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

Section 1 of 30Balancing equations

Symbol equation

• Symbol equation shows the number and type of each atom in reactants and products.

Conservation

• Conservation atoms cannot be created or destroyed, so each atom count must match on both sides.

Rules

• Rules never change a formula;
  • put balancing numbers in front of formulae;
  • keep polyatomic ions together if they appear on both sides.

Combustion of organics

• Combustion of organics 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.

Section 2 of 30Moles and reacting masses

Moles and reacting masses

Moles

• Moles n = m / M, with 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)

Type of particle

• 1 mol of CaF₂ contains 1 mol of formula units, 1 mol of Ca²⁺ and 2 mol of F⁻.

Stoichiometry

• Stoichiometry the balanced equation gives the mole ratio of reactants and products.

Units

• Units any mass unit can be used if consistent, since reacting masses stay in proportion.

1

Write the balanced equation.

2

Convert the known mass to moles.

3

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

4

Convert to mass using m = n × M.

Section 3 of 30Gas volumes and Avogadro's law

Avogadro's law

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

Volume ratio

• Volume ratio the volume ratio of gases in an equation equals the mole ratio.

Molar volume

• Molar volume at STP (0 °C, 100 kPa) one mole of any gas occupies 22.7 dm³.

Equations

• V = n × 22.7 and n = V / 22.7, with cm³ converted to dm³.

Limiting gas

• Limiting gas if gas volumes are not in the equation ratio, divide each by its coefficient;
  • the smallest value is limiting.

Liquid products

• Liquid products water formed as a liquid is not counted in gas volumes.

Section 4 of 30Concentration and titration

Volumetric analysis

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

Apparatus

• Apparatus a volumetric or graduated pipette and a burette.

Concordant results

• Concordant results repeat until titres agree.

Concentration

• Concentration c = n / V,
  • so 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)

Shortcut

• Shortcut for monoprotic acid and base, C₁V₁ = C₂V₂, and cm³ can be used directly.

Averaging

• Averaging burette averages are rounded to two decimal places, such as 23.225 rounded to 23.23 cm³.

1

Measure a known volume (20 or 25 cm³) into a conical flask with a pipette.

2

Fill the burette with the other solution.

3

Add a few drops of indicator.

4

Add from the burette until the indicator just changes colour.

5

Repeat until concordant results are obtained.

Section 5 of 30Four-step method for solution calculations

1

Write the balanced equation.

2

Find moles of the known substance.

3

Use the equation to deduce the mole ratio.

4

Calculate the unknown mass, moles, concentration or volume.

1Back titration

• Back titration a known excess of reagent reacts with the sample, the leftover excess is titrated, and subtraction gives the amount that reacted.

2Scaling

• Scaling when only an aliquot is titrated, scale up to the full volume made.

Section 6 of 30Limiting and excess reactants

Limiting reactant

• Limiting reactant the reactant used up first; it decides the amount of product.

Excess reactant

• Excess reactant the reactant left over.

Method

• Method find moles of each reactant and divide by its coefficient; the smallest value is limiting.

Why it matters

• Why it matters theoretical yield and percentage yield depend on the limiting reactant.

Section 7 of 30Percentage yield

Theoretical yield

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

Experimental yield

Experimental yield the mass actually collected.

Equation

• Equation percentage yield = actual yield ÷ theoretical yield × 100.

Lower yield

• Lower yield because of side reactions, losses in transfers or purification, and incomplete or reversible reactions.

Over 100 %

• Over 100 % usually means an impure or wet product.

Units

• Units actual and theoretical yields must use the same unit.

Section 8 of 30Atom economy

1Definition

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

2Equation

• Equation atom economy = M_r of desired product ÷ sum of M_r of all reactants × 100, from the balanced equation.

3Addition reactions

• Addition reactions they have 100 % atom economy because there is only one product.

4Low atom economy

• Low atom economy much of the reactant mass becomes waste, adding environmental and economic cost.

5Yield versus atom economy

• Yield versus atom economy a reaction can have a high yield but low atom economy, or the reverse.

