Learning Objectives
1 objectiveBy the end of this note, you should be able to:
- 1.2.AExplain the quantitative relationship between the mass spectrum of an element and the masses of the element’s isotopes.
Isotope Identification from Mass Spectra
A mass spectrum displays the isotopes of an element as distinct peaks, each at a specific mass-to-charge ratio (m/z), with heights proportional to relative abundance.
Because the AP exam considers only singly charged monatomic ions, the m/z value on the horizontal axis equals the isotopic mass in atomic mass units (amu). Each peak therefore represents one isotope, and the number of peaks reveals how many stable isotopes the element possesses.
The vertical axis reports relative abundance, typically as a percentage. The tallest peak corresponds to the most common isotope in nature.
For example, a mass spectrum of neon shows peaks at m/z = 20, 21, and 22. The peak at 20 is by far the tallest, indicating that ²⁰Ne is the most abundant isotope. The peak at 22 is much smaller, and the peak at 21 is barely visible.
Out of ScopeInterpreting mass spectra of samples containing multiple elements or peaks from species other than singly charged monatomic ions. Do not memorize this for the exam.
MisconceptionStudents sometimes assume every peak on a mass spectrum represents a different element. Each peak represents a different isotope of the same element — same number of protons, different number of neutrons.
Calculating Average Atomic Mass
The average atomic mass of an element equals the weighted average of its isotopic masses, where each isotope’s mass is multiplied by its fractional relative abundance. This value — not the mass of any single isotope — appears on the periodic table.
The calculation requires two pieces of data for every isotope: its exact isotopic mass (in amu) and its relative abundance expressed as a decimal fraction. An isotope that constitutes 75% of all atoms of that element contributes more heavily than one at 25%.
Equation used
$$\text{Average atomic mass}=\sum\limits_{i}({\text{mass}}_{i}\times {\text{fractional abundance}}_{i})$$
Where: mass_i = isotopic mass of isotope i (amu), fractional abundance_i = relative abundance of isotope i expressed as a decimal (unitless).
Worked Example
Chlorine has two stable isotopes: ³⁵Cl with a mass of 34.97 amu and 75.76% abundance, and ³⁷Cl with a mass of 36.97 amu and 24.24% abundance. Calculate the average atomic mass of chlorine.
Equation used
$$\text{Average atomic mass}=({m}_{1}\times {f}_{1})+({m}_{2}\times {f}_{2})$$
Given
$${m}_{1}=34.97\text{ amu}, {f}_{1}=0.7576$$
$${m}_{2}=36.97\text{ amu}, {f}_{2}=0.2424$$
Working
Contribution of ³⁵Cl:
$$34.97\times 0.7576=26.50\text{ amu}$$
Contribution of ³⁷Cl:
$$36.97\times 0.2424=8.961\text{ amu}$$
Sum the contributions:
$$26.50+8.961=35.46\text{ amu}$$
$$\text{Average atomic mass of Cl}=35.46\text{ amu}$$
This matches the value on the periodic table, confirming the calculation.
Practice Problem
Silicon has three stable isotopes:
| Isotope | Isotopic Mass (amu) | Relative Abundance (%) |
|---|---|---|
| ²⁸Si | 27.98 | 92.22 |
| ²⁹Si | 28.98 | 4.67 |
| ³⁰Si | 29.97 | 3.09 |
Calculate the average atomic mass of silicon.
Answer
Equation used
$$\text{Average atomic mass}=({m}_{1}\times {f}_{1})+({m}_{2}\times {f}_{2})+({m}_{3}\times {f}_{3})$$
Given
$${m}_{1}=27.98\text{ amu}, {f}_{1}=0.9222$$
$${m}_{2}=28.98\text{ amu}, {f}_{2}=0.0467$$
$${m}_{3}=29.97\text{ amu}, {f}_{3}=0.0309$$
Working
Contribution of ²⁸Si:
$$27.98\times 0.9222=25.80\text{ amu}$$
Contribution of ²⁹Si:
$$28.98\times 0.0467=1.353\text{ amu}$$
Contribution of ³⁰Si:
$$29.97\times 0.0309=0.9261\text{ amu}$$
Sum the contributions:
$$25.80+1.353+0.9261=28.08\text{ amu}$$
$$\text{Average atomic mass of Si}=28.08\text{ amu}$$
Examiner InsightAP free-response questions frequently give a mass spectrum (graph) and expect you to extract masses and abundances, then compute the weighted average. Practice going from spectrum → data → calculation in one smooth workflow.
MisconceptionStudents sometimes average isotopic masses without weighting (e.g., (35 + 37) ÷ 2 = 36 for Cl). This gives the wrong answer because it ignores that ³⁵Cl is far more abundant than ³⁷Cl. The average atomic mass always sits closer to the mass of the more abundant isotope.
QUICK RECAP
Key Points
- A mass spectrum of one element shows one peak per isotope.
- Peak position (m/z) gives isotopic mass in amu.
- Peak height gives relative abundance of that isotope.
- Only singly charged monatomic ions are assessed on the AP exam.
- The number of peaks equals the number of stable isotopes.
- The tallest peak corresponds to the most abundant isotope.
- Average atomic mass = sum of (mass × fractional abundance) for all isotopes.
- Convert percent abundance to a decimal before multiplying.
- The weighted average sits closer to the more abundant isotope’s mass.
- The periodic table value is a weighted average, not a single isotope’s mass.
- Averaging without weighting produces incorrect results.
- Estimate: the average is near the tallest peak, shifted toward other peaks.
CAN I…? PROGRESS CHECK
Self-Assessment
- Identify each isotope of an element from its mass spectrum by reading peak positions and heights.
- Determine the most abundant isotope from the tallest peak on a mass spectrum.
- Convert percent relative abundances to decimal fractions for use in calculations.
- Calculate the average atomic mass from isotopic masses and relative abundances using a weighted average.
- Predict which isotope’s mass the average atomic mass will be closest to based on relative peak heights.
- Explain why the average atomic mass on the periodic table is not a whole number.