IB Chemistry · Unit 2 · Foundations

Counting
in moles.

A mole is six hundred and two billion trillion of something. Get this counting trick right and quantitative chemistry falls into place.

Lessons Key terms
11Lessons
0HL extensions
42Key terms
SL+HLLevel
6.022 × 10²³ PER MOLE
Unit 2 · Standard & Higher Level

11 lessons to work through.

The required syllabus content for Unit 2, in order. Each card is one lesson-sized checkpoint.

Lesson 1

The mole

Conceptual understanding: The mole makes it possible to correlate the number of particles with the mass that can be measured.

Lesson 2

Empirical Formula

What is the meaning behind the term empirical formula?

Lesson 3

Water of crystallization

Complete the following practical investigation to determine the water of crystallization of hydrated copper sulphate.

Lesson 4

Reacting Masses

Lesson 4 of Unit 2.

Lesson 5

Solutions and dilution

Lesson 5 of Unit 2.

Lesson 6+7

Making a standard solution and calibration curve

Watch the videos provided on how to graph data on excel

Lesson 8

Titrations

Lesson 8 of Unit 2.

Lesson 9

Limiting Reagents

Lesson 9 of Unit 2.

Lesson 10

Percentage Yield and Atom Economy

Lesson 10 of Unit 2.

Lesson 11

Reacting Gases

Lesson 11 of Unit 2.

Lesson 12+13

Ideal Gases

Conceptual understanding: Mole ratios in chemical equations can be used to calculate reacting ratios by mass and gas volume.

Lessons in detail

The unit, lesson by lesson.

Each lesson card below mirrors the original teacher deck — syllabus refs, content, worked examples and practice questions in order.

Lesson 1

The mole

Structure 1.4.1Structure 1.4.2Structure 1.4.3

Lesson outcomes

The mole (mol) is the SI unit of amount of substance. One mole = exactly Avogadro's number of elementary entities — 6.022 × 10²³ — which is the same number of carbon atoms there are in exactly 12 g of ¹²C.

Three conversions you will use constantly

n = m / M   ·   amount from mass and molar mass
n = N / NA   ·   amount from particle count
N = n × NA   ·   particle count from amount

Relative formula / molecular mass (Mr)

Sum of relative atomic masses of all atoms in the formula unit. Mr(H₂O) = 2(1.01) + 16.00 = 18.02. Mr(Na₂SO₄·10H₂O) = 2(22.99) + 32.07 + 4(16.00) + 10(18.02) = 322.25.

Molar mass M in g mol⁻¹ is numerically equal to Mr.

Worked example

Moles in a sample of lead atoms

Problem. How many moles are there in a sample of 2.9 × 10²¹ atoms of Pb? How many mmol does this correspond to?
Solution. n = N / NA = 2.9 × 10²¹ / 6.022 × 10²³ = 4.8 × 10⁻³ mol = 4.8 mmol.
Worked example

Sucrose cube

Problem. A sugar cube is 2.80 g of sucrose C₁₂H₂₂O₁₁. Calculate (a) molar mass, (b) moles in the cube, (c) number of oxygen atoms in the cube.
Solution. (a) M = 12(12.01) + 22(1.01) + 11(16.00) = 144.12 + 22.22 + 176.00 = 342.34 g mol⁻¹.
(b) n = m/M = 2.80 / 342.34 = 8.18 × 10⁻³ mol.
(c) Each molecule has 11 O atoms, so N(O) = 11 × n × NA = 11 × 8.18 × 10⁻³ × 6.022 × 10²³ = 5.42 × 10²² atoms.

Try these

  1. Name the 7 SI base units.
    Show answer
    Metre (m, length), kilogram (kg, mass), second (s, time), ampere (A, electric current), kelvin (K, temperature), mole (mol, amount of substance), candela (cd, luminous intensity).
  2. Define a mole.
    Show answer
    The SI unit of amount of substance. One mole contains exactly 6.022 × 10²³ elementary entities (Avogadro's number) — the same number as there are carbon atoms in 12 g of ¹²C.
  3. Calculate the number of atoms in 2.5 mol of copper metal.
    Show answer
    N = n × NA = 2.5 × 6.022 × 10²³ = 1.51 × 10²⁴ atoms.
  4. How many molecules are in 0.25 mol of water? How many atoms are there in 0.25 mol of water?
    Show answer
    Molecules: N = 0.25 × 6.022 × 10²³ = 1.51 × 10²³ H₂O molecules. Atoms: each H₂O has 3 atoms, so 3 × 1.51 × 10²³ = 4.52 × 10²³ atoms.
  5. How does atomic mass compare to relative atomic mass Ar? What is the relative molecular mass Mr?
    Show answer
    Atomic mass: actual mass of a single atom in kg or u. Ar: weighted average mass of all isotopes of an element, relative to 1/12 of the mass of ¹²C — a dimensionless number. Mr is the sum of Ar values for all atoms in a formula unit.
  6. Calculate the Mr for H₂O, CaCl₂, NH₃, H₂SO₄ and Na₂SO₄·10H₂O.
    Show answer
    H₂O: 2(1.01)+16.00 = 18.02. CaCl₂: 40.08+2(35.45) = 110.98. NH₃: 14.01+3(1.01) = 17.04. H₂SO₄: 2(1.01)+32.07+4(16.00) = 98.09. Na₂SO₄·10H₂O: 2(22.99)+32.07+4(16.00)+10(18.02) = 322.25.
  7. What is the relationship linking molar mass (M), mass (m) and amount (n)?
    Show answer
    n = m / M. Equivalently m = n × M or M = m / n.
  8. How many moles of CO₂ are in 36 g of CO?
    Show answer
    Trick! 36 g of CO (not CO₂). M(CO) = 28.01. n = 36/28.01 = 1.28 mol of CO. There is no CO₂.
  9. How many carbon atoms are in 36.55 g of graphite?
    Show answer
    n(C) = 36.55 / 12.01 = 3.043 mol. N = 3.043 × 6.022 × 10²³ = 1.83 × 10²⁴ atoms.
  10. How many moles of Cl⁻ ions in 0.33 g of CaCl₂?
    Show answer
    M(CaCl₂) = 40.08 + 2(35.45) = 110.98. n(CaCl₂) = 0.33/110.98 = 2.97 × 10⁻³ mol. n(Cl⁻) = 2 × n(CaCl₂) = 5.95 × 10⁻³ mol.
  11. What is the mass, in grams, of 0.20 mol of C₃H₈?
    Show answer
    M(C₃H₈) = 3(12.01) + 8(1.01) = 44.11. m = n × M = 0.20 × 44.11 = 8.82 g.
  12. Which contains more particles: 50 g of water or 50 g of mercury?
    Show answer
    n(H₂O) = 50/18.02 = 2.77 mol. n(Hg) = 50/200.59 = 0.249 mol. Water has ~11× more particles.
  13. Put in descending order of mass: 1.0 mol N₂H₄, 2.0 mol N₂, 3.0 mol NH₃, 25.0 mol H₂.
    Show answer
    Masses: N₂H₄ = 32.05 g, N₂ = 2(28.02) = 56.04 g, NH₃ = 3(17.04) = 51.12 g, H₂ = 25(2.02) = 50.50 g. Order: N₂ > NH₃ > H₂ > N₂H₄.
Lesson 2