6Other measures

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

Section 9 of 30More from the specification

More from the specification

Limiting reactant stoichiometry

• Determining reactant moles to calculate theoretical mass yields.
  • (e.g. reacting magnesium or copper with hydrochloric acid)

Back titration analysis

• Reacting an impure solid with known excess hydrochloric acid,
  • (like calcium carbonate in marble)
• then titrating unreacted acid with standard sodium hydroxide.

Percentage purity calculations

• Found by dividing the determined mass of pure compound by the total mass of the sample and multiplying by 100%.

Gravimetric analysis of oxides

• Thermal decomposition of carbonates or synthesis of metal oxides to deduce empirical formulas.
  • (e.g. magnesium oxide)

Magnesium oxide

• White powder made when magnesium burns.

Section 10 of 30Avogadro's law and gas volume ratios

1Avogadro's law (1811)

• Avogadro's law (1811) equal volumes of gases at the same temperature and pressure contain the same number of molecules.

2Consequence

• Consequence the mole ratio of reacting gases equals their volume ratio.

3STP

• 0 °C (273 K) and 100 kPa; one mole of any gas occupies 22.7 dm³ mol⁻¹.

4Propane combustion

• C₃H₈ : O₂ : CO₂ is 1 : 5 : 3; water is liquid, so it is not counted.

5Not in ratio

• Not in ratio if the volumes are not in the coefficient ratio, the limiting reactant decides the product;
  • divide each volume by its coefficient and the smallest is limiting.

Section 11 of 30Calculations with molar gas volume

Volume

• Volume volume (dm³) = moles × 22.7 dm³ mol⁻¹.

Moles

• Moles moles = volume (dm³) ÷ 22.7 dm³ mol⁻¹.

Units

• Units convert cm³ to dm³ by dividing by 1000.

Formula triangle

• Formula triangle cover the wanted quantity and read the operation.

Section 12 of 30Titration method

Volumetric analysis

• Volumetric analysis uses a standard solution of known concentration to find the concentration of another solution.

Apparatus

• Apparatus a volumetric pipette and a burette.

1

Measure a known volume (usually 20 or 25 cm³) with a pipette into a conical flask.

2

Put the other solution in the burette.

3

Add a few drops of indicator.

4

Add from the burette portion by portion until the indicator just changes colour.

5

Repeat until concordant results are obtained and average them.

Section 13 of 30Four steps for concentration calculations

1

Write the balanced equation.

2

Find moles of the known substance, as concentration × volume in dm³ or mass ÷ M.

3

Use the equation to deduce the mole ratio.

4

Calculate the unknown: concentration = moles ÷ volume, or volume = moles ÷ concentration.

Section 14 of 30Shortcuts and precision

1Monoprotic acid and base

• Monoprotic acid and base moles at equivalence are equal, so C₁V₁ = C₂V₂.

2Volume units

• Volume units they cancel in this shortcut, so cm³ can be used directly.

3Rounding burette averages

• Rounding burette averages they are recorded to two decimal places, so 23.225 cm³ is rounded to 23.23 cm³.

4Significant figures

• Significant figures leave answers with the number of significant figures of the data.

Section 15 of 30Strategy for stoichiometry and gas questions

Pattern

• Pattern write the balanced equation, convert the known quantity to moles, use the mole ratio, then convert to the required quantity.

Gas ratio

• Gas ratio volume ratios replace mole ratios only when all gases are at the same temperature and pressure.

Molar volume

• 22.7 dm³ mol⁻¹ applies at STP; use the ideal gas equation for other conditions.

Units

• Units use the same unit when using ratios, and dm³ with molar volume or concentration.

Working

• Working show units at each stage to earn method marks.

Section 16 of 30More from the specification

1Covalent bonding between non-metal atoms

• Covalent bonding between non-metal atoms electrostatic attraction between shared valence electron pairs and positively charged nuclei binds non-metallic atoms stably.

