AP Chemistry · Handsworth Secondary 2026–27

Unit 0 · Chemistry 11 Refresher

Dr. Ras Mulinta
Handsworth Secondary
Notes Package

This is the ground floor. AP Chemistry does not re-teach BC Chemistry 11, it assumes it, from the first class. So before we touch Unit 1 we make the prerequisite skills rock solid, because everything above is built on them. Every section below names the BC Chemistry 11 learning standard it rests on, quoted word for word, so you can see exactly where you have already met this material.

How to use this:  work through A–J before or alongside our first week  ·  the paper Prerequisite Refresher and its answer key are handed out in class  ·  anything that does not click, bring it to me early, not in October.

Name:Block:Date:

A · Significant Figures & Scientific Notation BC Chem 11

AP graders take marks for a number reported to the wrong precision. Fix this now and it never costs you again.

BC Chemistry 11 learning standard (verbatim): “using significant figures”. And under Curricular Competencies, Planning and conducting: “Apply the concepts of accuracy and precision to experimental procedures and data: significant figures; uncertainty; scientific notation”.

Counting rules: non-zero digits always count · zeros between non-zeros count · leading zeros never count · trailing zeros count only when a decimal point is written.
Carrying rules: in × and ÷, the answer takes the fewest significant figures of any input. In + and −, the answer takes the fewest decimal places. Round once, at the very end, never mid-calculation.
Worked example
How many significant figures in 0.00250, and how is it written in scientific notation?
Leading zeros do not count. The digits are 2, 5, 0, and the trailing zero counts because a decimal point is written.
3 significant figures  ·  2.50 × 10⁻³

Practice A

  1. Significant figures in each: 1200, 1200. 0.0104, 8.00 × 10⁵.
  2. Evaluate to the correct precision: 4.20 × 3.1.
  3. Evaluate to the correct precision: 12.11 + 0.9.
  4. Write 0.000 042 8 and 93 500 000 in scientific notation.
  5. A balance reads 2.0 g and a second reads 2.000 g. What does the difference tell you about the two instruments?

B · Dimensional Analysis BC Chem 11

This is the workhorse of AP Chemistry. If you carry units through every line, the units tell you whether the setup is right before you ever check the arithmetic.

BC Chemistry 11 learning standard (verbatim): “dimensional analysis” “factor-label method (unit-analysis method)”; “calculation of mass and molar quantities (using significant figures)”.

The method: write what you are given, multiply by conversion factors written as fractions, and arrange each fraction so the unit you want to lose is on the bottom. Cancel units on the page. The unit that survives must be the unit you were asked for.
Worked example
A liquid has density 1.50 g/cm³. What is the mass of 2.50 L of it?
2.50 L × (1000 cm³ / 1 L) × (1.50 g / 1 cm³)   L cancels, then cm³ cancels, leaving g.
3.75 × 10³ g  (3 sig figs)

Practice B

  1. Convert 45.0 km/h to m/s.
  2. Convert 250. mL to L, then to m³.
  3. A reaction produces 0.0250 mol of gas. How many millimoles is that?
  4. SI prefixes: express 3.4 × 10⁻⁵ m in µm, and 2.5 × 10⁶ Pa in kPa.
  5. Set up (do not evaluate) the conversion from grams of NaCl to number of chloride ions. Which two conversion factors do you need?

C · The Mole BC Chem 11

You cannot count atoms, so you weigh them. The mole is the bridge from the mass on the balance to the number of particles reacting, and AP Chemistry crosses that bridge in almost every question.

BC Chemistry 11 learning standard (verbatim): “the mole”.

Core relationships: n = m / M  ·  N = n × N_A  ·  N_A = 6.022 × 10²³ mol⁻¹. Molar mass M is read off the periodic table and summed over the formula.
Worked example
How many moles, and how many molecules, are in 9.80 g of H₂SO₄? (M = 98.09 g/mol)
n = m / M = 9.80 / 98.09 = 0.0999 mol
N = 0.0999 × 6.022 × 10²³
0.0999 mol  ·  6.02 × 10²² molecules

Practice C

  1. Molar mass of Ca(NO₃)₂. Show the arithmetic.
  2. Moles in 25.0 g of CO₂.
  3. Mass of 0.350 mol of NaOH.
  4. Atoms of oxygen in 1.00 mol of Al₂(SO₄)₃. (Careful, count the oxygens in the formula first.)
  5. A sample contains 3.01 × 10²³ molecules of water. What is its mass?

