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General Chemistry I · Atoms, Molecules & Ions

Atomic Mass and Isotopic Abundance

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Key takeaway
  6. Study tools
  7. Sources & references

In 30 seconds

An element's atoms are not all identical in mass because most elements are mixtures of isotopes. The atomic mass printed on the periodic table is therefore a weighted average of the masses of all naturally occurring isotopes, with each isotope weighted by how abundant it is in nature. The unit is the atomic mass unit (u, also called the dalton, Da), defined so that carbon-12 has a mass of exactly 12 u. Because the weighting accounts for how much of each isotope actually exists — not a simple average of the isotope masses — the value for an element like chlorine (35.45 u) falls between its two isotope masses (35 u and 37 u) and is not a whole number.

Why this matters

The average atomic mass is the bridge from the invisible atomic world to the lab bench: it tells you that one mole (6.022 × 10²³ atoms) of any element has a mass in grams numerically equal to its average atomic mass. Every mass measurement in stoichiometry — every recipe, yield, and concentration — depends on this weighted number. Understanding abundance also explains real-world facts, from why atomic masses aren't integers to how mass spectrometry identifies elements and how isotopic dating (e.g., carbon-14) works.

The college version

Key Ideas

  • Atomic mass unit (u): 1 u = 1/12 the mass of one carbon-12 atom; 1 u ≈ 1.6605 × 10⁻²⁴ g.
  • Isotopic mass: the mass of one specific isotope (e.g., ³⁵Cl ≈ 34.97 u, ³⁷Cl ≈ 36.97 u).
  • Fractional abundance: the fraction (0 to 1) of atoms that are a given isotope; percent abundance = fractional abundance × 100%.
  • Average atomic mass: the abundance-weighted mean of isotopic masses.
  • The carbon-12 isotope (exactly 12 u) is the reference standard for the whole atomic-mass scale.
  • The Avogadro constant (Nₐ ≈ 6.022 × 10²³ mol⁻¹) links atomic-scale mass (u) to macroscopic molar mass (g/mol): 1 u × Nₐ = 1 g/mol.

Equations and Variables

Average atomic mass = Σ (isotopic mass × fractional abundance)

or, using percent abundances:

Average atomic mass = Σ (isotopic mass × % abundance / 100)

Variables:

  • isotopic mass = mass of one isotope in u (or g/mol for molar mass)
  • fractional abundance = dimensionless, between 0 and 1
  • % abundance = fractional abundance × 100

The abundances of all naturally occurring isotopes must sum to 1 (or to 100%).

How It Works or Problem-Solving Method

  1. List each isotope with its mass and its abundance.
  2. Convert abundances to fractions (divide percents by 100).
  3. Multiply each isotopic mass by its fractional abundance.
  4. Add the products.
  5. Sanity-check: the answer must lie between the lightest and heaviest isotope masses, and it will be closest to the most abundant isotope.

When the abundance is the unknown, set the lighter isotope's fraction to x and the heavier's to (1 − x), substitute into the weighted-average equation, and solve for x.

Worked Example

Problem 1: Chlorine has two stable isotopes: ³⁵Cl (34.969 u, 75.78%) and ³⁷Cl (36.966 u, 24.22%). Calculate the average atomic mass.

Convert: 0.7578 and 0.2422.

Average = (34.969 u × 0.7578) + (36.966 u × 0.2422) = 26.500 u + 8.954 u = 35.45 u.

This matches the periodic-table value for chlorine, confirming the weighting is essential — a simple mean (35.97 u) would be wrong.

Problem 2: Boron consists of ¹⁰B (10.013 u) and ¹¹B (11.009 u). If the average atomic mass of boron is 10.81 u, find the percent abundance of each isotope.

Let x = fractional abundance of ¹⁰B, so (1 − x) = abundance of ¹¹B.

10.81 = (10.013)(x) + (11.009)(1 − x) 10.81 = 10.013x + 11.009 − 11.009x 10.81 = 11.009 − 0.996x 0.996x = 11.009 − 10.81 = 0.199 x = 0.199/0.996 = 0.200 → 20.0% ¹⁰B, and 80.0% ¹¹B.

Problem 3 (conceptual): Why is the average atomic mass of hydrogen 1.008 u rather than exactly 1.000 u?

Because natural hydrogen is a mixture of ~99.99% ¹H (≈1.008 u) and a trace of ²H (deuterium, ≈2.014 u). The tiny contribution of the heavier isotope nudges the weighted average above 1.

Common Confusions

  • "Atomic mass is the mass of the most common isotope." Wrong — it is a weighted average of all stable isotopes; for chlorine it is 35.45 u, which matches no single isotope.
  • "Just average the isotope masses." Wrong — you must weight by abundance; a simple mean ignores how rare one isotope may be.
  • "Mass number and atomic mass are the same thing." Wrong — mass number A is a whole number for one isotope (³⁵Cl has A = 35); atomic mass is the abundance-weighted average across isotopes (35.45 u).
  • "Adding abundances to more than 100% is fine if I divide later." Wrong — the fractions must sum to exactly 1 (or 100%); exceeding 100% signals a data or arithmetic error.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine your class is taking a vote, but everyone's vote isn't equal — it's weighted by how many people are in each group. If 20 kids scored a "5" and 2 kids scored a "10," the average score isn't (5 + 10)/2 = 7.5; it's much closer to 5 because there are way more 5s. Atomic mass works the same way: the periodic table doesn't average the isotope masses equally — it lets each isotope's "vote" count according to how common it is. Limit of the analogy: the "weight" here is natural abundance measured by instruments, not an opinion — and the masses being averaged are exact physical quantities, not scores on a test.

Key takeaways

  • Periodic-table atomic mass = abundance-weighted average of isotope masses.
  • 1 u = 1/12 of a ¹²C atom ≈ 1.6605 × 10⁻²⁴ g.
  • Average atomic mass = Σ (isotopic mass × fractional abundance).
  • Abundances sum to 100% (or 1.0 as fractions).
  • The weighted average lies between the lightest and heaviest isotope masses.
  • Nₐ ≈ 6.022 × 10²³ mol⁻¹; 1 u × Nₐ = 1 g/mol.
  • Chlorine's 35.45 u comes from ~76% ³⁵Cl + ~24% ³⁷Cl.
  • Atomic mass = weighted average of isotope masses (not a simple mean).
  • 1 u = 1/12 of ¹²C; the ¹²C isotope sets the scale.
  • Weighted average = Σ (isotopic mass × fractional abundance).
  • Fractional abundance = percent/100; abundances total 1.0.
  • Nₐ ≈ 6.022 × 10²³ mol⁻¹; u ↔ g/mol via 1 u = 1 g/mol per mole.
  • Solve for unknown abundance by letting one isotope = x and the other = 1 − x.

Keep learning

Ready to build on this? Continue to the next lesson.

Practice General Chemistry I

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Related

You’ll learn to

  • Explain why the atomic masses on the periodic table are not whole numbers.
  • Define atomic mass unit (u) and relate it to the carbon-12 standard.
  • Calculate the average atomic mass of an element from isotopic masses and fractional abundances.
  • Convert between fractional abundance and percent abundance and use both in weighted averages.

Sources & references

  1. OpenStax, *Chemistry 2e*, Ch. 2.3, "Atomic Structure and Symbolism" (atomic mass and isotopes).
  2. Commission on Isotopic Abundances and Atomic Weights (CIAAW), "Atomic Weights."
  3. NIST, "CODATA value: Avogadro constant."

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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