2Limiting reactants and theoretical yield

• When calculating the amount of products formed, the limiting reactant is completely consumed while excess reactants remain unreacted, dictating the maximum theoretical yield.

3Percentage yield in synthesis

• Percentage yield compares the actual experimental mass obtained to the theoretical yield: Percentage Yield = * 100%, accounting for side reactions and incomplete recoveries.
  • (actual yield / theoretical yield)

4Atom economy calculations

• Atom economy evaluates synthetic efficiency: Atom Economy = * 100%, promoting sustainable green chemistry with minimal chemical waste.
  • (molar mass of desired product / total molar mass of all reactants)

5Stoichiometry of metal carbonates

• Heating metal carbonates (e.g. copper carbonate, calcium carbonate,
• or magnesium carbonate) leads to thermal decomposition, reacting with acid or decomposing into metal oxides and carbon dioxide in presence of oxygen.

6Limiting reactants and theoretical yield

• When calculating the amount of products formed, the limiting reactant is completely consumed while excess reactants remain unreacted, dictating the maximum theoretical yield.

7Percentage yield in synthesis

• Percentage yield compares the actual experimental mass obtained to the theoretical yield: Percentage Yield = * 100%, accounting for side reactions and incomplete recoveries.
  • (actual yield / theoretical yield)

8Atom economy calculations

• Atom economy evaluates synthetic efficiency: Atom Economy = * 100%, promoting sustainable green chemistry with minimal chemical waste.
  • (molar mass of desired product / total molar mass of all reactants)

9Stoichiometry of metal carbonates

• Heating metal carbonates (e.g. copper carbonate, calcium carbonate,
• or magnesium carbonate) leads to thermal decomposition, reacting with acid or decomposing into metal oxides and carbon dioxide in presence of oxygen.

10Percentage yield

• Actual yield divided by theoretical yield, times 100.

11Magnesium oxide

• White powder made when magnesium burns.

12Symbol equation

• Reaction written with formulae.

13Theoretical yield

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

14Excess

• More than enough of a reactant.

15Carbonate

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

16Oxygen

• A gas that relights a glowing splint.

Section 17 of 30Molar mass from titration data

1

Write the equation, for example HA + NaOH → NaA + H₂O.

2

Find moles of NaOH from volume (dm³) × concentration.

3

Use the mole ratio to get moles of HA in the aliquot.

4

Scale up to the whole volume of solution made.

5

Calculate molar mass = mass ÷ moles.

• Aliquot
  • a measured portion of the solution; scale up by the ratio of the total volume to the aliquot volume.

Section 18 of 30Back titration

Purpose

• Purpose finds the amount of an unknown substance indirectly.

Principle

• Principle the unknown reacts with a known excess of reagent, and the leftover excess is titrated.

Useful for

• Useful for insoluble or slowly reacting solids such as marble or antacid tablets, or when the direct reagent has no sharp end point.

Excess reagent

• Excess reagent it must be measured accurately with a pipette or burette.

1

Calculate moles of the excess reagent added originally.

2

Calculate moles of the excess left over from the titration.

3

Subtract to get moles that reacted with the sample.

4

Use the mole ratio to find moles of the unknown, then mass and percentage.

Section 19 of 30Limiting and excess reactants

Limiting reactant

• Limiting reactant the reactant used up first, which determines how much product forms.

Excess reactant

• Excess reactant left over after the reaction.

Limiting and excess reactants

Calculate moles of each reactant.

Divide each by its coefficient in the balanced equation.

The smallest value is the limiting reactant.

Section 20 of 30Theoretical and percentage yield

Theoretical yield

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

Experimental yield

• Experimental yield the mass actually collected, lower because of side reactions, losses in transfers or purification, and incomplete reaction.

Percentage yield

• × 100.
  • (actual yield ÷ theoretical yield)

Requirements

• Requirements both yields must have the same unit, and the theoretical yield must use the limiting reactant.

Section 21 of 30Choosing and judging titrations

Direct titration

• Direct titration suitable when the unknown reacts quickly and completely with the titrant.

