ACT Science: Understand Isotopes and Calculate Weighted Average Atomic Mass

Isotopes are atoms of the same element (same number of protons) with different numbers of neutrons (different mass numbers). Example: Carbon-12 (6 protons, 6 neutrons, mass 12) and Carbon-13 (6 protons, 7 neutrons, mass 13) are isotopes. Atomic mass on the periodic table is the weighted average of all isotopes, weighted by their natural abundance. Example: Boron has two stable isotopes: Boron-10 (10.013 u, 19.9% abundance) and Boron-11 (11.009 u, 80.1% abundance). Atomic mass=(10.013×0.199)+(11.009×0.801)=1.993+8.818=10.811 u≈10.81 u. The atomic mass is a weighted average, not the mass of a single atom. Understanding this explains why atomic masses are decimals.

Why it matters: Atomic mass affects molar mass, stoichiometry, and chemical behavior. Understanding isotopes helps you calculate average atomic masses and predict how isotope ratios change in nuclear reactions or radioactive decay.

Mistake 1: Assuming the atomic mass on the periodic table is the mass of the most abundant isotope. Chlorine's atomic mass is 35.45, but no single Chlorine isotope has this mass. Cl-35 (35.0) is 76% abundant, and Cl-37 (37.0) is 24% abundant. The weighted average is 35.45. Mistake 2: Forgetting to convert percentages to decimals. If an isotope has 75% abundance, use 0.75, not 75, in the calculation. Using 75 gives an answer 100 times too large. Always convert percentages to decimals (divide by 100) before calculating the weighted average.

After calculating atomic mass, check: Is the result between the masses of the lightest and heaviest isotopes? Is it closer to the more abundant isotope's mass? If yes, your calculation is likely correct.

Problem 1: Chlorine has two isotopes: Cl-35 (35.0 u, 75.8% abundant) and Cl-37 (37.0 u, 24.2% abundant). Calculate the average atomic mass. Setup: (35.0×0.758)+(37.0×0.242)=26.53+8.95=35.48 u. Problem 2: Copper has two isotopes: Cu-63 (63.0 u, 69.2% abundant) and Cu-65 (65.0 u, 30.8% abundant). Calculate the average atomic mass. Setup: (63.0×0.692)+(65.0×0.308)=43.6+20.0=63.6 u. Problem 3: An element has an atomic mass of 24.31 and two isotopes: Mass A (24.0 u) and Mass B (25.0 u). What is the % abundance of each isotope? Setup: Let x=abundance of A. (24.0×x)+(25.0×(1-x))=24.31. 24x+25-25x=24.31. -x=24.31-25=-0.69. x=0.69 or 69%. So A is 69% abundant, B is 31% abundant. For each problem, set up the weighted average equation and solve carefully, converting percentages to decimals first.

After solving, verify by checking that the result is between the two isotope masses and closer to the more abundant one.

Isotopes appear in chemistry questions about molar mass and stoichiometry, and in physics/earth science questions about radioactive decay and nuclear reactions. Understanding isotopes and atomic mass helps you solve problems involving molar calculations and predict properties of elements. Once you master isotope calculations, you unlock understanding of why atomic masses are decimals and how natural abundance affects measurable properties.

Spend 15 minutes this week calculating average atomic masses for 10 elements (use real isotope data from the periodic table or practice problems). Verify your calculations by checking that the result falls between the isotope masses. By test day, isotope calculations will be automatic, and you will answer questions about atomic mass with confidence grounded in understanding the underlying chemistry.