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1.5Unit 1 · Principles of Chemistry

Chemical formulae, equations, calculations — Edexcel IGCSE Chemistry Modular - UNIT 1 Revision Notes & Topical Questions

25 min read 27 parts Lesson 5 of 5 1 quiz

📚 Edexcel IGCSE Chemistry Modular - UNIT 1 Hub / Topical Revision
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Lesson summary

Write word and balanced chemical equations with state symbols.Mole equationmoles = mass (g) ÷ Mr.A balanced equation gives the mole ratio of reactants and products.Percentage yield is below 100% because of incomplete reactions, side reactions and losses in transfer.Molecular form…

  • Word and chemical equations
  • Relative mass
  • Moles, mass and RFM
  • Reacting masses
  • Empirical and molecular formulae
  • Experiment – formulae of simple compounds

Part 1 of 7Key ideas

Mole equation
Mole equation moles = mass (g) ÷ Mr.
A balanced equation gives the mole ratio of reactants and products.
Percentage yield is below 100% because of incomplete reactions, side reactions and losses in transfer.
Molecular formula = empirical formula × (Mr ÷ empirical formula mass).
One mole of any gas occupies 24 dm3 at room temperature and pressure.
Atoms are conserved in a reaction, so total mass of reactants = total mass of products.
Copper(II) oxide can be reduced by hydrogen or methane to find the formula of the oxide.

Part 2 of 7Learning objectives

1
Write word and balanced chemical equations with state symbols.
2
Calculate relative formula masses and use the mole concept.
3
Calculate reacting masses and percentage yield.
4
Determine empirical and molecular formulae from experimental data.

Part 3 of 7Essential knowledge

Essential knowledge
Word and chemical equations
Word and chemical equations a word equation names reactants and products; a balanced symbol equation has equal numbers of each type of atom on both sides. Balance using large numbers in front of formulae only, never by changing subscripts. Add state symbols (s), (l), (g), (aq).
Relative mass
Ar values come from the Periodic Table. Mr is the sum of the Ar values of all atoms, e.g. Mr of CaCO3 = 40 + 12 + (3 × 16) = 100.
Moles, mass and RFM
Moles, mass and RFM a mole is 6.02 × 1023 particles and the mass of one mole in grams equals Mr. Moles = mass ÷ Mr.
Reacting masses
(1) find moles of the known substance, (2) use the equation ratio to find moles of the unknown, (3) convert to mass using Mr. If quantities of two reactants are given, identify the limiting reactant first.
Percentage yield
Percentage yield actual yield ÷ theoretical yield × 100. Yields are less than 100% because reactions may be reversible or incomplete, other reactions may occur, and product is lost during filtering or transferring.
Empirical and molecular formulae
Empirical and molecular formulae divide each mass (or percentage) by Ar, divide by the smallest value, and turn into whole numbers. The molecular formula is a whole-number multiple of the empirical formula found using Mr.
Experiment – formulae of simple compounds
Experiment – formulae of simple compounds magnesium oxide is made by heating magnesium in a crucible; copper(II) oxide can be reduced by heating in a stream of hydrogen or methane, weighing the copper left, so the mass of oxygen lost is found.
Practical – formula of a metal oxide
Practical – formula of a metal oxide heat to constant mass to make sure the reaction is complete. Errors include loss of MgO smoke when the lid is lifted and magnesium reacting with nitrogen in air.
Elements written as molecules
Hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine and iodine exist as diatomic molecules, so write them as H2, N2, O2, F2, Cl2, Br2 and I2 in equations.
Balancing with groups
If a group of atoms such as NO3 or SO4 appears unchanged on both sides, count it as one unit. Balance elements that appear on their own (e.g. O2) last.
Equations from reacting masses
Convert each mass to moles and simplify to a whole-number ratio. If 2.4 g Mg reacts with 1.6 g O2 to form 4.0 g MgO, moles are 0.10 : 0.05 : 0.10 = 2 : 1 : 2, so 2Mg + O2 → 2MgO.
Awkward empirical ratios
If a ratio ends in .5 (e.g. 1 : 1.5), multiply all values by 2 to get whole numbers (2 : 3). Formulae of ionic compounds such as NaCl or MgO are always empirical formulae.
Water of crystallisation
Heat a weighed hydrated salt to constant mass and reweigh. Example: 5.00 g of MgSO4·xH2O leaves 2.44 g. Moles MgSO4 = 2.44 ÷ 120 = 0.0203; moles water = 2.56 ÷ 18 = 0.142; ratio 1 : 7, so MgSO4·7H2O.
Concentration
Concentration (mol/dm3) = moles of solute ÷ volume of solution in dm3. Divide cm3 by 1000 to get dm3. Example: 4.0 g NaOH (0.10 mol) in 250 cm3 gives 0.10 ÷ 0.250 = 0.40 mol/dm3.
g/dm3 and mol/dm3
Concentration can also be given in g/dm3. Convert using Mr: g/dm3 = mol/dm3 × Mr. So 0.40 mol/dm3 NaOH (Mr 40) is 16 g/dm3.
Titration calculations
Find moles of the known solution, use the equation ratio, then divide by volume.
  • 25.0 cm3 of 0.100 mol/dm3 NaOH (0.00250 mol) needs 20.0 cm3 H2SO4.
  • Ratio 2 : 1 gives 0.00125 mol acid, so concentration = 0.00125 ÷ 0.0200 = 0.0625 mol/dm3.
Gas volume from mass
Convert mass to moles first, then multiply by 24 dm3 (or 24 000 cm3) at rtp. 8.8 g CO2 (Mr 44) is 0.20 mol, which occupies 0.20 × 24 = 4.8 dm3.
Reacting gas volumes
Equal volumes of gases at the same temperature and pressure contain equal numbers of moles,
  • so gas volumes react in the same ratio as the equation.
  • CH4 + 2O2 → CO2 + 2H2O: 40 cm3 of methane needs 80 cm3 of oxygen and makes 40 cm3 of CO2.

