AP Chemistry · Handsworth Secondary 2026–27

Unit 1 · Atomic Structure & Properties

Dr. Ras Mulinta
Handsworth Secondary
Notes Package

CED Unit 1, 7–9% of the AP exam, taught in 4 blocks. You already know moles, formulas, and periodic trends from BC Chemistry 11, so we move quickly there and spend our time on the AP depth: mass-spectrum math, electron configuration, photoelectron spectroscopy, and Coulomb-law reasoning.

Blocks:  1 → A–D  ·  2 → E–F  ·  3 → G–H  ·  4 → review + Unit 1 test

Name:Block:Date:

A · The Mole & Molar Mass CED 1.1 · LO 1.1.A

We can't count atoms one at a time, so we weigh them. The mole is the bridge between the mass you measure and the number of particles reacting.

What you must be able to do (College Board learning objective 1.1.A): “Calculate quantities of a substance or its relative number of particles using dimensional analysis and the mole concept.”

Study this outside class. On AP Classroom, this topic’s daily videos cover:

Core: Avogadro's number N_A = 6.022×10²³ /mol  ·  n = m / M  ·  an atom's mass in amu equals its molar mass in g/mol.
Worked example
How many molecules are in 18.0 g of CO?
M(CO) = 12.01 + 16.00 = 28.01 g/mol → n = 18.0 / 28.01 = 0.643 mol
molecules = 0.643 × 6.022×10²³ = 3.87 × 10²³

Practice A

  1. How many moles are in 9.80 g of H₂SO₄? (M = 98.09 g/mol)
  2. How many oxygen atoms are in 2.0 mol of CO₂?
  3. A sample has 3.01×10²³ formula units of NaCl. Find its mass. (M = 58.44 g/mol)

B · Mass Spectra & Average Atomic Mass CED 1.2 · LO 1.2.A

A mass spectrometer separates an element's isotopes by mass. Each peak is an isotope, its position is the isotopic mass, its height is the relative abundance.

What you must be able to do (College Board learning objective 1.2.A): “Explain the quantitative relationship between the mass spectrum of an element and the masses of the element’s isotopes.”

Study this outside class. On AP Classroom, this topic’s daily videos cover:

Not assessed, don't spend time here: “Interpreting mass spectra of samples containing multiple elements or peaks arising from species other than singly charged monatomic ions will not be assessed on the AP Exam.” (CED exclusion statement)

Core: average atomic mass = Σ (isotope mass × fractional abundance). That's why periodic-table masses aren't whole numbers.
Worked example
Cl is 75.78% ³⁵Cl (34.97 amu) and 24.22% ³⁷Cl (36.97 amu). Find the average atomic mass.
M̄ = (0.7578)(34.97) + (0.2422)(36.97) = 26.50 + 8.954
M̄ = 35.45 amu ✓ (matches the periodic table)

Not tested: spectra of multiple elements, or of anything other than singly-charged atoms.

Practice B

  1. Cu is 69.17% ⁶³Cu (62.93 amu) and 30.83% ⁶⁵Cu (64.93 amu). Find the average atomic mass.
  2. Boron (average 10.81 amu) is ¹⁰B (10.01) and ¹¹B (11.01). Find each isotope's percent abundance.
  3. In a mass spectrum, what does a taller peak tell you about an isotope?

C · Empirical & Molecular Formulas CED 1.3 · LO 1.3.A

By the law of definite proportions, a pure compound always has the same mass ratio of elements. The empirical formula is the lowest whole-number atom ratio.

What you must be able to do (College Board learning objective 1.3.A): “Explain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance.”

Study this outside class. On AP Classroom, this topic’s daily videos cover:

Recipe: % → grams (assume 100 g) → moles → divide by the smallest → whole numbers. Molecular formula: multiply by n = (molar mass)/(empirical mass).
Worked example
A compound is 40.00% C, 6.71% H, 53.29% O; molar mass 180.2 g/mol. Find both formulas.
C 40.00/12.01 = 3.331 · H 6.71/1.008 = 6.66 · O 53.29/16.00 = 3.331 → ÷3.331 → 1 : 2 : 1
empirical = CH₂O (mass 30.03); n = 180.2/30.03 = 6.00
molecular = C₆H₁₂O₆ (glucose)

Practice C

  1. A compound is 52.14% C, 13.13% H, 34.73% O. Find the empirical formula.
  2. A hydrocarbon is 85.63% C, 14.37% H, molar mass 56.1 g/mol. Find the empirical and molecular formulas.
  3. Why do CH₂O and C₆H₁₂O₆ share the same percent composition?

D · Composition of Mixtures CED 1.4 · LO 1.4.A

A pure substance has one kind of particle; a mixture has two or more in variable proportions. Elemental analysis gives composition and purity.

What you must be able to do (College Board learning objective 1.4.A): “Explain the quantitative relationship between the elemental composition by mass and the composition of substances in a mixture.”

Study this outside class. On AP Classroom, this topic’s daily videos cover:

Core: work from the mass of each component → moles → mass or mole fraction. Purity = (mass of interest)/(total mass) × 100%.
Worked example
A 5.00 g road-salt sample is 60.0% KCl by mass (rest inert). How many moles of KCl?
mass KCl = 0.600 × 5.00 = 3.00 g; M(KCl) = 39.10 + 35.45 = 74.55 g/mol
n = 3.00 / 74.55 = 0.0402 mol

Practice D

  1. A 10.0 g ore sample yields 4.2 g of iron. What is the percent iron?
  2. A 2.50 g sample is 18.0% chloride by mass. What mass of chloride does it contain?
  3. Why can a mixture's percent composition vary while a pure compound's cannot?

