AP Chemistry

Complete course reading notes

Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.

1. Moles are a counting bridge, not a mass

Learning goal: Convert between particles, moles and grams, and read a mass spectrum or photoelectron spectrum as evidence about atoms.

Unit 1 starts with a counting unit. One mole is 6.022×10²³ particles, and the molar mass in g/mol links a sample you can weigh to the number of particles you cannot see. Every stoichiometry problem in the course uses this bridge: grams to moles with molar mass, moles to moles with a balanced equation, then back to grams or particles. Writing the unit on every number prevents most arithmetic slips.

Average atomic mass on the periodic table is a weighted average of isotope masses. Chlorine is about 35.45 u because roughly three-quarters of its atoms are ³⁵Cl and one-quarter are ³⁷Cl, not because any atom has mass 35.45. A mass spectrum shows this directly: each peak is an isotope, its position is mass and its height is relative abundance. Weighted average means multiplying each mass by its fraction and adding.

Photoelectron spectroscopy gives evidence about electron energies. Each peak is a subshell. The position, in energy, shows how tightly those electrons are held. The height shows how many electrons are in that subshell. Peaks at higher binding energy are closer to the nucleus. Across a period the same subshell peaks move to higher energy because effective nuclear charge increases while the electrons stay in the same shell. Down a group a new shell is added, so the outer electrons sit farther out and are easier to remove.

Coulomb's law is the master explanation for the whole unit: attraction grows with the charges and weakens with distance. When an exam asks you to compare ionisation energies, name the charge and the distance. A reason that says only that an atom is bigger is incomplete. A reason that says the outer electron is farther from the nucleus and more shielded, so the attraction is weaker, earns the point.

Worked example

Chlorine has two isotopes: ³⁵Cl (mass 34.97 u, 75.8%) and ³⁷Cl (mass 36.97 u, 24.2%). Estimate the average atomic mass.

Convert each percentage to a fraction: 0.758 and 0.242.
Multiply mass by fraction: 34.97×0.758 = 26.51 and 36.97×0.242 = 8.95.
Add: 26.51 + 8.95 = 35.46 u.
This matches the periodic table value of about 35.45 u within rounding.

Practice problem and solution

A sample of 36.0 g of water is weighed. How many moles of water is that? (molar mass 18.0 g/mol)

36.0 g ÷ 18.0 g/mol = 2.00 mol.

Mental model: Moles connect grams to particles. Spectra show isotopes or subshells. Coulomb's law explains trends.

Common trap: Giving 'bigger atom' as a reason without naming distance, shielding or charge.

2. Bonds are a tug of war over electrons

Learning goal: Predict bond type, draw Lewis structures with formal charge, and use VSEPR to predict geometry and polarity.

Unit 2 asks why atoms bond and what shape the result takes. Ionic bonding is the attraction between ions formed when electrons transfer, typically between a metal and a nonmetal. Covalent bonding shares electrons between nonmetals. Metallic bonding is a lattice of cations in a sea of delocalised electrons. Real bonds sit on a spectrum, and the electronegativity difference tells you where.

Lewis structures count valence electrons first, then place bonds, then complete octets. Formal charge, which is valence electrons minus nonbonding electrons minus half the bonding electrons, helps choose among possible structures. Structures with formal charges closest to zero are usually preferred, and negative formal charge belongs on the more electronegative atom. Resonance means the true structure is an average of several valid drawings, so bond lengths in the nitrate ion are all equal, not one short and two long.

VSEPR says electron domains repel and spread as far apart as possible. A domain is a bond (single, double or triple counts as one) or a lone pair. Two domains give linear, three give trigonal planar, four give tetrahedral. Lone pairs change the molecular shape: water has four domains but is bent, ammonia is trigonal pyramidal. Lone pairs repel more than bonding pairs, which is why bond angles in water are smaller than 109.5°.

Polarity needs two checks. First, is each bond polar? Second, do the bond dipoles cancel? Carbon dioxide has polar bonds but is linear, so the dipoles cancel and the molecule is nonpolar. Water is bent, the dipoles add, and the molecule is polar. Hybridisation is a bookkeeping label that follows the domain count: four domains is sp³, three is sp², two is sp.

Worked example

Draw the shape of SO₂ and say whether it is polar.

