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1.1.11Chemical Substances, Reactions and Essential Resources

The Avogadro Constant and the Mole

15 min read 14 parts Lesson 12 of 46 1 quiz

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Section 1 of 14Part 1 — Amount of substance and particles

mole avogadro constant
mole avogadro constant

  1. A mole is a unit of amount of substance. One mole contains 6.02 × 1023 specified particles. Always name the particles: atoms, molecules, ions or electrons. The Avogadro constant has units mol−1.

    Amount, n (mol)

    Start with the number of moles.

    → Multiply by NA

    N = n × 6.02 × 1023

    Number of particles, N

    To reverse the process, divide by NA.

    Particles must be specified
    SampleParticles countedAmount of atoms
    1 mol He6.02 × 1023 helium atoms1 mol He atoms
    1 mol O26.02 × 1023 oxygen molecules2 mol O atoms
    1 mol H2O6.02 × 1023 water molecules2 mol H atoms and 1 mol O atoms

    Example 2.0 mol of water contains 2.0 × 6.02 × 1023 = 1.20 × 1024 molecules, to three significant figures.

Section 2 of 14Part 2 — Mass and molar mass

moles mass molar mass
moles mass molar mass

  1. Molar mass is the mass of one mole, in g mol−1. Its numerical value equals the relative atomic mass or relative formula mass. Relative mass has no unit; molar mass does.

    Choose the required rearrangement
    FindRelationshipUnits
    Amountn = m / Mg divided by g mol−1 gives mol
    Massm = n × Mmol × g mol−1 gives g
    Molar massM = m / ng mol−1

    Worked calculation For 8.0 g of O2, M = 2 × 16 = 32 g mol−1. Therefore n = 8.0 / 32 = 0.25 mol. Using 16 would count oxygen atoms instead of oxygen molecules.

Section 3 of 14Part 3 — Gas volumes and solution concentrations

concentration gas volume rates
concentration gas volume rates
molar concentration moles volume
molar concentration moles volume
gas temperature volume constant pressure
gas temperature volume constant pressure

  1. At room temperature and pressure, use a molar gas volume of 24 dm3 mol−1 for these IGCSE calculations. This value applies to gases at the stated conditions; it is not the volume of one mole of a solid or liquid.

    Volume calculations and conversions
    SituationEquationEssential check
    Gas at r.t.p.n = V / 24V must be in dm3
    Gas measured in cm3n = V / 24000Do not mix this volume with 24
    Solutionn = c × Vc in mol dm−3, V in dm3
    Concentrationc = n / VUse total solution volume
    Mass concentrationg dm−3 = c × MMultiply molar concentration by molar mass

    Example 50.0 cm3 = 0.0500 dm3. A 0.200 mol dm−3 NaOH solution contains n = 0.200 × 0.0500 = 0.0100 mol in this volume.

Section 4 of 14Part 4 — Reaction ratios and the limiting reactant

Effect of Limiting Factors on the Rate of Photosynthesis — 2D Simulation
period3 oxygen reactions
period3 oxygen reactions

  1. The coefficients in a balanced equation give mole ratios. They do not directly give mass ratios. Convert the known quantity into moles before using the ratio.

    1 · Known quantity

    Mass, gas volume or concentration and volume.

    2 · Known moles

    Use n = m/M, V/24 or cV.

    3 · Equation ratio

    Multiply by required coefficient / known coefficient.

    4 · Required quantity

    Convert moles to mass, volume or concentration.

    For 2H2 + O2 → 2H2O, compare each available amount divided by its coefficient.

    • With 4 mol H2 and 1 mol O2, the values are 4/2 = 2 and 1/1 = 1.
      • Oxygen gives the smaller value and is limiting.
    • It forms 2 mol water, uses 2 mol hydrogen and leaves 2 mol hydrogen unreacted.

    The limiting reactant determines the theoretical maximum yield. The excess reactant remains after the limiting reactant is consumed.

Section 5 of 14Part 5 — Formulae, yield and purity

percentage yield 8g of 10g
percentage yield 8g of 10g

  1. An empirical formula gives the simplest whole-number atom ratio. A molecular formula gives the actual numbers of atoms in a molecule. When percentages are supplied, imagine a 100 g sample so each percentage becomes a mass in grams.

