Open a periodic table and chlorine reads 35.45. No chlorine atom has that mass. Not one. Every chlorine atom in the universe is either about 35 or about 37, and 35.45 is an average of the two — weighted by how common each is.

That gap between what the table prints and what any single atom weighs is where a surprising number of exam points live, and a mass spectrum is the instrument that closes it.

Isotopes: same element, different mass

Isotopes of an element have the same number of protons — that is what makes them the same element — and different numbers of neutrons. Chemically they behave essentially identically, because chemistry is done by electrons and the electron count has not changed.

The mass number (protons + neutrons) is a whole number and belongs to one isotope. The average atomic mass on the periodic table is not a whole number and belongs to the element as it is found on Earth. Confusing the two is the single most common error in this topic.

	Protons	Neutrons	Electrons (neutral)	Mass number
³⁵Cl	17	18	17	35
³⁷Cl	17	20	17	37
³⁵Cl⁻	17	18	18	35

Change the neutrons and you get an isotope. Change the electrons and you get an ion. Change the protons and you have a different element.

Reading a mass spectrum

A mass spectrometer ionizes a sample, accelerates the ions, and separates them by mass-to-charge ratio. The output is a bar chart: position along the x-axis is the isotope’s mass, and height is its relative abundance.

So a mass spectrum answers two questions at a glance — which isotopes are present, and in what proportion. Everything else is arithmetic.

[Diagram — see the figure in the print workbook.]

Peak position = mass. Peak height = abundance. The average atomic mass always sits nearer the taller peak, which is a free sanity check on any answer you calculate.

The weighted average, and how to run it backwards

Multiply each isotope’s mass by its fractional abundance and add. For chlorine: (34.969)(0.7576) + (36.966)(0.2424) = 26.49 + 8.96 = 35.45 — the number on the table.

AP asks this in reverse at least as often. Given the average and the two isotope masses, let x be the fraction of the lighter isotope, write the other as (1 − x), and solve one linear equation. There is never a second unknown to worry about, because the fractions must sum to 1.

$\text{average atomic mass}=\sum_i (\text{mass}_i)(\text{fractional abundance}_i)$

Worked example 1

Boron has two isotopes: ¹⁰B (10.013 u, 19.9%) and ¹¹B (11.009 u, 80.1%). Calculate the average atomic mass.

1.Convert to fractions: 0.199 and 0.801.
2.(10.013)(0.199) = 1.993.
3.(11.009)(0.801) = 8.818.
4.Sum: 1.993 + 8.818 = 10.811 u.

The answer sits much nearer 11 than 10, which is what an 80/20 split predicts. That agreement is the check — run it before you move on.

Answer: 10.81 u

Worked example 2

Copper has two stable isotopes, ⁶³Cu (62.930 u) and ⁶⁵Cu (64.928 u). The average atomic mass is 63.546 u. Find the percent abundance of each.

1.Let x be the fraction of ⁶³Cu; then (1 − x) is the fraction of ⁶⁵Cu.
2.62.930x + 64.928(1 − x) = 63.546.
3.Expand: 64.928 − 1.998x = 63.546.
4.So 1.998x = 1.382, giving x = 0.6915.
5.Convert: 69.15% ⁶³Cu and 30.85% ⁶⁵Cu.

63.546 is nearer 62.930 than 64.928, so the lighter isotope had to be the more abundant one. If your algebra had returned 31% for ⁶³Cu you would know to check it without re-reading the question.

Answer: 69.15% ⁶³Cu, 30.85% ⁶⁵Cu

Worked example 3

A mass spectrum of magnesium shows peaks at 24 (78.99%), 25 (10.00%) and 26 (11.01%). Calculate the average atomic mass to four significant figures.

1.Three isotopes work exactly like two — there are just three terms.
2.(23.985)(0.7899) = 18.945.
3.(24.986)(0.1000) = 2.4986.
4.(25.983)(0.1101) = 2.8607.
5.Sum: 18.945 + 2.499 + 2.861 = 24.305 u.

Note how little the two heavy isotopes move the answer: together they are only about 21% of the sample, so the average stays close to 24. The tallest peak dominates, always.

Answer: 24.31 u

Worked example 4

An element has two isotopes of mass 68.926 u and 70.925 u, and an average atomic mass of 69.723 u. Identify the element and give the abundance of the lighter isotope.

1.Let x be the fraction of the 68.926 u isotope.
2.68.926x + 70.925(1 − x) = 69.723.
3.70.925 − 1.999x = 69.723, so x = 1.202/1.999 = 0.6011.
4.The average atomic mass 69.72 identifies the element as gallium.

Identifying the element is done from the average mass against the periodic table, not from either isotope mass. Students who look up 68.926 find nothing and stall.

Answer: Gallium; the ⁶⁹Ga isotope is 60.11% abundant.

Worked example 5

A student claims that because chlorine’s average atomic mass is 35.45, a sample of chlorine gas must contain some atoms weighing 35.45 u. Explain what is wrong, and what the 35.45 actually describes.

1.Identify the claim: that the average is a property of an individual atom.
2.State the fact: every chlorine atom is ³⁵Cl or ³⁷Cl (or a rarer isotope). None has a mass of 35.45 u.
3.State what the number is: a weighted mean over the natural abundances of the isotopes, so it describes a sample, not an atom.
4.Say why it is still useful: any real sample contains enough atoms that the natural ratio holds, so the average predicts the sample’s mass exactly.

This is a Science Practice 6 question — argumentation. The mark is not for saying ‘wrong’; it is for naming what the quantity actually is. Answers that stop at ‘it is an average’ usually score half.

Answer: No atom has that mass. 35.45 u is the abundance-weighted mean over ³⁵Cl and ³⁷Cl, and it is a property of a macroscopic sample rather than of an individual atom.

Common trap: percentages that were never converted to fractions

An abundance quoted as 75.76% has to enter the sum as 0.7576. Leaving it as 75.76 produces an answer near 3545, which is not a plausible atomic mass for anything — and a student under time pressure writes it down anyway.

The second version of this trap is subtler. If the average you calculate does not land between the two isotope masses, you have made an error, always. An average of two numbers cannot fall outside them, and checking that takes one second.

Hack: the seesaw check

Picture the two isotope masses as the ends of a seesaw and the average as its balance point. The balance point always sits closer to the heavier crowd — the more abundant isotope.

Chlorine averages 35.45, which is much nearer 35 than 37, so Cl-35 must be the common one. You now know the answer’s shape before you calculate, and any arithmetic slip that flips the abundances is caught instantly.

Mass Spectra and Average Atomic Mass