E · Electron Configuration & Coulomb's Law CED 1.5 · LO 1.5.A

An atom is a small positive nucleus surrounded by electrons in shells and subshells. The ground-state electron configuration says where every electron sits.

What you must be able to do (College Board learning objective 1.5.A): “Represent the ground-state electron configuration of an atom of an element or its ions using the Aufbau principle.”

Study this outside class. On AP Classroom, this topic’s daily videos cover:

Not assessed, don't spend time here: “The assignment of quantum numbers to electrons in subshells of an atom will not be assessed on the AP Exam.” (CED exclusion statement)

Aufbau: fill lowest energy first, 1s 2s 2p 3s 3p 4s 3d 4p … (s holds 2, p 6, d 10). Core = inner [noble-gas] electrons; valence = outer electrons that do the chemistry.
Coulomb's law F ∝ q₁q₂ / r² explains the rest: an electron closer to the nucleus, or feeling a larger effective charge, is held more tightly → harder to remove → higher ionization energy.
Worked example
Write the ground-state configuration of Fe (Z = 26), then Fe²⁺ and Fe³⁺.
Fe = [Ar] 4s² 3d⁶
To make cations, remove the outer 4s electrons first:
Fe²⁺ = [Ar] 3d⁶  ·  Fe³⁺ = [Ar] 3d⁵

Not tested: assigning quantum numbers (n, ℓ, mℓ, ms), you need the configuration, not the quantum numbers.

Practice E

  1. Write the configuration of sulfur (Z = 16) in [noble-gas] shorthand.
  2. Write the configuration of Ca²⁺ (Ca is Z = 20). Which noble gas is it isoelectronic with?
  3. Use Coulomb's law to explain why removing a 1s electron takes more energy than removing a 2s electron from the same atom.

F · Photoelectron Spectroscopy (PES) CED 1.6 · LO 1.6.A

PES is the AP-signature way to "see" electron configuration: shine high-energy light on atoms, eject electrons, and measure the energy needed to remove them.

What you must be able to do (College Board learning objective 1.6.A): “Explain the relationship between the photoelectron spectrum of an atom or ion and: i. The ground-state electron configuration of the species. ii. The interactions between the electrons and the nucleus.”

Study this outside class. On AP Classroom, this topic’s daily videos cover:

Reading a spectrum: each peak is one subshell. Peak position (binding energy) shows how tightly electrons are held, 1s electrons need the most energy, so they sit on the high-energy side. Peak height ∝ number of electrons in that subshell.
Worked example
A neutral atom's PES shows four peaks; from highest binding energy to lowest the heights are 2, 2, 6, 2. Identify it.
1s² → 2s² → 2p⁶ → 3s² = 2+2+6+2 = 12 electrons
magnesium (Mg)

Practice F

  1. Sketch the expected PES peaks (position + relative height) for nitrogen (Z = 7).
  2. A PES spectrum has peaks 2, 2, 6, 1 (high → low energy). Identify the element.
  3. Atom X's 1s peak is at higher binding energy than atom Y's. Which atom has the greater nuclear charge? Explain with Coulomb's law.

G · Periodic Trends CED 1.7 · LO 1.7.A

The table is arranged so properties recur. Every trend comes from three ideas: nuclear charge, distance (shell), and shielding, Coulomb's law again.

What you must be able to do (College Board learning objective 1.7.A): “Explain the relationship between trends in atomic properties of elements and electronic structure and periodicity.”

Study this outside class. On AP Classroom, this topic’s daily videos cover:

Not assessed, don't spend time here: “Writing the electron configuration of elements that are exceptions to the aufbau principle will not be assessed on the AP Exam.” (CED exclusion statement)

Effective nuclear charge (Z_eff) = nuclear charge − shielding by inner electrons; bigger Z_eff pulls electrons in tighter.  Across a period → radius shrinks, ionization energy & electronegativity rise.  Down a group ↓ radius grows, ionization energy & electronegativity fall.  Ions: cations are smaller than the parent atom, anions larger.
Worked example
Rank Na, Mg, Cl (all period 3) by atomic radius and by first ionization energy.
Left → right, Z_eff rises (11 → 12 → 17 protons, similar shielding).
radius: Na > Mg > Cl  ·  ionization energy: Cl > Mg > Na

Not tested: writing the aufbau exceptions (Cr, Cu).

Practice G

  1. Rank K, Ca, Br by atomic radius and explain.
  2. Place Na⁺, Ne, F⁻ (all 10 electrons) in order of increasing radius; explain with nuclear charge.
  3. Why does ionization energy generally rise across a period?

H · Valence Electrons & Ionic Compounds CED 1.8 · LO 1.8.A

Whether two elements bond (and the formula they make) is set by their valence electrons and their positions on the table.

What you must be able to do (College Board learning objective 1.8.A): “Explain the relationship between trends in the reactivity of elements and periodicity.”

Study this outside class. On AP Classroom, this topic’s daily videos cover:

Typical ion charges from the group: 1 → +1, 2 → +2, 13 → +3, 15 → −3, 16 → −2, 17 → −1. Elements in the same column form analogous compounds.
Worked example
Predict the ionic compound of calcium and chlorine.
Ca → Ca²⁺ (group 2) · Cl → Cl⁻ (group 17); balance charge: one Ca²⁺ needs two Cl⁻
CaCl₂

Practice H

  1. Predict the compound of aluminum and oxygen.
  2. Potassium reacts vigorously with water; argon (next to it) doesn't react at all. Explain using valence electrons.
  3. Magnesium forms MgO. Predict the analogous compound formed by barium.
AP Chemistry · Unit 1, Atomic Structure & Properties · Dr. Ras Mulinta · Handsworth Secondary 2026–27. Pegged to the College Board AP Chemistry CED (topics 1.1–1.8) and BC Chemistry 11. Atomic masses from the IUPAC periodic table.