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Verify that the average atomic mass of lithium is 6.941 , given this information: \({ }^{6} \mathrm{Li},\) exact mass \(=6.015121 \mathrm{u}\) percent abundance \(=7.500 \%\) \({ }^{7} \mathrm{Li},\) exact mass \(=7.016003 \mathrm{u}\) percent abundance \(=92.50 \%\)

Short Answer

Expert verified
The average atomic mass of lithium, 6.941 u, is correct based on isotope data.

Step by step solution

01

Understanding Percent Abundance

The percentage abundance of each isotope indicates how much of that isotope is present in a typical sample of lithium. For example, \(^{6} Li\) has an abundance of 7.500%, which means that out of 100 lithium atoms, approximately 7.5 atoms will be \(^{6} Li\). Similarly, \(^{7} Li\) has an abundance of 92.50%.
02

Convert Percent to Decimal

Convert the percent abundance into a decimal for calculation. For \(^{6} Li\), the abundance is \(7.500\%\) or \(0.0750\) in decimal form. For \(^{7} Li\), the abundance is \(92.50%\) or \(0.9250\) in decimal form.
03

Calculate Weighted Mass Contribution

Multiply the exact mass of each isotope by its decimal abundance to find its contribution to the average atomic mass. For \(^{6} Li\):\[6.015121 \, \text{u} \times 0.0750 = 0.4511341 \\]For \(^{7} Li\):\[7.016003 \, \text{u} \times 0.9250 = 6.4888028\\]
04

Calculate Average Atomic Mass

Add up the contributions of each isotope to get the average atomic mass of lithium:\[0.4511341 \, \text{u} + 6.4888028 \, \text{u} = 6.9399369 \, \text{u}\\]Rounding this to three decimal places gives 6.940 u.
05

Verification

Compare the calculated average atomic mass (6.940 u) with the given average ( 6.941 u). The calculated mass is very close to the provided data, verifying that the average atomic mass provided is accurate given the isotope data.

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Key Concepts

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

Isotope Abundance
Isotope abundance, often expressed in percentages, tells us the distribution of different isotopes of an element found in nature. For example, if we look at lithium, it primarily exists as two isotopes:
  • ^{6}Li with an abundance of 7.500%
  • ^{7}Li with an abundance of 92.50%
This percent abundance helps us understand how common each isotope is within a given sample.
Every sample of lithium will have about 7.5% of ^{6}Li and 92.5% of ^{7}Li, highlighting the dominance of ^{7}Li in nature. This distribution is crucial when calculating the average atomic mass of lithium or any other element.
Weighted Average
The weighted average is essential when calculating the average atomic mass. This concept ensures that each isotope's mass is properly proportioned based on its abundance.
To compute the weighted average, convert percent abundances into their decimal forms by dividing by 100. This allows precise calculations:
  • ^{6}Li abundance as decimal: 0.0750
  • ^{7}Li abundance as decimal: 0.9250
Next, multiply the mass of each isotope by its decimal abundance:
  • ^{6}Li: 6.015121 u × 0.0750 = 0.4511341 u
  • ^{7}Li: 7.016003 u × 0.9250 = 6.4888028 u
Summing these products provides the average atomic mass:
6.9399369 u, rounded to 6.940 u. This demonstrates how significant isotope abundances are in calculating weighted averages.
Lithium Isotopes
Lithium, one of the lightest metals, exists naturally in two stable isotopes:
  • ^{6}Li
  • ^{7}Li
These isotopes vary by the number of neutrons. ^{6}Li contains three protons and three neutrons, while ^{7}Li contains three protons and four neutrons.
Despite the small difference in their mass numbers, ^{7}Li is significantly more abundant. It contributes more heavily to the average atomic mass of lithium.
This heavier isotope's predominance impacts physical properties and is crucial for applications such as nuclear fusion research and battery technology.
Mass Spectrometry
Mass spectrometry is an analytical technique that helps us determine isotope abundance and natural atomic masses. It separates isotopes based on their mass-to-charge ratio, allowing us to quantify and identify isotopes present in a sample.
For elements like lithium, this method can accurately measure the relative abundance of different isotopes such as ^{6}Li and ^{7}Li.
During mass spectrometry, the element's ions are generated, sorted, and detected, resulting in a spectrum that displays isotope abundance. This technology is integral in verifying isotope distribution and calculating precise average atomic masses of elements.

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Most popular questions from this chapter

There are four binary potassium compounds of oxygen. They contain these mass percents of potassium: Compound I, \(83.0 \%\); Compound II, \(55.0 \%\); Compound III, \(44.9 \%\); and Compound IV, \(71.0 \%\). One compound has molar mass equal to \(110.2 \mathrm{~g} / \mathrm{mol}\). Use this information to determine the chemical formula of each compound.

Ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), has a density of \(0.789 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Water weighs \(1.00 \mathrm{~kg} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\). Calculate the volume of ethanol that contains the same number of molecules as \(1.00 \mathrm{~L}\) water.

The present average concentration (mass percent) of magnesium ions in seawater is \(0.13 \%\). A chemistry textbook estimates that if \(1.00 \times 10^{8}\) tons \(\mathrm{Mg}\) were taken out of the sea each year, it would take one million years for the Mg concentration to drop to \(0.12 \% .\) Do sufficient calculations to either verify or refute this statement. Assume that Earth is a sphere with a diameter of \(8000 \mathrm{mi}, 67 \%\) of which is covered by oceans to a depth of \(1 \mathrm{mi}\), and that no \(\mathrm{Mg}\) is washed back into the oceans at any time.

Three oxygen-containing compounds of iron are \(\mathrm{FeCO}_{3}\), \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4} \cdot\) Calculate the percent iron by mass in each iron compound.

For each pair of elements, (i) through (vii), (a) Determine whether an ionic compound, a molecular compound, or no compound would form. (b) Write an appropriate formula for each compound you expect to form and name the compound. (i) \(\quad\) Chlorine and bromine (ii) \(\quad\) Lithium and tellurium (iii) Sodium and argon (iv) \(\quad\) Magnesium and fluorine (v) Nitrogen and bromine (vi) \(\quad\) Indium and sulfur (vii) Selenium and bromine

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