Unit 1 is where AP Chemistry starts. It’s also where many students quietly lose points they never needed to lose.
I’ve taught this course for years. The same slips show up every May: a forgotten unit, a sloppy electron configuration, a PES peak read backwards. This guide fixes those slips with real practice, written the way the College Board asks questions.
This guide follows the nine-topic outline of Unit 1 in the current AP Chemistry Course and Exam Description (CED). Colleges and the College Board can update details from year to year, so confirm the latest on [Insert link to College Board AP Chemistry Course Description here].
Expert Tips From an AP Teacher
How much is Unit 1 worth?
The CED lists Unit 1 (Atomic Structure and Properties) at about 7–9% of the exam score. That sounds small. It isn’t. Moles, mass percent, and periodic-trend reasoning show up in nearly every other unit. Weak Unit 1 skills drag down your stoichiometry, equilibrium, and thermodynamics scores later.
What’s in Unit 1?
- Moles and molar mass
- Mass spectra of elements
- Elemental composition of pure substances
- Composition of mixtures
- Atomic structure and electron configuration
- Photoelectron spectroscopy (PES)
- Periodic trends
- Valence electrons and ionic compound formulas
The exam format (check AP Central for the current year): the exam has a multiple-choice section and a free-response section, each worth half your score. The free-response section includes both long and short questions. Unit 1 ideas can turn up in either.
Five traps I see every year:
- Dropping units. Write g, mol, and g/mol in every line. Units catch your errors for you.
- Treating the average atomic mass like a real peak. A mass spectrum shows isotope masses. The weighted average appears nowhere on it.
- Removing 3d electrons before 4s when writing transition-metal cations. The 4s electrons leave first.
- Reading PES backwards. The peak farthest to the left is the highest binding energy, which means the 1s electrons.
- Explaining trends with “because it wants a full shell.” The College Board wants Coulombic attraction, nuclear charge, shielding, and distance. Use those words.
Want more structured practice across every unit? The study resources at LearnMinto are a good place to keep building.
Quick Formula Reference for Unit 1
| Concept | Formula / Relationship | Notes |
|---|---|---|
| Moles from mass | n = m ÷ M | n = moles, m = mass (g), M = molar mass (g/mol) |
| Particles in a sample | N = n × 6.022 × 10²³ | Avogadro’s number |
| Average atomic mass | Σ (isotope mass × fractional abundance) | Convert percent to decimal first |
| Mass percent | (mass of part ÷ mass of whole) × 100 | Use molar masses for a formula |
| Empirical formula | Mass → moles → divide by smallest → whole-number ratio | Multiply if you get .5, .33, or .25 |
| Photon energy | E = hν | h = 6.626 × 10⁻³⁴ J·s |
| Wave relationship | c = λν | c = 3.00 × 10⁸ m/s |
| PES energy balance | Photon energy = binding energy + kinetic energy | Higher binding energy means a more tightly held electron |
| Coulombic reasoning | Force increases with charge, decreases with distance | Use for every periodic-trend explanation |
You get a periodic table and a formula sheet on the exam. Know what’s on them, and know what isn’t. Check the current versions on link to College Board AP Chemistry Exam Reference Materials.
Part 1: Multiple-Choice Practice (8 Questions)
Try each one before reading the solution.
Question 1: Moles and Particles
How many aluminum atoms are in a 5.00 g sample of Al? (Molar mass of Al = 26.98 g/mol)
- (A) 1.85 × 10⁻¹
- (B) 1.12 × 10²³
- (C) 8.12 × 10²⁵
- (D) 3.01 × 10²⁴
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Answer: (B)
Step 1: Convert grams to moles.
n = 5.00 g ÷ 26.98 g/mol = 0.1853 mol
Step 2: Convert moles to atoms.
0.1853 mol × 6.022 × 10²³ atoms/mol = 1.12 × 10²³ atoms
Why the others fail:
- (A) is the number of moles. You stopped one step early.
- (C) comes from multiplying by the molar mass instead of dividing.
- (D) comes from skipping the molar mass entirely (5.00 × 6.022 × 10²³).
Teacher tip: the answer has to be less than one mole of atoms. A 5 g sample of a 27 g/mol element is well under a mole. That alone eliminates (C) and (D).
