Chemical Formulae
and Equations

Mole Concept
and Stoichiometry

IP Chemistry (6092), Sec 3

By the end of these topics, you can answer

1

What does the formula H₂SO₄ tell you about sulfuric acid?

2

Why is the formula of magnesium chloride MgCl₂ and not MgCl?

3

How do you balance an equation without changing any formula?

4

How can chemists count atoms that are far too small to see?

5

What mass of iron can be made from 2 kg of iron ore?

6

Why do real reactions make less product than the equation predicts?

Starter

PredictA recipe for a cake lists 2 eggs for every 1 cup of flour. You have 6 eggs and 2 cups of flour.

How many cakes can you make, and what is left over?

NoteChemical formulae and equations are recipes for chemicals. By the end, you will use them to work out exactly how much to use and how much you will get.
1

Chemical Formulae

1.1What a formula tells you

A chemical formula is like a recipe. It shows the ratio of elements in one unit of a substance.
  • The symbols show which elements are present. The subscripts show how many atoms of each.
  • The prefix di- means two. Carbon dioxide has two oxygen atoms.
1

Chemical Formulae

1.2Formulae of elements

MonoatomicSingle, uncombined atoms. The Group 18 elements, such as neon (Ne) and argon (Ar), have full outer shells.
DiatomicTwo atoms joined: H2, N2, O2, F2, Cl2, Br2, I2.
PolyatomicThree or more atoms joined, such as phosphorus and ozone (O3).
NoteThe seven diatomic elements are hydrogen, nitrogen, oxygen and the four Group 17 halogens. Learn them as a set: H, N, O, F, Cl, Br, I.
1

Chemical Formulae

1.3Formulae of compounds

  • Compounds have fixed formulae, while mixtures do not.
  • A subscript outside brackets multiplies everything inside them.

Ca(NO3)2 has 1 calcium, 2 nitrogen and 6 oxygen atoms.

1

Chemical Formulae

Quick checkHow many oxygen atoms are there in one unit of aluminium hydroxide, Al(OH)₃?

A1
B2
C3
D4
NoteC: the subscript 3 outside the bracket multiplies the O and the H inside, giving 3 O and 3 H.
2

Constructing Formulae

2.1Valency

Valency is the number of electrons an atom must lose, gain or share to get a noble gas configuration. The valency of an ion is its charge without the sign.
Group121314151617
ExamplesLi, Na, KBe, Mg, CaB, AlC, SiN, PO, SF, Cl, Br, I
Valency1234321
NoteA Roman numeral shows the valency of a transition metal: iron(II) is Fe2+, iron(III) is Fe3+, copper(II) is Cu2+.
2

Constructing Formulae

2.2Polyatomic ions

NameFormulaValencyNameFormulaValency
ammoniumNH4+1sulfateSO42−2
hydroxideOH−1carbonateCO32−2
nitrateNO3−1phosphatePO43−3
NoteA polyatomic ion is a group of atoms joined by covalent bonds that carries an overall charge. Learn the five marked ions by heart, as the O Level syllabus uses them throughout.
2

Constructing Formulae

2.3The crossing method

  1. Write the symbols and their valencies below.
  2. Cross the valencies: each valency becomes the subscript of the other element.
  3. Simplify the ratio if possible. C2O4 becomes CO2.
2

Constructing Formulae

2.4Worked examples

Worked exampleWrite the formulae of sodium phosphate and magnesium nitrate.

Step 1Na+ (valency 1) and PO43− (valency 3): three Na+ cancel one PO43−, so Na3PO4.

Step 2Mg2+ (valency 2) and NO3− (valency 1): two NO3− are needed.

Step 3Put the polyatomic ion in brackets before the subscript: Mg(NO3)2.

2

Constructing Formulae

2.5Finding a charge from a formula

Worked exampleFind the charge on the Al ion in KAl(SO4)2.

Step 1The compound is neutral, so the total charge is 0.

Step 2Let the charge on Al be x: 0 = (+1) + x + 2(−2).

