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Mass Spectra of Elements

Read isotope mass and relative abundance separately, normalize intensities and calculate a weighted mass.

AP Chemistry Topic 1.2 · CED printed p. 31 / PDF p. 38. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.

Before you start
  • Protons and neutrons
  • Weighted averages

Read the two jobs of a mass spectrum

Isotopes have the same proton number but different neutron numbers, so they are the same element with different masses. In the AP elemental mass-spectrum model, the detected particles are singly charged monatomic ions. The horizontal coordinate, m/z, therefore corresponds numerically to an isotope mass: z is +1. The vertical signal gives relative isotope abundance. A peak at a larger mass is not automatically a taller peak; mass and abundance are separate information.

Check the vertical scale. If peaks are labeled as percent abundance, their percentages should add to 100% apart from rounding. If the tallest peak is set to 100, the values are relative intensities, not percentages of the whole. Normalize by dividing each intensity by their sum. This topic does not ask you to interpret molecular fragments or multiply charged ions.

Average means weighted average

Use m‾=∑ifimi\overline{m}=\sum_i f_i m_i, where each fraction fi is the fraction of atoms, not a mass fraction. A 75% isotope counts three times as much as a 25% isotope. The periodic-table atomic mass describes a sample average, not an isotope that must physically exist. It must fall between the isotope masses and be closer to the more abundant isotope. Isotope mass is not exactly the integer mass number; use measured masses when supplied.

You can work backward too. For two isotopes, let the first fraction be x and the second be 1 − x. Substitute the known average into x m1 + (1 − x)m2. An answer outside 0 to 1 signals inconsistent assumptions or a reversed setup.

Let's work through it

Work it through

An original simplified spectrum has peaks at 62.0 u and 64.0 u with relative intensities 100 and 40. What average atomic mass does it imply?

Isotope mass / uRelative intensityFraction of atoms
62.0100100/140
64.04040/140
  1. The total intensity is 140; normalize to fractions 0.7142857 and 0.2857143. Treat the stated intensities as an exact ratio for this model.
  2. Weight the masses: 62.0(100/140) + 64.0(40/140) = 62.5714 u.
  3. Report 62.6 u at the precision of the supplied masses. It lies closer to 62.0 u, matching the taller low-mass peak.

Make the picture move

Interactive concept model

A weighted mean is a property of the whole sample.

Predict first. Predict which way the mean moves when the heavier isotope becomes more abundant. Can the mean represent a single atom?

m̄ = f₃₅m₃₅ + (1 − f₃₅)m₃₇

A weighted mean is a property of the whole sample.Two ideal mass-spectrum sticks are placed at the chlorine isotope masses, with heights equal to selected percent abundances. This selected mixture has mean atomic mass 35.453 u. Increasing chlorine-37 abundance increases the mean. The accessible data table and explanation give the same relationships.34353637380255075100³⁵Cl⁺³⁷Cl⁺Fraction of total ion signal (%)Approximate m/z for singly charged chlorine ions
  • Peak heights: fraction of total ion signal
Two ideal mass-spectrum sticks are placed at the chlorine isotope masses, with heights equal to selected percent abundances. This selected mixture has mean atomic mass 35.453 u. Increasing chlorine-37 abundance increases the mean.

On a small screen, scroll inside the graph or diagram to read its labels. The values and explanation also appear below.

Mean mass of this mixture35.453 u
Chlorine-3575.76%
Chlorine-3724.24%

This selected mixture has mean atomic mass 35.453 u. Increasing chlorine-37 abundance increases the mean.

Read the data and comparisons
Isotope weighting: model values and comparisons (numbers rounded for display)
IsotopeMass (u)FractionContribution to mean (u)
³⁵Cl34.9688530.757626.4924
³⁷Cl36.9659030.24248.9605
Sum—1.000035.4529

Macroscopic / data. The sample mean changes smoothly with isotope proportions.

