Problem 22
Question
An element \(X\) has the following isotopic composition \({ }^{200} \mathrm{X}=90 \%,{ }^{199} \mathrm{X}=8 \%,{ }^{202} \mathrm{X}=2 \%\) The weighed average atomic mass of the naturally occurring element ' \(X^{\prime}\) is closest to (a) \(201 \mathrm{amu}\) (b) 202 amus (e) 109 amu (d) 200 amu
Step-by-Step Solution
Verified Answer
The weighted average atomic mass is closest to 200 amu (option d).
1Step 1: Write Down Given Information
We are given the isotopic composition of element \(X\). The isotopes and their respective abundances are \(^{200}\mathrm{X} = 90\%,\) \(^{199}\mathrm{X} = 8\%,\) \(^{202}\mathrm{X} = 2\%\). We will use these values to calculate the weighted average atomic mass.
2Step 2: Convert Percentages to Decimals
Convert the percentage abundances to decimal form for easier calculations. - \(0.90\) for \(^{200}\mathrm{X}\)- \(0.08\) for \(^{199}\mathrm{X}\)- \(0.02\) for \(^{202}\mathrm{X}\)
3Step 3: Multiply Isotope Mass by Abundance
Multiply each isotope mass by its decimal abundance: - For \(^{200}\mathrm{X}\), the contribution to the atomic mass is \(200 \times 0.90 = 180\). - For \(^{199}\mathrm{X}\), the contribution is \(199 \times 0.08 = 15.92\). - For \(^{202}\mathrm{X}\), the contribution is \(202 \times 0.02 = 4.04\).
4Step 4: Sum the Contributions
To find the weighted average atomic mass, sum the contributions from all isotopes:\[180 + 15.92 + 4.04 = 199.96\] amu.
5Step 5: Compare with Given Options
Compare the calculated atomic mass, \(199.96\) amu, with the provided options:(a) 201 amu, (b) 202 amu, (e) 109 amu, (d) 200 amu. The closest value is \(d) 200\) amu.
Key Concepts
Isotopic CompositionWeighted AverageIsotope AbundancesElement X
Isotopic Composition
When talking about isotopic composition, we are referring to the different forms or variants an element can have based on the number of neutrons in its nucleus. Although isotopes of the same element have the same number of protons, they have different numbers of neutrons. This difference gives isotopes unique atomic masses. For instance, consider the element X. It has three isotopes identified by their atomic masses:
- \({ }^{200}\text{X}\) with an abundance of 90%.
- \({ }^{199}\text{X}\) with an 8% abundance.
- \({ }^{202}\text{X}\) with a mere 2% abundance.
Weighted Average
In many areas, especially in chemistry, we encounter the concept of a weighted average. A weighted average combines values with varying levels of importance or contribution. In the context of atomic mass calculations, it is used to determine the average atomic mass of an element based on the abundances of its isotopes.
For each isotope, you will:
- Multiply its atomic mass by the fractional abundance (percentage expressed as a decimal).
- Add all these products together to get the weighted average atomic mass.
Isotope Abundances
Isotope abundances provide essential insight into the natural occurrence and distribution of an element's isotopes. Represented as percentages, they indicate how much of each isotope exists compared to the total amount of the element. By knowing these abundances, we can understand how different isotopes add up to the average atomic weight we observe on the periodic table.For example:
- \({ }^{200}\text{X}\) predominates with 90% abundance.
- \({ }^{199}\text{X}\) comes next but only makes up 8%.
- \({ }^{202}\text{X}\) has the smallest presence with only 2%.
Element X
Element X refers to a hypothetical or unspecified element being analyzed. In this context, we're using it to understand basic principles of isotopic composition and atomic mass calculations. By examining Element X's isotopes, you can apply calculated practices in a real-world scenario, albeit with fictional data.
Considering Element X:
- Helps us explore how isotopes affect an element's property such as atomic mass.
- Provides a foundation for practicing calculations like weighted averages based on isotope abundances.
- Enables understanding of how periodic table entries often report average atomic masses derived from all isotopes of an element.
Other exercises in this chapter
Problem 20
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If \(30 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(20 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) react to form water. What is left at the end of the reaction? (a) \(1
View solution Problem 24
A.gas is found to have the formula (CO) \(_{x}\). Its V.D. is 70 . The value of \(x\) must be (a) 7 (b) 4 (c) 5 (d) 6
View solution