Mass spectrometry of elements

AP Chemistry· AP Chemistry CED — Atomic Structure and Properties· 14 min read

1. Core Concepts and Instrument Operation★★☆☆☆⏱ 3 min

Mass spectrometry is an experimental analytical technique that separates charged particles by mass to measure the mass and relative abundance of an element's isotopes. It appears regularly in both multiple-choice and free-response sections of the AP Chemistry exam.

📘 Definition

Mass-to-charge ratio

$m / z$

The ratio of an ion's mass (in amu) to its charge (in elementary charge units). For +1 ions, this equals the isotope's mass.

Example:

A 69 amu +1 gallium ion has $m / z = 69$

The core principle is that charged particles moving through a magnetic field deflect based on their $m / z$ ratio: lighter ions (or ions with higher charge) deflect more than heavier ions. A mass spectrum plots ion intensity (proportional to relative abundance) on the y-axis, and $m / z$ on the x-axis.

📐 Worked Example

A mass spectrometer analyzes two gallium isotopes: gallium-69 ( $m = 69$ amu, +1 charge) and gallium-71 ( $m = 71$ amu, +1 charge). Which isotope deflects more, and what is the $m / z$ of each peak?

1
Recall the deflection rule: for ions of equal charge, deflection is inversely proportional to mass.
2
Both ions have a charge of +1, so $m / z = \frac{mass}{1} = mass$ .
3
$m/z = 69 \text{ for Ga-69}, \quad m/z = 71 \text{ for Ga-71}$
4
Ga-69 has lower mass (and lower $m / z$ ), so it experiences greater deflection.
5
Final result: Gallium-69 deflects more, with peaks at $m / z = 69$ and $m / z = 71$ .

Exam tip:

Always check for stated ion charge. The AP exam can trick you with +2 ions; if charge is not +1, divide mass by charge to get $m / z$ , do not use mass directly.

2. Calculating Average Atomic Mass from Mass Spectra★★★☆☆⏱ 5 min

The most common AP exam question on this topic asks you to calculate average atomic mass from mass spectrum data. Average atomic mass is a weighted average, where each isotope contributes to the final value proportional to its relative abundance. The general formula is:

A_r = \sum (\text{mass of isotope } i \times \text{fractional abundance of isotope } i)

If given percentages, convert to fractions by dividing by 100. If given peak intensities, calculate total intensity then divide each individual peak intensity by the total to get fractional abundance. Always confirm the sum of all abundances equals 1 before starting your calculation.

📐 Worked Example

The mass spectrum of naturally occurring strontium has four isotopes with the following data: $^{84} S r$ (83.91 amu, 0.56%), $^{86} S r$ (85.91 amu, 9.86%), $^{87} S r$ (86.91 amu, 7.00%), $^{88} S r$ (87.91 amu, 82.58%). Calculate the average atomic mass of strontium.

1
Confirm the sum of percentages equals 100%: $0.56 + 9.86 + 7.00 + 82.58 = 100%$ , so convert to fractional abundances by dividing by 100.
2
Multiply each isotope mass by its fractional abundance:
3
$(83.91 \times 0.0056) = 0.470$
4
$(85.91 \times 0.0986) = 8.471$
5
$(86.91 \times 0.0700) = 6.084$
6
$(87.91 \times 0.8258) = 72.596$
7
Sum the products: $0.470 + 8.471 + 6.084 + 72.596 = 87.621$ amu.
8
Round to four significant figures, matching input data, to get 87.62 amu.

Exam tip:

Always check that the sum of fractional abundances equals 1 before you start calculating. This catches addition errors early.

3. Calculating Isotope Abundance from Average Atomic Mass★★★★☆⏱ 4 min

A common free-response question reverses the calculation: you are given the average atomic mass from the periodic table and the mass of each isotope, and asked to solve for the relative abundance of each isotope. For an element with two isotopes (the most common case on the AP exam), this is a simple one-variable algebra problem:

Let $x$ = fractional abundance of the first isotope, so the abundance of the second isotope is $1 - x$ , because the total abundance must equal 1. Substitute into the average atomic mass formula to get:

A_r = m_1 x + m_2 (1-x)

Rearrange this equation to solve for $x$ , then convert to percent abundance. For elements with three or more isotopes, you will usually be given all abundances except one, so you just subtract the sum of known abundances from 1 to get the missing abundance.

📐 Worked Example

Chlorine has two stable isotopes: Cl-35 (34.969 amu) and Cl-37 (36.966 amu). The average atomic mass of chlorine is 35.453 amu. Calculate the percent abundance of each isotope.