Step 3So x = +3

2

Constructing Formulae

2.6Practice with formulae

CompoundIons and valenciesFormula
calcium chlorideCa2+ (2) and Cl− (1)CaCl2
silver phosphateAg+ (1) and PO43− (3)Ag3PO4
iron(III) sulfateFe3+ (3) and SO42− (2)Fe2(SO4)3
aluminium oxideAl3+ (3) and O2− (2)Al2O3
NoteTitanium dioxide is TiO2. Two O2− give −4, so the titanium ion is Ti4+.
3

Chemical Equations

3.1Word and chemical equations

sodium + chlorine → sodium chloride
2Na(s) + Cl2(g) → 2NaCl(s)
3

Chemical Equations

3.1Reading an equation

  • Reactants go on the left of the arrow and products on the right.
  • The number in front of a formula is the number of units reacting or formed.
  • State symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous, which means dissolved in water.
NoteSodium chloride is (s) when it forms from sodium and chlorine, but (aq) when it forms in a solution.
3

Chemical Equations

3.2Balancing an equation

  1. Write the correct formulae, then count the atoms of each element on both sides.
  2. Add numbers in front of formulae to balance. Never change a subscript.
  3. Add the state symbols: 2H2(g) + O2(g) → 2H2O(l).
3

Chemical Equations

3.2From a word equation to a full equation

3

Chemical Equations

3.3Balancing with a solid and a gas

Worked exampleIron(III) oxide reacts with carbon monoxide to form iron and carbon dioxide. Write the balanced equation with state symbols.

Step 1Formulae: Fe2O3 + CO → Fe + CO2.

Step 2Two iron atoms need 2Fe. Three oxygen atoms leave as CO2, so use 3CO and 3CO2.

Step 3Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

3

Chemical Equations

3.4A harder equation

Worked exampleAqueous iron(III) sulfate reacts with aqueous barium nitrate to form aqueous iron(III) nitrate and solid barium sulfate. Write the equation with state symbols.

Step 1Formulae: Fe2(SO4)3 + Ba(NO3)2 → Fe(NO3)3 + BaSO4.

Step 2Balance the ions. Two Fe3+ need 2Fe(NO3)3. Three SO42− need 3BaSO4, so 3Ba(NO3)2.

Step 3Fe2(SO4)3(aq) + 3Ba(NO3)2(aq) → 2Fe(NO3)3(aq) + 3BaSO4(s)

3

Chemical Equations

3.5Ionic equations

  1. Write the balanced equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
  2. Split the aqueous substances into ions.
  3. Cancel the spectator ions, which appear unchanged on both sides.
3

Chemical Equations

3.5What to split

  • Solids, gases and liquids such as water are not split into ions.
  • Write the quantity in front of an ion: H2SO4(aq) splits into 2H+(aq) + SO42−(aq).
  • Aqueous covalent substances such as aqueous chlorine are not split: Cl2(g) + 2I−(aq) → 2Cl−(aq) + I2(aq).
AnalogySpectator ions are like fans at a football match: they are in the stadium at the start and at the end, but they never play. Only the players go into the ionic equation.
3

Chemical Equations

3.6An ionic equation with a solid and a gas

Worked exampleWrite the ionic equation for ZnCO3(s) + 2HNO3(aq) → Zn(NO3)2(aq) + CO2(g) + H2O(l).

Step 1Split only the aqueous substances. Keep ZnCO3(s), CO2(g) and H2O(l) whole.

Step 2Cancel the spectator ions, 2NO3−(aq).

Step 3ZnCO3(s) + 2H+(aq) → Zn2+(aq) + CO2(g) + H2O(l)

3

Chemical Equations

Quick checkWhat is the ionic equation for AgNO₃(aq) + HCl(aq) → AgCl(s) + HNO₃(aq)?

AH⁺(aq) + NO₃⁻(aq) → HNO₃(aq)
BAg⁺(aq) + Cl⁻(aq) → AgCl(s)
CAgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)
DAg⁺(aq) + HCl(aq) → AgCl(s) + H⁺(aq)
NoteB: H⁺ and NO₃⁻ are spectator ions. The solid AgCl, a precipitate, is not split into ions.
3

Chemical Equations

3.8Common mistakes to avoid

  • Changing a subscript to balance an equation. Only change the number in front.
  • Forgetting the brackets in a formula such as Mg(NO3)2.
  • Writing hydrogen as H instead of H2.
  • Splitting a solid, a gas or water into ions in an ionic equation.
  • Leaving the state symbols out when the question asks for them.

Mole Concept
and Stoichiometry

Counting particles by weighing them

1

Relative Masses

1.1Relative atomic mass, Ar

DefinitionThe relative atomic mass (Ar) 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 standard is carbon-12: one twelfth of a 12C atom is 1 unit.

