Chapter 4, Problem 78A

Interpretation Introduction

Interpretation : The percentage of naturally occurring lithium-7 is to be calculated.

Concept Introduction : The atomic mass is the weighted average mass of the atoms in a naturally occurring sample of an element. The formula to calculate atomic mass is given as

  $\text{Atomic mass}\text{=}\text{Isotope atomic mass}\times \text{Natural percent abundance}$

Answer to Problem 78A

The percentage of naturally occurring lithium-7 is $92.5\%$ .

Explanation of Solution

Given information :

Lithium-6 has an atomic mass of 6.015amu.

Lithium-7 has an atomic mass of 7.016 amu.

The atomic mass of lithium is 6.941 amu.

The atomic mass is the weighted average mass of the atoms in a naturally occurring sample of an element. The formula to calculate atomic mass is given as

  $\text{Atomic mass}\text{=}\text{Isotope atomic mass}\times \text{Natural percent abundance}$

Consider the natural percent abundance of Lithium-7 to be $\text{A}$ and Lithium-6 to be $1-\text{A}$ .

To calculate the natural occurring lithium-7, substitute the values in above formula:

  $\begin{array}{c}\text{Atomic mass}=\text{Isotope atomic mass×Natural percent abundance}\\ \text{6}\text{.941}=\left[\text{6}\text{.015×}\left(\text{1}-\text{A}\right)\right]\text{+}\left[\text{7}\text{.016×A}\right]\\ \text{6}\text{.941}-\text{6}\text{.015}=\text{7}\text{.016A}-\text{6}\text{.015A}\\ \text{A}=\text{0}\text{.925}\end{array}$

The value of A in percentage is $92.5\%$ .

So, the percentage of naturally occurring lithium-7 is $92.5\%$ .