AP Chemistry · Handsworth Secondary 2026–27

Unit 4 · Chemical Reactions & Stoichiometry

Dr. Ras Mulinta
Handsworth Secondary
Notes Package

CED Unit 4, 7–9% of the AP exam, taught in 7 blocks. This unit is not given a separate unit test, but it is on the AP exam, so we still front-load it fully. You already own the mole, balancing, and mass–mole–mass stoichiometry from BC Chemistry 11, so we move fast through those and spend real time on the AP-specific depth: net ionic equations, classifying reaction types, limiting reactant by moles, titration, and a first taste of redox (finished off in Unit 9).

Blocks:  1 → A–B  ·  2 → C  ·  3 → D  ·  4 → E  ·  5 → F  ·  6 → G  ·  7 → H + error analysis

Name:Block:Date:

A · Physical vs. Chemical Change CED 4.1 / 4.4 · LO 4.1.A / 4.4.A

A 15-minute warm-up: chemistry is the study of change, so first name the change. The dividing line is whether chemical bonds are made or broken.

Core: a physical change alters properties but not composition (phase changes, making/separating mixtures, only intermolecular forces shift). A chemical change makes new substances. Evidence of a chemical change: heat or light, a gas forming, a precipitate, or a colour change.
The honest grey zone: dissolving a salt in water can be argued either way, ionic bonds break, but new ion–dipole interactions form. Don't treat "physical vs. chemical" as always black-and-white; justify with bonds.
Worked example
Classify each, and give the deciding reason: (i) ice melting, (ii) iron rusting, (iii) mixing AgNO₃(aq) with NaCl(aq).
(i) phase change only, H₂O molecules stay H₂O, just intermolecular forces loosen → physical.
(ii) Fe + O₂ → new compound Fe₂O₃; bonds form → chemical (colour change is the evidence).
(iii) a white solid (AgCl) appears, a precipitate is direct evidence of a chemical change.

Practice A

  1. Classify: (a) sugar dissolving in tea, (b) a candle wick burning, (c) water boiling. Give the deciding reason for each.
  2. You mix two clear solutions and the test tube gets noticeably warm. What does that suggest, and why?
  3. Argue both sides: is dissolving NaCl in water a physical or a chemical change?

B · Writing, Balancing & Net Ionic Equations CED 4.2 / 4.3 · LO 4.2.A / 4.3.A

An equation is bookkeeping for atoms. Because atoms only rearrange, both mass and charge are conserved: equal atoms of each element on both sides, and equal total charge. The AP twist beyond Chem 11 is the net ionic equation.

Three forms of the same reaction: molecular (full formulas) → complete ionic (split every strong electrolyte / soluble ionic compound into its ions) → net ionic (cancel spectator ions that appear unchanged on both sides). Ag⁺(aq) + Cl⁻(aq) → AgCl(s) is the whole story of many precipitations.
Soluble for sure: all Na⁺, K⁺, NH₄⁺, and NO₃⁻ salts dissolve, so they are almost always spectators. That's the only solubility set the AP expects you to lean on.
Worked example
Aqueous lead(II) nitrate is mixed with aqueous potassium iodide; PbI₂ is insoluble. Give all three equations.
Molecular: Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq)
Complete ionic: Pb²⁺ + 2 NO₃⁻ + 2 K⁺ + 2 I⁻ → PbI₂(s) + 2 K⁺ + 2 NO₃⁻
Spectators K⁺ and NO₃⁻ cancel (unchanged both sides):
Net ionic: Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s)  (charge: +2 − 2 = 0 each side ✓)

Not tested: rote memorization of solubility rules beyond "all Na⁺, K⁺, NH₄⁺, and NO₃⁻ salts are soluble." If a problem needs more, it will tell you what's insoluble.

Practice B

  1. Balance: __ C₃H₈ + __ O₂ → __ CO₂ + __ H₂O.
  2. Write the net ionic equation for mixing BaCl₂(aq) with Na₂SO₄(aq) (BaSO₄ is insoluble).
  3. Mixing NaCl(aq) with KNO₃(aq) produces no reaction. Use the soluble-salts rule to explain why there is no net ionic equation.

C · Classifying Reaction Types CED 4.7 · LO 4.7.A

On the AP exam you must name the type and justify it from the equation. There are three to know: precipitation, acid–base, and oxidation–reduction (redox), with combustion as a redox subclass.

Three tests: Precipitation: two aqueous solutions form an insoluble solid (a (s) appears). Acid–base: one or more protons (H⁺) transfer between species. Redox: electrons transfer, shown by oxidation numbers changing. Combustion (a hydrocarbon + O₂ → CO₂ + H₂O) is redox.
Worked example
Classify each and justify: (i) HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l); (ii) Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s); (iii) AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).
(i) H⁺ moves from HCl to OH⁻ to make water → acid–base.
(ii) Zn goes 0 → +2 (loses e⁻), Cu goes +2 → 0 (gains e⁻) → redox.
(iii) insoluble AgCl(s) forms → precipitation.

