AP Chemistry · Unit 1 · Topic 1.2
Mass Spectra of Elements
An element does not necessarily produce a single peak in a mass spectrum. Atoms with the same proton number can carry different neutron numbers. A spectrum turns that invisible population into evidence, provided you read both the horizontal axis and the intensity scale carefully.
What the instrument separates
In a simplified elemental mass-spectrometry experiment, atoms are ionized and separated according to mass-to-charge ratio, m/z. For singly charged ions, numerical peak positions approximately match isotopic masses in atomic mass units. The detector counts ion signals; it is not weighing the entire sample at each point. Isotopic chemistry remains similar because the electron arrangements are nearly identical.
Two axes answer different questions
Peak position identifies a mass-to-charge ratio. Relative intensity describes how much signal occurs there. For an appropriately calibrated simple elemental spectrum, normalized signals estimate isotope fractions. Check charge state: an ion with twice the charge appears at half the m/z of a singly charged ion of the same mass. A low m/z peak is not automatically a lighter isotope.
An original model spectrum
| Approximate isotopic mass / u | Signal units | Normalized fraction |
|---|---|---|
| 62.0 | 84 | 84/120 = 0.700 |
| 64.0 | 30 | 30/120 = 0.250 |
| 65.0 | 6 | 6/120 = 0.050 |
These are invented teaching data for element X, not a claim about an actual element. The intensities add to 120 rather than 100, so normalization must precede the weighted average.
Worked example: a population average
Multiply each isotopic mass by the fraction of atoms having that mass, then add the contributions.
The answer lies between the smallest and largest masses and is closest to the dominant 62.0 u isotope. No individual atom needs a mass of 62.65 u. The periodic-table value summarizes a population; it is not a fractional neutron count inside each nucleus.
Recovering an unknown abundance
Suppose a second model element has only 20.0 u and 22.0 u isotopes and an average of 20.6 u. Let x be the fraction of the lighter isotope. Then 20.0x + 22.0(1 − x) = 20.6, giving x = 0.70. The fractions must sum to one. Two unknown fractions become solvable because conservation of total probability supplies another equation.
Judging evidence quality
Repeated spectra with the same peak positions but slightly different intensities suggest uncertainty in abundance, not changing proton numbers. Contamination or unequal instrument response can bias an apparent fraction. State the assumption that the selected peaks belong to the same element and charge state before treating their areas as an isotope census.
What not to conclude
The tallest peak is the most abundant detected isotope under the model, not necessarily the average atomic mass. Mass number is a whole-number nucleon count, whereas measured isotopic mass is not exactly that integer. Do not silently replace precise isotope masses with mass numbers if the question supplies measured masses.
Practice, then check your reasoning
If the 65.0 u isotope becomes more abundant at the expense of the 62.0 u isotope, what happens to the average?
Answer and reasoning: It increases: more of the population's weight moves toward the heavier mass.
A 64 u ion has charge 2+. Where is its approximate m/z peak?
Answer and reasoning: At 32, showing why charge state must be known before assigning isotope mass.
Scientific references: OpenStax: isotopes and weighted atomic masses; S. E. Van Bramer: isotope abundance. OpenStax Chemistry 2e authors: Paul Flowers, Klaus Theopold, Richard Langley, and William R. Robinson. Consult source pages for their license terms. Accessed September 13, 2026. Explanations, comparison tables, and practice tasks were independently written for this site; numerical teaching datasets are illustrative unless explicitly identified otherwise.
Official curriculum reference: College Board AP Chemistry course page. The explanation and worked example are independently written for this study site.