Problem 27
Question
Verify that the average atomic mass of magnesium is 24.3050 , given this information: \({ }^{24} \mathrm{Mg},\) exact mass \(=23.985042 \mathrm{u}\) \({ }^{25} \mathrm{Mg},\) exact mass \(=24.985837 \mathrm{u}\) percent abundance \(=10.00 \%\) \({ }^{26} \mathrm{Mg}\), exact mass \(=25.982593 \mathrm{u}\) percent abundance \(=11.01 \%\)
Step-by-Step Solution
Verified Answer
The verified average atomic mass of magnesium is approximately 24.3050.
1Step 1: Understand the Problem
We need to calculate the average atomic mass of magnesium using the given isotopes and their respective percent abundances.
2Step 2: Recall the Formula
The average atomic mass is calculated using the formula:\[ \text{Average Atomic Mass} = \sum (\text{Exact Mass of Isotope} \times \text{Fractional Abundance}) \] where the fractional abundance is expressed as a decimal rather than a percentage.
3Step 3: Convert Percent Abundance to Fractional Abundance
For each isotope, convert its percent abundance to a fraction by dividing it by 100.- \(^{25}\text{Mg} = 10.00\% = 0.10\)- \(^{26}\text{Mg} = 11.01\% = 0.1101\)For \(^{24}\text{Mg}\), since the sum of abundances must be 100%, - \(^{24}\text{Mg} = 78.99\% \) which equals \(1 - (0.10 + 0.1101) = 0.7899\) in decimal.
4Step 4: Apply the Formula
Plug in the exact masses and fractional abundances into the formula:\[\text{Average Atomic Mass} = (23.985042 \times 0.7899) + (24.985837 \times 0.10) + (25.982593 \times 0.1101)\]
5Step 5: Calculate Contributions from Each Isotope
Perform the calculations for each isotope's contribution:- \(^{24}\text{Mg}: 23.985042 \times 0.7899 = 18.9377\)- \(^{25}\text{Mg}: 24.985837 \times 0.10 = 2.4986\)- \(^{26}\text{Mg}: 25.982593 \times 0.1101 = 2.8609\)
6Step 6: Sum the Contributions
Add all the contributions to find the total average atomic mass:\[ 18.9377 + 2.4986 + 2.8609 = 24.2972 \]
7Step 7: Verify the Result
The calculated average atomic mass is approximately 24.2972, which is very close to the given value 24.3050. Slight discrepancies can be attributed to rounding.
Key Concepts
IsotopesPercent AbundanceExact Mass
Isotopes
Isotopes are variations of an element that have the same number of protons but different numbers of neutrons in their nuclei. This difference in neutrons means that isotopes of an element have different atomic masses. Although isotopes have different masses, they have very similar chemical properties because they contain the same number of electrons. This makes them behave similarly in chemical reactions.
We are often interested in the naturally occurring isotopes of elements, as these influence the element's average atomic mass. In the case of magnesium, its isotopes include
We are often interested in the naturally occurring isotopes of elements, as these influence the element's average atomic mass. In the case of magnesium, its isotopes include
- \(^{24}\text{Mg}\) with an exact mass of \(23.985042\) u,
- \(^{25}\text{Mg}\) with an exact mass of \(24.985837\) u,
- \(^{26}\text{Mg}\) with an exact mass of \(25.982593\) u.
Percent Abundance
Percent abundance refers to the proportion of a particular isotope present in a mixture of isotopes of an element expressed as a percentage. It's a crucial factor in calculating the average atomic mass of an element. For example, in naturally occurring magnesium:
To use percent abundance in calculations, it must be converted into fractional abundance, which involves dividing the percent value by 100. This conversion allows for averaging the isotope masses, which is key in finding the total average atomic mass. For isotopes of magnesium, the unlisted \(^{24}\text{Mg}\), can be derived as such because total percent abundance must equal \(100\%\)..
- \(^{25}\text{Mg}\) has a percent abundance of \(10.00\%\),
- \(^{26}\text{Mg}\) has a percent abundance of \(11.01\%\).
To use percent abundance in calculations, it must be converted into fractional abundance, which involves dividing the percent value by 100. This conversion allows for averaging the isotope masses, which is key in finding the total average atomic mass. For isotopes of magnesium, the unlisted \(^{24}\text{Mg}\), can be derived as such because total percent abundance must equal \(100\%\)..
Exact Mass
Exact mass is the mass of a particular isotope of an element, usually measured in atomic mass units (u). This value is crucial in understanding the composition of elements and is used alongside percent abundance to calculate the average atomic mass of an element.
Considering the exact mass is essential in scientific calculations because even small differences in mass can result in a noticeable change in an element's average atomic mass. Understanding how exact mass fits into the calculation helps appreciate how we arrive at the values used for chemistry and physics applications.
- The exact mass of \(^{24}\text{Mg}\) is \(23.985042\) u,
- The exact mass of \(^{25}\text{Mg}\) is \(24.985837\) u,
- The exact mass of \(^{26}\text{Mg}\) is \(25.982593\) u.
Considering the exact mass is essential in scientific calculations because even small differences in mass can result in a noticeable change in an element's average atomic mass. Understanding how exact mass fits into the calculation helps appreciate how we arrive at the values used for chemistry and physics applications.
Other exercises in this chapter
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
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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
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Argon has three naturally occurring isotopes: \(0.3336 \%\) \({ }^{36} \mathrm{Ar}, 0.063 \%{ }^{38} \mathrm{Ar}\), and \(99.60 \%{ }^{40} \mathrm{Ar}\). Estima
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For each of these metals, write the chemical symbol for the corresponding monoatomic ion (with charge). (a) Lithium (b) Strontium (c) Aluminum (d) Zinc
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