Elements in nature often exist as a blend of different atomic forms. Percent abundance quantifies the natural ratio of these forms, indicating the relative amount of each specific type of atom in a sample. This measurement is important for understanding the average mass of an element, the value typically seen on the periodic table.
The Building Blocks: Isotopes
Atoms of the same element share an identical number of protons, which defines their atomic number and chemical identity. However, these atoms can differ in their number of neutrons, leading to variations in their atomic mass. These differing forms are known as isotopes.
For example, all carbon atoms have six protons, but some have six neutrons (carbon-12), while others might have seven (carbon-13) or eight (carbon-14). This difference in neutron count means that isotopes of an element have distinct masses.
Most elements found in nature are a consistent mixture of several isotopes. The percent abundance describes the proportion of each of these isotopic varieties within that natural mixture. This natural ratio of isotopes remains largely constant across the Earth.
The Calculation Method
Determining the percent abundance of isotopes relies on the concept of the average atomic mass, which is a weighted average of the masses of an element’s naturally occurring isotopes. This average atomic mass is the value listed on the periodic table for each element. The calculation uses a formula that accounts for each isotope’s mass and its relative presence.
The formula for average atomic mass is: Average Atomic Mass = (Isotopic Mass₁ × Fractional Abundance₁) + (Isotopic Mass₂ × Fractional Abundance₂) + …
In this equation, “Isotopic Mass” refers to the precise mass of a particular isotope, while “Fractional Abundance” is the decimal equivalent of the percent abundance (percent abundance divided by 100). The sum of all fractional abundances for an element’s isotopes must always equal 1, representing 100% of the element.
Worked Example: Putting It All Together
To illustrate, consider chlorine (Cl), with an average atomic mass of 35.45 atomic mass units (amu). Chlorine exists as two stable isotopes: chlorine-35 (34.96885 amu) and chlorine-37 (36.96590 amu).
Let ‘x’ represent the fractional abundance of chlorine-35. Since the total fractional abundance must be 1, the fractional abundance of chlorine-37 will be (1 – x). Set up the equation: 35.45 amu = (34.96885 amu x) + (36.96590 amu (1 – x)).
Distribute the terms: 35.45 = 34.96885x + 36.96590 – 36.96590x. Combine ‘x’ terms and isolate ‘x’: 35.45 – 36.96590 = 34.96885x – 36.96590x, which simplifies to -1.5159 = -1.99705x.
Dividing by -1.99705 yields x ≈ 0.7590. To convert this fractional abundance to percent, multiply by 100. Chlorine-35 has an abundance of approximately 75.90%. Chlorine-37’s abundance is 1 – 0.7590 = 0.2410, or 24.10%.