Problem 19 - Fundamentals of Physics

Question

A periodic table might list the average atomic mass of magnesium as being \(24.312 \mathrm{u},\) which is the result of weighting the atomic masses of the magnesium isotopes according to their natural abundances on Earth. The three isotopes and their masses are \({ }^{24} \mathrm{Mg}(23.98504 \mathrm{u}),{ }^{25} \mathrm{Mg}(24.98584 \mathrm{u}),\) and \({ }^{26} \mathrm{Mg}(25.98259 \mathrm{u})\) The natural abundance of \({ }^{24} \mathrm{Mg}\) is \(78.99 \%\) by mass (that is, \(78.99 \%\) of the mass of a naturally occurring sample of magnesium is due to the presence of \({ }^{24} \mathrm{Mg}\) ). What is the abundance of (a) \({ }^{25} \mathrm{Mg}\) and (b) \({ }^{26} \mathrm{Mg} ?\)

Step-by-Step Solution

Answer

The abundance of \(^{25} \text{Mg}\) is 10.5%, and \(^{26} \text{Mg}\) is 10.5%.

Step 1: Understand the Problem

We are given the average atomic mass of magnesium and need to find the natural abundance of two isotopes: \(^{25} \text{Mg}\) and \(^{26} \text{Mg}\). We know the natural abundance of \(^{24} \text{Mg}\).

Step 2: Set Up the Equations

First, we express the average atomic mass equation. If \(x\) is the natural abundance of \(^{25} \text{Mg}\) and \(y\) of \(^{26} \text{Mg}\), the equation becomes:\[ 23.98504 \times 0.7899 + 24.98584 \times x + 25.98259 \times y = 24.312 \]Additionally, since the total natural abundance must be 100%, we have:\[ 0.7899 + x + y = 1 \]

Step 3: Solve for the Abundance of \(^{25} \text{Mg}\) and \(^{26} \text{Mg}\)

We solve the second equation first for \(y\):\[ y = 1 - 0.7899 - x = 0.2101 - x \]Substitute \(y\) in the first equation:\[ 23.98504 \times 0.7899 + 24.98584 \times x + 25.98259 \times (0.2101 - x) = 24.312 \]

Step 4: Simplify and Solve for \(x\)

Calculate the left side:\[ 18.953348 + 24.98584x + 5.4647959 - 25.98259x = 24.312 \]\[ ?0.99675x = 24.312 - 18.953348 - 5.4647959 \]\[ -0.99675x = -0.1061439 \]

Step 5: Solve for \(x\)

Solve the equation for \(x\):\[ x = \frac{-0.1061439}{-0.99675} \approx 0.105 \]

Step 6: Find \(y\) Using \(x\)

Substitute \(x\) back into the equation for \(y\):\[ y = 0.2101 - 0.105 \approx 0.105 \]

Step 7: Final Results

The natural abundance of \(^{25} \text{Mg}\) is approximately 10.5%, and for \(^{26} \text{Mg}\), it is also approximately 10.5%.

Key Concepts

Average Atomic Mass

The concept of average atomic mass is central to understanding the weighted contributions of an element's isotopes. Isotopes are variations of the same element, each having a different number of neutrons. While individual isotopes have specific atomic masses, the average atomic mass takes the weighted mean based on natural abundance. This ensures that the mass listed on the periodic table is representative of the element as it naturally occurs.
The average atomic mass is calculated using the formula:

• Given isotopes:
  • ext{Isotope 1}: mass \( m_1 \), abundance \( x_1 \)
  • ext{Isotope 2}: mass \( m_2 \), abundance \( x_2 \)
  • ext{Isotope 3}: mass \( m_3 \), abundance \( x_3 \)

The formula becomes:\[ \text{Average Atomic Mass} = m_1 \times x_1 + m_2 \times x_2 + m_3 \times x_3 \]This simple yet powerful equation compiles the diverse isotopic data into a singular value that reflects the element's presence on Earth. Calculating the average atomic mass helps chemists understand isotope distribution and anticipate how elements behave in different chemical reactions.

Periodic Table

The periodic table is an iconic symbol in chemistry, organizing elements based on atomic number, electron configuration, and recurring chemical properties. Each element is listed with crucial data, including its symbol, atomic number, and average atomic mass. This table is fundamental for scientists to quickly understand elemental relationships and properties.
Each slot in the periodic table represents an element and provides keys to understanding its behavior:

• Atomic Number: The number of protons in an atom's nucleus.
• Element Symbol: A one or two-letter designation representing the element.
• Average Atomic Mass: Reflects the weighted average of all isotope masses based on their natural abundance.

Understanding how elements are grouped, such as in columns called groups or families and rows termed periods, allows for predicting many of their physical or chemical properties. The periodic table thus serves as a comprehensive guide to not only identify an element but also forecast its behavior in compounds, reactions, and experimental settings.

Natural Abundance

Natural abundance is the percentage of an isotope compared to all isotopes of that element found on Earth. Different isotopes exist due to variations in the number of neutrons, while the atomic number remains constant.
Consider magnesium, for example. It has multiple isotopes, including

• \(^{24} \text{Mg} \) with a natural abundance of 78.99%
• \(^{25} \text{Mg} \) and \(^{26} \text{Mg} \) both approximately at 10.5%

These percentages indicate how commonly each isotope appears in naturally occurring samples. This abundance data is vital:

• It influences the calculated average atomic mass, as more abundant isotopes contribute more to the average.
• Provides insights into an element’s stability and its tendency to participate in nuclear reactions.
• Impacts economic considerations, such as sourcing and utilizing specific isotopes for industrial applications.

Therefore, knowing the natural abundance of isotopes is essential for chemists working in fields ranging from analytical chemistry to environmental science. It aids in making informed decisions in both laboratory and industrial settings.