Chapter 4, Problem 114A

Interpretation Introduction

Interpretation:

The identity and percent abundance of other isotope of indium needs to be determined.

Concept introduction:

The isotope contains the different numbers of neutron and same number of protons. So, the atomic mass of the isotope is different. The calculation of the atomic mass of elements is termed as the average atomic mass.

$\text{Average}\text{atomic}\text{mass}\text{=}\text{(}\text{mass}\text{of}\text{isotope}\text{×}\text{relative}\text{abundance)}$

Answer to Problem 114A

The other isotope of Indium is In $-114$ and the percentage of abundance is $95.7\%$.

Explanation of Solution

Given information:

The atomic mass of indium is $114.818amu$ and the abundance is present in the fraction form,

For the calculation of percentage abundance −

$\text{Average}\text{atomic}\text{mass}\text{=}(\text{mass}{\text{of In}}^{\text{114}}\text{×relative}\text{abundance)+(mass}\text{of}{\text{In}}^{\text{113}}\text{×relative}\text{abundance}\text{)}$

$\begin{array}{l}114.82=(114.82\times X/100)+(112.9\times 0.043)\\ =(114.82\times X/100)+(4.8547)\end{array}$

$\begin{array}{l}114.82-4.8547=(114.82\times X/100)\\ 109.9653÷114.82=X/100\end{array}$

$X=95.7\%$

Conclusion

The abundance is $100\%$ for every element, so when one isotope is $4.3\%$ then the other isotope is $95.7\%$.