D · Percent Composition, Empirical & Molecular Formulas BC Chem 11

This is the same mole arithmetic run backwards: from masses to a formula.

BC Chemistry 11 learning standard (verbatim): “calculation of mass and molar quantities (using significant figures)” under “dimensional analysis”; built on “the mole”.

The route: percent → assume a 100 g sample so percents become grams → divide each by its molar mass to get moles → divide every mole value by the smallest → scale to whole numbers. That gives the empirical formula. Then molecular = k × empirical where k = M(molecular) / M(empirical).
Worked example
A compound is 26.7% C, 2.24% H, 71.1% O by mass. Find the empirical formula.
Assume 100 g: n(C) = 26.7/12.01 = 2.223   n(H) = 2.24/1.008 = 2.222   n(O) = 71.1/16.00 = 4.444
Divide by the smallest (2.222): C 1.00, H 1.00, O 2.00
CHO₂  (if M = 90.0 g/mol, then k = 2 and the molecular formula is C₂H₂O₄)

Practice D

  1. Percent by mass of nitrogen in NH₄NO₃.
  2. Empirical formula of a compound that is 52.2% C, 13.0% H, 34.8% O.
  3. A compound has empirical formula CH₂ and molar mass 42.08 g/mol. Molecular formula?
  4. A 5.00 g sample of a metal oxide contains 3.22 g of the metal, which has molar mass 55.85 g/mol. Find the empirical formula.
  5. Why does assuming exactly 100 g not change the answer?

E · Writing & Balancing Reactions BC Chem 11

An unbalanced equation makes every number after it wrong. Balance first, always.

BC Chemistry 11 learning standard (verbatim): “reactions” “predicting products, reactants and energy changes (ΔH)”.

Balance by conservation of atoms: change only coefficients, never subscripts. Leave free elements (like O₂) until last. Check every element, then check the total charge if the equation is ionic.
Reaction types to recognise on sight: synthesis · decomposition · single replacement · double replacement (including precipitation and neutralisation) · combustion. AP will later add net ionic equations and redox on top of these.
Worked example
Balance the combustion of propane: C₃H₈ + O₂ → CO₂ + H₂O
Carbon: 3 on the left → 3 CO₂. Hydrogen: 8 on the left → 4 H₂O. Oxygen now needed on the right: (3×2) + 4 = 10, so 5 O₂.
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Practice E

  1. Balance: Al + O₂ → Al₂O₃
  2. Balance: Fe₂O₃ + CO → Fe + CO₂
  3. Balance the combustion of ethanol, C₂H₅OH.
  4. Write and balance the reaction of aqueous silver nitrate with aqueous sodium chloride. Which product is the precipitate?
  5. Write and balance the neutralisation of sulfuric acid by sodium hydroxide.

F · Stoichiometry, Limiting Reactant & Yield BC Chem 11

Every stoichiometry problem is the same three moves: to moles → across the mole ratio → out of moles. The mole ratio comes from the balanced coefficients, and from nowhere else.

BC Chemistry 11 learning standard (verbatim): “stoichiometric calculations” “mass”; “number of molecules”; “gas volumes”; “molar quantities”; “excess and limiting reactants”.