    Formula calculation: 40.0% C, 6.7% H, 53.3% O
    StepCHO
    Mass in 100 g40.0 g6.7 g53.3 g
    Divide by Ar40.0/12 = 3.336.7/1 = 6.753.3/16 = 3.33
    Divide by smallest121

    The empirical formula is CH2O, with relative formula mass 30. If molecular Mr = 60, the multiplier is 60/30 = 2, giving C2H4O2.

    Different percentage calculations
    QuantityCalculationInterpretation
    Percentage yieldactual yield / theoretical yield × 10015 g obtained from a theoretical 20 g gives 75%
    Percentage puritymass of pure substance / sample mass × 1008 g pure substance in a 10 g sample gives 80%
    Element compositiontotal Ar contribution / Mr × 100Use the number of atoms in the formula

    Yield can be reduced by incomplete reaction, side reactions or product loss during transfer and separation. An apparent yield above 100% indicates a problem such as wet or impure product, or a measurement or calculation error.

Section 6 of 14Part 6 — Titration: measurement to concentration

titration results concordance
titration results concordance
titration acid added 25ml
titration acid added 25ml
naoh titration 15ml example
naoh titration 15ml example
ester equilibrium concentrations worked
ester equilibrium concentrations worked

  1. A titration measures the volume of one solution needed to react with a measured volume of another. Use a suitable indicator and approach the end-point dropwise while swirling. Repeat to obtain closely agreeing titres and calculate a mean from the concordant results, excluding the rough trial.

    Worked titration calculation
    StageReasoning and calculation
    Balanced equation2NaOH + H2SO4 → Na2SO4 + 2H2O
    Known acid20.0 cm3 of 0.100 mol dm−3: n = 0.100 × 0.0200 = 0.00200 mol
    Apply 2:1 ration(NaOH) = 2 × 0.00200 = 0.00400 mol
    Unknown alkali25.0 cm3 = 0.0250 dm3
    Find concentrationc(NaOH) = 0.00400 / 0.0250 = 0.160 mol dm−3

    Check the equation ratio before calculating concentration: acid and alkali do not always react in a 1:1 ratio. Keep extra digits in intermediate calculations and round the final answer appropriately.

Section 7 of 14The Avogadro Constant and the Mole

oom heavy dna pani light
Essential knowledge
Core
Relative atomic mass, Ar
  • the average mass of the isotopes of an element compared with 1/12 of the mass of a carbon-12 atom.
  • It is a ratio, so it has no units;
    • e.g. Ar(Mg) = 24 means a magnesium atom is on average twice as heavy as a carbon-12 atom.
Core
Relative molecular / formula mass, Mr
  • the sum of the Ar values of all the atoms in the formula ("relative formula mass" is used for ionic compounds).
  • E.g.
  • Mg(OH)2 = 24 + 2 × (16 + 1) = 58.
  • Remember to multiply everything inside brackets.
Core
Conservation of mass
  • atoms are only rearranged in a reaction, so total mass of reactants = total mass of products, and reacting masses stay in a fixed ratio.
  • E.g.
  • 2Mg + O2 → 2MgO: 48 g Mg + 32 g O2 → 80 g MgO, so 12 g Mg gives 20 g MgO (same ratio in kg or tonnes).
Supplement
Equal gas volumes, equal molesat the same temperature and pressure, equal volumes of any gases contain the same number of moles (molecules).
  • So gas volume ratios equal the mole ratios in the equation: in 2H2(g) + O2(g) → 2H2O(l), 60 cm3 of H2 needs exactly 30 cm3 of O2.
Supplement
Non-whole-number ratios in empirical formula work: if dividing by the smallest gives values like 1 : 1.5 or 1 : 1.33, multiply all values by 2 or 3 to get whole numbers. E.g. Fe : O = 1 : 1.5 becomes 2 : 3, so the empirical formula is Fe2O3. Never round 1.5 to 2.
Supplement
Finding x in a hydrated salt
  • heat a weighed sample to constant mass;
    • mass lost = water.
  • Find moles of anhydrous salt and of water, then divide by the smaller.
  • E.g.
  • 2.50 g CuSO4·xH2O leaves 1.60 g CuSO4 (0.0100 mol);
    • 0.90 g water = 0.050 mol, so x = 5.
Core
State symbols show the physical state of each substance in an equation: (s) solid, (l) liquid, (g) gas, (aq) dissolved in water. E.g. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g). Water as a product is (l), and a precipitate is (s).
Core
Balancing a symbol equationwrite correct formulae first,
  • then change only the large numbers in front of formulae until each element has the same number of atoms on both sides.
    • (never the subscripts)
  • E.g.
  • Al + O2 → Al2O3 becomes 4Al + 3O2 → 2Al2O3.
Supplement
Writing an ionic equationsplit soluble ionic compounds (aq) into their ions, cancel the spectator ions that appear unchanged on both sides,
  • and check that atoms and charges balance.
  • E.g.
  • AgNO3(aq) + KBr(aq) → AgBr(s) + KNO3(aq) gives Ag+(aq) + Br−(aq) → AgBr(s).