ExampleA magnesium atom weighs as much as 24 of those units, so Ar(Mg) = 24.
1

Relative Masses

1.1What the unit means

  • Atoms are far too light to weigh in grams, so we compare each atom with a tiny standard unit: 1/12 of the mass of a carbon-12 atom.
  • This unit is about the mass of one proton or one neutron. So Ar is close to the number of nucleons: magnesium, with 12 protons and 12 neutrons, has Ar = 24.
  • The Ar values are in the Periodic Table. They have no units.
1

Relative Masses

1.1Averages from isotopes

Chlorine is 75 % chlorine-35 and 25 % chlorine-37.

A simple average of 35 and 37 is 36, but there are three times as many 35Cl atoms, so the average is pulled towards 35.

Worked exampleFind the Ar of chlorine.

Ar = 75100 × 35 + 25100 × 37 = 35.5
1

Relative Masses

1.2Relative molecular and formula mass, Mr

DefinitionMr is the sum of the relative atomic masses of all the atoms in the formula. For an ionic compound it is the relative formula mass.
SubstanceFormulaWorkingMr
waterH2O2(1) + 1618
carbon dioxideCO212 + 2(16)44
calcium carbonateCaCO340 + 12 + 3(16)100
iron(II) nitrateFe(NO3)256 + 2(14) + 6(16)180
1

Relative Masses

1.3Percentage mass of an element

percentage mass = number of atoms × Ar of the elementMr of compound × 100 %

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

Step 1Mr = 2(14 + 4) + 32 + 4(16) = 132.

Step 2Percentage 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.
number of moles = number of particles6.02 × 1023
2

The Mole

2.2Particles in a mole

AmountParticlesNumber
1 mol of NeNe atoms6.02 × 1023
1 mol of H2H2 molecules6.02 × 1023
1 mol of H2H atoms (2 mol)1.204 × 1024
2 mol of Mg(NO3)2NO3− ions (4 mol)2.408 × 1024
NoteDo 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

The Mole

2.3Moles and mass

DefinitionThe mass of one mole is the molar mass. It equals the Ar or Mr in grams. The unit is g/mol.
Analogy6.02 × 1023 of the tiny atomic mass units (each about the mass of one proton) add up to 1 g. The mole is the exchange rate between the atom scale and the gram scale, just as 100 cents make 1 dollar. So 1 mol of carbon-12, at 12 units per atom, has a mass of 12 g.
number of moles = mass in gmolar mass in g/mol
2

The Mole

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

Step 1Molar mass = 207 + 32 + 4(16) = 303 g/mol.

Step 2Mass = moles × molar mass = 4.5 × 303 = 1360 g (3 s.f.)

2

The Mole

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.
2

The Mole

Quick checkWhat volume does 0.5 mol of oxygen gas occupy at r.t.p.?

A8 dm³
B12 dm³
C16 dm³
D48 dm³
NoteB: volume = moles × 24 = 0.5 × 24 = 12 dm³. One mole of any gas occupies 24 dm³ at r.t.p.
2

The Mole

2.5Concentration of a solution

DefinitionThe concentration of a solution is the amount of solute dissolved in a unit volume of solution, in g/dm3 or 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 = 1 litre. Always change cm3 to dm3 by dividing by 1000.
2

The Mole

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

Step 1Molar mass = 2(39) + 32 + 4(16) = 174 g/mol.

Step 2Moles = 1.74174 = 0.01 mol

Step 3Concentration = 0.01 mol0.200 dm3 = 0.05 mol/dm3

2

The Mole

2.6The formula triangles

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

3

Empirical and Molecular Formulae

3.1Two kinds of formula

Empirical formulaThe simplest whole-number ratio of the atoms of each element in a compound.
Molecular formulaThe exact number of atoms of each element in one molecule.
CompoundMolecularEmpirical
waterH2OH2O
phosphorus(V) oxideP4O10P2O5
etheneC2H4CH2
3

Empirical and Molecular Formulae

3.2Empirical formula from percentage mass

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

StepCHO
1 Assume 100 g, so mass in g40.06.7053.3
2 Divide by Ar to get moles40.012 = 3.336.701 = 6.7053.316 = 3.33
3 Divide by the smallest121

The empirical formula is CH2O

3

Empirical and Molecular Formulae

3.3Molecular formula

n = Mr of compoundMr of empirical formula

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

Step 1Mr of CH2O = 12 + 2(1) + 16 = 30.

Step 2n = 6030 = 2

Step 3Molecular formula = (CH2O)2 = C2H4O2

3

Empirical and Molecular Formulae

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.

Step 1Moles: C = 59.912 = 4.99, H = 8.11 = 8.1, O = 32.016 = 2.00

Step 2Divide by the smallest (2.00): C = 2.5, H = 4.05, O = 1.

