Journey Inside the Atom · Lesson 12 of 14
Average Atomic Mass
“Natural isotope abundance determines weighted average atomic mass”
• Explain why a simple average can be misleading. • Define relative abundance. • Calculate weighted average atomic mass. • Interpret chlorine's average atomic mass. • Explain why a fractional average is not an individual atom mass. • Solve a bromine weighted-average problem.
A natural element may contain more than one isotope. To represent the mass of the element as it occurs in nature, we must consider both the isotope masses and how common each isotope is.
Average Atomic Mass
Why a Simple Average Is Not Enough
Chlorine occurs mainly as isotopes with masses 35 u and 37 u. An ordinary mean gives 36 u only if the two isotopes are treated as equally common. In nature the lighter isotope forms about three quarters of chlorine while the heavier isotope forms about one quarter.
The proportion or percentage in which an isotope occurs in a natural sample.
Weighted Average Atomic Mass
The average obtained by multiplying each isotope mass by its relative abundance and adding the contributions.
Problem
Chlorine-35 is about 75% and chlorine-37 about 25%. Find the average.
- 1.Convert to decimals: 0.75 and 0.25.
- 2.35 × 0.75 = 26.25 u.
- 3.37 × 0.25 = 9.25 u.
- 4.Add: 26.25 + 9.25 = 35.5 u.
- 5.The weighted average is 35.5 u.
What a Fractional Average Means
The value 35.5 u does not mean one chlorine atom has a mass halfway between 35 u and 37 u. It is the average across a large natural sample containing many atoms of the two isotopes in unequal proportions.
A Reliable Weighted-Average Method
List each isotope and abundance, convert percentages to decimals, multiply mass by abundance, add all contributions, and check that the final average lies between the smallest and largest isotope masses.
Problem
Bromine-79 is 49.7% and bromine-81 is 50.3%. Find the average.
- 1.Convert abundances: 0.497 and 0.503.
- 2.79 × 0.497 = 39.263.
- 3.81 × 0.503 = 40.743.
- 4.Add: 39.263 + 40.743 = 80.006 u.
- 5.The average is approximately 80.0 u.
- 6.The answer lies between 79 u and 81 u, so the result is reasonable.
Do not use an ordinary mean unless isotopes occur equally. The more abundant isotope pulls the weighted average closer to its mass.
Quiz
Which description best matches Relative Abundance?
Which description best matches Weighted Average Atomic Mass?
Which term matches this description: The proportion or percentage in which an isotope occurs in a natural sample.
Which term matches this description: The average obtained by multiplying each isotope mass by its relative abundance and adding the contributions.
Which statement is a key takeaway from this lesson?
Practice Problems
- Why is 36 u not the best natural value for chlorine in a three-to-one mixture?
- Find the weighted average of masses 10 u and 11 u occurring at 20% and 80%.
- Why can an average atomic mass be decimal even though isotope mass numbers are whole numbers?
- What range check should you perform after a weighted-average calculation?
Key Takeaways
• Many elements occur naturally as mixtures of isotopes. • Average atomic mass depends on both isotope mass and relative abundance. • It is calculated as a weighted average rather than a simple arithmetic mean. • The resulting average atomic mass lies between the masses of the isotopes present.