Problem 134 - Chemistry The Molecular Science

Question

There are three naturally occurring isotopes of potassium: \({ }^{39} \mathrm{~K} 38.963707 \mathrm{u} ;{ }^{40} \mathrm{~K} 39.963999 \mathrm{u}\); and \({ }^{41} \mathrm{~K}\) \(40.961825 \mathrm{u}\). The average atomic mass of potassium is \(39.0983 \mathrm{u}\) and the natural abundance of the lightest isotope is \(93.2581 \% .\) Calculate the natural abundances of the other two isotopes.

Step-by-Step Solution

Answer

Natural abundances: \({}^{40}\mathrm{K} = 1.0369\%\), \({}^{41}\mathrm{K} = 5.705\%\).

Step 1: Assign Variables

Assign variables to the unknown natural abundances of the isotopes. Let \( x \) be the natural abundance of \( {}^{40}\mathrm{K} \) and \( y \) be the natural abundance of \( {}^{41}\mathrm{K} \). Since the total abundance must be 100%, we can state that \( 93.2581\% + x + y = 100\% \).

Step 2: Set Up the Equation for Average Atomic Mass

The average atomic mass of an element can be calculated using the formula: \[\text{Average atomic mass} = \sum (\text{isotope mass} \times \text{natural abundance})\]Substitute the given values:\[39.0983 = (38.963707 \times 0.932581) + (39.963999 \times x) + (40.961825 \times y)\]

Step 3: Solve the Total Abundance Equation

Simplify the total abundance equation from Step 1: \[x + y = 100\% - 93.2581\% = 6.7419\%\]So, \(x + y = 0.067419\) when expressed as a decimal.

Step 4: Solve the Average Atomic Mass Equation

Calculate the contribution of the \({ }^{39}\mathrm{K}\) isotope to the average atomic mass: \[38.963707 \times 0.932581 = 36.344256\]Insert this into the equation:\[39.0983 = 36.344256 + (39.963999x) + (40.961825y)\]Rearrange the equation:\[2.754044 = 39.963999x + 40.961825y\]

Step 5: Solve the System of Equations

We now have two equations:1. \( x + y = 0.067419 \)2. \( 39.963999x + 40.961825y = 2.754044 \)Solve equation 1 for \( x \):\[ x = 0.067419 - y\]Substitute \( x \) in equation 2:\[ 39.963999(0.067419 - y) + 40.961825y = 2.754044 \]This simplifies to:\[ 2.697135 - 39.963999y + 40.961825y = 2.754044 \]Combine like terms:\[ 0.997826y = 2.754044 - 2.697135 \]Solve for \( y \):\[ y = \frac{0.056909}{0.997826} = 0.05705\]

Step 6: Find x using y

Now, substitute \( y = 0.05705 \) back into the equation \( x = 0.067419 - y \):\[ x = 0.067419 - 0.05705 = 0.010369 \]

Step 7: Convert to Percentages

Convert \( x \) and \( y \) back to percentages by multiplying by 100:- \( x = 0.010369 \times 100 = 1.0369\% \)- \( y = 0.05705 \times 100 = 5.705\% \)

Key Concepts

Average Atomic Mass

The concept of "average atomic mass" is an essential aspect of understanding how elements behave in nature. Atoms of the same element can have different numbers of neutrons, resulting in isotopes. Each isotope has a different atomic mass.
To calculate the average atomic mass of an element, you use the formula:

• Average atomic mass = sum of (mass of each isotope × its natural abundance)

This calculation involves summing up all the products of the individual isotopes' masses and their respective natural abundances.
The average atomic mass is not simply the average of the masses of the isotopes, but a weighted average, where each mass contributes according to its abundance in nature. This is why the average atomic mass often does not equal the mass of any particular isotope, but rather represents a kind of "center of mass" for the natural mix of isotopes.

Natural Abundance

Natural abundance refers to the proportion of a particular isotope that occurs naturally relative to all the isotopes of that element. It is an indication of how frequently you would encounter that specific isotope if you were to sample the element in nature.
Natural abundance is usually expressed as a percentage, representing out of 100% the likelihood of finding each specific isotope.

• For example, if an isotope's natural abundance is said to be 93.2581%, it means that approximately 93 out of every 100 atoms of that element found in nature will be of that isotope.

This information is crucial when calculating the average atomic mass of an element because each isotope contributes to the element's overall atomic mass in proportion to its natural abundance.
In the exercise example, we are given the natural abundance of one isotope ( ^{39}K = 93.2581% ), and we use this along with the atomic masses to find the abundances of the other isotopes that make up the element.

Potassium Isotopes

Potassium (symbol K) is an element that has multiple isotopes with different atomic masses. In nature, three isotopes of potassium occur naturally, namely ^{39}K, ^{40}K, and ^{41}K.

Isotope	Mass
^{39}K	38.963707 u
^{40}K	39.963999 u
^{41}K	40.961825 u

Each of these isotopes contributes to potassium's average atomic mass within nature based on its own mass and its natural abundance.
The occurrence and behavior of these isotopes have practical applications in scientific fields, like geology and archaeology, due to ^{40}K's usefulness in dating rocks through its radioactive decay properties.

• ^{39}K: Most abundant isotope with significant impact on average atomic mass.
• ^{40}K: Known for slight radioactivity and use in radioisotope dating.
• ^{41}K: Less common, contributing less to overall mass calculation.

Understanding isotopes and their distribution helps scientists interpret data related to both elemental and isotopic compositions.