Sulfur has 6 valence electrons, each oxygen has 6, total 18.
Sulfur has two bonds and one lone pair, so three electron domains.
Three domains is trigonal planar for electron arrangement; with one lone pair the molecular shape is bent.
The bond dipoles do not cancel in a bent shape, so SO₂ is polar.

Practice problem and solution

How many electron domains does the central atom of NH₃ have?

Three N–H bonds plus one lone pair is four electron domains.

Mental model: Bond polarity plus shape decides molecular polarity. Count domains, then place lone pairs.

Common trap: Naming the shape from the domain count and forgetting that lone pairs are not atoms.

3. Why liquids evaporate and solutions mix

Learning goal: Rank intermolecular forces, connect them to boiling point and solubility, and use gas laws, Beer's law and chromatography.

Unit 3 carries the heaviest weighting on the exam, 18% to 22% of the score, so it deserves time. Intermolecular forces, or IMFs, are attractions between separate molecules. London dispersion forces exist in every substance and grow with the number of electrons and with surface contact. Dipole-dipole forces act between polar molecules. Hydrogen bonding is a strong dipole-dipole attraction when H is bonded to N, O or F and is attracted to a lone pair on another N, O or F.

Stronger IMFs mean higher boiling point, higher melting point, higher viscosity and lower vapour pressure. When the exam asks you to compare two substances, name the strongest force in each, compare them, and then connect to the property. Do not confuse IMFs with covalent bonds: boiling water breaks IMFs between molecules, not the O–H bonds inside them.

Solubility follows 'like dissolves like'. Polar solvents dissolve polar and ionic solutes because new solute-solvent attractions compensate for those broken. Ideal gases are modelled with PV = nRT; real gases deviate at high pressure and low temperature, where molecular volume and attractions matter. Dalton's law adds partial pressures, and the partial pressure of a gas is its mole fraction times the total pressure.

Beer's law, A = εbc, says absorbance is proportional to concentration when path length and wavelength are fixed. A calibration curve of known concentrations lets you read an unknown. Chromatography separates a mixture because components spend different fractions of time in the mobile and stationary phases; stronger attraction to the stationary phase means slower travel. Molarity is moles of solute per litre of solution, and dilution follows M₁V₁ = M₂V₂.

Worked example

You have 50.0 mL of 2.00 M NaCl and dilute it to 200.0 mL. What is the new molarity?

Use M₁V₁ = M₂V₂.
(2.00 M)(50.0 mL) = M₂(200.0 mL).
M₂ = 100 ÷ 200.0 = 0.500 M.
The volume rose by a factor of four, so concentration fell by a factor of four.

Practice problem and solution

A calibration curve gives absorbance 0.40 for a 0.20 M solution. Assuming Beer's law with the same cell, what absorbance would you expect for a 0.30 M solution?

Absorbance is proportional to concentration: 0.40 × (0.30/0.20) = 0.60.

Mental model: Name the IMF, then connect it to the property. Beer's law is a straight line.

Common trap: Saying covalent bonds break when a molecular substance boils.

4. Writing what actually happens in a reaction

Learning goal: Write net ionic equations, balance reactions, find limiting reagents and classify reactions.

Unit 4 turns observations into equations. A molecular equation shows all formulas. A complete ionic equation breaks strong electrolytes into ions. The net ionic equation removes spectator ions, those that appear unchanged on both sides, and keeps only the species that react. Strong acids, strong bases and soluble ionic compounds are written as ions; weak acids, weak bases, solids, liquids and gases stay as formulas.

Precipitation reactions form an insoluble solid when two ionic solutions mix. Acid-base reactions transfer H⁺. Redox reactions transfer electrons; oxidation is loss of electrons, reduction is gain, and oxidation numbers track the change. The oxidising agent is reduced and the reducing agent is oxidised. A common error is to swap these labels, so assign oxidation numbers to every atom before naming anything.

Stoichiometry uses the balanced equation as a recipe. The limiting reagent is the reactant that runs out first and sets the maximum product. Compare moles divided by coefficient for each reactant; the smallest value is limiting. Percent yield is actual yield divided by theoretical yield times 100. Theoretical yield always comes from the limiting reagent.

