How much? The amount of chemical change — IB Chemistry SL Revision Notes & Topical Questions

Section 1 of 30Balancing equations

• Conservation
  • atoms are neither created nor destroyed, so each element has the same number of atoms on both sides.

1

Write the formulae of the reactants and products.

2

Count the atoms of each element on each side.

3

Balance one element at a time using coefficients only.

4

Add state symbols.

Balancing equations

Rules

• Rules never change a formula, and keep polyatomic ions such as SO₄²⁻ and NO₃⁻ together when they appear on both sides.

Combustion of organic compounds

• Combustion of organic compounds balance carbon first, then hydrogen, then oxygen.

State symbols

• (s), (l), (g) and (aq).

Diatomic elements

• H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂ are written with subscript 2.

Meaning of coefficients

• Meaning of coefficients they give the mole ratio, not the mass ratio.

Section 2 of 30Reacting mass calculations

Reacting mass calculations

Amount

• Amount n = m/M, with n in mol, m in g and M in g mol⁻¹.
  What the symbols mean: n = amount of substance (mol) · m = mass (g) · M or M_r = molar mass (g mol⁻¹, numerically equal to M_r)

Particle type

• Particle type state which particle is meant, e.g. 1 mol of CaF₂ contains 1 mol of formula units, 1 mol of Ca²⁺ and 2 mol of F⁻.

1

Write the balanced equation.

2

Convert the given mass to moles.

3

Use the mole ratio from the equation to find moles of the wanted substance.

4

Convert moles to mass using mass = n × M.

Units

• Units masses can be in grams or tonnes as long as they are consistent, because reacting masses stay in proportion to the equation.

Conservation of mass

• Conservation of mass total mass of reactants equals total mass of products.

Section 3 of 30Gas volumes and Avogadro's law

1

Avogadro's law

• Avogadro's law equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

2

Consequence

• Consequence the volume ratio of reacting gases equals the mole ratio, so no mole conversion is needed at the same conditions.

3

Molar volume

• 22.7 dm³ mol⁻¹ at STP (273 K, 100 kPa).

4

Volume from moles

• V = n × 22.7 dm³ mol⁻¹.
• Moles from volume — n = V ÷ 22.7 dm³ mol⁻¹.

5

Ideal gas equation

• Ideal gas equation pV = nRT with p in Pa, V in m³, T in K and R = 8.31 J K⁻¹ mol⁻¹ for non-standard conditions.

6

Liquids

• Liquids water formed as a liquid is not included in the gas volume ratio.

7

Limiting gas

• Limiting gas divide each volume by its coefficient and the smallest value is limiting.

Section 4 of 30Concentration and titration calculations

Concentration

• Concentration n = cV with V in dm³ (divide cm³ by 1000).

Volumetric analysis

• Volumetric analysis uses a standard solution to find the concentration of an unknown.

1

Pipette a known volume (usually 20 or 25 cm³) into a conical flask.

2

Fill the burette with the other solution and add a few drops of indicator.

3

Run in the burette solution until the indicator just changes colour.

4

Repeat until titres are concordant.

1

Write the balanced equation.

2

Calculate the moles of the known substance.

3

Use the mole ratio to find moles of the unknown.

4

Calculate the unknown mass, concentration or volume.

1Shortcut

• Shortcut for monoprotic acid and base reacting 1 : 1, c₁V₁ = c₂V₂ and cm³ can be used directly.

2Rounding burette averages

• Rounding burette averages do not give an average more precisely than the 2-decimal-place readings.

Section 5 of 30Back titration

• Purpose
  • finds the amount of an unknown indirectly, useful for insoluble, impure or slow-reacting substances.

1

React the sample with a measured excess of a reagent, e.g. standard HCl.

2

Titrate the leftover excess reagent with a standard solution, e.g. NaOH.

3

Subtract the excess from the amount originally added to find the amount that reacted.

4

Use the mole ratio of the sample reaction to find the sample amount, e.g. 2HCl : 1 CaCO₃.

5

Convert to mass and percentage by mass of the sample.

Section 6 of 30Limiting and excess reactants

Limiting reactant

• Limiting reactant the reactant that runs out first and determines the amount of product.

Excess reactant

• Excess reactant remains after the reaction, and its leftover amount equals initial moles minus moles reacted.

1

Calculate the moles of each reactant.

2

Divide each by its coefficient.

