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Chapter 1: Problem 22

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

Short Answer

Expert verified
The weighted average atomic mass is closest to 200 amu (option d).

Step by step solution

01

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.
02

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 \(^{202}\mathrm{X}\)
03

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 \(202 \times 0.02 = 4.04\).
04

Sum the Contributions

To find the weighted average atomic mass, sum the contributions from all isotopes:\[180 + 15.92 + 4.04 = 199.96\] amu.
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.
The isotopic composition of an element essentially describes the relative amounts or abundance of each of its isotopes. Understanding these compositions is crucial, as they directly affect the overall atomic weight of an element.
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.
Weighted averages allow us to obtain a single value that accurately represents the entire set of different isotopic contributions. It reflects both the mass and its significance based on abundance, ensuring precise calculations in elemental chemistry.
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%.
These abundances are necessary for accurately calculating any element's atomic mass. They tell us how these isotopes influence the total average mass that characterizes that element.
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.
By mastering the example of Element X, learners can confidently approach similar exercises involving real elements and their isotopes.

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Most popular questions from this chapter

For the reaction \(A+2 B \rightarrow C, 5\) moles of \(A\) and 8 moles of \(B\) will produce (a) 5 moles of \(\bar{C}\) (b) 4 moles of \(\vec{C}\) (c) 8 moles of \(\mathrm{C}\) (d) 13 moles of \(C\)

The density in grams per litre of a mixture containing an equal number of moles of methane and ethane at STP is (a) \(1.03\) (b) \(1.10\) (c) \(0.94\) (d) \(1.20\)

A \(0.242 \mathrm{~g}\) sample of potassium is heated in oxygen. The result is \(0.440 \mathrm{~g}\) of a crystalline compound. What is the formula of this compound? (a) \(\mathrm{KO}\) (b) \(\mathrm{K}_{2} \mathrm{O}\) (c) \(\mathrm{KO}_{2}\) (d) \(\mathrm{KO}_{3}\)

A \(3 \mathrm{~g}\) of oxide of a metal is converted to chloride completely and it yields \(5 \mathrm{~g}\) of chloride. The equivalent weight of the metal is (a) \(33.25\) (b) \(3.325\) (c) 12 (d) 20

The decomposition of a certain mass of \(\mathrm{CaCO}_{3}\) gave \(11.2 \mathrm{dm}^{-3}\) of \(\mathrm{CO}_{2}\) gas at STP. The mass of \(\mathrm{KOH}\) required to completely neutralize the gas is (a) \(56 \mathrm{~g}\) (b) \(28 \mathrm{~g}\) (c) \(42 \mathrm{~g}\) (d) \(20 \mathrm{~g}\)
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