Titration is stoichiometry in a flask. At the equivalence point, moles of acid equal moles of base for a 1:1 reaction. The indicator changes colour near that point; the colour change is the endpoint. Gravimetric analysis weighs a precipitate to find the amount of an ion in the original sample. In each method, keep the chain going: measure, convert to moles, use the ratio, convert to the requested unit.

Worked example

Silver nitrate solution is mixed with sodium chloride solution. Write the net ionic equation.

AgNO₃ and NaCl are soluble, so they are written as ions.
Silver chloride is insoluble and forms a solid.
Na⁺ and NO₃⁻ appear unchanged on both sides, so they are spectators.
Net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s).

Practice problem and solution

2.0 mol of H₂ reacts with 2.0 mol of O₂ in 2H₂ + O₂ → 2H₂O. Which reactant is limiting? Type H2 or O2.

H₂: 2.0/2 = 1.0. O₂: 2.0/1 = 2.0. Hydrogen has the smaller value, so it is limiting.

Mental model: Net ionic equations drop spectators. Limiting reagent sets yield.

Common trap: Using moles of reactant directly without dividing by the coefficient.

5. Rates come from collisions and mechanisms

Learning goal: Determine rate laws from data, use integrated rate laws, and judge a mechanism against a rate law.

Unit 5 asks how fast a reaction goes and why. The rate law, rate = k[A]ᵐ[B]ⁿ, links rate to concentration. The orders m and n must be found by experiment; they are not the coefficients from the balanced equation. The method of initial rates compares two trials where one concentration changes and the rest stay fixed. If doubling [A] doubles the rate, the order in A is 1. If it quadruples the rate, the order is 2. If the rate does not change, the order is 0.

The units of k reveal the overall order. For a first-order reaction k has units of s⁻¹, for second order M⁻¹s⁻¹, and for zero order M·s⁻¹. Integrated rate laws turn that into straight lines: ln[A] versus time is linear for first order, 1/[A] versus time for second order, and [A] versus time for zero order. A first-order half-life is constant and equals ln 2 divided by k, so it does not depend on the starting concentration.

Collision theory says that a reaction needs collisions with enough energy and the right orientation. Raising temperature increases the fraction of collisions that exceed the activation energy, which is why the rate constant climbs quickly with temperature. A catalyst provides a different pathway with lower activation energy and is not consumed. It speeds up forward and reverse reactions and does not change the equilibrium position.

A mechanism is a sequence of elementary steps. The rate law of an elementary step follows its coefficients, and the slow step sets the overall rate. An intermediate is made and then used; a catalyst is used first and then regenerated. For a proposed mechanism to be acceptable, the steps must add to the overall equation and the slow step must give the observed rate law. A mechanism can be consistent with data without being proven.

Worked example

Trial 1: [A]=0.10 M, rate 2.0×10⁻³ M/s. Trial 2: [A]=0.30 M, rate 1.8×10⁻² M/s. Find the order in A.

The concentration increased by a factor of 3.
The rate increased by 1.8×10⁻² ÷ 2.0×10⁻³ = 9.
3ⁿ = 9, so n = 2.
The reaction is second order in A.

Practice problem and solution

A first-order reaction has k = 0.0347 s⁻¹. Estimate the half-life in seconds (ln 2 = 0.693).

t½ = 0.693 ÷ 0.0347 ≈ 20 s.

Mental model: Orders come from data. Slow step sets the rate. Catalysts change pathway, not equilibrium.

Common trap: Reading orders from the coefficients of the overall equation.

6. Heat flow, enthalpy and calorimetry

Learning goal: Track energy as heat, calculate with q = mcΔT, apply Hess's law and interpret bond energy and energy diagrams.

Unit 6 is about energy changes. The system is the reaction; the surroundings are everything else. Heat flows from hot to cold. If the surroundings get warmer, the reaction is exothermic and releases heat to them, and ΔH is negative. If the surroundings cool, it is endothermic and ΔH is positive. Heat lost by one object equals heat gained by another when the system is isolated, which is the basis of calorimetry.

For a temperature change with no phase change, q = mcΔT, where m is mass, c is specific heat capacity and ΔT is final minus initial temperature. During a phase change temperature stays constant and the heat is q = n·ΔH of fusion or vaporisation. Heating curves show flat sections for phase changes and sloped sections for single-phase heating. A common error is to use q = mcΔT while ice melts, when temperature is constant.