Empirical formula

Structure 1.4.4

Lesson outcomes

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual count. They are related by a whole-number multiplier:

molecular formula = (empirical formula) × k where k = Mr(molecular) / Mr(empirical).

Method from % composition

  1. Treat each percentage as grams (assume a 100 g sample).
  2. Divide each mass by its Ar → moles of each element.
  3. Divide every result by the smallest → ratio.
  4. Multiply up to whole numbers if needed.

Method from combustion analysis (C, H, O)

Burn a known mass in excess O₂. Measure the masses of CO₂ and H₂O produced. All the C ends up as CO₂; all the H ends up as H₂O. Any O in the original is found by subtracting (mass of C + mass of H) from the original sample mass.

Worked example

Combustion analysis: C, H, O compound

Problem. A 4.406 g sample of a compound containing only C, H and O was burnt in excess oxygen, producing 8.802 g CO₂ and 3.604 g H₂O. Determine the empirical formula.
Solution. n(CO₂) = 8.802 / 44.01 = 0.200 mol → n(C) = 0.200 mol → m(C) = 0.200 × 12.01 = 2.402 g.
n(H₂O) = 3.604 / 18.02 = 0.200 mol → n(H) = 0.400 mol → m(H) = 0.400 × 1.01 = 0.404 g.
m(O) = 4.406 − 2.402 − 0.404 = 1.600 g → n(O) = 1.600 / 16.00 = 0.100 mol.
Ratio C : H : O = 0.200 : 0.400 : 0.100 = 2 : 4 : 1 → empirical formula C₂H₄O.
Worked example

Empirical → molecular formula

Problem. A compound has empirical formula CH₂O and molar mass 180.16 g mol⁻¹. Determine the molecular formula.
Solution. Mr(CH₂O) = 12.01 + 2(1.01) + 16.00 = 30.03. k = 180.16 / 30.03 = 6.0 → molecular formula = (CH₂O)₆ = C₆H₁₂O₆ (glucose).

Try these

  1. What is the meaning of the term 'empirical formula'? What is the relationship between empirical and molecular formula?
    Show answer
    Empirical formula: the simplest whole-number ratio of atoms in a compound. Molecular formula: the actual count of each atom in a molecule. They are related by a whole-number multiplier k: molecular = (empirical)k, where k = Mr(molecular) / Mr(empirical).
  2. A compound has empirical formula CH₂O and molar mass 180.16 g mol⁻¹. Determine the molecular formula.
    Show answer
    Mr(CH₂O) = 30.03. k = 180.16 / 30.03 = 6. Molecular = (CH₂O)₆ = C₆H₁₂O₆ (glucose).
  3. Name three practical ways to calculate an empirical formula.
    Show answer
    (1) Burn a metal in excess O₂ to form a metal oxide; weigh before and after. (2) Reduce a metal oxide back to the metal; mass loss = O removed. (3) Burn a volatile organic compound through combustion; trap and weigh CO₂ and H₂O products to find C and H.
  4. What is the percentage by mass of Ca, P and O in calcium phosphate Ca₃(PO₄)₂?
    Show answer
    M = 3(40.08) + 2(30.97) + 8(16.00) = 310.18. %Ca = (120.24/310.18)×100 = 38.76%. %P = (61.94/310.18)×100 = 19.97%. %O = (128.00/310.18)×100 = 41.27%.
  5. Deduce the empirical formula of an oxide of iron containing 72.36% Fe.
    Show answer
    Assume 100 g sample → 72.36 g Fe, 27.64 g O. n(Fe) = 72.36/55.85 = 1.296. n(O) = 27.64/16.00 = 1.727. Divide by smallest: Fe = 1.000, O = 1.333 → multiply by 3 → Fe₃O₄.
  6. Crocetin contains C, H, O. 1.00 g forms 2.68 g CO₂ and 0.657 g H₂O on combustion. Find the empirical formula.
    Show answer
    n(C) = 2.68/44.01 = 0.0609 → m(C) = 0.731 g. n(H) = 2 × 0.657/18.02 = 0.0729 → m(H) = 0.0737 g. m(O) = 1.00 − 0.731 − 0.0737 = 0.195 g → n(O) = 0.0122. Ratio = 0.0609 : 0.0729 : 0.0122 = 5.0 : 6.0 : 1.0 → C₅H₆O.
  7. Deduce the empirical formula of an oxide of manganese containing 36.81% oxygen.
    Show answer
    %Mn = 63.19. n(Mn) = 63.19/54.94 = 1.150. n(O) = 36.81/16.00 = 2.301. Ratio Mn:O = 1.000 : 2.001 → MnO₂.
  8. A hydrocarbon has molar mass 42.09 g mol⁻¹. Combustion gives 5.501 g CO₂ and 2.253 g H₂O. Deduce the empirical and molecular formulae.
    Show answer
    n(C) = 5.501/44.01 = 0.1250 → m(C) = 1.500 g. n(H) = 2(2.253)/18.02 = 0.2500 → m(H) = 0.252 g. Total sample mass = 1.500 + 0.252 = 1.752 g (no O present, since hydrocarbon).
    Ratio C:H = 0.125:0.250 = 1:2 → empirical = CH₂.
    Mr(CH₂) = 14.03. k = 42.09/14.03 = 3 → molecular = C₃H₆ (propene).
Lesson 3