1
Assign variables: Let $x$ = fractional abundance of Cl-35, so $1 - x$ = fractional abundance of Cl-37.
2
Substitute into the average atomic mass formula:
3
$35.453 = 34.969x + 36.966(1-x)$
4
Expand and simplify the right-hand side:
5
$35.453 = 34.969x + 36.966 - 36.966x = 36.966 - 1.997x$
6
Rearrange to solve for $x$ :
7
$1.997x = 36.966 - 35.453 = 1.513 \implies x = \frac{1.513}{1.997} \approx 0.758$
8
Calculate the second abundance: $1 - 0.758 = 0.242$ . Convert to percentages: 75.8% Cl-35, 24.2% Cl-37.

Exam tip:

Always sanity-check your result: the average atomic mass should be closer to the mass of the more abundant isotope. If your result is not, you swapped your variables.

4. AP-Style Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding with this AP-style multiple-choice question:

An element has two stable isotopes: Q-63 (62.93 amu) and Q-65 (64.93 amu). The average atomic mass of Q is 63.55 amu. What is the approximate percent abundance of the heavier isotope?
- 31%
- 45%
- 55%
- 69%
Reveal answer
31% —
Correct. Letting $x$ = abundance of the heavier isotope, solving gives $x = 0.31 = 31%$ . If you got a different answer, check your variable assignment and algebra.

5. Common Pitfalls

Wrong move:

Using percent abundances directly in the average atomic mass formula without converting to fractional abundances. For example, calculating $A_{r} = (24 \times 78) + (25 \times 10) + (26 \times 12)$ instead of converting percentages to fractions.

Why:

Students rush the problem and forget the formula uses fractions of the total, not percentages. The result is ~100x too large, which is often not caught.

Correct move:

Always convert all percentages to fractional abundances by dividing by 100 before multiplying by mass.

Wrong move:

Confusing peak height (y-axis) with peak position (x-axis), assigning a larger mass to the tallest peak.

Why:

Students assume the largest peak corresponds to the largest mass, mixing up axis labels.

Correct move:

Label the x-axis as $m / z$ (mass for +1 ions) and y-axis as abundance before starting any calculation, and double-check axis labels.

Wrong move:

For a +2 charged ion, using the isotope mass as the $m / z$ value.

Why:

Students are used to +1 ions for elemental mass spectrometry, so they automatically assume $z = 1$ even when the problem states a different charge.

Correct move:

Always check the problem statement for ion charge before calculating $m / z$ , and divide mass by charge to get the correct $m / z$ value.

Wrong move:

When calculating abundance for two isotopes, assigning $x$ as the abundance of both isotopes, leading to $A_{r} = m_{1} x + m_{2} x$ .

Why:

Students forget that total abundance must equal 1, so they incorrectly use two independent variables.

Correct move:

For two unknown abundances, always assign the first as $x$ and the second as $1 - x$ to ensure total abundance sums to 1.

Wrong move:

Rounding intermediate products when calculating average atomic mass, leading to a final value outside the acceptable tolerance for the correct answer.

Why:

Students round to clean numbers early, which introduces cumulative rounding error.

Correct move:

Keep all extra significant figures in intermediate steps, and only round the final answer to match the significant figures of the input data.

6. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Mass-to-charge ratio	$m / z = \frac{ion mass (amu)}{ion charge}$	Equal to isotope mass for +1 ions
Deflection rule	Deflection $\propto \frac{1}{m / z}$	Lower $m / z$ = more deflection
Average atomic mass	$A_{r} = \sum (m_{i} \times a_{i})$	$m_{i}$ = isotope mass, $a_{i}$ = fractional abundance
Percent to fraction conversion	$a_{i} = \frac{percent abundance}{100}$	Always do this before plugging into the $A_{r}$ formula
Two-isotope abundance	$A_{r} = m_{1} x + m_{2} (1 - x)$	$x$ = fractional abundance of first isotope
Mass spectrum axes	X-axis = $m / z$ ; Y-axis = relative abundance	Peak position = mass, peak height = abundance
Total abundance rule	$\sum a_{i} = 1$ (fractional) = 100% (percent)	Use to find missing abundances and check work

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Calculate average atomic mass from spectrum
2022 · FRQ
Solve for unknown isotope abundance

What's Next

Mass spectrometry of elements is the foundational experimental technique that confirms the existence of isotopes, which is core to all subsequent work in atomic structure and chemical calculations. Mastery of weighted average atomic mass from mass spectrometry is required for nearly every calculation-based topic in AP Chemistry, including molar mass and stoichiometry, so errors here will propagate through other problems. This topic also sets the foundation for more advanced mass spectrometry of molecules, used in organic chemistry to identify compound structures. After completing this sub-topic, you will move on to core atomic structure topics where the concept of isotopes is foundational.