3

The smallest value identifies the limiting reactant.

4

Calculate the product from the limiting reactant.

Section 7 of 30Percentage yield

Percentage yield

Theoretical yield

• Theoretical yield the maximum product from the limiting reactant, assuming perfect conversion.

Actual (experimental) yield

• Actual (experimental) yield the amount of product actually collected.

Formula

Formula percentage yield = (actual yield ÷ theoretical yield) × 100, with both yields in the same unit.

Causes of low yield

• Causes of low yield incomplete or reversible reactions, side reactions, losses in transfer, filtering and purification, and impure reactants.

Yield above 100%

• Yield above 100% usually shows the product is impure or wet.

Section 8 of 30Atom economy

Atom economy

Definition

Definition the proportion of the total mass of reactants that ends up in the desired product.

Formula

Formula atom economy = (M_r of desired product ÷ sum of M_r of all reactants) × 100, using coefficients from the balanced equation.

Basis

• Basis it is calculated from the equation, not from experimental data.

Addition reactions

• Addition reactions have 100% atom economy because only one product forms.

Green chemistry

• Green chemistry a high atom economy means less waste and a more sustainable process.

Yield versus atom economy

• Yield versus atom economy a reaction can have a high yield but low atom economy, so both are considered.

Other measures

• Other measures rate, amounts of catalysts and solvents, energy use and economic efficiency.

Section 9 of 30More from the specification

More from the specification

Limiting reactant calculations

• Determining stoichiometric amounts when active metals like magnesium react with hydrochloric acid to form hydrogen and metal chlorides.

Percentage yield determination

• The ratio of actual isolated product mass to theoretical product mass expressed as a percentage: x 100%.
  • (actual yield / theoretical yield)

Back titration methodology

• Determining the purity of calcium carbonate samples by reacting with excess standardized hydrochloric acid and back-titrating with sodium hydroxide.

Synthesis of magnesium oxide

• Burning magnesium in air under crucible lids to calculate empirical formulas from measured masses before and after oxidation.

Limiting reactant calculations

• Determining stoichiometric amounts when active metals like magnesium react with hydrochloric acid to form hydrogen and metal chlorides.

Percentage yield determination

• The ratio of actual isolated product mass to theoretical product mass expressed as a percentage: x 100%.
  • (actual yield / theoretical yield)

Back titration methodology

• Determining the purity of calcium carbonate samples by reacting with excess standardized hydrochloric acid and back-titrating with sodium hydroxide.

Synthesis of magnesium oxide

• Burning magnesium in air under crucible lids to calculate empirical formulas from measured masses before and after oxidation.

Magnesium oxide

• White powder made when magnesium burns.

Section 10 of 30Avogadro's law

Statement

Statement equal volumes of all gases at the same temperature and pressure contain the same number of molecules (Avogadro, 1811).

Proportionality

• Proportionality gas volume is directly proportional to the amount in moles.

Mole ratio

• Mole ratio the mole ratio of reacting gases follows from their volumes.

No conversion

• No conversion no mole conversion is needed when all gases are measured under the same conditions.

Section 11 of 30Reacting gas volumes

1

Deduce the mole ratio of the gases from the balanced equation, leaving out liquids such as water.

2

Find the limiting gas by dividing each volume by its coefficient.

3

Apply the volume ratio to the limiting gas to find the volume of product and of reactant used.

4

Add any unreacted excess gas to the product volume to find the total gas remaining.

Ammonia and oxygen

• Ammonia and oxygen in 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l) the gas ratio NH₃ : O₂ : NO is 4 : 5 : 4.

Same conditions

• Same conditions the volume ratio applies only if the gases are measured at the same T and p.

Section 12 of 30Molar volume

Definition

Definition the volume of one mole of any gas at STP (273 K, 100 kPa), which is 22.7 dm³ mol⁻¹.

Volume

• V (dm³) = n (mol) × 22.7 dm³ mol⁻¹.

Amount

• Amount n = V ÷ 22.7.

Unit conversion

• 1000 cm³ = 1 dm³.

Formula triangle

• Formula triangle put n at the top and V and 22.7 below, then cover the quantity wanted.

Section 13 of 30Using gas volumes with masses

Chain

• Chain mass → moles (m/M) → moles of gas from the equation ratio → volume (n × 22.7).

Reverse chain

• Reverse chain volume → moles → moles of other substance → mass.