Section 8 of 14Process & sequence

indicator equal colour concentrations
indicator equal colour concentrations
    • Write the balanced equation for the reaction.
    • Calculate moles of the substance you know about (n = m/M, n = V/24 or n = c × V).
      What the symbols mean: n = amount of substance (mol) · m = mass (g) · M or Mr = molar mass (g mol−1, numerically equal to Mr)
    • If two reactant amounts are given, compare mole ratios to find the limiting reactant.
    • Use the mole ratio from the balanced equation to find moles of the required substance.
    • Convert moles into the quantity asked for (mass = n × M, volume = n × 24 dm3, concentration = n ÷ V).
    • Give the answer with a unit and to a sensible number of significant figures (usually 3).

    Finding concentration by acid–alkali titrationwrite these steps in this order; each step is a marking point.

    1. Rinse and fill a burette with acid; pipette 25.0 cm3 of alkali into a conical flask.
    2. Add a few drops of indicator (e.g. methyl orange or thymolphthalein).
    3. Add acid from the burette, swirling, dropwise near the end-point until the indicator just changes colour.
    4. Record the titre; repeat until concordant results (within 0.10 cm3) and average them.
    5. Calculate moles of known solution (c × V), use the mole ratio, then c = n ÷ V for the unknown.

Section 9 of 14Compare

Key mole relationships
To findEquationUnits
moles from massn = m ÷ Mn in mol, m in g, M in g/mol
moles of gas at r.t.p.n = V ÷ 24V in dm3 (or V ÷ 24 000 for cm3)
moles in solutionn = c × Vc in mol/dm3, V in dm3
number of particlesN = n × 6.02 × 1023N has no unit
concentration in g/dm3mass concentration = c (mol/dm3) × Mg/dm3

Convert cm3 to dm3 by dividing by 1000.

Section 10 of 14Worked examples

Worked examples
What volume of CO2 at r.t.p. is produced when 25.0 g of CaCO3 reacts with excess hydrochloric acid? (Mr CaCO3 = 100)
  • CaCO3 + 2HCl → CaCl2 + H2O + CO2. n(CaCO3) = 25.0 ÷ 100 = 0.250 mol.
    • Ratio 1:1 so n(CO2) = 0.250 mol.
    • V = 0.250 × 24 = 6.00 dm3.
25.0 cm3 of NaOH(aq) is neutralised by 20.0 cm3 of 0.100 mol/dm3 H2SO4. Find the concentration of NaOH.
  • 2NaOH + H2SO4 → Na2SO4 + 2H2O. n(H2SO4) = 0.100 × 20.0/1000 = 0.00200 mol. n(NaOH) = 2 × 0.00200 = 0.00400 mol. c = 0.00400 ÷ (25.0/1000) = 0.160 mol/dm3.
A compound contains 40.0% C, 6.7% H and 53.3% O by mass; Mr = 60. Find the empirical and molecular formulae.
  • C: 40.0/12 = 3.33;
    • H: 6.7/1 = 6.7;
    • O: 53.3/16 = 3.33.
    • Divide by 3.33: 1 : 2 : 1 → CH2O (Mr 30).
    • 60 ÷ 30 = 2 → molecular formula C2H4O2.

