Isotopes and average atomic mass: define isotopes, and calculate the weighted average atomic mass of an element from the masses and natural abundances of its isotopes.

What this topic is asking

The Core Curriculum requires you to explain that the elements occur as isotopes and to calculate an element's average atomic mass as a weighted average of its isotopes. This is one of the most reliable Part B-2 and Part C calculation questions on the Regents exam, and it explains why the atomic masses on the Periodic Table are decimals rather than whole numbers.

What isotopes are

For example, the three natural isotopes of hydrogen are protium ($^{1}_{1}\text{H}$, one proton, no neutrons), deuterium ($^{2}_{1}\text{H}$, one proton, one neutron) and tritium ($^{3}_{1}\text{H}$, one proton, two neutrons). Because they have the same number of electrons and the same nuclear charge, isotopes of an element behave almost identically in chemical reactions; they differ measurably only in mass and, for unstable isotopes, in radioactivity. The Regents notation for an isotope is either the nuclide symbol $^{A}_{Z}\text{X}$ or hyphen notation such as carbon-14.

Why the periodic-table mass is a decimal

This is why you never need to do isotope averaging just to find a molar mass for stoichiometry: the periodic-table value has already done the weighting for you. You only carry out the weighted-average calculation when a question gives you isotope masses and abundances directly.

The weighted-average calculation

The method is the same every time:

where each $m$ is an isotope mass and each $f$ is that isotope's abundance written as a decimal fraction (so $25\%$ becomes $0.25$). The fractions add up to $1$. A plain average (just adding the masses and dividing) is wrong unless the isotopes happen to be equally abundant, which is a common trap.

Reading the question carefully

Regents questions phrase abundance as a percentage; convert it to a decimal before multiplying. Sometimes a question gives the number of atoms out of a sample (for example "$3$ atoms of mass $10$ and $1$ atom of mass $11$"), in which case the fractions are $\frac{3}{4}$ and $\frac{1}{4}$. Either way, the principle is mass times fractional abundance, summed.

Try this

Q1. An element has two isotopes: mass $6.02$ at $7.5\%$ and mass $7.02$ at $92.5\%$. Show the setup for the average atomic mass. [1 point]

• Cue. $(6.02)(0.075) + (7.02)(0.925)$.

Q2. Why is the atomic mass of most elements not a whole number? [1 point]

• Cue. It is a weighted average of isotopes of different mass numbers, so it falls between whole-number masses.