Water of crystallisation

Structure 1.4.4

Lesson outcomes

Many ionic compounds crystallise with water molecules locked into the lattice. These are hydrated salts, written with a centred dot: CuSO₄·5H₂O is "copper(II) sulfate pentahydrate" — five waters per formula unit.

Heating drives off the water (the anhydrous salt remains). Measuring the mass before and after heating gives the ratio of moles of salt to moles of water, and hence the value of x in MX·xH₂O.

Practical · determine x for hydrated CuSO₄

  1. Weigh an empty crucible: m₀.
  2. Add the hydrated salt, reweigh: m₁. Mass of hydrate = m₁ − m₀.
  3. Heat gently to constant mass (the blue colour fades to white). Reweigh: m₂.
  4. Mass of anhydrous CuSO₄ = m₂ − m₀. Mass of water lost = m₁ − m₂.
  5. n(CuSO₄) = (m₂ − m₀) / 159.62 · n(H₂O) = (m₁ − m₂) / 18.02 · ratio = x.
Worked example

Hydrated salt formula

Problem. 2.500 g of a hydrated salt of copper(II) sulfate is heated to constant mass, yielding 1.598 g of anhydrous CuSO₄. Determine x in CuSO₄·xH₂O.
Solution. n(CuSO₄) = 1.598 / 159.62 = 0.01001 mol.
Mass of water lost = 2.500 − 1.598 = 0.902 g → n(H₂O) = 0.902 / 18.02 = 0.05006 mol.
Ratio H₂O : CuSO₄ = 0.05006 / 0.01001 = 5.00.
Formula = CuSO₄·5H₂O (pentahydrate). ✓

Try these

  1. What is meant by 'water of crystallisation'? Show the dot notation for hydrated copper sulfate pentahydrate.
    Show answer
    Water molecules locked into the crystal lattice of an ionic compound. Written with a centred dot: CuSO₄·5H₂O = five water molecules per formula unit.
  2. Why must the sample be heated to constant mass and not just for a fixed time?
    Show answer
    Until the mass stops changing, water of crystallisation may still be present. Constant mass is the experimental evidence that all the water has been driven off.
  3. 5.00 g of hydrated magnesium sulfate MgSO₄·xH₂O is heated to constant mass yielding 2.44 g of anhydrous MgSO₄. Calculate x.
    Show answer
    M(MgSO₄) = 120.37. n(MgSO₄) = 2.44/120.37 = 0.02028 mol. m(H₂O lost) = 5.00 − 2.44 = 2.56 g. n(H₂O) = 2.56/18.02 = 0.1421 mol. Ratio = 0.1421/0.02028 = 7.01. x = 7. Formula: MgSO₄·7H₂O (Epsom salt).
  4. What would happen to your calculated value of x if you didn't heat the sample for long enough? Justify.
    Show answer
    x would be too small. Residual water would be counted as anhydrous mass, so 'anhydrous mass' is overstated and 'water lost' is understated. The ratio H₂O / salt comes out lower than the true value.
Lesson 4

Reacting masses and volumes

Reactivity 2.1.1Reactivity 2.1.2

Lesson outcomes

A balanced equation gives the ratios in which substances react. The five-step method works for every stoichiometry problem:

  1. Write a balanced equation (with state symbols).
  2. Convert the given quantity (mass, volume, concentration) into moles.
  3. Apply the mole ratio from the equation to find moles of the unknown.
  4. Convert back to the unit required (mass, volume, concentration).
  5. Check significant figures and units.

Golden rule: always convert to moles first.

Worked example

Complete combustion of propane

Problem. Calculate the mass of CO₂ produced when 4.40 g of propane (C₃H₈) is completely burned in oxygen.
Solution. 1. Equation: C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l).
2. Moles: M(C₃H₈) = 44.11 g mol⁻¹. n = 4.40 / 44.11 = 0.0997 mol.
3. Ratio: 1 mol C₃H₈ gives 3 mol CO₂. n(CO₂) = 3 × 0.0997 = 0.299 mol.
4. Mass: M(CO₂) = 44.01. m = 0.299 × 44.01 = 13.2 g.