Section 11 of 14Keep going

  1. The Mole & the Avogadro Constant IGCSE Chemistry 0620: what to do next

    These The Mole & the Avogadro Constant IGCSE Chemistry 0620 notes covered the key ideas, worked examples and exam technique for this topic. Check your The Mole & the Avogadro Constant IGCSE Chemistry 0620 understanding with the practice questions below, then reread any The Mole & the Avogadro Constant IGCSE Chemistry 0620 point you were unsure about before moving on.

    The Mole & the Avogadro Constant IGCSE Chemistry 0620: revision checklist

    • Revise the main The Mole & the Avogadro Constant IGCSE Chemistry 0620 definitions until you can write them from memory.
    • Practise The Mole & the Avogadro Constant IGCSE Chemistry 0620 exam questions under timed conditions.
    • Link each The Mole & the Avogadro Constant IGCSE Chemistry 0620 idea to a diagram or a worked example.
    • List the common The Mole & the Avogadro Constant IGCSE Chemistry 0620 mistakes and how to avoid them.
    • Explain one The Mole & the Avogadro Constant IGCSE Chemistry 0620 idea out loud as if teaching a friend.
    • Summarise The Mole & the Avogadro Constant IGCSE Chemistry 0620 in five key points on a revision card.
    • Check which command words appear most often in The Mole & the Avogadro Constant IGCSE Chemistry 0620 questions.
    • Come back to The Mole & the Avogadro Constant IGCSE Chemistry 0620 after a week for spaced practice.
    • Mark your answers to The Mole & the Avogadro Constant IGCSE Chemistry 0620 questions against the key ideas above.
    • Highlight the Mole & the Avogadro Constant IGCSE Chemistry 0620 keywords that examiners expect to see.
    • Draw a mind map that connects The Mole & the Avogadro Constant IGCSE Chemistry 0620 to earlier lessons.
    • Write a short The Mole & the Avogadro Constant IGCSE Chemistry 0620 recap before you start the next lesson.
    • Turn each The Mole & the Avogadro Constant IGCSE Chemistry 0620 heading into a question and answer it without notes.

Section 12 of 14Key terms

Section 13 of 14Command terms

Command terms
State / Name
Give a short answer only; no explanation is needed (1 mark = 1 fact).
Define
Give the precise meaning. Learn the syllabus definitions word-for-word: every keyword in the definition is a marking point.
Describe
Say what happens / what the data show, with key features and figures. Do not give reasons.
Explain
Give reasonswhat happens AND why/how, linked with “because”, “so”, “therefore”.
Suggest
Apply knowledge to an unfamiliar situation; any sensible, scientifically-based answer can score.
Compare
Give similarities AND differences, each as a direct comparison (“X is … whereas Y is …”).
Outline
Set out the main points briefly.
Calculate
Work out from given facts; show working, give the unit and the requested significant figures.
Identify
Name/select from given information.
Predict
Give an expected result without a full justification.
Evaluate
Judge the importance or quality of the evidence/method, with reasons.

Section 14 of 14Exam technique

Keywords the mark scheme looks for in this lesson

Keywords the mark scheme looks for in this lesson
Mole · Avogadro constant · Molar mass · Molar gas volume · Room temperature and pressure (r.t.p.) · Concentration · Solute / solvent / solution · Limiting reactant · Excess reactant · Percentage yield · Percentage purity · Percentage composition by mass · Titration
Learn each definition in the Key words tab exactly; a missing keyword in a definition usually loses the mark.

How to get full marks (examiner guidance)

1
Use the exact scientific term and learn syllabus definitions word-for-word.
2
Write “rate of … increases”, not “… increases”.
3
Say “concentration gradient”, never just “gradient”; name particles precisely (atoms, molecules, ions) — never “it”.
4
Give units with every numerical answer and use the significant figures asked for; show all working.
5
Compare = a similarity AND a difference; describe = what happens; explain = why it happens.
6
Repeats let you identify anomalies and calculate a mean — they do not make an experiment “valid”.

Words to avoid

1
a mole is the mass of a substancea mole is the amount of substance containing 6.02 × 1023 particles
2
n = c × 25 using 25 cm3convert to dm3 first: 25 ÷ 1000 = 0.025 dm3
3
the gas volume is 24 dm3 for any conditions24 dm3 per mole at room temperature and pressure
4
yield is low because some product was destroyedyield is low because of incomplete reaction, side reactions or loss during transfer/purification
5
mass ratio from the equationuse the mole ratio from the balanced equation
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