Finding the limiting reactant: convert each reactant to moles, then divide each by its own coefficient. The smallest result is the limiting reactant. It sets the yield; the other reactant is in excess and some of it survives.
Yield: % yield = (actual / theoretical) × 100. Theoretical yield always comes from the limiting reactant.
Worked example
10.0 g of H₂ reacts with 100.0 g of O₂: 2 H₂ + O₂ → 2 H₂O. What mass of water forms, and what is left over?
n(H₂) = 10.0/2.016 = 4.960 mol → ÷2 = 2.480  ·  n(O₂) = 100.0/32.00 = 3.125 mol → ÷1 = 3.125
2.480 < 3.125, so H₂ is limiting. Water: 4.960 mol × 18.02 g/mol.
89.4 g of H₂O, with 20.6 g of O₂ left unreacted

Practice F

  1. Mass of CO₂ from the complete combustion of 12.0 g of CH₄.
  2. 4.00 g of H₂ reacts with 28.0 g of N₂ to make NH₃. Which is limiting, and what mass of NH₃ forms?
  3. A reaction has a theoretical yield of 15.0 g and gives 12.3 g. Percent yield?
  4. How many molecules of O₂ are needed to burn 0.500 mol of C₂H₆?
  5. Explain in one sentence why doubling the excess reactant does not change the yield.

G · Solutions: Molarity, Dilution & Ions BC Chem 11

AP Chemistry lives in aqueous solution. Titrations, equilibrium, acids and bases and electrochemistry all assume you are fluent here.

BC Chemistry 11 learning standards (verbatim): “solubility” “dissociation of ions, dissociation equation”; and “stoichiometric calculations in aqueous solutions” “molarity”; “dilution effect”; “concentration of ions in solution when two solutions are mixed”.

Core relationships: c = n / V with V in litres  ·  dilution: c₁V₁ = c₂V₂ (moles are conserved; only the water changes).
Ion concentrations: write the dissociation equation first. 0.100 M Na₂SO₄ gives 0.200 M Na⁺ and 0.100 M SO₄²⁻ the subscript multiplies the concentration. This single step is the most common lost mark in solution chemistry.
Worked example
What volume of 12.0 M stock HCl is needed to make 500.0 mL of 0.150 M HCl?
c₁V₁ = c₂V₂ → V₁ = c₂V₂/c₁ = (0.150 × 500.0) / 12.0
6.25 mL of stock, then add water to a final volume of 500.0 mL

Practice G

  1. Molarity of a solution made from 5.85 g of NaCl in 250.0 mL of solution.
  2. Moles of solute in 35.0 mL of 0.200 M KMnO₄.
  3. Write the dissociation equation for Al₂(SO₄)₃ and give every ion concentration in a 0.050 M solution.
  4. 25.0 mL of 0.100 M HCl is diluted to 100.0 mL. New concentration?
  5. 50.0 mL of 0.100 M NaCl is mixed with 50.0 mL of 0.100 M AgNO₃. Which ions remain in solution, and at what concentrations, once the precipitate forms?

H · Electron Configuration & Periodic Trends BC Chem 11

Unit 1 of AP Chemistry starts here and immediately goes deeper. Arrive fluent.

BC Chemistry 11 learning standards (verbatim): “quantum mechanical model and electron configuration”; “valence electrons and Lewis structures”.

Filling order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p. Read it straight off the periodic table by block, do not memorise a diagonal diagram you cannot reconstruct.
Ions: anions gain electrons into the next available orbital. Cations of transition metals lose the outer s electrons first, before any d electrons.
Trends across a period → and down a group ↓: atomic radius decreases → increases ↓ · ionisation energy and electronegativity increase → decrease ↓. The reason is always the same: nuclear charge, shielding, and distance.
Worked example
Write the ground-state electron configuration of Fe (Z = 26) and of Fe³⁺.
Fe: [Ar] 4s² 3d⁶. For the ion, remove the two 4s electrons first, then one 3d.
Fe = [Ar]4s²3d⁶  ·  Fe³⁺ = [Ar]3d⁵

Practice H

  1. Full electron configurations of S, Ca, and Br.
  2. Configurations of S²⁻ Ca²⁺ and Cu²⁺.
  3. How many valence electrons in P, in Se, in Al?
  4. Rank by atomic radius: Na, Mg, K. Justify with nuclear charge and shell number.
  5. Which has the higher first ionisation energy, Li or Be? Why?

I · Bonding, Lewis Structures & VSEPR BC Chem 11

AP Unit 2 assumes you can draw a correct Lewis structure in seconds and read a shape off it.

