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Chapter 2: Problem 98

Potassium has three naturally occurring isotopes \(\left(^{39} \mathrm{K}\right.\) \(\left.^{40} \mathrm{K}, \text { and }^{41} \mathrm{K}\right),\) but \(^{40} \mathrm{K}\) has a very low natural abundance. Which of the other two isotopes is more abundant? Briefly explain your answer.

Short Answer

Expert verified
The isotope \(^{39}\mathrm{K}\) is more abundant because its atomic weight is closer to the average atomic weight of potassium.

Step by step solution

01

Identify Atomic Weights

The approximate atomic weights for the potassium isotopes are as follows: \(^{39}\mathrm{K}\) has an atomic weight of about 39 u (unified atomic mass units), and \(^{41}\mathrm{K}\) has an atomic weight of about 41 u.
02

Consider Average Atomic Weight

The average atomic weight of potassium that you would find on the periodic table is about 39.10 u. This average should closely reflect the relative abundance of each isotope.
03

Compare Isotope Weights to Average

Compare the isotopic atomic weights to the average atomic weight of potassium. \(^{39}\mathrm{K}\) is closer to the average atomic weight of 39.10 u than \(^{41}\mathrm{K}\) is.
04

Conclusion on Abundance

Given that \(^{39}\mathrm{K}\) is closer to the average atomic weight of potassium, it indicates that \(^{39}\mathrm{K}\) is more abundant than \(^{41}\mathrm{K}\), because the average atomic weight is skewed towards the more abundant isotope.

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Key Concepts

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

Average Atomic Weight
The average atomic weight of an element gives us a clue about the relative abundance of its isotopes. This value is a weighted average, not simply a mean. It considers both the mass of each isotope and its relative abundance. For potassium, the average atomic weight is approximately 39.10 u.
This means that isotopes closer to this number are more prevalent. The average atomic weight reflects the natural distribution of isotopes in a sample.
Here鈥檚 how it works:
  • If an isotope has a higher percentage in nature, the average atomic weight will be closer to its atomic weight.
  • This concept helps us predict which isotopes are more common based on their contributions to the average value found on the periodic table.
Potassium Isotopes
Potassium is an element with three naturally occurring isotopes:
  • \(^{39}\mathrm{K}\) - Atomic weight of about 39 u.
  • \(^{40}\mathrm{K}\) - Present in low abundance, not a focus for common calculations.
  • \(^{41}\mathrm{K}\) - Atomic weight of about 41 u.
These isotopes differ in the number of neutrons they have, but they all share the same number of protons. In nature, these isotopes occur in varying amounts.
Since \(^{40}\mathrm{K}\) has a negligible natural abundance, between \(^{39}\mathrm{K}\) and \(^{41}\mathrm{K}\), it is \(^{39}\mathrm{K}\) that is more abundant because its atomic weight is closer to the average atomic weight of potassium, 39.10 u. This proximity suggests it is the dominant isotope by mass.
Atomic Weights
Atomic weights are a critical part of understanding isotopes and their behavior. An atomic weight is essentially the mass of an atom, expressed in unified atomic mass units (u).
For isotopes, individual atomic weights are not integers because they take into consideration the binding energy of nuclei and other quantum effects. Each isotope of an element has its unique atomic weight:
  • \(^{39}\mathrm{K}\) - approximately 39 u, which is why it strongly influences the average atomic weight of potassium.
  • \(^{41}\mathrm{K}\) - about 41 u, contributes less to the average due to lower abundance.
Understanding atomic weights helps chemists and scientists predict chemical behaviors and interactions of elements, determine molecular weights, and perform quantitative chemical analyses.

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Most popular questions from this chapter

Analysis of a 10.0 -g sample of apatite (a major component of tooth enamel) showed that it was made up of \(3.99 \mathrm{g} \mathrm{Ca}, 1.85 \mathrm{g} \mathrm{P}, 4.14 \mathrm{g} \mathrm{O},\) and \(0.020 \mathrm{g} \mathrm{H} .\) I ist these elements based on relative amounts (moles), from smallest to largest.

Calculate the molar mass and the mass percent of each element in the blue solid compound \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O} .\) What is the mass of copper and the mass of water in \(10.5 \mathrm{g}\) of the compound?

Direct reaction of iodine \(\left(I_{2}\right)\) and chlorine \(\left(C l_{2}\right)\) produces an iodine chloride, \(I_{x} C l_{y},\) a bright yellow solid. If you completely consume \(0.678 \mathrm{g}\) of \(\mathrm{I}_{2}\) in a reaction with excess \(\mathrm{Cl}_{2}\) and produce \(1.246 \mathrm{g}\) of \(\mathrm{I}_{x} \mathrm{Cl}_{y}\) what is the empirical formula of the compound? A later experiment showed that the molar mass of \(\mathrm{I}_{x} \mathrm{Cl}_{y}\) was \(467 \mathrm{g} / \mathrm{mol} .\) What is the molecular formula of the compound?

Write the formula for each of the following compounds and indicate which ones are best described as ionic: (a) sodium hypochlorite (b) boron triiodide (c) aluminum perchlorate (d) calcium acetate (e) potassium permanganate (f) ammonium sulfite (g) potassium dihydrogen phosphate (h) disulfur dichloride (i) chlorine trifluoride (j) phosphorus trifluoride

Write the formulas for the four ionic compounds that can be made by combining the cations \(\mathrm{Mg}^{2+}\) and \(\mathrm{Fe}^{3+}\) with the anions \(\mathrm{PO}_{4}^{3-}\) and \(\mathrm{NO}_{3}^{-} .\) Name each compound formed.
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