Not tested: the terms "oxidizing agent" and "reducing agent" are not assessed on the AP exam, you must identify what is oxidized/reduced, but you won't be graded on the "agent" labels.

Practice C

  1. Classify and justify: 2 H₂(g) + O₂(g) → 2 H₂O(l).
  2. Classify and justify: CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g).
  3. Classify and justify: Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s).

D · Acid–Base Reactions (Brønsted–Lowry) CED 4.8 · LO 4.8.A

Zoom in on the acid–base type. The AP definition is all about the proton: who donates H⁺ and who accepts it.

Brønsted–Lowry: an acid is a proton (H⁺) donor; a base is a proton acceptor. After the transfer you get a conjugate pair: the acid loses H⁺ to become its conjugate base; the base gains H⁺ to become its conjugate acid. Water can do either job (it is amphiprotic).
Worked example
In NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), label each species and pair the conjugates.
H₂O donates an H⁺ to NH₃ → H₂O is the acid, NH₃ is the base.
NH₃ + H⁺ → NH₄⁺ (its conjugate acid); H₂O − H⁺ → OH⁻ (its conjugate base).
Conjugate pairs: NH₃ / NH₄⁺  and  H₂O / OH⁻.

Not tested: Lewis acid–base concepts are not assessed here; AP Chemistry's emphasis is on proton transfer in aqueous solution.

Practice D

  1. In HF(aq) + H₂O(l) ⇌ F⁻(aq) + H₃O⁺(aq), identify the acid, the base, and both conjugate pairs.
  2. Show how water acts as a base in HCl(aq) + H₂O(l) → Cl⁻(aq) + H₃O⁺(aq).
  3. What is the conjugate base of H₂SO₄? The conjugate acid of CO₃²⁻?

E · Mole-Ratio Stoichiometry CED 4.5 · LO 4.5.A

This is your Chem 11 home turf, sharpened for AP. The coefficients of a balanced equation are a mole ratio: the only legitimate bridge from one substance to another. Always travel through moles.

The road map: given → moles → (× mole ratio) → moles wanted → answer. n = m / M for mass, n = M × V for solutions (molarity × litres), PV = nRT for gases. Never compare mass to mass directly.
Worked example
For N₂(g) + 3 H₂(g) → 2 NH₃(g), what mass of NH₃ forms from 14.0 g of N₂ with excess H₂?
M(N₂) = 2 × 14.01 = 28.02 g/mol → n(N₂) = 14.0 / 28.02 = 0.4997 mol
mole ratio NH₃ : N₂ = 2 : 1 → n(NH₃) = 0.4997 × 2 = 0.9993 mol
M(NH₃) = 14.01 + 3(1.008) = 17.03 g/mol
m(NH₃) = 0.9993 × 17.03 = 17.0 g

Practice E

  1. For 2 H₂ + O₂ → 2 H₂O, how many moles of water form from 5.0 mol of O₂ (excess H₂)?
  2. What mass of CO₂ (M = 44.01) forms when 25.0 g of CaCO₃ (M = 100.09) decomposes: CaCO₃ → CaO + CO₂?
  3. How many mL of 0.500 M HCl are needed to react completely with 0.0250 mol of NaOH (1 : 1)?

F · Limiting Reactant & Percent Yield CED 4.5 · LO 4.5.A

When amounts of both reactants are given, one runs out first and caps the product, the limiting reactant. The AP-classic mistake is comparing masses; you must compare moles against the mole ratio.

Method: convert each reactant to moles, divide each by its coefficient, and the smaller quotient is the limiting reactant. Build product from the limiter. % yield = (actual / theoretical) × 100%.
Worked example
2 Al + 3 Cl₂ → 2 AlCl₃. React 5.40 g Al (M = 26.98) with 12.0 g Cl₂ (M = 70.90). Find the limiting reactant and the theoretical mass of AlCl₃ (M = 133.33).
n(Al) = 5.40 / 26.98 = 0.2001 mol → ÷2 = 0.1001
n(Cl₂) = 12.0 / 70.90 = 0.1693 mol → ÷3 = 0.05642 → smaller, so Cl₂ limits.
n(AlCl₃) = n(Cl₂) × (2 AlCl₃ / 3 Cl₂) = 0.1693 × 2/3 = 0.1128 mol
m(AlCl₃) = 0.1128 × 133.33 = 15.0 g (theoretical)

Practice F

  1. N₂ + 3 H₂ → 2 NH₃. Mix 2.0 mol N₂ with 3.0 mol H₂. Which is limiting, and how many mol NH₃ form?
  2. In the Al/Cl₂ example above, the lab actually isolated 12.8 g of AlCl₃. What is the percent yield?
  3. Why is comparing the masses 5.40 g Al and 12.0 g Cl₂ directly a wrong way to find the limiting reactant?

