AP Chemistry Mass Spectrometry Practice Problems
Three fully worked examples, from a simple two-isotope case to solving in reverse.
Mass spectrometry problems on the AP exam are almost entirely about one skill: turning isotope masses and relative abundances into an average atomic mass, or working backward from a known average to find a missing abundance. Here are three worked examples covering the real range of how this gets asked.
What's actually tested: you don't need to know the instrument's internal mechanics (ionization, acceleration, magnetic deflection, detection). You need to be comfortable with the overall purpose of the technique and, specifically, reading a spectrum to calculate average atomic mass.
Reading the Spectrum
A mass spectrum plots mass (x-axis) against relative abundance (y-axis). Each peak is one isotope: its position gives the isotope's mass, its height gives how abundant it is compared to the other peaks.
Example 1: Two Isotopes
Copper has two isotopes: 69.2% at mass 63 and 30.8% at mass 65. Find copper's average atomic mass.
Average = (0.692 × 63) + (0.308 × 65) = 43.60 + 20.02 = 63.62
(Copper's real average atomic mass is 63.55, this problem's rounded abundances land close to that real value.)
Example 2: Three Isotopes
Magnesium has three isotopes: 78.99% at mass 24, 10.00% at mass 25, and 11.01% at mass 26. Find magnesium's average atomic mass.
Average = (0.7899 × 24) + (0.1000 × 25) + (0.1101 × 26)
= 18.96 + 2.50 + 2.86 = 24.32
Same method as the two-isotope case, just one more term added to the sum.
Example 3: Solving in Reverse for an Unknown Abundance
An element has two isotopes, one at mass 63 and one at mass 65. Its average atomic mass is 63.55. What is the percent abundance of each isotope?
Let x = fractional abundance of the mass-63 isotope. Then (1 − x) is the fractional abundance of the mass-65 isotope, since the two must add to 100%.
63x + 65(1 − x) = 63.55
63x + 65 − 65x = 63.55
−2x = −1.45
x = 0.725
72.5% at mass 63, and 27.5% at mass 65. Check: (0.725 × 63) + (0.275 × 65) = 45.68 + 17.88 = 63.55. ✓
Common Mistakes
- Using percentages directly instead of converting to decimals first. 69.2% has to become 0.692 before multiplying, not 69.2.
- Forgetting a term when there are three or more isotopes. Every isotope present contributes one term to the sum, none get skipped.
- Setting up the reverse problem with two unknowns instead of one. Since the abundances must add to 100%, one variable (x and 1−x) is enough; introducing a second unknown just adds unnecessary algebra.
- Trying to reason about the instrument's mechanics (ionization, deflection) when the question is really just asking for the weighted-average calculation.
Related Resources
- Unit 1 Review: Atomic Structure and Properties
- AP Chemistry Isotope Notation
- How to Write Electron Configurations
- AP Chemistry Significant Figures Rules
- AP Chemistry Study Guide
Frequently Asked Questions
Do I need to know how a mass spectrometer physically works for AP Chemistry?
No. You need to be comfortable with the overall idea and purpose of the technique, separating isotopes and measuring their relative abundance, not the instrument's internal mechanics (ionization, acceleration, magnetic deflection, detection). The actual tested skill is reading a spectrum and calculating average atomic mass from it.
What does a mass spectrum actually show?
A graph with mass (or mass-to-charge ratio) on the x-axis and relative abundance on the y-axis. Each peak represents one isotope of the element; its position tells you that isotope's mass, and its height (relative to the other peaks) tells you how abundant it is.
How do you handle three or more isotopes instead of two?
The same weighted-average method, just with more terms. Multiply each isotope's mass by its fractional abundance, then add all of the products together, whether there are two isotopes or five.
How do you solve for an unknown abundance instead of the average mass?
Set up the same weighted-average equation, but let the unknown abundance be a variable (commonly x) instead of the average. Since the two abundances must add to 100%, you only need one variable, then solve the resulting linear equation.
Are the percentages in these problems always given as exact real values?
On the real exam, yes, isotope masses and abundances are always given in the problem. The point being tested is whether you can set up and execute the weighted-average calculation correctly, not whether you've memorized any element's actual isotopic composition.
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