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Question 2: Mass Spectrometry
A mass spectrum of element X shows two peaks. One is at mass 63 with 69.2% relative abundance. The other is at mass 65 with 30.8% relative abundance. What is the average atomic mass of X?
- (A) 63.00
- (B) 63.62
- (C) 64.00
- (D) 65.00
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Answer: (B)
Step 1: Convert percentages to decimals. 0.692 and 0.308.
Step 2: Weight each isotope.
(63 × 0.692) + (65 × 0.308) = 43.596 + 20.020 = 63.616 ≈ 63.62 amu
Why not (C)? 64.00 is the simple average of 63 and 65. It ignores abundance. Since the lighter isotope is more abundant, the answer must sit closer to 63 than to 65.
Teacher tip: do this sanity check every time. The weighted average always lands closer to the more abundant isotope.
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Question 3: Mass Percent
What is the mass percent of nitrogen in ammonium sulfate, (NH₄)₂SO₄?
- (A) 10.6%
- (B) 21.2%
- (C) 22.6%
- (D) 42.4%
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Answer: (B)
Step 1: Count atoms. (NH₄)₂SO₄ has 2 N, 8 H, 1 S, and 4 O.
Step 2: Find the molar mass.
- N: 2 × 14.01 = 28.02
- H: 8 × 1.008 = 8.064
- S: 1 × 32.07 = 32.07
- O: 4 × 16.00 = 64.00
- Total = 132.15 g/mol
Step 3: Divide and multiply by 100.
(28.02 ÷ 132.15) × 100 = 21.2%
Why the others fail:
- (A) counts only one nitrogen.
- (C) leaves out the hydrogen mass in the denominator.
- (D) double-counts the nitrogen.
Teacher tip: the subscript outside the parentheses multiplies everything inside. That’s where most errors start.
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Question 4: Empirical Formula
A compound is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. What is its empirical formula?
- (A) CHO
- (B) CH₂O
- (C) C₂H₄O
- (D) C₃H₆O₃
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Answer: (B)
Step 1: Assume a 100 g sample. Percentages become grams: 40.0 g C, 6.7 g H, 53.3 g O.
Step 2: Convert to moles.
- C: 40.0 ÷ 12.01 = 3.33 mol
- H: 6.7 ÷ 1.008 = 6.65 mol
- O: 53.3 ÷ 16.00 = 3.33 mol
Step 3: Divide by the smallest (3.33).
C = 1.00, H = 2.00, O = 1.00
Empirical formula: CH₂O.
Why not (D)? C₃H₆O₃ is a valid molecular formula with the same ratio. The empirical formula is the simplest whole-number ratio, so it must be reduced.
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Question 5: Photoelectron Spectroscopy
A PES spectrum of a neutral atom shows three peaks with these approximate binding energies and relative heights:
| Binding Energy (MJ/mol) | Relative Height |
|---|---|
| 39.6 | 2 |
| 3.6 | 2 |
| 1.40 | 3 |
Which element produced this spectrum?
- (A) Boron
- (B) Carbon
- (C) Nitrogen
- (D) Oxygen
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Answer: (C)
Step 1: Match each peak to a subshell. The highest binding energy belongs to the electrons closest to the nucleus, the 1s electrons. The next peak is 2s, and the lowest is 2p.
Step 2: Use the heights as electron counts. 2 : 2 : 3 gives 1s² 2s² 2p³.
Step 3: Count the total. 2 + 2 + 3 = 7 electrons. A neutral atom with 7 electrons has 7 protons: nitrogen.
Teacher tip: relative peak height equals the number of electrons in that subshell. The peak position tells you which subshell. Keep the two ideas separate.
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Question 6: Periodic Trends (Ionization Energy)
Which statement best explains why the first ionization energy of aluminum is lower than that of magnesium?
- (A) Aluminum has a smaller nuclear charge than magnesium.
- (B) Aluminum’s outermost electron is in a 3p subshell, which is higher in energy and farther from the nucleus on average than a 3s electron.
- (C) Aluminum has a larger atomic radius because it has more energy levels.
- (D) Aluminum has a completely filled 3s subshell.