Try these

  1. What are the advantages of chemical equations over word equations?
    Show answer
    (1) Universal language — readable by chemists worldwide regardless of native language. (2) Show ratios and stoichiometry directly. (3) Show state symbols and explicit number of each species. (4) Concise — no ambiguity from how words are phrased.
  2. Balance these equations and add state symbols: (a) ethane + O₂ → CO₂ + H₂O. (b) C₆H₁₂O₆ → C₂H₅OH + CO₂ (anaerobic respiration). (c) Fe + O₂ → Fe₂O₃. (d) Mg(OH)₂ + HNO₃ → Mg(NO₃)₂ + H₂O.
    Show answer
    (a) 2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(l). (b) C₆H₁₂O₆(aq) → 2 C₂H₅OH(aq) + 2 CO₂(g). (c) 4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s). (d) Mg(OH)₂(aq) + 2 HNO₃(aq) → Mg(NO₃)₂(aq) + 2 H₂O(l).
  3. Why must you always convert to moles before applying stoichiometric ratios?
    Show answer
    Balanced equations give ratios in moles, not grams or volumes. 1 mol Mg reacts with 1 mol H₂SO₄ — but 24.31 g Mg reacts with 98.09 g H₂SO₄. You must convert from g (or mL) → mol via M (or via molar volume for gases) before using the mole ratio, then convert back.
  4. What mass of CaCO₃ is needed to produce 4.40 g of CO₂ on heating? CaCO₃ → CaO + CO₂.
    Show answer
    n(CO₂) = 4.40/44.01 = 0.1000 mol. From 1:1 ratio, n(CaCO₃) = 0.1000 mol. m = 0.1000 × 100.09 = 10.0 g.
Lesson 5

Solutions and dilution

Structure 1.4.5

Lesson outcomes

A solution is a homogeneous mixture of a solute (dissolved substance) and a solvent (substance doing the dissolving — usually water in chemistry).

Concentration formulas

c = n / V  mol dm⁻³  (molar concentration; square brackets [X] mean concentration of X)
ρ = m / V  g dm⁻³  (mass concentration)
c = ρ / M  (links the two via molar mass)

Unit conversion: 1 dm³ = 1000 cm³ = 1 L. Divide cm³ by 1000 to get dm³.

Dilution

c1V1 = c2V2 · moles are conserved when only solvent is added.

Worked example

Concentration of dissolved NaCl

Problem. 3.00 g NaCl is dissolved in deionised water to make a 150 cm³ solution. Calculate the concentration in mol dm⁻³ and g dm⁻³.
Solution. V = 150 / 1000 = 0.150 dm³.
ρ = 3.00 / 0.150 = 20.0 g dm⁻³.
M(NaCl) = 58.44. n = 3.00 / 58.44 = 0.0513 mol.
c = 0.0513 / 0.150 = 0.342 mol dm⁻³.
Worked example

Mass of solute from concentration

Problem. Calculate the mass of H₂SO₄ in 50 cm³ of a 1.50 mol dm⁻³ solution.
Solution. V = 0.050 dm³. n = c × V = 1.50 × 0.050 = 0.075 mol. M(H₂SO₄) = 98.09. m = 0.075 × 98.09 = 7.36 g.
Worked example

Mass concentration of CuSO₄

Problem. Calculate the mass concentration (g dm⁻³) of a 2.00 mol dm⁻³ solution of CuSO₄.
Solution. M(CuSO₄) = 63.55 + 32.07 + 4(16.00) = 159.62. ρ = c × M = 2.00 × 159.62 = 319 g dm⁻³.

Try these

  1. [Na₂CO₃] = 0.100 mol dm⁻³. Calculate the concentration in g cm⁻³.
    Show answer
    M(Na₂CO₃) = 105.99. ρ = 0.100 × 105.99 = 10.6 g dm⁻³ = 0.0106 g cm⁻³.
  2. What volume of 2.00 mol dm⁻³ NaOH must be diluted to make 500 cm³ of 0.400 mol dm⁻³ NaOH?
    Show answer
    c₁V₁ = c₂V₂. V₁ = (0.400 × 500)/2.00 = 100 cm³. Take 100 cm³ of stock and dilute with water to 500 cm³ total.
  3. Why are square brackets [X] used in chemistry?
    Show answer
    [X] denotes the molar concentration of species X in mol dm⁻³. Saves writing 'concentration of'. So [HCl] = 0.10 mol dm⁻³ means a 0.10 M solution of HCl.
  4. Convert 0.150 mol dm⁻³ to g dm⁻³ for CuSO₄·5H₂O.
    Show answer
    M(CuSO₄·5H₂O) = 249.69. ρ = 0.150 × 249.69 = 37.5 g dm⁻³.
Lesson 6+7

Standard solutions and calibration curves

Structure 1.4.5Tool 2 · Tool 3

Lesson outcomes

A standard solution is one whose concentration is precisely known. They are needed for quantitative analysis — for titrations, calibration of spectrophotometers and determination of unknowns.

Making a standard solution

  1. Accurately weigh the solute on a balance.
  2. Transfer to a volumetric flask using a funnel and rinsing.
  3. Add solvent and dissolve, then make up to the graduation mark.
  4. Stopper and invert several times to mix.

Serial dilution + calibration curve

Dilute the stock solution by known factors (e.g. 1:2, 1:5, 1:10). Measure absorbance of each at a chosen λ in a spectrophotometer. Plot absorbance (y) vs concentration (x) — Beer-Lambert predicts a straight line through the origin (A = ε·c·l). Read unknown concentrations by matching their absorbance to the line.

