College Board AP 2020 (style)-style practice: Section I (multiple choice). The periodic-table atomic mass of chlorine is 35.45 amu, yet chlorine has only two stable isotopes, ^35Cl and ^37Cl. Which statement best explains this? (A) Chlorine atoms have a mass of 35.45 amu. (B) ^35Cl is much more abundant than ^37Cl. (C) The two isotopes are equally abundant. (D) Chlorine has a third undetected isotope. Justify your choice.

A 1-point conceptual MCQ. The answer is (B). The periodic-table value is a weighted average. Because 35.45 is much closer to 35 than to 37, the lighter isotope ^35Cl must be far more abundant (about 76\% versus 24\%). No individual atom has a mass of 35.45 amu (A is wrong), equal abundance would give an average near 36 (C is wrong), and no third isotope is needed (D is wrong).

United StatesChemistrySyllabus dot point

How does a mass spectrum reveal the isotopes of an element and let us calculate its average atomic mass?

Topic 1.2 Mass Spectra of Elements: interpret a mass spectrum to identify the isotopes of an element and their relative abundances, and calculate the average atomic mass from the data.

A focused answer to AP Chemistry Topic 1.2, covering isotopes, the mass spectrum, mass-to-charge ratio, relative abundance, and the weighted-average calculation of atomic mass, with full worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Isotopes
How a mass spectrum is produced
Calculating average atomic mass
Reading peaks carefully
Try this

What this topic is asking

The College Board (Topic 1.2) wants you to read a mass spectrum of an element, identify its isotopes and their relative abundances, and calculate the average atomic mass as a weighted average. You should also be able to explain why the periodic-table atomic mass is usually not a whole number.

Isotopes

Because the chemical behavior of an atom is set by its electrons (and so by its proton count), isotopes of an element are chemically almost identical; they differ mainly in mass. For example $^{12}\text{C}$ and $^{13}\text{C}$ both have 6 protons but 6 and 7 neutrons respectively.

How a mass spectrum is produced

In a mass spectrometer, atoms of a sample are ionized (electrons are knocked off to make positive ions), accelerated, and deflected by a magnetic field. Lighter ions and more highly charged ions are deflected more, so the instrument separates ions by their mass-to-charge ratio ( $m/z$ ). A detector records how many ions arrive at each $m/z$ value.

Calculating average atomic mass

The average atomic mass is a weighted average: each isotope contributes in proportion to its abundance.

\bar{A} = \sum_i f_i\, m_i

where $f_i$ is the fractional abundance (the percentage divided by 100) of isotope $i$ and $m_i$ is its mass. Because heavier isotopes are often less abundant, the average sits between the isotope masses but is pulled toward the most common one. This is exactly why the periodic-table atomic mass is almost never a whole number, and why you can estimate which isotope dominates just by seeing which whole number the average is closest to.

The same logic runs in reverse. If you are told the average atomic mass and the two isotope masses, you can solve for the abundances. Let the fractional abundance of the lighter isotope be $x$ ; then the heavier one is $1 - x$ , and you set up $\bar{A} = x\,m_1 + (1-x)\,m_2$ and solve the single linear equation for $x$ . This "given the average, find the abundances" version is a favorite AP twist, so practice it both ways.

Reading peaks carefully

A common spectrum shows the relative abundances as percentages that add to 100. If instead the tallest peak is set to 100 (a relative scale), convert to true percentages by dividing each peak height by the sum of all peak heights before weighting. Always check whether the abundances already sum to 100 or need normalizing.

Average atomic mass of magnesium

A mass spectrum of magnesium shows three peaks: $^{24}\text{Mg}$ at $78.99\%$ , $^{25}\text{Mg}$ at $10.00\%$ , and $^{26}\text{Mg}$ at $11.01\%$ . Calculate the average atomic mass.

step 1 Convert percentages to fractional abundances

$f_{24} = 0.7899$ , $f_{25} = 0.1000$ , $f_{26} = 0.1101$ . (They sum to $1.000$ , so no normalizing is needed.)

step 2 Multiply each isotope mass by its fraction

$0.7899 \times 24 = 18.96$ , $\quad 0.1000 \times 25 = 2.500$ , $\quad 0.1101 \times 26 = 2.863$ .

step 3 Add the contributions

$\bar{A} = 18.96 + 2.500 + 2.863 = 24.32$ amu.

step 4 Check against the periodic table

The periodic-table value for magnesium is $24.31$ amu, which matches, confirming the calculation.

Try this

Q1. An element has two isotopes: mass $10$ ( $19.9\%$ ) and mass $11$ ( $80.1\%$ ). Calculate its average atomic mass and identify the element. [2 points]

Cue. $\bar{A} = (0.199)(10) + (0.801)(11) = 1.99 + 8.81 = 10.8$ amu, which is boron.

Q2. Explain why isotopes of an element have nearly identical chemical properties. [1 point]

Cue. They have the same number of protons and therefore the same number and arrangement of electrons, which determine chemical behavior.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)3 marksSection II (short FRQ). The mass spectrum of an element shows two peaks: one at mass

63

with relative abundance

69.0\%

and one at mass

65

with relative abundance

31.0\%

. (a) Identify the two isotopes shown. (b) Calculate the average atomic mass of the element. (c) Identify the element.

Show worked answer →

A 3-point quantitative FRQ on reading a spectrum and weighting masses.

(a) Identify (1 point): the isotopes are mass-63 and mass-65 versions of the same element (copper-63 and copper-65); they have the same number of protons but different numbers of neutrons.
(b) Average atomic mass (1 point): $\bar{A} = (0.690)(63) + (0.310)(65) = 43.47 + 20.15 = 63.6$ amu.
(c) Identify (1 point): an average atomic mass near $63.6$ amu matches copper (Cu), whose periodic-table value is $63.55$ .

Markers reward converting percentages to fractions, a correct weighted sum, and matching the result to the periodic table.

AP 2020 (style)1 marksSection I (multiple choice). The periodic-table atomic mass of chlorine is

35.45

amu, yet chlorine has only two stable isotopes,

^{35}\text{Cl}

and

^{37}\text{Cl}

. Which statement best explains this? (A) Chlorine atoms have a mass of

35.45

amu. (B)

^{35}\text{Cl}

is much more abundant than

^{37}\text{Cl}

. (C) The two isotopes are equally abundant. (D) Chlorine has a third undetected isotope. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

The periodic-table value is a weighted average. Because $35.45$ is much closer to $35$ than to $37$ , the lighter isotope $^{35}\text{Cl}$ must be far more abundant (about $76\%$ versus $24\%$ ). No individual atom has a mass of $35.45$ amu (A is wrong), equal abundance would give an average near $36$ (C is wrong), and no third isotope is needed (D is wrong).

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)