Question

The atomic masses of \({ }^{6} \mathrm{Li}\) and \({ }^{7} \mathrm{~L}_{\mathrm{i}}\) are \(6.0151 \mathrm{amu}\) and 7.0160 amu, respectively. Calculate the natural abundances of these two isotopes. The average atomic mass of \(\mathrm{Li}\) is 6.941 amu.

Short answer

The natural abundances are 7.49% for \( ^6 \text{Li} \) and 92.51% for \( ^7 \text{Li} \).

Step-by-step solution

Understand the Problem

We need to find the natural abundances of \( ^6 \text{Li} \) and \( ^7 \text{Li} \). This involves using their atomic masses and the given average atomic mass of lithium, which is 6.941 amu.

Define Variables and Equation

Let \( x \) be the fraction of \( ^6 \text{Li} \), then \( 1-x \) will be the fraction of \( ^7 \text{Li} \). The equation relating atomic masses and abundances is:\[ x \times 6.0151 + (1-x) \times 7.0160 = 6.941. \]

Expand and Simplify the Equation

Expand the equation:\[ 6.0151x + 7.0160 - 7.0160x = 6.941. \]Simplify it to:\[ -1.0009x + 7.0160 = 6.941. \]

Solve for x

Rearrange the equation to solve for \( x \):\[ -1.0009x = 6.941 - 7.0160, \]\[ -1.0009x = -0.075. \]Divide both sides by \(-1.0009\):\[ x = \frac{-0.075}{-1.0009} \approx 0.0749. \]

Solve for 1-x

Since \( x \) represents the fraction of \( ^6 \text{Li} \), \( 1-x \) represents the fraction of \( ^7 \text{Li} \):\[ 1-x = 1 - 0.0749 = 0.9251. \]

Convert Fractions to Percentages

Convert fractions to percentages for natural abundances:\( ^6 \text{Li} \) abundance is \( 0.0749 \times 100\% = 7.49\% \).\( ^7 \text{Li} \) abundance is \( 0.9251 \times 100\% = 92.51\% \).

Key concepts

Isotopic Abundance

Isotopic abundance refers to the percentage of a particular isotope present in a sample of an element. Isotopes are atoms of the same element with different numbers of neutrons, resulting in different atomic masses. Understanding isotopic abundance is crucial because elements in nature usually exist as a mix of isotopes.
Consider lithium (Li) as an example. It primarily exists as two isotopes: \(^6 ext{Li}\) and \(^7 ext{Li}\). When you weigh a sample of lithium, its mass will largely be influenced by the relative amounts of each isotope present. These isotopic abundances determine the average atomic mass of the element.
Because isotopic abundance affects the chemical properties and reactions of elements, it's a significant aspect of many fields, including chemistry, physics, and Earth science.

Average Atomic Mass

The average atomic mass is a weighted average of the atomic masses of an element's isotopes, based on their natural abundances. This value is what you often see on the periodic table. It reflects the average mass of an atom of that element, considering the varying contributions from each isotope.
Calculating the average atomic mass involves multiplying the mass of each isotope by its fractional abundance and summing these values. For lithium (Li), the average atomic mass is given as 6.941 amu, which incorporates the masses of \(^6 ext{Li}\) and \(^7 ext{Li}\) and their respective abundances.
The formula for average atomic mass can be represented as follows:

• If \(x\) is the fraction for \(^6 ext{Li}\), then \(1-x\) represents \(^7 ext{Li}\).
• The formula becomes: \[x(6.0151) + (1-x)(7.0160) = 6.941\].

Utilizing this formula gives us insights into how the isotopic masses and their abundances contribute to the element's average atomic mass.

Mass Spectrometry

Mass spectrometry is an analytical technique that measures the mass-to-charge ratio of ions. It is essential for identifying the isotopic composition of an unknown sample by revealing the individual isotopes present and their abundances.
In this context, mass spectrometry aids in calculating isotopic abundances by providing detailed mass distribution data for an element. When analyzing lithium (Li) through mass spectrometry, the machine would output peaks corresponding to the isotopes \(^6 ext{Li}\) and \(^7 ext{Li}\). The area of these peaks reflects their relative abundances.
Mass spectrometry is crucial because it allows chemists to accurately determine the average atomic mass of elements, verify the presence of isotopes, and carry out various chemical calculations with precision. This technique is widely used in chemistry and is instrumental in drug testing, environmental analysis, and biochemical studies.

Chemical Calculations

Chemical calculations involve various mathematical procedures to determine quantities such as the amount of substance, concentration, yield, and, in this case, isotopic abundances.
In the exercise provided, chemical calculations were used to derive the natural abundances of lithium isotopes. This process involved setting up an equation based on the weighted average formula, which is applied to the known atomic masses and average atomic mass. After defining variables, students solve the equation to find the abundance of each isotope.
Such calculations are integral to chemistry, enabling scientists to predict reactions, understand elemental properties, and develop new materials. Mastering these calculations empowers one to approach various scientific challenges with confidence and creativity.

• The calculation follows these steps:
  • Define variables for isotopic fractions.
  • Construct an equation based on isotopic masses.
  • Solve the equation for each isotope's fraction and convert to percentages.

Understanding these critical elements of chemical calculations is foundational for any budding chemist.