Worked example

Standard CuSO₄·5H₂O for calibration

Problem. 2.497 g of CuSO₄·5H₂O is dissolved in deionised water in a 100 cm³ volumetric flask. A 5.00 cm³ aliquot is diluted to 50.0 cm³. Calculate [Cu²⁺] in the diluted solution.
Solution. M(CuSO₄·5H₂O) = 249.69. n = 2.497 / 249.69 = 0.01000 mol.
[stock] = 0.01000 / 0.100 = 0.1000 mol dm⁻³.
Dilution: c1V1 = c2V2 → c2 = (0.1000 × 5.00) / 50.0 = 0.0100 mol dm⁻³.
Since each CuSO₄ gives one Cu²⁺, [Cu²⁺] = 0.0100 mol dm⁻³.

Try these

  1. What is a stock solution? Why are stock solutions considered standard solutions?
    Show answer
    Stock solution: a concentrated solution of accurately known concentration, prepared in advance and stored. It's a standard solution because its concentration is precisely defined; it can be reliably diluted to any chosen concentration via c₁V₁ = c₂V₂.
  2. Why are standard solutions necessary?
    Show answer
    Quantitative analysis (titrations, calibration curves, determining unknowns) requires a reference of known concentration. Without a standard, you cannot calibrate or compare.
  3. How does spectrophotometry determine the concentration of an unknown using a calibration curve?
    Show answer
    Make standards of known concentration; measure each absorbance at a chosen λ. Plot A (y) vs c (x) — straight line through origin (Beer-Lambert). Measure A of unknown; read off c from the curve.
  4. Why is the absorbance vs concentration line expected to be straight and pass through the origin?
    Show answer
    Beer-Lambert law A = ε·c·l. With ε and l constant, A ∝ c. Zero concentration → zero absorbance, so the line passes through the origin.
  5. What is serial dilution and why is it used to prepare a calibration curve?
    Show answer
    Repeatedly diluting a stock solution by a known factor (e.g. 1:2, 1:5, 1:10) to create a sequence of standards. Allows you to span a range of concentrations from a single stock solution using volumetric pipettes — much more accurate than weighing many small masses.
  6. If a sample's absorbance falls above the highest point on your calibration curve, what should you do?
    Show answer
    Dilute the sample by a known factor and re-measure. Reading A above the calibrated range is unreliable (Beer-Lambert can break down at high concentration). Multiply your read concentration by the dilution factor to get the original.
Lesson 8

Titrations

Structure 1.4.5Reactivity 2.1.2

Lesson outcomes

A titration determines the concentration of one solution by reacting it with a measured volume of another of known concentration, until the reaction is complete.

The setup

The calculation

For a balanced reaction aA + bB → products, at the endpoint:

n(A) / a = n(B) / b  or equivalently  cAVA / a = cBVB / b

Worked example

HCl + NaOH titration

Problem. 25.0 cm³ of 0.10 mol dm⁻³ HCl requires 21.50 cm³ of NaOH to reach the endpoint. Calculate the concentration of NaOH.
Solution. HCl + NaOH → NaCl + H₂O. 1:1 mole ratio.
n(HCl) = 0.10 × 25.0/1000 = 2.50 × 10⁻³ mol.
n(NaOH) = n(HCl) = 2.50 × 10⁻³ mol.
c(NaOH) = 2.50 × 10⁻³ / (21.50/1000) = 0.116 mol dm⁻³.

Try these

  1. What does the term 'endpoint' mean in a titration? How does it differ from 'equivalence point'?
    Show answer
    Equivalence point: stoichiometric amounts of reactants have been mixed (where the reaction is exactly complete). Endpoint: the point at which the indicator changes colour, indicating the experimenter judges the equivalence point has been reached. A well-chosen indicator gives endpoint ≈ equivalence point.
  2. What are the four uses of titration listed?
    Show answer
    (1) Determine concentrations of acids/bases. (2) Determine concentrations of other reactants. (3) Following the rate of a reaction. (4) Determining equilibrium constants.
  3. 20.0 cm³ of 0.250 mol dm⁻³ Ca(OH)₂ is titrated with HCl. The titre is 24.50 cm³. Calculate [HCl]. Equation: Ca(OH)₂ + 2 HCl → CaCl₂ + 2 H₂O.
    Show answer
    n(Ca(OH)₂) = 0.250 × 0.020 = 5.00 × 10⁻³ mol. From 1:2 ratio, n(HCl) = 2 × 5.00 × 10⁻³ = 1.00 × 10⁻² mol. [HCl] = 1.00 × 10⁻² / 0.02450 = 0.408 mol dm⁻³.
  4. Why do we repeat titrations until we get concordant titres?
    Show answer
    To minimise random error. Concordant titres (within ±0.10 cm³) are averaged to give a reliable mean — averaging an outlier with good values would bias the result.
  5. How would your calculated concentration of NaOH change if the conical flask had been rinsed with NaOH solution before titration?
    Show answer
    The flask would contain extra NaOH, so more HCl would be needed at the endpoint and you would calculate [NaOH] as too high. The flask should be rinsed only with deionised water.
  6. Why should the burette be rinsed with the titrant before being filled?
    Show answer
    Water clinging to the burette walls would dilute the titrant, lowering its effective concentration → larger titre → overestimated [analyte]. Rinsing with the titrant displaces water, leaving the correct concentration throughout.
  7. Sketch what happens to pH as you add strong base to strong acid in a titration. Where is the equivalence point?
    Show answer
    pH starts low (~1), rises slowly as acid is consumed, then jumps almost vertically through pH 7 around the equivalence point, then levels off near pH 13. The equivalence point is the midpoint of the steep jump.
Lesson 9

Limiting reagents

Reactivity 2.1.3

Lesson outcomes

The limiting reactant is the one fully consumed first — it sets the maximum amount of product. The excess reactant has some left over.