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Answer: (B)
Step 1: Write the configurations.
- Mg: [Ne] 3s²
- Al: [Ne] 3s² 3p¹
Step 2: Compare the electron being removed. Mg loses a 3s electron. Al loses a 3p electron. The 3p electron sits at a higher energy, is on average farther from the nucleus, and is partly shielded by the 3s electrons. Less energy is needed to remove it.
Why the others fail:
- (A) is false. Aluminum has more protons (13 vs. 12).
- (C) is false. Both are in period 3, so they have the same number of energy levels.
- (D) is false in a way that matters: it’s magnesium’s 3s that is filled.
Teacher tip: this is a classic “exception to the trend” question. Memorize the reason: the 3s to 3p change in subshell.
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Question 7: Ionic Radius
Which species has the largest radius: Na⁺, Mg²⁺, F⁻, or O²⁻?
- (A) Na⁺
- (B) Mg²⁺
- (C) F⁻
- (D) O²⁻
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Answer: (D)
Step 1: Count electrons. All four species have 10 electrons (1s² 2s² 2p⁶). They are isoelectronic.
Step 2: Compare nuclear charge.
- O²⁻: 8 protons
- F⁻: 9 protons
- Na⁺: 11 protons
- Mg²⁺: 12 protons
Step 3: Apply Coulomb’s law. With the same number of electrons and the same shielding, more protons pull the electron cloud in tighter. Fewer protons means a larger radius, so O²⁻ is the largest.
Teacher tip: for isoelectronic series, ignore periods and groups. Just count protons.
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Question 8: Electron Configuration of an Ion
What is the ground-state electron configuration of Fe³⁺?
- (A) [Ar] 4s² 3d³
- (B) [Ar] 3d⁵
- (C) [Ar] 4s¹ 3d⁴
- (D) [Ar] 3d⁶
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Answer: (B)
Step 1: Write neutral iron. Fe has 26 electrons: [Ar] 4s² 3d⁶.
Step 2: Remove electrons from the outermost shell first. That’s the 4s subshell (n = 4). Remove both 4s electrons: [Ar] 3d⁶.
Step 3: Remove one more electron from 3d: [Ar] 3d⁵.
Result: [Ar] 3d⁵. This half-filled configuration is also why Fe³⁺ is so common.
Teacher tip: electrons fill 4s before 3d, but they leave 4s first. Students who forget this lose easy points on every exam.
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Part 2: Free-Response Practice (3 Questions)
Write full answers before checking. Free-response points come from your reasoning, not just the final number.
FRQ 1: Mass Spectrometry and Average Atomic Mass
A mass spectrum of a sample of magnesium shows three peaks:
| Isotope | Mass (amu) | Relative Abundance |
|---|---|---|
| ²⁴Mg | 24 | 78.99% |
| ²⁵Mg | 25 | 10.00% |
| ²⁶Mg | 26 | 11.01% |
(a) Calculate the average atomic mass of magnesium.
(b) Explain why no peak appears at 24.3 on the spectrum.
(c) A student says the tallest peak shows the isotope with the most protons. Is this correct? Justify your answer.
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(a)
- ²⁴Mg: 24 × 0.7899 = 18.958
- ²⁵Mg: 25 × 0.1000 = 2.500
- ²⁶Mg: 26 × 0.1101 = 2.863
Sum = 24.32 amu
(b) The spectrum shows the masses of individual isotopes, and no single atom has a mass of 24.3. The average atomic mass is a calculated weighted value across the whole sample, so it never appears as a peak.
(c) Incorrect. All magnesium isotopes have 12 protons, which is what makes them magnesium. The peak height shows relative abundance. The tallest peak is the most common isotope (²⁴Mg). The isotopes differ in neutrons, not protons.
Scoring note: for (b) and (c), name the specific idea (weighted average, relative abundance, or same atomic number). Vague answers earn nothing.
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FRQ 2: Elemental Composition and Mixtures
A 3.00 g sample of an iron oxide contains 2.10 g of iron. The rest is oxygen.
(a) Calculate the mass percent of iron in the sample.
(b) Determine the empirical formula of the oxide.