BC Chemistry 11 learning standards (verbatim): “chemical bonding” “Lewis structures of compounds, polarity based on electronegativity”; “molecular geometry, valence shell electron pair repulsion (VSEPR) theory”; “bonds/forces” “covalent bond” “hydrogen bond” “intra- and intermolecular forces” “impact on properties”.

Lewis method: count total valence electrons → connect atoms with single bonds → complete octets on the outer atoms → put any leftovers on the central atom → if the centre is short, form double or triple bonds.
VSEPR: electron domains repel and spread as far apart as possible. 2 domains linear · 3 trigonal planar · 4 tetrahedral. Lone pairs occupy a domain but are invisible in the shape name, which is why NH₃ is pyramidal and H₂O is bent.
Polarity: polar bonds come from an electronegativity difference; a polar molecule needs polar bonds that do not cancel by symmetry. CO₂ has polar bonds and is non-polar. This distinction is worth marks all year.
Worked example
Draw NH₃ and give its shape and polarity.
Valence electrons: 5 + (3 × 1) = 8. Three N–H bonds use 6, leaving one lone pair on N → 4 electron domains.
Trigonal pyramidal (bond angle just under 109.5° because the lone pair repels harder), and polar: the N–H dipoles do not cancel

Practice I

  1. Lewis structures for H₂O, CH₄ CO₂ and NH₄⁺.
  2. Give the electron-domain count and the molecular shape for each of the four above.
  3. Which of CH₄ CHCl₃ CCl₄ are polar? Explain using symmetry, not just electronegativity.
  4. Rank the intermolecular forces in CH₄ H₂O, and HCl, and predict the order of boiling points.
  5. Ionic or covalent: MgO, CO, KBr, Cl₂. Justify with electronegativity difference.

J · Algebra & Graphing Toolkit BC Chem 11

Almost every AP Chemistry mark lost to “maths” is lost here rather than in the chemistry. These are the exact moves the course will ask for.

BC Chemistry 11 Curricular Competencies (verbatim): “Seek and analyze patterns, trends, and connections in data, including describing relationships between variables, performing calculations, and identifying inconsistencies”; “Construct, analyze, and interpret graphs, models, and/or diagrams”.

Rearranging: whatever you do to one side you do to the other. Practise solving n = m/M for M, c = n/V for V, and q = mcΔT for ΔT until it is automatic, AP will hand you these rearranged.
Straight lines: y = mx + b. The slope has units, and in chemistry the slope is nearly always the answer, the rate constant, the molar absorptivity, the enthalpy. Read slope as (rise units)/(run units), never as a bare number.
Proportion: if two quantities are directly proportional the graph is a straight line through the origin; inversely proportional gives a straight line only when you plot against 1/x. AP uses this constantly in kinetics and gas laws.
Worked example
A graph of absorbance against concentration is a straight line through the origin, passing through (0.0200 M, 0.240).
Slope = Δy/Δx = 0.240 / 0.0200 M
slope = 12.0 M⁻¹ and note the unit came out of the arithmetic, it was not guessed

Practice J

  1. Solve q = mcΔT for c, and state the units of c.
  2. Solve PV = nRT for T.
  3. A line passes through (2.0, 5.0) and (6.0, 13.0). Find the slope and the intercept.
  4. If y is inversely proportional to x, what should you plot on the horizontal axis to get a straight line?
  5. A student reports a slope as “12”. What is missing, and why does it matter?

Where this goes next

A–D and F feed straight into AP Unit 1 (moles, mass spectra, formulas, mixtures) and Unit 4 (reactions and stoichiometry). H and I feed Unit 1 and Unit 2. G underpins Units 3, 4, 7 and 8. J runs through the entire course.

Not in this refresher, because AP does not assume it: organic nomenclature, green chemistry, and the environmental-analysis techniques from BC Chemistry 11. They are real learning standards and worth knowing, they are simply not prerequisites for the AP course.

Sources: BC Chemistry 11 Content and Curricular Competencies, quoted verbatim from the BC Ministry of Education curriculum (curriculum.gov.bc.ca/curriculum/science/11/chemistry), retrieved 2 September 2026. AP unit references follow the AP Chemistry Course and Exam Description, College Board. Practice questions are original.