Back titration

• Back titration chosen for insoluble substances, slow reactions or poor end points.

Precision

• Precision limited by the least precise value;
  • burette readings of ±0.05 cm³ on a titre of about 24 cm³ give a percentage uncertainty of about 0.4 %.

Pipette

• Pipette adds its own uncertainty.

Scaling factor

• Scaling factor a factor such as 100.0 / 25.0 is exact and does not change the percentage uncertainty.

Section 22 of 30More from the specification

1Percentage Atom Economy Defined

• Atom economy measures the proportion of reactant mass converted into desired products, calculated as × 100%.
  • (molar mass of desired product / total molar mass of all reactants)

2Atom Economy vs Percentage Yield

• A reaction may have a high percentage yield but a low atom economy if significant mass ends up as unwanted chemical by-products.
  • (actual product mass / theoretical mass)

3Industrial Reduction of Metal Oxides

• In the extraction of metals, blast furnaces reduce iron(III) oxide with carbon monoxide,
  • (Fe₂O₃ + 3CO → 2Fe + 3CO₂)
• while copper(II) oxide is reduced with hydrogen or carbon to obtain pure metal.

4Stoichiometric Calculations with Metal Carbonates

• Thermal decomposition of group 2 carbonates generates metal oxides and carbon dioxide,
  • (e.g. magnesium carbonate MgCO₃ → MgO + CO₂ or calcium carbonate)
• allowing theoretical yield calculations.

5Magnesium oxide

• White powder made when magnesium burns.

6Carbonate

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

7Formula

• Symbols and numbers showing the atoms in a compound.

8Oxygen

• A gas that relights a glowing splint.

Section 23 of 30Method for percentage yield

1

Write the balanced equation.

2

Calculate moles of the limiting reactant.

3

Use the mole ratio to get the maximum moles of product.

4

Convert to the theoretical mass using mol × M.

5

Calculate percentage yield = actual ÷ theoretical × 100.

Limiting reactant

• Limiting reactant always base the theoretical yield on it.

Comparison

• Comparison percentage yield compares the product obtained with the maximum possible.

Section 24 of 30Atom economy

Definition

• Definition how efficiently a reaction turns reactants into the desired product, as the proportion of total reactant mass ending up in it.

Equation

• Equation atom economy = M_r of desired product ÷ sum of M_r of all reactants × 100;
  • this equation is not in the data booklet.

Alternative

• Alternative mass of desired product ÷ total mass of all products × 100.

Source

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

High atom economy

• High atom economy means less waste and a more sustainable process.

Section 25 of 30Addition reactions and waste

Addition reactions

• Addition reactions always 100 % atom economy because only one product forms, as in CH₂=CH₂ + Br₂ → CH₂BrCH₂Br.

Working

• Working still show the calculation if the question asks for it.

Low atom economy

• Low atom economy much reactant mass becomes waste or by-products, needing more resources and waste management at greater cost.

Section 26 of 30Comparing routes

1Hydration of ethene

• C₂H₄ + H₂O → C₂H₅OH has 100 % atom economy, as all atoms are in the product.

2Substitution of bromoethane

• C₂H₅Br + NaOH → C₂H₅OH + NaBr has lower atom economy because NaBr is waste.

3Other measures of efficiency

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

Section 27 of 30Yield versus atom economy in industry

1Percentage yield

• Percentage yield shows how successful a particular experiment was, including losses and incomplete conversion.

2Atom economy

• Atom economy a property of the balanced equation showing how much reactant mass could ever end up in the useful product.

3High yield, poor atom economy

• High yield, poor atom economy substitution reactions that produce a salt as a by-product.

4Perfect atom economy, low yield

• Perfect atom economy, low yield an incomplete equilibrium reaction.

5Industry preference

• Industry preference high atom economy routes because by-products need disposal or markets, raw materials are wasted and extra separation costs energy.

6Green chemistry

• Green chemistry a guiding principle;
  • recycling unreacted material, using catalysts and continuous processes raises effective yield.