Enthalpy is a state function, which means ΔH depends on initial and final states and not on the path. Hess's law uses this: add known reactions to get the target, flip a reaction and the sign of ΔH flips, multiply a reaction and ΔH is multiplied. Standard enthalpy of formation gives ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants), with elements in their standard states set to zero.

Bond energy gives a rough estimate: ΔH ≈ energy of bonds broken minus energy of bonds formed. Breaking bonds absorbs energy and forming bonds releases it. An energy diagram shows reactants, products and the peak. The height from reactants to peak is the activation energy; the difference between reactants and products is ΔH. The same ΔH can come with a large or small barrier, so thermodynamics and kinetics are separate questions.

Worked example

How much heat is needed to warm 50.0 g of water from 20.0 °C to 30.0 °C? (c = 4.18 J/g·°C)

Use q = mcΔT.
ΔT = 30.0 − 20.0 = 10.0 °C.
q = 50.0 × 4.18 × 10.0 = 2090 J.
The sign is positive because the water absorbs heat.

Practice problem and solution

A reaction has ΔH = −90 kJ. What is ΔH for the reverse reaction, in kJ?

Reversing the reaction flips the sign, so +90 kJ.

Mental model: q = mcΔT without phase change. Hess's law adds ΔH. Energy diagrams show Eₐ and ΔH separately.

Common trap: Applying mcΔT during a phase change.

7. Equilibrium is a balance of rates, not a halt

Learning goal: Write K, use Q to predict direction, apply Le Châtelier's principle and solve ICE tables.

Unit 7 describes reactions that go both ways. At equilibrium the forward and reverse rates are equal, so concentrations stay constant, but both reactions are still happening. The equilibrium constant expression is products over reactants, each raised to its coefficient, using concentrations for aqueous species and pressures for gases (Kp). Pure solids and liquids are left out because their concentration does not change. K depends only on temperature.

The reaction quotient Q has the same form as K but uses current concentrations. If Q < K the reaction shifts toward products; if Q > K it shifts toward reactants; if Q = K it is at equilibrium. A large K, such as 10⁵, means products are favoured at equilibrium. A small K, such as 10⁻⁵, means reactants are favoured. K says nothing about how fast equilibrium is reached.

An ICE table organises a calculation: Initial, Change, Equilibrium. Write the change in terms of x using the coefficients, substitute the equilibrium row into the K expression and solve. When K is very small compared with the starting concentration, x is negligible next to the starting value and the approximation saves time; check it afterwards. Le Châtelier's principle predicts how a system responds when disturbed: adding reactant shifts right, removing product shifts right, compressing a gas shifts toward fewer moles of gas, and raising temperature shifts toward the endothermic direction. Only a temperature change alters K.

Solubility equilibrium uses Ksp. For AgCl, Ksp = [Ag⁺][Cl⁻]. Adding a common ion, such as Cl⁻ from NaCl, shifts the equilibrium toward the solid and lowers the solubility of AgCl. Comparing Q with Ksp predicts whether a precipitate forms: precipitation occurs when Q exceeds Ksp. Adding acid increases the solubility of salts of weak-acid anions because H⁺ removes the anion.

Worked example

For H₂ + I₂ ⇌ 2HI, K = 50 at some temperature. A mixture has [H₂]=0.10, [I₂]=0.10, [HI]=0.50 M. Which way does it shift?

Q = [HI]² / ([H₂][I₂]).
Q = 0.50² / (0.10 × 0.10) = 0.25 / 0.010 = 25.
Compare: Q = 25 is smaller than K = 50.
Q < K, so the reaction shifts toward products.

Practice problem and solution

For 2NO₂ ⇌ N₂O₄, K = 4 at some temperature and [N₂O₄] = 0.16 M. What equilibrium [NO₂] (M) gives that K?

K = [N₂O₄]/[NO₂]². 4 = 0.16/[NO₂]². [NO₂]² = 0.04, so [NO₂] = 0.20 M.

Mental model: Compare Q with K for direction. Only temperature changes K.

Common trap: Saying equilibrium means the reaction has stopped.