Finding the limiting reactant — two methods

Method 1 (ratio): for each reactant, divide moles by its stoichiometric coefficient. The smallest answer wins.

Method 2 (test): pick one reactant, calculate how much of the other is needed. If the actual amount is less, that reactant is limiting.

Once the limiting reactant is identified, all further calculations (theoretical yield, excess remaining) start from its moles, not from the moles in the chemical equation.

Worked example

Ammonia + oxygen (3.25 g + 3.50 g)

Problem. 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O. 3.25 g NH₃ reacts with 3.50 g O₂. (a) Which is limiting? (b) Mass of NO formed? (c) Excess remaining?
Solution. Moles: n(NH₃) = 3.25/17.04 = 0.1907. n(O₂) = 3.50/32.00 = 0.1094.
Divide by coefficient: NH₃ → 0.1907/4 = 0.04767. O₂ → 0.1094/5 = 0.02188. O₂ is limiting.
(b) n(NO) = (4/5) × n(O₂) = 0.8 × 0.1094 = 0.0875 mol. m(NO) = 0.0875 × 30.01 = 2.63 g.
(c) n(NH₃ reacted) = (4/5) × n(O₂) = 0.0875 mol. n(NH₃ excess) = 0.1907 − 0.0875 = 0.103 mol → m = 0.103 × 17.04 = 1.76 g.
Worked example

Benzene + bromine

Problem. C₆H₆ + Br₂ → C₆H₅Br + HBr. 42.1 g C₆H₆ reacts with 73.0 g Br₂. What is the theoretical yield of C₆H₅Br?
Solution. M(C₆H₆) = 78.12, M(Br₂) = 159.81, M(C₆H₅Br) = 157.02.
n(C₆H₆) = 42.1/78.12 = 0.539. n(Br₂) = 73.0/159.81 = 0.457.
1:1 ratio, so Br₂ is limiting (smaller moles).
n(C₆H₅Br) = n(Br₂) = 0.457 mol. m = 0.457 × 157.02 = 71.8 g.

Try these

  1. Why might the amounts of reactants in a real mixture not be in the ideal stoichiometric ratio?
    Show answer
    Often deliberate — one reactant kept in excess to push the other to completion. Also: cost (use less of the expensive reactant); safety; subsequent processing needs.
  2. 4.54 dm³ H₂ + 2.27 dm³ Cl₂ → HCl(g) at STP. Find limiting reactant, and final volumes of each gas.
    Show answer
    H₂ + Cl₂ → 2 HCl. Equal volume = equal moles (Avogadro). H₂:Cl₂ = 1:1, so Cl₂ is limiting (smaller amount, given the 1:1 ratio). All Cl₂ reacts; consumes 2.27 dm³ H₂. Final: H₂ = 4.54 − 2.27 = 2.27 dm³, Cl₂ = 0, HCl = 2 × 2.27 = 4.54 dm³.
  3. 1.00 dm³ of 0.500 mol dm⁻³ HCl is mixed with 1.00 dm³ of 0.200 mol dm⁻³ NaOH. Determine the final concentrations of all solutes (assume additive volumes).
    Show answer
    HCl + NaOH → NaCl + H₂O. n(HCl) = 0.500 mol. n(NaOH) = 0.200 mol. NaOH limiting. After reaction: n(NaOH) = 0, n(HCl) = 0.500 − 0.200 = 0.300 mol, n(NaCl) = 0.200 mol. Total V = 2.00 dm³.
    [HCl] = 0.300/2.00 = 0.150 mol dm⁻³. [NaCl] = 0.200/2.00 = 0.100 mol dm⁻³. [NaOH] = 0.
Lesson 10

Percentage yield and atom economy

Reactivity 2.1.4Reactivity 2.1.5

Lesson outcomes

Theoretical yield = maximum mass (or moles) of product calculated from the limiting reactant, assuming 100% conversion. Actual yield = mass actually obtained in the laboratory.

% yield = (actual / theoretical) × 100%

Reasons actual < theoretical: side reactions; equilibrium; product loss during transfer or purification; incomplete reaction; experimental error.

Atom economy

A different metric — looks at the percentage of atoms in the reactants that end up in the desired product:

AE = (Mr of desired product / Σ Mr of all reactants) × 100%

A high atom economy means little waste. A reaction can have 100% yield but only 30% atom economy if it produces lots of by-products. Atom economy is a key green-chemistry metric.

Worked example

Yield from aluminium + oxygen

Problem. 4 Al + 3 O₂ → 2 Al₂O₃. 9.443 g Al reacts with 7.945 dm³ O₂ (STP) and produces 17.3 g Al₂O₃. Calculate the theoretical and percentage yields.
Solution. n(Al) = 9.443 / 26.98 = 0.3500. n(O₂) = 7.945 / 22.7 = 0.3500.
Ratio Al:O₂ = 4:3. By coefficient: 0.3500/4 = 0.0875, 0.3500/3 = 0.1167. Al is limiting.
n(Al₂O₃) = (2/4) × n(Al) = 0.1750 mol. M(Al₂O₃) = 101.96. Theoretical = 0.1750 × 101.96 = 17.84 g.
% yield = (17.3 / 17.84) × 100% = 97.0%.
Worked example

Atom economy of dimethyl carbonate synthesis

Problem. Calculate the atom economy for: 4 CH₃OH + 2 CO + O₂ → 2 (CH₃O)₂CO + 2 H₂O. Assume 100% yield.
Solution. Desired product: 2 × M((CH₃O)₂CO) = 2 × 90.08 = 180.16.
Total reactants: 4(32.04) + 2(28.01) + 32.00 = 128.16 + 56.02 + 32.00 = 216.18.
AE = (180.16 / 216.18) × 100% = 83.3%.

