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The atomic masses of \({ }^{204} \mathrm{~Pb}\left(1.4\right.\) percent), \({ }^{, 06} \mathrm{P} \mathrm{b}\) (24.1 percent), \({ }^{207} \mathrm{~Pb}\) ( 22.1 percent), and \({ }^{20 \mathrm{c} \mathrm{s}} \mathrm{Pb}\) \((52.4\) percent) are \(203.973020 .205 .974440,206.975872,\) and 207.976627 amu, respectively. Calculate the average atomic mass of lead. The percentages in parentheses denote the relative abundances.

Short Answer

Expert verified
The average atomic mass of lead is 207.2053 amu.

Step by step solution

01

Understand the Problem

We need to calculate the average atomic mass of lead based on the given isotopes and their abundances. The average atomic mass is a weighted average, where each isotope's mass is multiplied by its abundance.
02

Set up the Formula

The formula to calculate the average atomic mass is:\[ \text{Average Atomic Mass} = \sum (\text{fractional abundance} \times \text{atomic mass}) \]For each isotope, convert the percentage to a fractional abundance. For instance, if an isotope has a 1.4% abundance, its fractional abundance is 0.014.
03

Calculate Fractional Abundances

Convert each percentage into a decimal (fractional abundance):- For \(^{204}\text{Pb}\): 1.4% becomes 0.014- For \(^{206}\text{Pb}\): 24.1% becomes 0.241- For \(^{207}\text{Pb}\): 22.1% becomes 0.221- For \(^{208}\text{Pb}\): 52.4% becomes 0.524
04

Multiply Each Mass by Its Fractional Abundance

Calculate the contribution of each isotope to the average atomic mass:- \(^{204}\text{Pb}\): \(203.973 \times 0.014 = 2.855622\)- \(^{206}\text{Pb}\): \(205.974 \times 0.241 = 49.641234\)- \(^{207}\text{Pb}\): \(206.976 \times 0.221 = 45.742296\)- \(^{208}\text{Pb}\): \(207.977 \times 0.524 = 108.966148\)
05

Sum the Contributions

Add up all the products from Step 4 to find the average atomic mass:\[2.855622 + 49.641234 + 45.742296 + 108.966148 = 207.2053 \text{ amu}\]

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Key Concepts

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

Understanding Isotopes
An isotope refers to one of two or more species of atoms of a chemical element that have the same number of protons in their nucleus but different numbers of neutrons. This results in varying atomic masses for isotopes of the same element. For example, lead has several isotopes like ^{204}Pb, ^{206}Pb, ^{207}Pb, and ^{208}Pb. Each isotope has a specific atomic mass, which contributes to the overall average atomic mass of the element found on the periodic table. Despite the differences in atomic mass, the chemical properties of isotopes of the same element remain very similar.
Fractional Abundance Explained
Fractional abundance is a way to express the abundance of a particular isotope compared to all isotopes of that element. To calculate fractional abundance, the percentage abundance of each isotope is converted into a fraction. For instance, if an isotope has an abundance of 24.1%, it is equivalent to a fractional abundance of 0.241. This value is essential as it represents how frequently the isotope occurs in nature relative to other isotopes of the element. This concept is important in calculating the average atomic mass, as it weighs how each isotope's atomic mass should contribute to the overall average.
The Concept of Weighted Average
The weighted average is a statistical method used to calculate the average of quantities with different relevancies, or weights. In the context of atomic masses, it involves multiplying the mass of each isotope by its fractional abundance and then summing these values. The formula is: \[ \text{Average Atomic Mass} = \sum (\text{fractional abundance} \times \text{atomic mass}) \]. This approach accounts for the fact that not all isotopes occur equally in nature, ensuring that more common isotopes contribute more significantly to the average atomic mass.
Atomic Mass Unit (amu) and Its Role
An atomic mass unit (amu) is a standard unit of measure for expressing atomic and molecular weights. It is defined as one twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state, and is equivalent to approximately 1.66 x 10^{-27} kilograms. The use of amu is crucial in chemistry and physics because it allows scientists to measure and compare the exceedingly small masses of atoms and molecules in a more manageable way. The average atomic mass of elements, like that of lead calculated as 207.2053 amu from its isotopes, is expressed in this unit on the periodic table.

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Most popular questions from this chapter

Describe the eontributions of the following scientists 10 our knowledge of atomic structure: J. J. Thomson. R. A Millikan, Ernest Rutherford, and James Chadwick:

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