G · Titration CED 4.6 · LO 4.6.A

A titration is stoichiometry done with a buret. You add a titrant of known concentration to an analyte until they react in exact stoichiometric proportion, the equivalence point.

Key terms: the equivalence point is where moles of titrant exactly consume the analyte (set by the mole ratio). The endpoint is what you actually see (the indicator's colour change) ideally right at equivalence. Work it as: moles titrant → mole ratio → moles analyte → concentration.
Worked example
25.00 mL of HCl is titrated to the equivalence point by 32.10 mL of 0.1000 M NaOH. Find [HCl]. (HCl + NaOH → NaCl + H₂O, ratio 1 : 1.)
n(NaOH) = 0.1000 mol/L × 0.03210 L = 3.210×10⁻³ mol
ratio 1 : 1 → n(HCl) = 3.210×10⁻³ mol
[HCl] = 3.210×10⁻³ mol / 0.02500 L = 0.1284 M

Practice G

  1. 20.00 mL of NaOH is neutralized by 18.50 mL of 0.150 M HCl (1 : 1). Find [NaOH].
  2. It takes 27.40 mL of 0.200 M NaOH to titrate 25.00 mL of H₂SO₄ (H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O). Find [H₂SO₄].
  3. Distinguish the equivalence point from the endpoint. Which one do you observe directly?

H · A First Look at Redox + Error Analysis CED 4.9 · LO 4.9.A

Redox is introduced here and finished in Unit 9 (electrochemistry). For now: assign oxidation numbers, spot what is oxidized/reduced, and balance a simple redox reaction from half-reactions. We close the unit with the lab skill that runs through all of it, error analysis.

Oxidation-number rules (in order): free element = 0; monatomic ion = its charge; O = −2 (peroxides −1); H = +1 with nonmetals; group 1 = +1, group 2 = +2; F = −1. The numbers in a neutral compound sum to 0; in an ion, to the ion's charge. Oxidation = loses e⁻ (number rises); reduction = gains e⁻ (number falls).
Half-reactions: split the reaction into an oxidation half and a reduction half, balance atoms, then balance charge with electrons, scale so electrons cancel, and add. Error analysis: % error = |measured − accepted| / accepted × 100%. Name specific sources (e.g. overshooting the endpoint, an uncalibrated buret, incomplete drying) and their direction of effect.
Worked example
Balance Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) by half-reactions, and confirm what is oxidized.
Oxidation: Zn → Zn²⁺ + 2 e⁻ (Zn: 0 → +2, loses electrons)
Reduction: Cu²⁺ + 2 e⁻ → Cu (Cu: +2 → 0, gains electrons)
Electrons already equal (2 = 2); add and cancel them:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s); Zn is oxidized, Cu²⁺ is reduced.
Worked example, error analysis
A redox titration finds an iron(II) sample to be 18.9% Fe; the accepted value is 20.1%. Find the percent error and give one plausible source.
% error = |18.9 − 20.1| / 20.1 × 100% = 1.2 / 20.1 × 100%
% error = 5.97% ≈ 6.0%; e.g. some Fe²⁺ air-oxidized to Fe³⁺ before titrating, lowering the measured Fe²⁺ (a determinate, low-biased error).

Heads-up, not "not tested": the full redox toolkit, assigning agents, electrochemical cells, balancing in acidic/basic solution, is developed in Unit 9. Here you only need oxidation numbers and a simple half-reaction balance.

Practice H

  1. Assign the oxidation number of: Mn in MnO₄⁻, S in SO₄²⁻, N in NH₄⁺.
  2. In Mg(s) + 2 H⁺(aq) → Mg²⁺(aq) + H₂(g), which species is oxidized and which is reduced? Write both half-reactions.
  3. A titration gives a molar mass of 41.5 g/mol for a compound whose accepted value is 40.0 g/mol. Find the percent error and name one source that would raise the measured value.
AP Chemistry · Unit 4, Chemical Reactions & Stoichiometry · Dr. Ras Mulinta · Handsworth Secondary 2026–27. Pegged to the College Board AP Chemistry CED (topics 4.1–4.9) and BC Chemistry 11 (stoichiometry). This unit is not separately tested but appears on the AP exam. Redox is introduced here and completed in Unit 9. Atomic masses from the IUPAC periodic table.