Try these

  1. What is the difference between theoretical, actual and percentage yield?
    Show answer
    Theoretical yield: maximum predicted from the limiting reactant assuming 100% conversion. Actual yield: mass (or mol) actually obtained in the lab. Percentage yield: (actual/theoretical) × 100%.
  2. Suggest reasons why the percentage yield is usually less than 100%.
    Show answer
    Side reactions producing unwanted products; loss of product during transfer / filtration / recrystallisation; reaction not reaching completion (equilibrium); impurities in starting materials; evaporation of volatile products.
  3. What is green chemistry? Why was it developed?
    Show answer
    Design of chemical products and processes that reduce or eliminate the use and generation of hazardous substances. Developed in the 1990s in response to growing environmental concerns — to make industrial chemistry sustainable, less polluting and less wasteful.
  4. What is atom economy and how does it help measure efficiency? Give the equation and an example.
    Show answer
    % of reactant atoms that end up in the desired product. AE = (Mr of desired product / Σ Mr of all reactants) × 100%. Example: HCl + NaOH → NaCl + H₂O. Mr(NaCl) = 58.44. Total reactants = 36.46 + 40.00 = 76.46. AE = (58.44/76.46) × 100 = 76.4% — i.e. 23.6% of reactant mass is 'wasted' as H₂O.
  5. Dimethyl carbonate (CH₃O)₂CO is a green solvent. Calculate the atom economy for: 4 CH₃OH + 2 CO + O₂ → 2 (CH₃O)₂CO + 2 H₂O. Assume 100% yield.
    Show answer
    Desired: 2 × 90.08 = 180.16. Reactants: 4(32.04) + 2(28.01) + 32.00 = 216.18. AE = (180.16/216.18) × 100 = 83.3%.
  6. How can the atom economy of the synthesis of N-methylphenylamine (from phenylamine, using dimethyl carbonate, producing methanol and CO₂ as by-products) be improved?
    Show answer
    Recycle the methanol by-product back into the production of dimethyl carbonate (recall the 83.3% AE reaction above). Coupling the two processes turns a by-product into a starting material, raising overall atom economy.
Lesson 11

Reacting gases · Avogadro's law

Structure 1.4.6

Lesson outcomes

Avogadro's law: at the same temperature and pressure, equal volumes of any gases contain equal numbers of molecules. Equivalently, the volume per mole of any gas under the same conditions is the same — the molar volume.

At STP (273 K, 100 kPa): molar volume = 22.7 dm³ mol⁻¹.

For reactions involving only gases, you can work entirely in volume ratios — no need to convert to moles.

Worked example

Volume of H₂ from Li + HCl

Problem. What volume of H₂ gas is produced when 0.0500 mol of Li reacts with excess HCl at STP? 2 Li + 2 HCl → 2 LiCl + H₂.
Solution. n(H₂) = (1/2) × n(Li) = 0.0250 mol. V(H₂) = 0.0250 × 22.7 = 0.568 dm³.
Worked example

Combustion of H₂S — volume calculation

Problem. 2 H₂S(g) + 3 O₂(g) → 2 H₂O(l) + 2 SO₂(g). 0.908 dm³ of H₂S reacts. Calculate V(O₂) consumed and V(SO₂) produced.
Solution. Ratio H₂S : O₂ = 2 : 3, so V(O₂) = (3/2) × 0.908 = 1.36 dm³. Ratio H₂S : SO₂ = 1 : 1, so V(SO₂) = 0.908 dm³.

Try these

  1. State Avogadro's law in your own words.
    Show answer
    Equal volumes of any gas at the same temperature and pressure contain equal numbers of molecules (or moles). Equivalent: at fixed T and p, V ∝ n, regardless of which gas.
  2. What is the molar volume of an ideal gas at STP? At RTP?
    Show answer
    STP (273 K, 100 kPa): 22.7 dm³ mol⁻¹. RTP (298 K, 100 kPa): 24.5 dm³ mol⁻¹ (legacy — not used in IB).
  3. What is the minimum volume of H₂ required to fully reduce 10.0 g CuO at STP?
    Show answer
    CuO + H₂ → Cu + H₂O. n(CuO) = 10.0/79.55 = 0.1257 mol. n(H₂) = 0.1257 mol. V = 0.1257 × 22.7 = 2.85 dm³.
  4. In an airbag, 2 NaN₃ → 2 Na + 3 N₂. What mass of NaN₃ is needed to fully inflate a 60.0 dm³ airbag at STP?
    Show answer
    n(N₂) = 60.0/22.7 = 2.643 mol. n(NaN₃) = (2/3) × 2.643 = 1.762 mol. M(NaN₃) = 65.02. m = 115 g.
  5. 500 cm³ CH₄ + 600 cm³ O₂ → CO₂ + 2 H₂O. Final volumes of each gas?
    Show answer
    O₂ limiting (since CH₄ needs 1000 cm³ O₂ for full combustion but only 600 cm³ given). CH₄ used = 300 cm³ → CH₄ remaining = 200 cm³. CO₂ produced = 300 cm³ (1:1 with CH₄ used). H₂O is liquid, not counted as gas. CH₄: 200, O₂: 0, CO₂: 300 cm³.
  6. 2 CO + O₂ → 2 CO₂. The reaction produces 2.00 dm³ of CO₂. Calculate the volumes of CO and O₂ consumed, in dm³ and cm³.
    Show answer
    Ratio CO : O₂ : CO₂ = 2 : 1 : 2. V(CO) = 2.00 dm³ = 2000 cm³. V(O₂) = 1.00 dm³ = 1000 cm³.
Lesson 12+13