Part 4 of 7Process & sequence

1
Weigh an empty crucible with its lid.
2
Add a coiled strip of magnesium ribbon and reweigh.
3
Heat strongly, lifting the lid occasionally to let air in without losing smoke.
4
Continue until the magnesium no longer glows, then cool and reweigh.
5
Heat again and reweigh until the mass is constant.
6
Calculate masses of Mg and O, convert to moles and find the simplest ratio.

Finding a reacting masswrite these steps in this order; each step is a marking point.

1
Write balanced equation
2
Moles = mass ÷ Mr
3
Use mole ratio from equation
4
Mass = moles × Mr
5
Compare with actual for % yield

Part 5 of 7Worked examples

14.0 g
What mass of calcium oxide is formed by heating 25.0 g of calcium carbonate? CaCO3 → CaO + CO2 (Ca = 40, C = 12, O = 16)
  • Mr CaCO3 = 100; moles = 25.0 ÷ 100 = 0.250 mol.
    • Ratio 1:1, so 0.250 mol CaO.
    • Mr CaO = 56; mass = 0.250 × 56 = 14.0 g.
    • If only 12.6 g was collected, % yield = 12.6 ÷ 14.0 × 100 = 90%.
A compound contains 85.7% carbon and 14.3% hydrogen. Its Mr is 56. Find the empirical and molecular formulae.
  • C: 85.7 ÷ 12 = 7.14;
    • H: 14.3 ÷ 1 = 14.3.
    • Divide by 7.14: C 1, H 2.
    • Empirical formula CH2 (mass 14).
    • 56 ÷ 14 = 4, so molecular formula C4H8.

Part 6 of 7Summary

1
Moles = mass ÷ Mr.
2
Use the balanced equation ratio to find reacting masses.
3
% yield = actual ÷ theoretical × 100.
4
Empirical formula is the simplest whole-number ratio.

Part 7 of 7Compare: Compare

Key calculation formulae
QuantityFormulaUnits
Moles from massn = mass ÷ Mrmol
Mass from molesmass = n × Mrg
Percentage yieldactual ÷ theoretical × 100%
Percentage by mass(Ar × number of atoms) ÷ Mr × 100%
Gas volume (rtp)volume = n × 24dm3
Empirical formula table method (magnesium oxide example)
StepMgO
Mass (g)0.240.16
Divide by Ar0.24 ÷ 24 = 0.0100.16 ÷ 16 = 0.010
Divide by smallest11
Empirical formulaMgO

Key terms

Command terms

Command terms
Give / State / Name
Recall a fact; one idea per mark.
Define
Give the exact meaning using the specification keywords.
Describe
Give an account of what happens / the pattern in data, quoting values; no reasons needed.
Explain
Give reasons with linked statements (cause → effect). Each linked step is usually a separate marking point.
Suggest
Use knowledge in a new context; more than one answer may be accepted.
Compare
Similarities and differences; each point must refer to both items.
Calculate
Show working; include units; check significant figures. Method marks are given even if the final answer is wrong.
Discuss
Explore the issue from different angles and link the points.
Evaluate / Assess
Weigh strengths and weaknesses and reach a supported judgement.
Deduce
Draw a conclusion from the information given.
Justify
Support a statement/choice with evidence.

Exam technique

Keywords the mark scheme looks for in this lesson

Keywords the mark scheme looks for in this lesson
Relative atomic mass (Ar) · Relative formula mass (Mr) · Mole · Molar mass · Limiting reactant · Theoretical yield · Percentage yield · Empirical formula · Molecular formula · State symbols
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)

How to get full marks (examiner guidance)
Use specification keywords exactly (e.g. electronegativity, nucleophile, dynamic equilibrium, activation energy).
Link ideas
Link ideas for “explain” each marking point is usually a cause → consequence step; unlinked facts often score nothing.
Use the information in the question
“discuss/evaluate” answers must use the data provided and match the number of marks.
Convert units carefully (mg → g, cm³ → dm³) and show working for calculations.
Avoid vague words
“it”, “stuff”, “things”, “goes up”, “energy is made”. Name the substance and the process.
Do not extrapolate a graph line beyond the data or to zero unless asked.
Six-mark extended responses are marked on content AND a logical, linked line of reasoning.

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