Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Isotope Abundance Problem**
*Problem Statement:*
An unknown element is a mixture of isotopes \(^{129}\text{X}\) and \(^{132}\text{X}\). The average atomic mass of X is 130.35 amu. What is the percent abundance of \(^{132}\text{X}\)?
*Solution Approach:*
To solve this problem, you need to use the formula for the average atomic mass of isotopes:
\[
\text{Average Atomic Mass} = (f_1 \times m_1) + (f_2 \times m_2)
\]
where:
- \(f_1\) and \(f_2\) are the fractional abundances of isotopes \(^{129}\text{X}\) and \(^{132}\text{X}\), respectively.
- \(m_1\) and \(m_2\) are the masses of \(^{129}\text{X}\) and \(^{132}\text{X}\), which are 129 amu and 132 amu.
Also, \(f_1 + f_2 = 1\).
Let's set \(f_2 = x\). Then \(f_1 = 1 - x\).
Substitute these values into the equation:
\[
130.35 = (1-x) \times 129 + x \times 132
\]
Simplify and solve for \(x\) to find the percent abundance of \(^{132}\text{X}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0157a351-404e-45f5-af25-2ff50bf4117b%2Faf25f4b2-2368-4613-90f7-f76c57581f01%2Fcavpo3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Isotope Abundance Problem**
*Problem Statement:*
An unknown element is a mixture of isotopes \(^{129}\text{X}\) and \(^{132}\text{X}\). The average atomic mass of X is 130.35 amu. What is the percent abundance of \(^{132}\text{X}\)?
*Solution Approach:*
To solve this problem, you need to use the formula for the average atomic mass of isotopes:
\[
\text{Average Atomic Mass} = (f_1 \times m_1) + (f_2 \times m_2)
\]
where:
- \(f_1\) and \(f_2\) are the fractional abundances of isotopes \(^{129}\text{X}\) and \(^{132}\text{X}\), respectively.
- \(m_1\) and \(m_2\) are the masses of \(^{129}\text{X}\) and \(^{132}\text{X}\), which are 129 amu and 132 amu.
Also, \(f_1 + f_2 = 1\).
Let's set \(f_2 = x\). Then \(f_1 = 1 - x\).
Substitute these values into the equation:
\[
130.35 = (1-x) \times 129 + x \times 132
\]
Simplify and solve for \(x\) to find the percent abundance of \(^{132}\text{X}\).
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