(c) A 10.0 g mixture of this oxide and silicon dioxide (SiO₂) contains 4.20 g of iron. Calculate the mass percent of the iron oxide in the mixture.
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(a) Mass percent Fe = (2.10 g ÷ 3.00 g) × 100 = 70.0%
(b)
- Mass of O = 3.00 − 2.10 = 0.90 g
- Moles Fe = 2.10 ÷ 55.85 = 0.0376 mol
- Moles O = 0.90 ÷ 16.00 = 0.0563 mol
- Ratio O : Fe = 0.0563 ÷ 0.0376 = 1.50
A ratio of 1.5 means you multiply both by 2: Fe : O = 2 : 3.
Empirical formula: Fe₂O₃
(c) Silicon dioxide has no iron, so all 4.20 g of Fe comes from Fe₂O₃.
- Mass percent of Fe in Fe₂O₃ = (2 × 55.85) ÷ 159.70 = 111.70 ÷ 159.70 = 69.94%
- Mass of Fe₂O₃ = 4.20 g ÷ 0.6994 = 6.00 g
- Mass percent Fe₂O₃ in the mixture = (6.00 ÷ 10.0) × 100 = 60.0%
Teacher tip: in (b), a ratio like 1.5 is never “round to 2.” It’s a signal to multiply by 2. This is a favorite exam move.
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FRQ 3: PES, Periodic Trends, and Valence Electrons
Consider phosphorus (Z = 15) and sulfur (Z = 16).
(a) Write the ground-state electron configuration of phosphorus.
(b) Describe the PES spectrum of phosphorus: how many peaks, and what are their relative heights?
(c) The first ionization energy of sulfur is lower than that of phosphorus. Explain this.
(d) The 1s peak of chlorine appears at a higher binding energy than the 1s peak of phosphorus. Explain why.
(e) Predict the formula of the ionic compound formed between magnesium and phosphorus.
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(a) 1s² 2s² 2p⁶ 3s² 3p³ (or [Ne] 3s² 3p³)
(b) Five peaks, with relative heights 2 : 2 : 6 : 2 : 3. They correspond to 1s, 2s, 2p, 3s, and 3p, listed from highest binding energy to lowest.
(c)
- Phosphorus: 3p³. Each 3p electron is in its own orbital.
- Sulfur: 3p⁴. One 3p orbital now holds a pair of electrons.
- The paired electrons repel each other. That repulsion makes it easier to remove one electron from sulfur.
- Sulfur’s extra proton does not make up for this repulsion.
(d) Both 1s electrons are in the same subshell, so shielding and distance are similar. Chlorine has 17 protons and phosphorus has 15. A greater nuclear charge means a stronger Coulombic attraction on the 1s electrons, so more energy is needed to remove them.
(e)
- Mg loses 2 electrons → Mg²⁺
- P gains 3 electrons → P³⁻
- Balance charges: 3 Mg²⁺ (+6) and 2 P³⁻ (−6)
Formula: Mg₃P₂
Scoring note: in (c) and (d), always name charge, shielding, or distance, and say how each changes. “It’s more stable” earns zero.
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How to Use This Guide
- Do the problems cold. Cover the solutions. Struggle for a few minutes before you look.
- Track your mistakes by type. Math slip, concept gap, or misread question? Each one needs a different fix.
- Write explanations in full sentences. Free-response graders look for cause and effect.
- Time yourself. Aim for about 90 seconds per multiple-choice question.
For official practice materials and past free-response questions, go to College Board AP Central AP Chemistry past exam questions. For an interactive look at isotopes and atom building, try to PhET Interactive Simulations atom or isotope simulator.
Your Turn: What Unit Do You Want Next?
You just worked through the foundation of AP Chemistry. If those PES questions and mass-percent problems felt shaky, redo them tomorrow. Repetition is how this stuff sticks.
Tell me where you want to go next. Unit 2 (Molecular and Ionic Compound Structure)? Unit 3 (Intermolecular Forces)? Unit 4 (Chemical Reactions)? Drop your pick in the comments, and I’ll build the next practice set around the questions students miss most.
You’ve got this. Keep going.
Disclaimer:
This study guide is an independent educational resource and is not officially endorsed or affiliated with the College Board.