8. Acids, bases and buffers

Learning goal: Calculate pH, work with weak acid equilibria, buffers and titration curves.

Unit 8 carries 11% to 15% of the exam. pH = −log[H⁺], and pOH = −log[OH⁻], with pH + pOH = 14.00 at 25 °C. A strong acid dissociates completely, so [H⁺] equals the acid concentration for a monoprotic acid. A weak acid only partly dissociates, and its strength is expressed by Ka. A larger Ka means a stronger acid. A conjugate pair is linked by one proton, and Ka × Kb = Kw = 1.0×10⁻¹⁴ for a pair at 25 °C.

For a weak acid HA, set up an ICE table with HA ⇌ H⁺ + A⁻. If the acid is not too dilute and Ka is small, x is small relative to the initial concentration, so Ka ≈ x²/C. Then pH = −log x. Check that x is under about 5% of C. Salts of weak acids give basic solutions because the conjugate base reacts with water; salts of weak bases give acidic solutions. Salts of strong acid and strong base are neutral.

A buffer contains a weak acid and its conjugate base in comparable amounts. Added H⁺ is consumed by A⁻ and added OH⁻ is consumed by HA, so pH changes little. The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), gives the pH. A buffer works best when the ratio is near 1 and pH is within one unit of pKa. A buffer has limited capacity: it fails when one component is used up.

In a titration of a weak acid with strong base, the pH rises slowly in the buffer region, equals pKa at the half-equivalence point, and is above 7 at the equivalence point because the conjugate base is present. For a strong acid with strong base, the equivalence point is at pH 7. Choose an indicator whose colour change falls in the steep part of the curve. Molecular structure matters too: acid strength increases with a more stable conjugate base, which often means higher electronegativity or more resonance.

Worked example

A buffer has 0.20 M CH₃COOH and 0.20 M CH₃COONa. pKa of acetic acid is 4.74. What is the pH, and what is it after adding a small amount of strong acid?

Ratio [A⁻]/[HA] = 1, so log term = 0.
pH = pKa + 0 = 4.74.
Adding a small amount of H⁺ converts some A⁻ to HA.
The ratio changes slightly, so the pH falls only a little below 4.74.

Practice problem and solution

At the half-equivalence point of a weak acid titration, pH = 4.76. What is the pKa of the acid?

At half-equivalence [HA] = [A⁻], so pH = pKa = 4.76.

Mental model: pH = pKa + log ratio. Half-equivalence gives pKa.

Common trap: Treating a weak acid as if it dissociates completely.

9. Entropy, free energy and electrochemistry

Learning goal: Predict spontaneity with ΔG, link ΔG to K, and calculate cell potentials.

Unit 9 asks whether a process will happen and how to harness it. Entropy, S, measures the number of ways energy and matter can be arranged. It increases when a solid becomes a liquid or gas, when the number of gas moles increases and when a solute dissolves. The Gibbs free energy change, ΔG = ΔH − TΔS, combines enthalpy and entropy. A reaction is thermodynamically favourable when ΔG is negative.

The signs of ΔH and ΔS give four cases. Negative ΔH and positive ΔS is favourable at all temperatures. Positive ΔH and negative ΔS is never favourable. The other two cases depend on temperature: they are favourable at low temperature if ΔH is negative and ΔS is negative, and at high temperature if ΔH is positive and ΔS is positive. Favourable does not mean fast; a reaction with negative ΔG can have a very high activation energy.

ΔG° relates to K by ΔG° = −RT ln K. A negative ΔG° means K is greater than 1. In an electrochemical cell, ΔG° = −nFE°, with n moles of electrons, F = 96,485 C/mol and E° the standard cell potential. A positive E° means a galvanic (voltaic) cell that can do work. In a galvanic cell, oxidation happens at the anode, reduction at the cathode, electrons flow through the wire from anode to cathode, and the salt bridge lets ions move to keep the solutions neutral.

E°cell = E°cathode − E°anode, using reduction potentials. Do not multiply the potential by the coefficient when balancing electrons; potential is intensive. The Nernst equation, E = E° − (RT/nF) ln Q, shows that a cell's potential falls as Q rises, reaching zero at equilibrium. In electrolysis, an external source drives a nonspontaneous reaction. Faraday's law links charge to moles: moles of electrons = (current × time) ÷ F.

