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Mole Concept and Stoichiometry

1

Relative Masses

1.1Relative atomic mass, A_r

DefinitionThe relative atomic mass (A_r) of an element is the average mass of one atom of the element compared with 1/12 of the mass of a carbon-12 atom.

The A_r values are in the Periodic Table. They have no units.

Worked exampleChlorine is 75 % chlorine-35 and 25 % chlorine-37. Find its A_r.

A_r = (75/100 × 35) + (25/100 × 37) = 35.5

1.2Relative molecular and formula mass, M_r

DefinitionThe relative molecular mass (M_r) of a molecule is the sum of the relative atomic masses of all the atoms in its formula. For an ionic compound it is called the relative formula mass and has the same symbol, M_r.

Substance	Formula	Working	M_r
water	H₂O	2(1) + 16	18
carbon dioxide	CO₂	12 + 2(16)	44
calcium carbonate	CaCO₃	40 + 12 + 3(16)	100
iron(II) nitrate	Fe(NO₃)₂	56 + 2(14) + 6(16)	180

NoteMultiply the A_r by the number of atoms of that element, including the number outside any brackets.

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Relative Masses (continued)

1.3Percentage mass of an element

percentage mass = number of atoms × A_r of the elementM_r of compound × 100 %

Worked exampleFind the percentage mass of nitrogen in (NH₄)₂SO₄.

M_r = 2(14 + 4) + 32 + 4(16) = 132.Percentage of N = 2 × 14132 × 100 % = 21.2 %.

2

The Mole

2.1Counting particles

DefinitionOne mole of any substance contains 6.02 × 10²³ particles. This number is the Avogadro constant. The symbol for the mole is mol.

The particles can be atoms, molecules, ions or even electrons.

number of moles = number of particles6.02 × 10²³

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2

The Mole (continued)

2.2Particles in a mole

Amount	Particles	Number
1 mol of Ne	Ne atoms	6.02 × 10²³
1 mol of H₂	H₂ molecules	6.02 × 10²³
1 mol of H₂	H atoms (2 mol)	1.204 × 10²⁴
2 mol of Mg(NO₃)₂	NO₃⁻ ions (4 mol)	2.408 × 10²⁴

Exam tipDo not mix up the number of moles of a substance with the number of particles. Look at what the question asks you to count.

2.3Moles and mass

DefinitionThe mass of one mole is the molar mass. It equals the A_r or M_r in grams. The unit is g/mol.

number of moles = mass in gmolar mass in g/mol

Worked exampleFind the mass of 4.5 mol of lead(II) sulfate, PbSO₄.

Molar mass = 207 + 32 + 4(16) = 303 g/mol.Mass = 4.5 × 303 = 1360 g (3 s.f.).

2.4Moles and gas volume

DefinitionOne mole of any gas occupies 24 dm³ at r.t.p. (25 °C and 1 atmosphere).

number of moles of gas = volume in dm³24 dm³

Worked exampleFind the volume of 2 mol of oxygen at r.t.p.

Volume = 2 × 24 = 48 dm³.

34Topic 07 · Mole Concept

2

The Mole (continued)

2.5Concentration of a solution

DefinitionThe concentration of a solution is the amount of solute dissolved in a unit volume of solution. Two units are used: g/dm³ and mol/dm³.

concentration in g/dm³ = mass of solute in gvolume of solution in dm³

concentration in mol/dm³ = amount of solute in molvolume of solution in dm³

concentration in mol/dm³ = concentration in g/dm³molar mass in g/mol

Note1 dm³ = 1000 cm³. Change cm³ to dm³ by dividing by 1000.

Worked example500 cm³ of solution contains 2 mol of NaCl. Find the concentration in g/dm³.

Mass of NaCl = 2 × 58.5 = 117 g.Concentration = 117 g0.500 dm³ = 234 g/dm³.

Worked example1.74 g of K₂SO₄ is dissolved to make 200 cm³ of solution. Find the concentration in mol/dm³.

Molar mass = 2(39) + 32 + 4(16) = 174 g/mol. Moles = 1.74174 = 0.01 mol.Concentration = 0.01 mol0.200 dm³ = 0.05 mol/dm³.

2.6The formula triangles

NoteCover the quantity you want. A horizontal line means divide and a vertical line means multiply.

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3

Empirical and Molecular Formulae

3.1Two kinds of formula

Empirical formula

The simplest whole-number ratio of the atoms of each element in a compound.

Molecular formula

The exact number of atoms of each element in one molecule, a multiple of the empirical one.

Compound	Molecular	Empirical
water	H₂O	H₂O
phosphorus(V) oxide	P₄O₁₀	P₂O₅
ethene	C₂H₄	CH₂

3.2Empirical formula from percentage mass

A compound is 40.0 % carbon, 6.70 % hydrogen and 53.3 % oxygen by mass.

