1.2.6 · D5Atomic Structure (Classical)
Question bank — Calculation of atomic mass from isotopic abundance
True or false — justify
The relative atomic mass equals the mass of the most common isotope.
False. It is a weighted average, so it sits strictly between the lightest and heaviest isotope, only leaning toward the common one — never equal to it (unless one isotope is 100%).
A weighted average can be smaller than the smallest isotopic mass.
False. An average of numbers always lies between the minimum and maximum of those numbers, so can never fall below the lightest isotope's mass.
If a hypothetical element had all isotopes equally abundant, the weighted average would reduce to the ordinary (simple) mean.
True. Equal fractions turn into , which is exactly the plain average — the simple mean is the special case of equal weights.
Doubling the size of your chlorine sample changes the calculated .
False. In the derivation the total atom count cancels, so depends only on proportions, not on how much material you have.
Relative atomic mass has units of grams.
False. It is measured in atomic mass units (u); see Atomic Mass Unit (u) and the Carbon-12 standard. Only when you take in grams do you get the mass of one mole (see Mole Concept and Molar Mass).
For an element with a single stable isotope (e.g. fluorine), is essentially just that isotope's mass.
True. With one isotope , so — no averaging happens, and comes out very close to a whole number.
The fractional abundances of an element must always add to exactly 1.
True by definition — every atom in the sample is some isotope, so the fractions partition the whole and .
is closer to the heavier isotope whenever the heavier isotope has the larger abundance.
True. The average is "pulled" toward whichever mass carries more weight, so a dominant heavy isotope drags upward.
Spot the error
A student computes chlorine as . Find the mistake.
Mixed units: the first term uses a fraction (0.7577) but the second uses a percentage (24.23). Convert both to fractions, or both to percentages with a — never one of each.
"Boron's , so its most common isotope must be B because 10.811 rounds to 11."
The reasoning is sloppy but the conclusion is right for the wrong reason. is near 11 because B is abundant (), not because 10.811 "rounds" — rounding is coincidental and can mislead for other elements.
A student writes and divides by 2 for chlorine, getting 35.97.
This is the simple mean, which wrongly treats both isotopes as equally common. Weight by abundance: Cl is ~3× more common, so the true (35.45) leans hard toward 35.
Given Cl at 75.8% and Cl at 24.2% (sum 100.0%), a student divides by 99.9%. Error?
No error in principle — dividing by the actual sum of weights is always correct. Here the sum is 100.0%, so dividing by 100 or by the sum gives the same answer; the danger only appears when the given percentages genuinely fail to reach 100.
"I'll use mass numbers: ."
The method is fine but the inputs are approximations. Mass number is a whole-number nucleon count, not the measured isotopic mass; using it gives a close-but-not-exact 35.48 instead of 35.45. Use measured masses when the question provides them.
"Since neutrons are heavier than protons, adding a neutron adds exactly 1 u to the isotopic mass."
Not exactly. A neutron is ~1.0087 u and the actual isotopic mass differs from the integer count because of the mass defect (binding energy), so isotopic masses are close to — but never precisely — whole numbers. See Nuclear Binding Energy and Mass Defect.
A student solving Boron backwards writes using the same for both.
Wrong — that double-counts one fraction. The two abundances are not equal; use for B and for B so they sum to 1, then solve the single equation.
Why questions
Why is atomic mass on the periodic table almost never a whole number?
Because it averages isotopes of different mass numbers, weighted by abundance; a weighted mix of integers generally lands between them, not on one. (Plus each isotopic mass is itself slightly non-integer from the mass defect.)
Why does the total number of atoms cancel in the derivation?
Because and , so the 's divide out — the average is a property of the proportions of the mixture, blind to sample size. This is the heart of Weighted Average (mathematics).
Why must we divide by 100 (or by ) when using percentages?
A genuine average divides by the total weight. Percentages are "parts per hundred," so their weights sum to 100; dividing by 100 renormalises them back into fractions that sum to 1.
Why is the weighted average always trapped between the extreme isotopic masses?
It is a convex combination — a blend where the weights are positive and sum to 1. Any such blend of numbers lies within their range; you cannot average your way outside the smallest or largest value.
Why do we need experiments like Mass Spectrometry to do this calculation at all?
The formula needs two ingredients we cannot deduce from theory: the exact isotopic masses and the natural abundances. A mass spectrometer measures both by separating isotopes by mass and counting how many of each.
Why doesn't change if we scoop chlorine from Earth's ocean versus a lab bottle?
Because natural isotopic abundances are essentially fixed by the element's origin, the proportions are the same everywhere, and depends only on proportions — so both give ~35.45 u.
Why can two different elements never share the same but have the same isotopes?
Because isotopes of one element all share the same proton number ; a different is a different element by definition. Isotopes differ only in neutron count within one element.
Edge cases
An element has one isotope at 99.99% and another at 0.01%. Does the rare isotope affect ?
Yes, but only in the far decimal places. Its tiny weight nudges a hair toward its mass; the pull is real but negligible for most practical rounding.
What is for a pure, artificially isotope-separated sample (e.g. 100% Cl)?
It equals that single isotope's mass, 34.969 u — with there is nothing to average. This differs from the natural-abundance value of 35.45 u.
Can the abundance of an isotope be exactly zero in the formula?
Yes — a term simply contributes nothing (), which is why an element with only one natural isotope drops all other terms.
What happens to in the limiting case where one isotope's abundance approaches 100%?
approaches that isotope's mass. The average continuously slides toward whichever isotope dominates, reaching the pure-isotope value in the limit .
If a question gives abundances that sum to 97% (rest is a third unnamed isotope), what should you do?
You cannot just divide the two terms by 97 — that ignores the missing isotope's mass entirely and biases the result. You need the third isotope's mass and abundance too; an incomplete abundance table gives an incomplete average.
Two isotopes have abundances 50% and 50% but very different masses. Where does land?
Exactly at their midpoint — equal weights collapse the weighted average into the simple mean , regardless of how far apart the masses are.
Could an element's ever decrease over geological time?
In principle yes, if a radioactive isotope's abundance shifts as it decays — the mixture's proportions change, so the abundance-weighted average tracks that change. For stable-isotope elements, though, it holds constant.
Connections
- Calculation of atomic mass from isotopic abundance (index 1.2.6) — the parent this bank drills.
- Weighted Average (mathematics) — why the average is trapped between extremes and ignores sample size.
- Isotopes and Mass Number — what the "same element, different mass" ingredients actually are.
- Atomic Mass Unit (u) and the Carbon-12 standard — the unit behind every .
- Mass Spectrometry — where the masses and abundances come from.
- Nuclear Binding Energy and Mass Defect — why isotopic masses aren't whole numbers.
- Mole Concept and Molar Mass — in grams = mass of one mole.