Worked example

A cell uses Zn²⁺/Zn (E° = −0.76 V) and Cu²⁺/Cu (E° = +0.34 V). Find E°cell and the anode.

The higher reduction potential is the cathode: copper.
Zinc is the anode.
E°cell = 0.34 − (−0.76) = 1.10 V.
E° is positive, so the cell is spontaneous as written.

Practice problem and solution

Calculate ΔG in kJ for ΔH = −100 kJ, ΔS = −0.20 kJ/K at 300 K.

ΔG = −100 − (300)(−0.20) = −100 + 60 = −40 kJ.

Mental model: ΔG = ΔH − TΔS. Positive E°cell means spontaneous.

Common trap: Doubling E° when you double the reaction.

10. Exam day: reading the question like a grader

Learning goal: Use the exam format, skills and common scoring habits to turn chemistry knowledge into points.

The AP Chemistry Exam has two sections. Section I has 60 multiple-choice questions in 1 hour 30 minutes and counts for 50% of the exam score. Section II has seven free-response questions in 1 hour 45 minutes and also counts for 50%: three long-answer questions worth 10 points each and four short-answer questions worth 4 points each. Calculators are permitted, and reference information is provided. The College Board's exam page lists the current format; check it for your exam year.

The course framework names six skills: models and representations, question and method, representing data and phenomena, model analysis, mathematical routines and argumentation. In the free-response section, mathematical routines carry roughly 43% to 53% of the points and argumentation roughly 15% to 24%, according to the course and exam description. That split means calculations need setup, units and significant figures, and explanations need a claim plus evidence plus reasoning.

For a calculation, write the formula, substitute with units, and state the answer with units. Partial credit follows steps, so do not skip them. For an argument, use the pattern claim, evidence, reasoning: the claim answers the question, the evidence is a number or observation from the stimulus, and the reasoning names the chemical principle, such as Coulomb's law or IMFs. Avoid reasons that restate the claim in other words.

For particulate diagrams, count atoms, show lone pairs and keep spacing and proportions consistent with the substance's state. For experimental design, name the measurement, the variable you change, the variables you hold constant and how you will analyse the data. For graph questions, label axes with units and say what the slope means. Use the first minutes of each free-response question to underline what is asked, because 'explain' and 'justify' need a reason and 'calculate' needs numbers.

Worked example

Explain why a 0.10 M solution of acetic acid has a pH higher than a 0.10 M solution of HCl.

Claim: acetic acid solution has a higher pH.
Evidence: acetic acid is a weak acid and HCl is a strong acid.
Reasoning: HCl dissociates completely, so [H⁺] = 0.10 M; acetic acid only partly dissociates, so [H⁺] is lower.
Lower [H⁺] means higher pH.

Practice problem and solution

A free-response question is worth 4 points and asks you to 'justify' a claim. How many parts should a strong answer have? Enter a number.

At least two: the claim supported by evidence from the prompt, and the chemical principle connecting them. Some answers add a third part, comparison.

Mental model: Show steps and units. Argue with claim, evidence and reasoning.

Common trap: Writing a conclusion with no principle behind it.

Course outline

Moles are a counting bridge, not a mass

Bonds are a tug of war over electrons

Why liquids evaporate and solutions mix

Writing what actually happens in a reaction

Rates come from collisions and mechanisms

Heat flow, enthalpy and calorimetry

Equilibrium is a balance of rates, not a halt

Acids, bases and buffers

Entropy, free energy and electrochemistry

Exam day: reading the question like a grader

Sources and curriculum note

Complete course reading notes

1. Moles are a counting bridge, not a mass

Worked example

2. Bonds are a tug of war over electrons

Worked example

3. Why liquids evaporate and solutions mix

Worked example

4. Writing what actually happens in a reaction

Worked example

5. Rates come from collisions and mechanisms

Worked example

6. Heat flow, enthalpy and calorimetry

Worked example

7. Equilibrium is a balance of rates, not a halt

Worked example

8. Acids, bases and buffers

Worked example

9. Entropy, free energy and electrochemistry

Worked example

10. Exam day: reading the question like a grader

Worked example