Gas states

• Gas states only gases count towards a final gas volume, since liquids and solids do not contribute.

Non-standard conditions

• Non-standard conditions molar volume cannot be used, so pV = nRT is applied with p in pascals, V in cubic metres and T in kelvin.

Section 14 of 30Introduction to volumetric analysis

Volumetric analysis

• Volumetric analysis uses the volume and concentration of one reactant to find the concentration of another.
  • (a standard solution)

Titration

• Titration the most common form, in which a measured volume (usually 20 or 25 cm³) is pipetted into a conical flask and the other solution is added from a burette until the end point.

Standard solution

• Standard solution a solution of accurately known concentration.

Calculations

• Calculations are developed from n = cV and the equation ratio.

Section 15 of 30More from the specification

1Percentage Yield and Theoretical Calculations

• Percentage yield is determined using × 100%, quantifying the efficiency of a preparation by comparing the mass of isolated product against the maximum calculated stoichiometric mass.
  • (actual yield / theoretical yield)

2Atom Economy in Sustainable Chemical Synthesis

• Atom economy measures the proportion of reactant atoms converted into the desired product: × 100%, emphasizing green chemical manufacturing with minimal waste.
  • (Mr of desired product / sum of Mr of all reactants)

3Thermal Decomposition of Carbonates

• When metal carbonates such as magnesium carbonate or calcium carbonate are heated, they undergo endothermic thermal decomposition to form the metal oxide and carbon dioxide gas: CaCO₃(s) → CaO(s) + CO₂(g).

4Stoichiometry in Metal and Oxide Reductions

• In metallurgical reactions, stoichiometric ratios determine the exact amounts of reducing agent required to extract pure copper or iron from copper oxide or iron oxide ore.

5Magnesium oxide

• White powder made when magnesium burns.

6Symbol equation

• Reaction written with formulae.

7Carbonate

• A compound containing the CO₃ group, such as calcium carbonate.

8Percentage yield

• Actual yield divided by theoretical yield, times 100.

Section 16 of 30Titration calculations

• Amount of solute
  • n = c × V with V in dm³.

1

Write the balanced equation for the reaction.

2

Calculate the moles of the known substance from its concentration and volume.

3

Use the mole ratio to find the moles of the unknown.

4

Calculate the unknown concentration as c = n ÷ V.

Volume needed

• Volume needed rearrange to volume = moles ÷ concentration.

Shortcut

• Shortcut when both acid and base are monoprotic the amounts are equal at equivalence, so c₁V₁ = c₂V₂ and cm³ can be used directly.

Not 1 : 1

• Not 1 : 1 the shortcut fails for other ratios, and H₂SO₄ is diprotic.

Section 17 of 30Finding molar mass and unknown formulae

1Method

• Method a titration of a known mass of an unknown acid or base gives the moles, and M = m ÷ n.

2Diluted samples

• Diluted samples if only part of a made-up solution (e.g. 25.0 cm³ of 100.0 cm³) is titrated, multiply the result by the dilution factor.

3Stages

• Stages find moles of titrant, deduce moles of the sample, scale up to the original solution, then calculate the molar mass.

Section 18 of 30Back titration

• When used
  • when the substance is insoluble, impure or reacts slowly, such as calcium carbonate in an antacid tablet or eggshell.

1

React the sample with a measured excess of standard acid.

2

Titrate the unreacted acid with standard sodium hydroxide.

3

Subtract the moles left over from the moles of acid originally added to find the moles that reacted.

4

Use the equation 2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O (2 : 1) to find the moles of CaCO₃.

5

Convert to mass and percentage purity.

• Common error
  • do not use the moles left over as the moles reacted.

Section 19 of 30Limiting and excess reactants

Limiting reactant

• Limiting reactant the one used up first.

Excess reactant

• Excess reactant the others, which remain after the reaction.

1

Calculate the moles of each reactant.

2

Divide each by its coefficient in the balanced equation.

3

The smallest value is the limiting reactant.

4

Find the excess left as moles available minus moles used.

• Product yield
  • is calculated from the limiting reactant.

Section 20 of 30Presenting titration calculations

1Equation first

• Equation first write the balanced equation and show the mole ratio.

2Data lines

• Data lines list n = c × V with the volume converted to dm³.

3Separate steps

• Separate steps show the ratio step on its own line.

4Back titration layout

• Back titration layout use three labelled lines: moles added, moles left over and moles reacted.

