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Chapter 10: Problem 22

Silver has two stable isotopes. The nucleus, \(^{107}_{47} \mathrm{Ag},\) has atomic mass \(106.905095 \mathrm{g} / \mathrm{mol}\) with an abundance of \(51.83 \% ;\) whereas \(^{109}_{47} \mathrm{Ag}\) has atomic mass \(108.904754 \mathrm{g} /\)mol with an abundance of \(48.17 \% .\) Find the atomic mass of the element silver.

Short Answer

Expert verified
The atomic mass of the element silver is approximately \(107.881 \, \mathrm{g/mol}\).

Step by step solution

01

Identify given values

We are given the following information: - Isotope 1: - Atomic mass: 106.905095 g/mol - Abundance: 51.83% - Isotope 2: - Atomic mass: 108.904754 g/mol - Abundance: 48.17%
02

Convert percentages to decimals

To use the given percentages in further calculations, we need to convert them to decimals: - Abundance of isotope 1: \(51.83\% = \frac{51.83}{100} = 0.5183\) - Abundance of isotope 2: \(48.17\% = \frac{48.17}{100} = 0.4817\)
03

Calculate the weighted average

The atomic mass of silver can be found by calculating the weighted average of the atomic masses of its isotopes: \(Atomic \, mass \, of \, silver = (Atomic \, mass \, of \, isotope \, 1 \times Abundance \, of \, isotope \, 1) + (Atomic \, mass \, of \, isotope \, 2 \times Abundance \, of \, isotope \, 2)\) Plug in the values and perform the calculation: \(Atomic \, mass \, of \, silver = (106.905095 \times 0.5183) + (108.904754 \times 0.4817)\) \(Atomic \, mass \, of \, silver = 55.45041529 + 52.43043445\)
04

Find the atomic mass of silver

Combine the results from the previous step: \(Atomic \, mass \, of \, silver = 55.45041529 + 52.43043445 = 107.88084974 \, \mathrm{g/mol}\) The atomic mass of the element silver is approximately \(107.881 \, \mathrm{g/mol}\).

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Key Concepts

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

Isotopic Abundance

Isotopic abundance is a key concept in understanding the nature of elements. It refers to the relative percentage of each isotope of an element that occurs in nature. Since isotopes of an element vary only in the number of neutrons in their nuclei, they have almost identical chemical properties but different atomic masses. Isotopic abundance is crucial for calculating the average atomic mass of an element, which is what we observe on the periodic table and use for calculations in chemical reactions.

Significance in Calculations

The specific proportion in which different isotopes of an element are found naturally affects its overall atomic mass. To get an accurate understanding of an element's behavior in reactions, scientists and students must account for these natural variations through isotopic abundance.

Weighted Average

The concept of a weighted average is integral in various scientific fields, especially in chemistry when calculating atomic masses. A weighted average, unlike a simple arithmetic mean, accounts for the varying significance—or weight—of each value it includes. In terms of isotopes, the atomic mass of each isotope isn't equally representative of an element's overall atomic mass. Instead, we must weigh each isotopic mass by its abundance.

Calculating Atomic Mass

To find the average atomic mass of an element, multiply the mass of each isotope by its fractional abundance (percentage abundance as a decimal), adding these products together. This ensures that more abundant isotopes have a greater influence on the final result, more accurately reflecting the average atomic weight one would encounter.

Chemistry Education

Chemistry education strives to impart a deep understanding of not just chemical reactions and lab techniques, but foundational principles like atomic theory and properties of elements. Comprehension of isotopic abundance and weighted averages is crucial for students, as these concepts underpin the calculations that allow chemists to predict the outcomes of chemical reactions.

Effective Learning Approaches

In educational settings, breaking down complex topics into simpler components can improve student understanding. Interactive exercises, like calculating the atomic mass of an element, bridge theory with practical application, reinforcing the material through active learning. Emphasizing the real-world implications of these calculations also helps in retaining students’ interest and curiosity.

Nuclear Physics

Nuclear physics examines the components and behavior of atomic nuclei. Isotopic abundance relates to this field as it provides insights into nuclear stability and the processes by which isotopes form, such as radioactive decay or stellar nucleosynthesis. These variations in isotope distribution contribute to the diversity of atomic mass among elements.

Real-World Implications

Nuclear physics not only aids our understanding of fundamental aspects of matter but also has practical applications like energy production. The nuclear industry often exploits the differences between isotopes, such as using certain uranium isotopes for fuel in nuclear reactors.

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Most popular questions from this chapter

If two nuclei are to fuse in a nuclear reaction, they must be moving fast enough so that the repulsive Coulomb force between them does not prevent them for getting within \(R \approx 10^{-14} \mathrm{m}\) of one another. At this distance or nearer, the attractive nuclear force can overcome the Coulomb force, and the nuclei are able to fuse. (a) Find a simple formula that can be used to estimate the minimum kinetic energy the nuclei must have if they are to fuse. To keep the calculation simple, assume the two nuclei are identical and moving toward one another with the same speed \(v\). (b) Use this minimum kinetic energy to estimate the minimum temperature a gas of the nuclei must have before a significant number of them will undergo fusion. Calculate this minimum temperature first for hydrogen and then for helium. (Hint: For fusion to occur, the minimum kinetic energy when the nuclei are far apart must be equal to the Coulomb potential energy when they are a distance \(R\) apart.)

Explain why a bound system should have less mass than its components. Why is this not observed traditionally, say, for a building made of bricks?

A rare decay mode has been observed in which \(^{222} \mathrm{Ra}\) emits a \(^{14} \mathrm{C}\) nucleus. (a) The decay equation is \(^{222} \mathrm{Ra} \rightarrow^{A} \mathrm{X}+^{14} \mathrm{C} .\) Identify the nuclide \(^{A} \mathrm{X} .\) (b) Find the energy emitted in the decay. The mass of \(222 \mathrm{Ra}\) is 222.015353 u.

What is the difference between \(\gamma\) rays and characteristic X-rays and visible light?

(a) Calculate the energy released in the neutron-induced fission reaction \(n+^{235} \mathrm{U} \rightarrow^{92} \mathrm{Kr}+^{142} \mathrm{Ba}+2 n,\) given \(m\left(^{92} \mathrm{Kr}\right)=91.926269 \mathrm{u}\) and \(m\left(^{142} \mathrm{Ba}\right)=141.916361 \mathrm{u} .\) (b) Confirm that the total number of nucleons and total charge are conserved in this reaction.
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