Step	C	H	O
1 Assume 100 g, so mass in g	40.0	6.70	53.3
2 Divide by A_r to get moles	40.0 ÷ 12 = 3.33	6.70 ÷ 1 = 6.70	53.3 ÷ 16 = 3.33
3 Divide by the smallest	1	2	1

The empirical formula is CH₂O.

3.3Molecular formula

n = M_r of compoundM_r of empirical formula

Worked exampleA compound has M_r = 60 and empirical formula CH₂O. Find its molecular formula.

M_r of CH₂O = 12 + 2(1) + 16 = 30.n = 60 ÷ 30 = 2.Molecular formula = (CH₂O)₂ = C₂H₄O₂.

3.4When the ratio is not a whole number

Worked examplePlexiglass is 59.9 % carbon, 8.1 % hydrogen and 32.0 % oxygen by mass. Find its empirical formula.

Moles: C = 59.9 ÷ 12 = 4.99, H = 8.1 ÷ 1 = 8.1, O = 32.0 ÷ 16 = 2.00.Divide by the smallest (2.00): C = 2.5, H = 4.05, O = 1.2.5 is not a whole number, so multiply all by 2: C = 5, H = 8, O = 2. The formula is C₅H₈O₂.

36Topic 07 · Mole Concept

4

Stoichiometry

4.1The mole ratio

Stoichiometry is the ratio between the quantities of substances, in moles, in a chemical reaction.

Exam tipThe ratio of moles equals the ratio of the numbers in front of the formulae, not the ratio of masses.

4.2Reacting masses and gas volumes

Find the moles of the substance you know (from mass, gas volume or concentration and volume).
Use the mole ratio from the balanced equation to find the moles of the substance you want.
Change the moles to the mass, volume or concentration you want.

Worked exampleFind the mass of magnesium carbonate needed to produce 100 dm³ of carbon dioxide at r.t.p. MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + CO₂(g) + H₂O(l)

Moles of CO₂ = 100 ÷ 24 = 4.17 mol.1 mol of MgCO₃ gives 1 mol of CO₂, so moles of MgCO₃ = 4.17 mol.Mass = 4.17 × (24 + 12 + 48) = 350 g.

Worked exampleFind the mass of iron made from 2000 g of iron(III) oxide. Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

M_r of Fe₂O₃ = 160, so moles = 2000 ÷ 160 = 12.5 mol.1 mol of Fe₂O₃ gives 2 mol of Fe, so moles of Fe = 25.0 mol.Mass = 25.0 × 56 = 1400 g.

4.3Gas volumes in a reaction

NoteAvogadro's law: at the same temperature and pressure, equal volumes of gases contain the same number of particles. So the mole ratio of gases is also their volume ratio.

Worked exampleN₂(g) + O₂(g) → 2NO(g). Find the volume of nitrogen needed to make 800 cm³ of nitric oxide.

Volume ratio N₂ : NO = 1 : 2.Volume of N₂ = 800 ÷ 2 = 400 cm³.

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4

Stoichiometry (continued)

4.4Titration

The titrant has a known concentration. The analyte has an unknown concentration.
Add a fixed volume of analyte (usually 25.0 cm³, by pipette) and a few drops of indicator to the flask. Add titrant from the burette, swirling, until the colour changes permanently. This is the end-point.
Repeat until two titres agree to within 0.10 cm³, and use the average of those titres.

Worked example20.0 cm³ of H₂SO₄ reacts with 30.0 cm³ of 0.100 mol/dm³ KOH. 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O. Find the concentration of the acid.

Moles of KOH = 0.100 × 0.0300 = 0.003 mol.2 mol KOH react with 1 mol H₂SO₄, so moles of acid = 0.0015 mol.Concentration = 0.0015 ÷ 0.0200 = 0.075 mol/dm³.

4.5A gas reacting with a solution

Worked example250 cm³ of calcium hydroxide solution reacts completely with 6 dm³ of carbon dioxide at r.t.p. CO₂(g) + Ca(OH)₂(aq) → CaCO₃(s) + H₂O(l). Find the concentration of the calcium hydroxide and the mass of calcium carbonate formed.

Moles of CO₂ = 6 ÷ 24 = 0.25 mol, so moles of Ca(OH)₂ = 0.25 mol (ratio 1 : 1).Concentration = 0.25 ÷ 0.250 = 1.0 mol/dm³.Moles of CaCO₃ = 0.25 mol, so mass = 0.25 × 100 = 25 g.

38Topic 07 · Mole Concept

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Stoichiometry (continued)

4.6Limiting reactants

recipe2 bread+1 cheese→1 sandwich we have10 bread4 cheese4 sandwiches 2 left over: in excessall used: limitingset by the cheese

The reactant that is completely used up is the limiting reactant. The other is in excess.
The amount of product depends on the limiting reactant, so calculate from it.
Example: 5 mol of NaOH and 3 mol of HCl react 1 : 1. All 3 mol of HCl are used up and 2 mol of NaOH remain. HCl is limiting and 3 mol of NaCl form.

