The atomic masses of 6 Li and7Li are 6.0151 amu and 7.0160 amu, respectively. Calculate the natural abundances of the heaviest isotope. The average atomic mass of Li is 6.941 amu. Put your answer in three significant figures.
The atomic masses of 6 Li and7Li are 6.0151 amu and 7.0160 amu, respectively. Calculate the natural abundances of the heaviest isotope. The average atomic mass of Li is 6.941 amu. Put your answer in three significant figures.
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter2: Atoms, Molecules, And Ions
Section: Chapter Questions
Problem 15QAP
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Question
![### Problem Statement:
The atomic masses of \( \phantom{6}^6_3 \text{Li} \) and \( \phantom{7}^7_3 \text{Li} \) are 6.0151 amu and 7.0160 amu, respectively. Calculate the natural abundances of the heaviest isotope. The average atomic mass of Li is 6.941 amu. Put your answer in three significant figures.
**Explanation:**
To find the natural abundances of the isotopes, we need to solve the following system of equations based on the given data:
1. \( x + y = 1 \) (since the total abundance must equal 1)
2. \( 6.0151x + 7.0160y = 6.9410 \) (weighted average of the atomic masses)
Where:
- \( x \) = natural abundance of \( \phantom{6}^6_3 \text{Li} \)
- \( y \) = natural abundance of \( \phantom{7}^7_3 \text{Li} \)
Steps to solve:
1. From \( x + y = 1 \), we get \( y = 1 - x \).
2. Substitute \( y \) in the second equation:
\[
6.0151x + 7.0160(1 - x) = 6.9410
\]
3. Solve the equation for \( x \):
\[
6.0151x + 7.0160 - 7.0160x = 6.9410 \\
-1.0009x + 7.0160 = 6.9410 \\
-1.0009x = 6.9410 - 7.0160 \\
-1.0009x = -0.075 \\
x = \frac{-0.075}{-1.0009} \\
x ≈ 0.07496
\]
4. Calculate \( y \):
\[
y = 1 - x \\
y = 1 - 0.07496 \\
y ≈ 0.925
\]
Thus, the natural abundance of the heaviest isotope \( \phantom{7}^7_3 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5362cbea-9427-4b2b-99cb-57a208edd853%2F2f0b8e71-a6ce-4c16-8929-1051520b3691%2Fxeue6m8.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
The atomic masses of \( \phantom{6}^6_3 \text{Li} \) and \( \phantom{7}^7_3 \text{Li} \) are 6.0151 amu and 7.0160 amu, respectively. Calculate the natural abundances of the heaviest isotope. The average atomic mass of Li is 6.941 amu. Put your answer in three significant figures.
**Explanation:**
To find the natural abundances of the isotopes, we need to solve the following system of equations based on the given data:
1. \( x + y = 1 \) (since the total abundance must equal 1)
2. \( 6.0151x + 7.0160y = 6.9410 \) (weighted average of the atomic masses)
Where:
- \( x \) = natural abundance of \( \phantom{6}^6_3 \text{Li} \)
- \( y \) = natural abundance of \( \phantom{7}^7_3 \text{Li} \)
Steps to solve:
1. From \( x + y = 1 \), we get \( y = 1 - x \).
2. Substitute \( y \) in the second equation:
\[
6.0151x + 7.0160(1 - x) = 6.9410
\]
3. Solve the equation for \( x \):
\[
6.0151x + 7.0160 - 7.0160x = 6.9410 \\
-1.0009x + 7.0160 = 6.9410 \\
-1.0009x = 6.9410 - 7.0160 \\
-1.0009x = -0.075 \\
x = \frac{-0.075}{-1.0009} \\
x ≈ 0.07496
\]
4. Calculate \( y \):
\[
y = 1 - x \\
y = 1 - 0.07496 \\
y ≈ 0.925
\]
Thus, the natural abundance of the heaviest isotope \( \phantom{7}^7_3 \
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