Ideal gases

Structure 1.5.1Structure 1.5.2

Lesson outcomes

Five assumptions of the ideal gas model

The gas laws

Boyle's law (constant T, n): pV = constant → halve V, double p.
Charles' law (constant p, n): V / T = constant → double T, double V.
Combined: p1V1/T1 = p2V2/T2.

Ideal gas equation

pV = nRT

SI units: p in Pa (kPa × 1000), V in m³ (dm³ ÷ 1000, cm³ ÷ 10⁶), n in mol, T in K (°C + 273). R = 8.31 J K⁻¹ mol⁻¹.

Real gas deviations

Real gases deviate from ideal behaviour at low T (IMFs significant) and high p (particle volume non-negligible). Gases with strong IMFs (NH₃, H₂O) deviate more than non-polar gases (He, Ne).

Worked example

Boyle's law · weather balloon

Problem. A weather balloon contains 32.0 dm³ He at 100 kPa at sea level. It rises to where p = 58.0 kPa. Calculate the new volume (T constant).
Solution. p1V1 = p2V2 → V2 = (100 × 32.0) / 58.0 = 55.2 dm³.
Worked example

Molar mass from pV = nRT

Problem. 1.048 g of unknown gas A occupies 846 cm³ at 500 K and 101 kPa. Find its molar mass.
Solution. Convert: p = 1.01 × 10⁵ Pa, V = 8.46 × 10⁻⁴ m³.
n = pV / (RT) = (1.01 × 10⁵)(8.46 × 10⁻⁴) / (8.31 × 500) = 0.02055 mol.
M = m / n = 1.048 / 0.02055 = 51.0 g mol⁻¹.
Worked example

Pressure in a fire extinguisher

Problem. 45.4 mol CO₂ in a 3000 cm³ vessel at 50 °C. Find the pressure.
Solution. V = 3.00 × 10⁻³ m³. T = 323 K. p = nRT/V = (45.4 × 8.31 × 323)/3.00 × 10⁻³ = 4.06 × 10⁷ Pa.

Try these

  1. List the assumptions of the ideal gas model.
    Show answer
    (1) Molecules in ceaseless random motion. (2) All collisions are elastic (no kinetic energy loss). (3) Particle volume is negligible compared to container volume. (4) No intermolecular forces between particles. (5) Average kinetic energy proportional to absolute temperature T.
  2. Why is the kelvin scale used (and not Celsius)?
    Show answer
    The gas laws require an absolute temperature scale where 0 corresponds to zero kinetic energy. T must be a true ratio. The Kelvin scale starts at absolute zero (0 K = −273.15 °C).
  3. State Boyle's law and derive how doubling V at constant T and n changes p.
    Show answer
    At constant T and n: pV = constant. So if V doubles, p halves (inverse proportionality).
  4. Which gas is more likely to behave ideally: (a) at low or high pressure? (b) at low or high temperature? Explain.
    Show answer
    (a) Low pressure — particles far apart, so their own volume is negligible. (b) High temperature — kinetic energy dominates over IMF attractions; collisions effectively elastic.
  5. Predict which is closer to ideal behaviour: He or NH₃, at the same T and p. Justify.
    Show answer
    He is closer to ideal. He is non-polar with the weakest possible London forces; NH₃ is polar with hydrogen bonding. NH₃'s strong IMFs cause significant deviation.
  6. Why do strong intermolecular forces cause deviation from ideal behaviour?
    Show answer
    IMFs reduce the effective velocity during collisions (attraction slows approach) — collisions are no longer perfectly elastic, and measured pressure drops below the ideal prediction. Also: at low T, attractions can hold molecules together → liquefaction.
  7. A weather balloon at sea level holds 32.0 dm³ He at 100 kPa, 293 K. At altitude it experiences 58 kPa and 250 K. Calculate the new volume.
    Show answer
    Combined gas law: p1V1/T1 = p2V2/T2. V2 = (p1V1T2)/(T1p2) = (100 × 32.0 × 250)/(293 × 58) = 800 000/17 000 = 47 dm³.
  8. Practical: explain how to derive Boyle's law from a simulation of a piston with masses 10 kg, 20 kg, 50 kg ... up to 500 kg on a fixed-T gas.
    Show answer
    Adding masses increases the pressure on the gas. Record p (from m × g / A) and V at each step. Convert p to Pa, V to m³. Plot V vs p — inverse curve. Plot V vs 1/p — straight line through origin. The straight line confirms pV = constant at fixed T (Boyle's law).
Vocabulary

42 terms to own.

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elementcompoundmixturehomogeneousatomrelative atomic massmolemolar massrelative formula massempirical formulamolecular formulastoichiometrylimiting reactantexcess reactanttheoretical yieldactual yieldpercentage yieldatom economyideal gasideal gas equationmolar volumestpconcentrationsolutesolventsolutiondilutiontitrationbalanced equationpercentage compositiongraphcalibration curvepressuretemperatureequilibriumequilibrium constantkpoxideintermolecular forcehydrogen bondhydrocarbonamine