4.7Finding the limiting reactant

Worked example2.00 g of calcium carbonate powder is added to 50.0 cm³ of 2.00 mol/dm³ hydrochloric acid. Find the volume of carbon dioxide at r.t.p. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Moles of HCl = 2.00 × 0.0500 = 0.100 mol.Moles of CaCO₃ = 2.00 ÷ 100 = 0.0200 mol.0.100 mol of HCl needs 0.0500 mol of CaCO₃, but only 0.0200 mol is present. CaCO₃ is limiting.1 mol of CaCO₃ gives 1 mol of CO₂, so volume = 0.0200 × 24 = 0.480 dm³.

4.8Common mistakes in mole calculations

Forgetting to change cm³ to dm³: divide by 1000.
Using the ratio of masses instead of the ratio of moles from the equation.
Using the M_r of the wrong substance, or leaving out the number in front of a formula.
Using 24 dm³ for a solid, a liquid or a solution. It is only for gases at r.t.p.

Topic 07 · Mole Concept39

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5

Percentage Yield and Purity

5.1Percentage yield

percentage yield = actual yieldtheoretical maximum yield × 100 %

The theoretical maximum is worked out from the mole ratio in the equation.

Worked example100 g of CaCO₃ reacts with excess HNO₃: CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂. 129.6 g of Ca(NO₃)₂ is collected. Find the percentage yield.

Moles of CaCO₃ = 100 ÷ 100 = 1 mol, so 1 mol of Ca(NO₃)₂ is the theoretical maximum.M_r of Ca(NO₃)₂ = 164, so the theoretical mass is 164 g.Percentage yield = 129.6 ÷ 164 × 100 % = 79.0 %.

5.2Percentage purity

percentage purity = mass of pure substance in the sampletotal mass of the sample × 100 %

Worked example50.0 g of impure aspirin, C₉H₈O₄, contains 0.20 mol of pure aspirin. Find the percentage purity.

Molar mass = 9(12) + 8(1) + 4(16) = 180 g/mol.Mass of pure aspirin = 0.20 × 180 = 36.0 g.Percentage purity = 36.0 ÷ 50.0 × 100 % = 72.0 %.

5.3Purity from a reaction

Worked example200 g of impure zinc is added to excess copper(II) chloride solution and 171 g of copper is obtained. Zn + CuCl₂ → ZnCl₂ + Cu. Find the percentage purity of the zinc.

Moles of Cu = 171 ÷ 64 = 2.67 mol, so moles of Zn = 2.67 mol (ratio 1 : 1).Mass of pure zinc = 2.67 × 65 = 174 g.Percentage purity = 174 ÷ 200 × 100 % = 86.8 %.

40Topic 07 · Mole Concept

6

Summary

6.1Moles at the centre

To go between any two quantities, change to moles first, then change to the quantity you want.

6.2Formulae to know

To find	Use
M_r	add the A_r of every atom in the formula
moles from mass	mass in g ÷ molar mass in g/mol
moles of a gas	volume in dm³ ÷ 24
moles from particles	number of particles ÷ 6.02 × 10²³
moles in solution	concentration in mol/dm³ × volume in dm³
concentration in g/dm³	concentration in mol/dm³ × M_r
molecular formula	(empirical formula)_n, n = M_r ÷ M_r of empirical formula
percentage yield	actual ÷ theoretical maximum × 100 %
percentage purity	pure mass ÷ sample mass × 100 %

Exam tipWrite the equation, then moles, then the mole ratio, then the answer. Show every step and give the units.

Topic 07 · Mole Concept41

Relative Masses

1.1Relative atomic mass, Ar

1.2Relative molecular and formula mass, Mr

Relative Masses (continued)

1.3Percentage mass of an element

The Mole

2.1Counting particles

The Mole (continued)

2.2Particles in a mole

2.3Moles and mass

2.4Moles and gas volume

The Mole (continued)

2.5Concentration of a solution

2.6The formula triangles

Empirical and Molecular Formulae

3.1Two kinds of formula

3.2Empirical formula from percentage mass

3.3Molecular formula

3.4When the ratio is not a whole number

Stoichiometry

4.1The mole ratio

4.2Reacting masses and gas volumes

4.3Gas volumes in a reaction

Stoichiometry (continued)

4.4Titration

4.5A gas reacting with a solution

Stoichiometry (continued)

4.6Limiting reactants

4.7Finding the limiting reactant

4.8Common mistakes in mole calculations

Percentage Yield and Purity

5.1Percentage yield

5.2Percentage purity

5.3Purity from a reaction

Summary

6.1Moles at the centre

6.2Formulae to know

1.1Relative atomic mass, A_r

1.2Relative molecular and formula mass, M_r