Step 32.5 is not a whole number, so multiply all by 2: C5H8O2

4

Stoichiometry

4.1The mole ratio

Stoichiometry is the ratio between the quantities of substances, in moles, in a chemical reaction.
NoteThe ratio of moles equals the ratio of the numbers in front of the formulae, not the ratio of masses.
4

Stoichiometry

4.2Reacting masses and gas volumes

  1. Find the moles of the substance you know.
  2. Use the mole ratio from the balanced equation to find the moles of the substance you want.
  3. Change the moles to the mass, volume or concentration you want.
4

Stoichiometry

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)

Step 1Moles of CO2 = 10024 = 4.17 mol

Step 21 mol of MgCO3 gives 1 mol of CO2, so moles of MgCO3 = 4.17 mol.

Step 3Mass = 4.17 × (24 + 12 + 48) = 350 g

4

Stoichiometry

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

Step 1Mr of Fe2O3 = 160, so moles = 2000160 = 12.5 mol

Step 21 mol of Fe2O3 gives 2 mol of Fe, so moles of Fe = 25.0 mol.

Step 3Mass = 25.0 × 56 = 1400 g

4

Stoichiometry

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 = 8002 = 400 cm3
4

Stoichiometry

4.4Titration

  • The titrant has a known concentration. The analyte has an unknown concentration.
  • Add 25.0 cm3 of analyte (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 at least twice and use the average volume.
4

Stoichiometry

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.

Step 1Moles of KOH = 0.100 × 0.0300 = 0.003 mol.

Step 22 mol KOH react with 1 mol H2SO4, so moles of acid = 0.0015 mol.

Step 3Concentration = 0.0015 mol0.0200 dm3 = 0.075 mol/dm3

4

Stoichiometry

4.6Limiting reactants

A sandwich needs 2 slices of bread and 1 slice of cheese. With 10 bread and 4 cheese, how many sandwiches?

Only 4: the cheese runs out first, and 2 slices of bread are left over.

  • 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.
4

Stoichiometry

4.7Finding the limiting reactant

Worked example2.00 g of calcium carbonate 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

Step 1Moles of HCl = 2.00 × 0.0500 = 0.100 mol. Moles of CaCO3 = 2.00100 = 0.0200 mol

Step 20.100 mol of HCl needs 0.0500 mol of CaCO3, but only 0.0200 mol is present. CaCO3 is limiting.

Step 31 mol of CaCO3 gives 1 mol of CO2, so volume = 0.0200 × 24 = 0.480 dm3

4

Stoichiometry

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 Mr 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.
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.
  • A higher yield means a more efficient reaction. A low yield means reactant was wasted, which costs more.

Worked example100 g of CaCO3 reacts with excess HNO3 and 129.6 g of Ca(NO3)2 (Mr 164) is collected. Find the percentage yield.

1 mol of CaCO3 gives 1 mol of Ca(NO3)2, so the theoretical mass is 164 g.Percentage yield = 129.6164 × 100 % = 79.0 %
5

Percentage Yield and Purity

5.2Percentage purity

percentage purity = mass of pure substance in the sampletotal mass of the sample × 100 %
  • 24-carat gold is 99.9 % pure. 18-carat gold is 75.0 % pure, and its impurities make it harder, so it lasts longer in jewellery.
  • Impurities in food and medicines can be harmful, so they are made as pure as possible.

Worked example50.0 g of impure aspirin, C9H8O4 (molar mass 180 g/mol), contains 0.20 mol of pure aspirin. Find the percentage purity.

Mass of pure aspirin = 0.20 × 180 = 36.0 g.Percentage purity = 36.050.0 × 100 % = 72.0 %
5

Percentage Yield and Purity

Quick checkThe theoretical yield of a reaction is 40 g but only 30 g of product is collected. What is the percentage yield?

A30 %
B40 %
C75 %
D133 %
NoteC: 30 ÷ 40 × 100 % = 75 %. A yield can never be above 100 %.

Exit ticket

1Write the formula of ammonium sulfate.

(NH₄)₂SO₄[1]

2Balance: Al + O₂ → Al₂O₃

4Al + 3O₂ → 2Al₂O₃[1]

3How many moles are in 8.8 g of carbon dioxide?

8.8 ÷ 44 = 0.20 mol[1]

4What volume does 0.25 mol of any gas occupy at r.t.p.?

0.25 × 24 = 6.0 dm³[1]