amu}$ and $161.8326\text{amu}$.

(b)

Interpretation Introduction

To determine: The natural abundance of each of the molecules of the predicted masses of ${\text{Br}}_2$.

Answer to Problem 2.120AP

Solution:

The natural abundance possible masses of ${\text{Br}}_2$ is $25\%$ $\left({}^{79}\text{Br}{-}^{79}\text{Br}\right)$, $50\%$ $\left({}^{81}\text{Br}{-}^{79}\text{Br}\right)$ and $25\%$ $\left({}^{81}\text{Br}{-}^{81}\text{Br}\right)$.

Explanation of Solution

The possible masses of ${\text{Br}}_2$ is $157.8366\text{amu}$ $\left({}^{79}\text{Br}{-}^{79}\text{Br}\right)$, $159.8346\text{amu}$ $\left({}^{81}\text{Br}{-}^{79}\text{Br}\right)$ and $161.8326\text{amu}$ $\left({}^{81}\text{Br}{-}^{81}\text{Br}\right)$.

Therefore, the average mass of ${\text{Br}}_2$ is,

$\frac{157.8366\text{amu}+159.8346\text{amu}+161.8326\text{amu}}{3}=159.8346\text{amu}$

The average mass of ${\text{Br}}_2$ is same as that of ${}^{81}\text{Br}{-}^{79}\text{Br}$, therefore, the natural abundance of ${}^{81}\text{Br}{-}^{79}\text{Br}$ is $50\%$ and that of the others is $50\%$.

The natural abundance of ${}^{81}\text{Br}{-}^{81}\text{Br}$ is assumed to be $\frac{x}{100}$.

Therefore, the natural abundance of ${}^{79}\text{Br}{-}^{79}\text{Br}$ is $\frac{50-x}{100}$.

The average mass is calculated by the formula,

$\text{Average}\text{mass}={\displaystyle \sum _{i=1}^n{\left(\text{Natural}\text{abundance}\text{of}\text{isotope}\right)}_i\times {\left(\text{Mass}\text{of}\text{isotope}\right)}_i}$

Therefore, the average mass of bromine molecule is calculated by the formula,

$\text{Average}\text{mass}=\left(\begin{array}{l}\text{Abundance}\text{of}{}^{79}\text{Br}{-}^{79}\text{Br}\times \text{Mass}\text{of}{}^{\text{79}}{\text{Br-}}^{\text{79}}\text{Br}\\ \text{+Abundance}\text{of}{}^{81}\text{Br}{-}^{79}\text{Br}\times \text{Mass}\text{of}{}^{81}\text{Br}{-}^{79}\text{Br}\\ \text{+Abundance}\text{of}{}^{81}\text{Br}{-}^{81}\text{Br}\times \text{Mass}\text{of}{}^{81}\text{Br}{-}^{81}\text{Br}\end{array}\right)$

Substitute the natural abundance and mass of ${}^{79}\text{Br}{-}^{79}\text{Br}$, ${}^{81}\text{Br}{-}^{79}\text{Br}$ and ${}^{81}\text{Br}{-}^{81}\text{Br}$ in the above equation.

$\begin{array}{c}159.8346\text{amu}=\left(\begin{array}{l}\left(\frac{50-x}{100}\right)\times 157.8366\text{amu}\\ +\frac{50}{100}\times 159.8346\text{amu}\\ \text{+}\frac{x}{100}\times 161.8326\text{amu}\end{array}\right)\\ 159.8346\times 100=7891.83-157.8366x+7991.73+161.8326x\\ 15983.46-7891.83-7991.73=3.996x\\ x=25\%\end{array}$

Therefore, the abundance of both ${}^{79}\text{Br}{-}^{79}\text{Br}$ and ${}^{81}\text{Br}{-}^{81}\text{Br}$ is $25\%$.

Conclusion:

1. a. The possible masses of ${\text{Br}}_2$ is $157.8366\text{amu}$ $\left({}^{79}\text{Br}{-}^{79}\text{Br}\right)$, $159.8346\text{amu}$ $\left({}^{81}\text{Br}{-}^{79}\text{Br}\right)$ and $161.8326\text{amu}$ $\left({}^{81}\text{Br}{-}^{81}\text{Br}\right)$.
2. b. The natural abundance possible masses of ${\text{Br}}_2$ is $25\%$ $\left({}^{79}\text{Br}{-}^{79}\text{Br}\right)$, $50\%$ $\left({}^{81}\text{Br}{-}^{79}\text{Br}\right)$ and $25\%$ $\left({}^{81}\text{Br}{-}^{81}\text{Br}\right)$