5Dilution

• Dilution state the factor and apply it at the end.

6Final answer

• Final answer give the right unit and significant figures, and check it is sensible, since a percentage purity above 100% signals an error.

Section 21 of 30More from the specification

More from the specification

Atom Economy Calculation

• Atom economy evaluates synthetic green efficiency: × 100%.
  • (molar mass of desired product / total molar mass of all reactants)

Empirical Formula Determination

• Using experimental masses of elements, dividing by relative atomic masses gives the simplest whole-number ratio of atoms.
  • (e.g. heating magnesium or copper in oxygen)

Avogadro Constant and Moles

• One mole of any chemical substance contains 6.02 × 10²³ elementary particles (atoms, molecules,
• or formula units), defined by the Avogadro constant (N_A).

Molar Volume of Ideal Gases

• At standard temperature and pressure, one mole of any ideal gas occupies an equal volume of 22.7 dm³,
• allowing stoichiometric calculations between reactant and product gases.

Theoretical yield

• The maximum mass of product predicted from the balanced equation.

Formula

• Symbols and numbers showing the atoms in a compound.

Oxygen

• A gas that relights a glowing splint.

Magnesium oxide

• White powder made when magnesium burns.

Symbol equation

• Reaction written with formulae.

Section 22 of 30Percentage yield

Theoretical yield

• Theoretical yield the maximum mass of product predicted from the limiting reactant and the balanced equation.

Actual yield

• Actual yield the mass actually obtained.

Formula

Formula percentage yield = (actual yield ÷ theoretical yield) × 100.

1

Write the balanced equation.

2

Identify the limiting reactant.

3

Calculate the moles of the limiting reactant.

4

Use the mole ratio to find the theoretical moles of product.

5

Convert to the theoretical mass.

6

Divide the actual mass by it and multiply by 100.

• Units
  • both yields must be in the same unit.

Section 23 of 30Why yields are below 100%

Reversible reactions

• Reversible reactions may not go to completion.

Side reactions

• Side reactions form other products.

Losses

• Losses material is lost during transfer, filtering and washing.

Impure reactants

• Impure reactants reduce the useful amount.

Above 100%

• Above 100% suggests the product is wet or impure.

Section 24 of 30Atom economy

Atom economy

Definition

Definition the proportion of the total mass of reactants that ends up in the useful product.

Formula

Formula atom economy = (molar mass of desired product ÷ sum of molar masses of all reactants) × 100.

Coefficients

• Coefficients include the coefficients in both the product mass and the reactant mass.

Mass form

• Mass form it can also be found as mass of desired product ÷ total mass of products × 100.

Basis

• Basis it comes from the balanced equation, not experimental masses.

Reduction of iron(III) oxide

• Fe₂O₃ + 3CO → 2Fe + 3CO₂ has a low atom economy because CO₂ is a by-product.

Showing working

• Showing working even for an addition reaction where it is obviously 100%, the working must be shown when asked.

Section 25 of 30Why atom economy matters

Low atom economy	High atom economy
More waste to manage	Little waste
Higher environmental and economic costs	Lower disposal costs and fewer raw materials
Typical of substitution and elimination with by-products	Typical of addition reactions (100%)

Green chemistry

• Green chemistry designs processes that reduce waste and the use of hazardous materials.

Ethanol example

• Ethanol example hydration of ethene has 100% atom economy, whereas substitution of bromoethane with NaOH produces NaBr waste.

Section 26 of 30Other measures of efficiency

1Reagents

• Reagents quantity and toxicity of catalysts and solvents.

2Energy

• Energy temperature and pressure needed.

3Catalyst recovery

• Catalyst recovery whether the catalyst can be reused.

4Number of steps

• Number of steps fewer steps usually waste less.

5By-products

• By-products whether they can be sold or reused.

6Caution

• Caution a high atom economy can still be wasteful if the yield is low or much solvent is used.

Section 27 of 30Applying yield and atom economy in industry

Waste handling

• Waste handling a process with low atom economy needs a plan for waste: by-products may be sold, recycled or neutralised.

Preferred reactions

• Preferred reactions addition reactions such as hydration of ethene are favoured for their 100% atom economy.

Raising yield

• Raising yield remove the product as it forms, use an excess of a cheap reactant, or choose conditions favouring the forward direction at equilibrium.

