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Chapter 2: Problem 93

The nucleus of \({ }^{6} \mathrm{Li}\) is a powerful absorber of neutrons. It exists in the naturally occurring metal to the extent of \(7.5 \%\). In the era of nuclear deterrence, large quantities of lithium were processed to remove \({ }^{6} \mathrm{Li}\) for use in hydrogen bomb production. The lithium metal remaining after removal of \({ }^{6} \mathrm{Li}\) was sold on the market. (a) What are the compositions of the nuclei of \({ }^{6} \mathrm{Li}\) and \({ }^{7} \mathrm{Li}\) ? (b) The atomic masses of \({ }^{6} \mathrm{Li}\) and \({ }^{7} \mathrm{Li}\) are \(6.015122\) and 7.016004 amu, respectively. A sample of lithium depleted in the lighter isotope was found on analysis to contain \(1.442 \%^{6} \mathrm{Li}\). What is the average atomic weight of this sample of the metal?

Short Answer

Expert verified
The compositions of the \({ }^{6}\mathrm{Li}\) and \({ }^{7}\mathrm{Li}\) nuclei are as follows: \({ }^{6}\mathrm{Li}\) has 3 protons and 3 neutrons, while \({ }^{7}\mathrm{Li}\) has 3 protons and 4 neutrons. Given a sample of lithium depleted in \({ }^{6}\mathrm{Li}\) with a concentration of \(1.442 \%\) of said isotope, the remaining \({ }^{7}\mathrm{Li}\) concentration is \(98.558\%\). The average atomic weight of this sample is calculated as \(7.00426 \) amu.

Step by step solution

01

Identify the compositions of \({ }^{6} \mathrm{Li}\) and \({ }^{7} \mathrm{Li}\) nuclei

The notation "\({ }^{A} \mathrm{X}\)" represents an isotope of element X with mass number A. The mass number (A) is the sum of the number of protons (Z) and the number of neutrons (N) in the nucleus of the atom. In this case, we need to determine the numbers of protons and neutrons for each isotope of lithium (\({ }^{6}\mathrm{Li}\) and \({ }^{7}\mathrm{Li}\)). The atomic number of lithium (Li) is 3, which means it has 3 protons in the nucleus. Number of protons (Z) for lithium isotopes: Z = 3 For \({ }^{6}\mathrm{Li}\): Number of neutrons (N) = Mass number (A) - Number of protons (Z) N = 6 - 3 = 3 For \({ }^{7}\mathrm{Li}\): N = 7 - 3 = 4
02

Calculate average atomic weight of the sample

Given that a sample of lithium depleted in the lighter isotope was found to contain \(1.442 \%^{6} \mathrm{Li}\), we can determine the percentage of \({ }^{7}\mathrm{Li}\) in the sample as follows: Percentage of \({ }^{7}\mathrm{Li} = 100\% - 1.442\% = 98.558\%\) Now we can calculate the average atomic weight of this sample by taking the weighted average of the atomic masses of \({ }^{6} \mathrm{Li}\) and \({ }^{7} \mathrm{Li}\): Average atomic weight = (Percentage of \({ }^{6}\mathrm{Li}\))(Atomic mass of \({ }^{6}\mathrm{Li}\)) + (Percentage of \({ }^{7}\mathrm{Li}\))(Atomic mass of \({ }^{7}\mathrm{Li}\)) = \((0.01442)(6.015122) + (0.98558)(7.016004)\)
03

Calculate the result

Now let's perform the calculations: Average atomic weight = \((0.01442)(6.015122) + (0.98558)(7.016004)\) = \(0.086744 + 6.91752\) = \(7.00426\) Thus, the average atomic weight of this sample of lithium depleted in the lighter isotope is approximately 7.004 amu.

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Key Concepts

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

Atomic Mass
Atomic mass is a fundamental concept in the study of elements and their isotopes. It represents the total mass of an atom, primarily due to the mass of its protons and neutrons. Electrons have very little mass compared to protons and neutrons and contribute negligibly to the atomic mass.
The atomic mass is usually expressed in atomic mass units (amu), where 1 amu is defined as one twelfth of the mass of a carbon-12 atom. This allows chemists to compare the masses of different atoms on a relative scale.
For isotopes, the atomic mass reflects the number of protons and neutrons in the nucleus. For example, the atomic mass of the isotope extsuperscript{6}Li is 6.015122 amu. This reflects that the nucleus of this isotope contains six nucleons in total鈥攖hree protons and three neutrons.
Nuclear Composition
The nuclear composition of an atom refers to the arrangement and number of protons and neutrons in its nucleus. This composition defines the identity and isotope of the element. Here, we'll examine the nuclear compositions of extsuperscript{6}Li and extsuperscript{7}Li isotopes.
  • extsuperscript{6}Li: This isotope has an atomic number (Z) of 3, meaning it has 3 protons. To determine the number of neutrons, subtract the number of protons from the isotope's mass number (A): extsuperscript{6}Li has 3 neutrons ( A = 6 - Z = 6 - 3 = 3).
  • extsuperscript{7}Li: Similarly, extsuperscript{7}Li has 3 protons and 4 neutrons ( A = 7 - Z = 7 - 3 = 4).
The composition of the nucleus affects not only the atomic mass but also the behavior of the isotope. For example, extsuperscript{6}Li is known for its ability to absorb neutrons, playing a role in nuclear reactions.
Average Atomic Weight
The average atomic weight of an element is a weighted mean of the masses of its naturally occurring isotopes, based on their relative abundances.
In the case of a sample containing less extsuperscript{6}Li, as described in the exercise, the average atomic weight needs to be recalculated. It must account for the different proportions of extsuperscript{6}Li and extsuperscript{7}Li.
To find the average atomic weight, you can use the following formula:
  • Determine the percentage of each isotope in the sample. For example, extsuperscript{6}Li at 1.442% and extsuperscript{7}Li at 98.558%.
  • Use a weighted average: \(\text{Average Atomic Weight} = (\text{Percentage of } \text{ extsuperscript{6}Li})\times(\text{Atomic Mass of } \text{ extsuperscript{6}Li}) + (\text{Percentage of } \text{ extsuperscript{7}Li})\times(\text{Atomic Mass of } \text{ extsuperscript{7}Li})\)
In our problem, this calculation results in an average atomic weight of approximately 7.004 amu for the lithium sample.
Lithium Isotopes
Lithium has two naturally occurring isotopes: extsuperscript{6}Li and extsuperscript{7}Li. These isotopes differ in the number of neutrons within their nuclei, giving them distinct nuclear compositions and properties.
  • extsuperscript{6}Li: Comprising about 7.5% of natural lithium, this isotope has an atomic mass of 6.015122 amu. It contains 3 protons and 3 neutrons. extsuperscript{6}Li is well-known for its neutron absorption capabilities, making it critical for certain nuclear applications including hydrogen bomb production.
  • extsuperscript{7}Li: This isotope is more abundant, making up around 92.5% of natural lithium. With an atomic mass of 7.016004 amu, it has 3 protons and 4 neutrons. This isotope is more stable and less reactive compared to extsuperscript{6}Li.
The different isotopes of lithium result in its varied applications in both commercial and scientific fields, highlighting their importance.

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Most popular questions from this chapter

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