Particle view. Every chlorine nucleus has 17 protons; these isotopes have 18 or 20 neutrons. No atom is assigned the sample mean mass.

Symbolic view. f₃₅ = 0.7576; f₃₇ = 0.2424; f₃₅ + f₃₇ = 1.

Why it works. Multiply each isotopic mass by the fraction of atoms of that isotope, then add. Peak positions identify mass-to-charge ratios; peak signal proportions supply the weighting. The mass numbers 35 and 37 are whole-number nucleon counts, not the measured masses.

Conditions and limits. The default abundances are the NIST representative composition (75.76% and 24.24%); natural samples vary. Moving the slider constructs a hypothetical isotope mixture. This ideal spectrum includes singly charged atomic ions with equal relative response and no fragments. The stick positions approximate ion m/z using neutral atomic masses; the electron mass difference is unresolved on the graph. The table gives neutral atomic masses for the mean calculation. Peaks are schematic sticks.

Try it from memory

Pause & retrieve

Try it before you peek.

Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.

Question 1 of 20 self-assessed

Question 1

A spectrum’s tallest peak is set to 100 and a second peak has intensity 50. Is the second isotope 50% abundant?

See every recall question with its answer

These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.

Retrieval 1

A spectrum’s tallest peak is set to 100 and a second peak has intensity 50. Is the second isotope 50% abundant?

Worked answer and reason

No. The fractions are 100/150 and 50/150; the second isotope is 33.3% abundant.

The scale is relative to the tallest peak, not relative to total counts.

Retrieval 2

Why can an element have an average atomic mass of 35.45 u while having no isotope at 35.45 u?

Worked answer and reason

The average combines populations of distinct isotopes; an average need not be an individual observation.

Weighted averages describe the sample distribution.

Catch the tempting mistake

The average is the midpoint of the two masses.
Only equal abundances justify the midpoint.
Isotopes differ in the number of protons.
Changing proton number changes the element; isotopes differ in neutron number.

Put it to work

Give the chemical reason as well as the calculation. Worked solutions are on a separate page.

Scientific calculator workspace

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Scientific calculator

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For the teacher
  • Make students mark what each axis measures before doing the average.
  • Offer both percent and base-peak-normalized spectra to test normalization.

Unit planning and formative guidance →

Source objectives and scope

Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →

1.2.A
Explain the quantitative relationship between the mass spectrum of an element and the masses of the element’s isotopes.
1.2.A.1
The mass spectrum of a sample containing a single element can be used to determine the identity of the isotopes of that element and the relative abundance of each isotope in nature.
1.2.A.2
The average atomic mass of an element can be estimated from the weighted average of the isotopic masses using the mass of each isotope and its relative abundance.
  • Interpreting mass spectra of samples containing multiple elements or peaks arising from species other than singly charged monatomic ions will not be assessed on the AP Exam.
Science references for this teaching
Glossary and deeper reading · 4 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Isotope

    Isotopes are atoms of the same element with different numbers of neutrons. They have the same proton number but different masses. For example, carbon-12 and carbon-13 each have six protons but have six and seven neutrons, respectively.

  • Ion

    An ion is an atom or group of atoms with a net electric charge. Ordinary ion formation changes electron count without changing the element's proton count. For example, a sodium atom losing one electron becomes a sodium ion with charge +1.

    Read more: Valence Electrons and Ionic Compounds →

  • Relative isotope abundance

    Relative isotope abundance is the fraction or percentage of an element's atoms belonging to a particular isotope. It concerns the number of atoms rather than their fraction of the sample's mass. For example, an isotope present in 25 percent of the atoms contributes a factor of 0.25 to the weighted average atomic mass.

  • Mass fraction

    Mass fraction is a component's mass divided by the total mass of the sample. Multiplying that fraction by 100 gives its mass percentage. For example, 5 grams of a solute in 100 grams of solution corresponds to a mass fraction of 0.05 and a mass percentage of 5 percent.

    Read more: Composition of Mixtures →

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