Catalyst

• Catalyst improves rate and allows a lower temperature, but not the equilibrium yield.

Judgement

• Judgement balance yield, atom economy, energy use and hazard.

Section 28 of 30More from the specification

1Electrostatic lattice of bonded atoms

• Electrostatic lattice of bonded atoms sharing delocalised valence electrons among metallic cations binds regular arrays of metallic atoms in robust lattices.

2Molar Volume of Gases at RTP

• Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules;
  • at room temperature and pressure (rtp), one mole of any gas occupies 24.0 dm³ (24,000 cm³).

3Gas Volume Stoichiometric Calculations

• In reactions of carbonates with acids, the volume of carbon dioxide gas evolved is calculated from moles using V = n × 24.0 dm³.
  • (e.g. calcium carbonate or magnesium carbonate)

4Solution Concentration Formula

• The concentration of a solute in aqueous solution is defined as the amount of substance in moles divided by solution volume in dm³ (c = n / V), expressed in mol dm⁻³.

5Stoichiometry of Acid-Carbonate Reactions

• Reacting metal carbonates with dilute acid produces a metal salt, water,
• and carbon dioxide (e.g. CuCO₃(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l) + CO₂(g)).

6Magnesium oxide

• White powder made when magnesium burns.

7Symbol equation

• Reaction written with formulae.

8Volume

• The space taken up by an object.

9Concentration

• Moles ÷ volume in dm³, or g/dm³.

10Carbonate

• A compound containing the CO₃ group, such as calcium carbonate.

11Oxygen

• A gas that relights a glowing splint.

12Substance

• A material with only one type of particle.

Section 29 of 30Key terms

Term	Meaning
Mole	The SI unit of amount of substance, containing 6.02 × 1023 entities.
Limiting reactant	The reactant that is used up first and limits the product.
Percentage yield	Actual yield divided by theoretical yield × 100.
Atom economy	The mass of desired product as a percentage of the total mass of reactants.
Standard solution	A solution of accurately known concentration.
Back titration	Titration of the excess reagent left after reaction with a sample.

Term	Meaning
Avogadro's law	Equal volumes of gases at the same T and p contain equal numbers of molecules.
Molar volume	The volume of one mole of gas, 22.7 dm3 mol−1 at STP.
STP	Standard temperature and pressure, 273 K and 100 kPa.
Volumetric analysis	Quantitative analysis by measuring volumes of solutions.
Standard solution	A solution of accurately known concentration.

Term	Meaning
Back titration	A titration of the excess reagent left after reaction with a sample.
Limiting reactant	The reactant that runs out first.
Monoprotic	Able to donate one proton per molecule.
Equivalence point	The point where exact stoichiometric amounts have mixed.
Percentage purity	The mass of pure substance divided by sample mass × 100.

Term	Meaning
Theoretical yield	The maximum product predicted from the balanced equation.
Percentage yield	Actual yield divided by theoretical yield × 100.
Atom economy	The mass of desired product as a percentage of the total mass of reactants.
Green chemistry	The design of processes that reduce waste and hazards.
By-product	An unwanted product of a reaction.

Section 30 of 30Exam technique

1

Using the mass ratio instead of the mole ratio — coefficients give moles.

2

Forgetting to convert cm³ to dm³ — divide by 1000 before n = cV.

3

Calculating the theoretical yield from the excess reactant — use the limiting reactant.

4

Calculating atom economy from experimental masses — use the balanced equation and include coefficients.

1

Using the molar volume of 22.7 dm³ mol⁻¹ at room temperature — it applies only at STP.

2

Including liquid water in the gas volume ratio — only gases count.

3

Applying Avogadro's law to volumes measured under different conditions — they must be the same.

4

Not converting cm³ to dm³ — divide by 1000.

1

Using c₁V₁ = c₂V₂ for reactions that are not 1 : 1 — use the mole ratio.

2

Forgetting that H₂SO₄ is diprotic — it reacts with two OH⁻.

3

In a back titration, using the moles left over as the moles reacted — subtract them from the total added.

4

Not applying the dilution factor — scale the titrated sample back to the original solution.

1

Calculating atom economy from experimental masses — use the balanced equation.

2

Forgetting stoichiometric coefficients in atom economy — multiply the M_r by the coefficient.

3

Assuming a high yield means a high atom economy — they measure different things.

4

Using the mass of the excess reactant for the theoretical yield — use the limiting reactant.