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3.3Stoichiometry

The mole and the Avogadro constant Core — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions

5 min read 11 parts Lesson 3 of 3 1 quiz

📚 Cambridge IGCSE Chemistry 0620 Hub / Topical Revision
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3.3 The mole and the Avogadro constant Core — Notes — IGCSE Chemistry notes: State and use the mole, Avogadro constant, molar gas volume and concentration in calculations.Mole — the SI unit of amount of substance, symbol molAvogadro constant — one mole contains 6.02 × 1023 partic…

  • Number of particles
  • Use of the constant
  • Molar mass
  • Moles from mass
  • Mass from moles
  • Avogadro's law

Part 1 of 9The mole and the Avogadro constant

The mole and the Avogadro constant
Mole
  • Mole the SI unit of amount of substance, symbol mol
Avogadro constant
  • Avogadro constant one mole contains 6.02 × 1023 particles, which can be atoms, molecules, ions or formula units
One mole of Na contains 6.02 × 1023 atoms.
One mole of H2 contains 6.02 × 1023 molecules.
One mole of NaCl contains 6.02 × 1023 formula units.
Number of particles
  • N = amount in moles × 6.02 × 1023
Use of the constant
  • Use of the constant the value need not be remembered, but the equation must be used and rearranged

Part 2 of 9Moles, mass and molar mass

Molar mass
  • Molar mass the mass of one mole, in g/mol
For an element it is the Ar in grams, such as carbon 12 g and helium 4 g.
For a compound it is the Mr in grams, such as carbon dioxide 44 g.
Moles from mass
  • Moles from mass n = m ÷ Mr, with the mass in grams
    What the symbols mean: n = amount of substance (mol) · m = mass (g) · M or Mr = molar mass (g mol−1, numerically equal to Mr)
Mass from moles
  • Mass from moles m = n × Mr
Elements
  • Elements use the same equation with Ar in place of Mr
Molar mass values
  • Molar mass values one mole of carbon is 12 g, of water is 18 g and of NaCl is 58.5 g.

Part 3 of 9Gas volumes

Avogadro's law
  • Avogadro's law at the same temperature and pressure, equal amounts of gases occupy the same volume
Molar gas volume
  • Molar gas volume one mole of any gas occupies 24 dm3 (24 000 cm3) at room temperature and pressure (r.t.p.)
r.t.p. is 20 °C and 1 atmosphere.
Moles of gas
  • Moles of gas n = V ÷ 24 with V in dm3, or n = V ÷ 24 000 with V in cm3
Volume of gas
  • V = n × 24 dm3
Volume ratio
  • Volume ratio gases react in volume ratios equal to the mole ratios in the equation.
In 2H2 + O2 → 2H2O, 100 cm3 of hydrogen reacts with 50 cm3 of oxygen.

Part 4 of 9Concentration of solutions

Concentration
  • Concentration the amount of solute in 1 dm3 of solution, measured in g/dm3 or mol/dm3
Volume units
  • 1 dm3 = 1000 cm3, so divide cm3 by 1000 to get dm3
Moles in solution
  • Moles in solution n = c × V, with V in dm3
    What the symbols mean: c = concentration (mol dm−3) · n = amount (mol) · V = volume (dm3; divide cm3 by 1000)
Concentration
  • Concentration c = n ÷ V
Mass concentration
  • Mass concentration g/dm3 = mol/dm3 × Mr
Converting units
  • Converting units mol/dm3 = g/dm3 ÷ Mr.
A 2.0 mol/dm3 NaOH solution contains 2 × 40 = 80 g in every dm3.

Part 5 of 9Reacting amounts and limiting reactants

1Mole ratio
  • Mole ratio the coefficients in the balanced equation give the ratio of the amounts of each substance
2Limiting reactant
  • Limiting reactant the reactant used up first, which limits the amount of product
3
The amount of product is directly proportional to the amount of limiting reactant.
4
The reactant that remains is the excess reactant.
5
Write the balanced equation and find the mole ratio.
6
Convert the mass of each reactant to moles by dividing by its molar mass.
7
Compare the moles with the ratio in the equation to find the limiting reactant.
8
Use the limiting reactant to find the maximum amount of product.

Part 6 of 9Titration calculations (Extended)

Titration calculations (Extended)
Titration
  • Titration analyses the concentration of a solution by finding the volume of one solution needed to react exactly with another
Acid–base titrations find how much alkali neutralises a quantity of acid.
Average titre
  • Average titre calculated from the titration results before using it in the calculation
Write the balanced equation and find the mole ratio.
Calculate the moles of the solution of known concentration using n = c × V.
Use the mole ratio to find the moles of the other reactant.
Calculate the unknown concentration (c
Calculate the unknown concentration (c n ÷ V) or volume (V = n ÷ c).

