Problem 26
Question
Verify that the average atomic mass of lithium is 6.941 , given this information: \({ }^{6} \mathrm{Li},\) exact mass \(=6.015121 \mathrm{u}\) percent abundance \(=7.500 \%\) \({ }^{7} \mathrm{Li},\) exact mass \(=7.016003 \mathrm{u}\) percent abundance \(=92.50 \%\)
Step-by-Step Solution
Verified Answer
The average atomic mass of lithium, 6.941 u, is correct based on isotope data.
1Step 1: Understanding Percent Abundance
The percentage abundance of each isotope indicates how much of that isotope is present in a typical sample of lithium. For example, \(^{6} Li\) has an abundance of 7.500%, which means that out of 100 lithium atoms, approximately 7.5 atoms will be \(^{6} Li\). Similarly, \(^{7} Li\) has an abundance of 92.50%.
2Step 2: Convert Percent to Decimal
Convert the percent abundance into a decimal for calculation. For \(^{6} Li\), the abundance is \(7.500\%\) or \(0.0750\) in decimal form. For \(^{7} Li\), the abundance is \(92.50%\) or \(0.9250\) in decimal form.
3Step 3: Calculate Weighted Mass Contribution
Multiply the exact mass of each isotope by its decimal abundance to find its contribution to the average atomic mass. For \(^{6} Li\):\[6.015121 \, \text{u} \times 0.0750 = 0.4511341 \\]For \(^{7} Li\):\[7.016003 \, \text{u} \times 0.9250 = 6.4888028\\]
4Step 4: Calculate Average Atomic Mass
Add up the contributions of each isotope to get the average atomic mass of lithium:\[0.4511341 \, \text{u} + 6.4888028 \, \text{u} = 6.9399369 \, \text{u}\\]Rounding this to three decimal places gives 6.940 u.
5Step 5: Verification
Compare the calculated average atomic mass (6.940 u) with the given average (
6.941 u). The calculated mass is very close to the provided data, verifying that the average atomic mass provided is accurate given the isotope data.
Key Concepts
Isotope AbundanceWeighted AverageLithium IsotopesMass Spectrometry
Isotope Abundance
Isotope abundance, often expressed in percentages, tells us the distribution of different isotopes of an element found in nature. For example, if we look at lithium, it primarily exists as two isotopes:
Every sample of lithium will have about 7.5% of ^{6}Li and 92.5% of ^{7}Li, highlighting the dominance of ^{7}Li in nature. This distribution is crucial when calculating the average atomic mass of lithium or any other element.
- ^{6}Li with an abundance of 7.500%
- ^{7}Li with an abundance of 92.50%
Every sample of lithium will have about 7.5% of ^{6}Li and 92.5% of ^{7}Li, highlighting the dominance of ^{7}Li in nature. This distribution is crucial when calculating the average atomic mass of lithium or any other element.
Weighted Average
The weighted average is essential when calculating the average atomic mass. This concept ensures that each isotope's mass is properly proportioned based on its abundance.
To compute the weighted average, convert percent abundances into their decimal forms by dividing by 100. This allows precise calculations:
6.9399369 u, rounded to 6.940 u. This demonstrates how significant isotope abundances are in calculating weighted averages.
To compute the weighted average, convert percent abundances into their decimal forms by dividing by 100. This allows precise calculations:
- ^{6}Li abundance as decimal: 0.0750
- ^{7}Li abundance as decimal: 0.9250
- ^{6}Li: 6.015121 u × 0.0750 = 0.4511341 u
- ^{7}Li: 7.016003 u × 0.9250 = 6.4888028 u
6.9399369 u, rounded to 6.940 u. This demonstrates how significant isotope abundances are in calculating weighted averages.
Lithium Isotopes
Lithium, one of the lightest metals, exists naturally in two stable isotopes:
Despite the small difference in their mass numbers, ^{7}Li is significantly more abundant. It contributes more heavily to the average atomic mass of lithium.
This heavier isotope's predominance impacts physical properties and is crucial for applications such as nuclear fusion research and battery technology.
- ^{6}Li
- ^{7}Li
Despite the small difference in their mass numbers, ^{7}Li is significantly more abundant. It contributes more heavily to the average atomic mass of lithium.
This heavier isotope's predominance impacts physical properties and is crucial for applications such as nuclear fusion research and battery technology.
Mass Spectrometry
Mass spectrometry is an analytical technique that helps us determine isotope abundance and natural atomic masses. It separates isotopes based on their mass-to-charge ratio, allowing us to quantify and identify isotopes present in a sample.
For elements like lithium, this method can accurately measure the relative abundance of different isotopes such as ^{6}Li and ^{7}Li.
During mass spectrometry, the element's ions are generated, sorted, and detected, resulting in a spectrum that displays isotope abundance. This technology is integral in verifying isotope distribution and calculating precise average atomic masses of elements.
For elements like lithium, this method can accurately measure the relative abundance of different isotopes such as ^{6}Li and ^{7}Li.
During mass spectrometry, the element's ions are generated, sorted, and detected, resulting in a spectrum that displays isotope abundance. This technology is integral in verifying isotope distribution and calculating precise average atomic masses of elements.
Other exercises in this chapter
Problem 24
Give the complete symbol \({ }_{Z}^{A} \mathrm{X}\) for each of these atoms: (a) sodium with 12 neutrons, (b) argon with 21 neutrons, and (c) gallium with 38 ne
View solution Problem 25
Give the complete symbol \({ }_{Z}^{A} \mathrm{X}\) for each of these atoms: (a) nitrogen with 8 neutrons, (b) zinc with 34 neutrons, and (c) xenon with 75 neut
View solution Problem 27
Verify that the average atomic mass of magnesium is 24.3050 , given this information: \({ }^{24} \mathrm{Mg},\) exact mass \(=23.985042 \mathrm{u}\) \({ }^{25}
View solution Problem 29
Argon has three naturally occurring isotopes: \(0.3336 \%\) \({ }^{36} \mathrm{Ar}, 0.063 \%{ }^{38} \mathrm{Ar}\), and \(99.60 \%{ }^{40} \mathrm{Ar}\). Estima
View solution