Section 28 of 30More from the specification

1Stoichiometric mole calculations

• Amount in moles is calculated from mass (n = m / M) or solution volume and molarity (n = c * V), relating reactant and product amounts via balanced equation ratios.
  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)

2Limiting reactants and percentage yield

• The limiting reactant is completely consumed first and determines theoretical product yield; percentage yield = * 100.
  • (actual yield / theoretical yield)

3Atom economy in green chemistry

• Atom economy assesses the synthetic efficiency of a reaction: atom economy = * 100.
  • (molar mass of desired product / total molar mass of all reactants)

4Molar volume of reacting gases

• Molar volume of reacting gases at standard temperature and pressure, equal gas volumes contain identical molar amounts according to Avogadro's gas law.

5Titration of acid and carbonate

• Titration of acid and carbonate measuring solution volume and concentration allows determination of reacting molar amounts of hydrochloric acid and metal carbonate.

6Excess reagent in copper reactions

• Excess reagent in copper reactions reacting copper carbonate with excess dilute acid ensures complete conversion of the limiting reactant into soluble product.

7Substance

• A material with only one type of particle.

8Magnesium oxide

• White powder made when magnesium burns.

9Molar volume

• Volume of one mole of gas24 dm³ at RTP.

10Molar ratio

• The ratio of the moles of the elements or substances in a formula or equation.

11Volume

• The space taken up by an object.

12Concentration

• Moles ÷ volume in dm³, or g/dm³.

13Excess

• More than enough of a reactant.

14Carbonate

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

15Formula

• Symbols and numbers showing the atoms in a compound.

16Oxygen

• A gas that relights a glowing splint.

Section 29 of 30Key terms

Term	Meaning
Stoichiometry	The mole ratio of reactants and products in a balanced equation.
Limiting reactant	The reactant that is used up first.
Percentage yield	Actual yield as a percentage of theoretical yield.
Atom economy	Mass of desired product as a percentage of total reactant mass.
Standard solution	A solution of accurately known concentration.
Back titration	Titrating the excess of a reagent left after reaction with the sample.

Term	Meaning
Avogadro's law	Equal volumes of gases at the same T and p hold equal numbers of molecules.
Molar volume	22.7 dm3 mol−1 at STP.
Standard solution	A solution of accurately known concentration.
Concordant titres	Titres that agree closely.
Titre	The volume added from the burette.

Term	Meaning
Aliquot	A measured portion of a solution.
Back titration	Titration of the excess reagent after reaction with a sample.
Excess reactant	The reactant remaining after the limiting reactant is used up.
Theoretical yield	The maximum possible mass of product.
Experimental yield	The mass of product actually obtained.

Term	Meaning
Percentage yield	Actual yield as a percentage of theoretical yield.
Atom economy	Mass of desired product as a percentage of total reactant mass.
By-product	A product other than the desired one.
Green chemistry	Chemistry designed to reduce waste and environmental harm.

Section 30 of 30Exam technique

1

Using the ratio of masses instead of the ratio of moles from the equation.

2

Forgetting to convert cm³ to dm³ in concentration calculations.

3

Basing the theoretical yield on the excess reactant.

4

Calculating atom economy from experimental masses rather than the balanced equation.

1

Using 22.7 dm³ mol⁻¹ at room temperature — it applies at STP.

2

Including liquid water in gas volume ratios.

3

Forgetting to convert cm³ to dm³ with molar volume or concentration.

4

Using C₁V₁ = C₂V₂ when the acid and base have different mole ratios.

1

Forgetting to scale from the aliquot to the whole solution.

2

Using moles of acid added as moles reacted — in back titration subtract the excess.

3

Dividing the moles by the wrong coefficient when finding the limiting reactant.

4

Basing percentage yield on the excess reactant.

1

Using the experimental masses to calculate atom economy.

2

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

3

Using the molar mass of one reactant instead of the sum of all reactants in atom economy.

4

Forgetting to show working for an addition reaction with 100 % atom economy.