Part 7 of 9Empirical and molecular formulae from data

Empirical and molecular formulae from data
Write the mass or percentage of each element.
Divide each by its Ar to get the moles.
Divide all the moles by the smallest number.
If a ratio is not whole, such as 1.5, multiply all the ratios to get whole numbers.
Write the empirical formula from the whole-number ratio.
Molecular formula
  • Molecular formula the actual number of atoms of each element in a molecule
Find the relative formula mass of the empirical formula.
Divide the Mr of the molecule by this to get a number, then multiply each atom count in the empirical formula by it.
Empirical and molecular
  • Empirical and molecular methane CH4 and CH4, ethane CH3 and C2H6, ethene CH2 and C2H4, benzene CH and C6H6
Hydrated salts
  • Hydrated salts find the moles of the anhydrous salt and of the water lost on heating, then divide by the smaller to get the formula of the hydrate
Combustion or reduction data
  • Combustion or reduction data start from the masses found by weighing, then follow the same method.

Part 8 of 9Percentage yield, composition and purity

1Yield
  • Yield the amount of product obtained from a reaction
2Percentage yield
  • Percentage yield actual yield ÷ theoretical yield × 100
3
Actual yield is the recorded amount of product.
4
Theoretical yield is the amount under perfect conditions, from a reacting mass calculation.
5Why yield is below 100%
  • Why yield is below 100% reactants left behind in the apparatus, reversible reactions, loss during filtration or distillation, side reactions,
    • and loss during transfers
6Percentage composition
  • Percentage composition total mass of the element in the formula ÷ Mr of the compound × 100
7
Use the total mass of the element, not one atom.
8
The percentages of all the elements add up to 100%.
9Percentage purity
  • Percentage purity mass of pure substance ÷ total mass of substance × 100
10
Impurities include unreacted reactants and catalysts.
11Percentage composition equation
  • Ar × number of atoms of the element ÷ Mr × 100.
12Fertilisers
  • Fertilisers the percentage of nitrogen compares how much nitrogen each provides per kilogram, such as NH4NO3 35%, (NH4)2SO4 21% and urea CO(NH2)2 47%.
13Ores
  • Ores the percentage of metal compares ores, such as Fe2O3 70% iron and Fe3O4 72.4% iron.
14Hydrated salts
  • Hydrated salts the water of crystallisation in CuSO4.5H2O is 90 ÷ 249.5 = 36.1%.

Part 9 of 9More from the specification

1The Mole Equation Triangle
  • The fundamental stoichiometric relationship links mass (m), number of moles (n), and molar mass (M) via: Mass = Moles x Molar mass (m = n x Mr).
2The Avogadro Constant as a Counting Unit
  • One mole of any substance contains exactly 6.02 x 1023 elementary entities, with mass in grams numerically equal to its relative atomic or formula mass.
    • (atoms, molecules, or formula units)
3Stoichiometric Mole Ratios in Neutralisation Reactions
  • Neutralising nitric acid with aqueous ammonia yields ammonium nitrate fertilizer (NH3 + HNO3 → NH4NO3) in a precise 1:1 equimolar ratio.
4Ammonium nitrate
  • NH4NO3, a fertiliser with 35% nitrogen.

Key terms

TermMeaning
MoleThe amount of substance containing 6.02 × 10²³ particles.
Avogadro constant6.02 × 10²³ per mole.
Molar massThe mass of one mole, in g/mol.
Molar gas volume24 dm³ at r.t.p.
ConcentrationMoles ÷ volume in dm³, or g/dm³.
Limiting reactantthe reactant that is used up first and so limits the amount of product.
Percentage yieldactual yield divided by theoretical yield, times 100.
Percentage compositionthe percentage by mass of each element in a compound.

Avoid these mistakes

1
Saying a mole is a mass — it is an amount containing 6.02 × 10²³ particles
2
Using cm³ directly in n = c × V — convert to dm³ by dividing by 1000
3
Saying 24 dm³ applies in all conditions — it applies at r.t.p., and 22.4 dm³ is the s.t.p. value
4
Using the mass ratio from the equation or an unbalanced equation — balance first and use the mole ratio
5
Accepting a percentage yield above 100% — it signals an error or an impure product

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