Amount of Substance

Section 1 of 32Unified atomic mass unit

Reason

• Reason atoms are too small to weigh directly, so masses are compared with a standard

Definition

• Definition the unified atomic mass unit, u (or Da), is one twelfth of the mass of a carbon-12 atom

Value

• 1 u = 1.66 × 10^-27 kg

Section 2 of 32Relative atomic mass

Relative atomic mass, A_r = the ratio of the average mass of the atoms of an element to the unified atomic mass unit

Basis

• Basis it uses the average mass of the isotopes of the element

Units

• Units it has none because it is a ratio

Formula

• A_r = average mass of one atom of X ÷ one twelfth of the mass of a carbon-12 atom

Calculation

• Calculation it can also be found from isotopic abundances

Common definition

• Common definition the mass of an atom compared with one twelfth of a carbon-12 atom

Section 3 of 32Relative molecular and formula mass

Relative molecular mass, M_r = the ratio of the weighted average mass of a molecule to the unified atomic mass unit, with no units

Calculation

• Calculation add the relative atomic masses of all atoms in one molecule

Simplest formula

• Simplest formula for giant structures the formula unit is used, such as SiO₂

Relative formula mass

• Relative formula mass the term used for ionic compounds, calculated in the same way

Substance	Calculation	Mr
H2	2 × 1.0	2.0
H2O	(2 × 1.0) + 16.0	18.0
SiO2	28.1 + (2 × 16.0)	60.1
K2CO3	(2 × 39.1) + 12.0 + (3 × 16.0)	138.2
Ca(OH)2	40.1 + (2 × 16.0) + (2 × 1.0)	74.1
(NH4)2SO4	(2 × 14.0) + (8 × 1.0) + 32.1 + (4 × 16.0)	132.1

Section 4 of 32The mole and the Avogadro constant

The mole and the Avogadro constant

Avogadro constant, NA

• Avogadro constant, NA the number of particles in a mole, 6.02 × 10²³ mol^-1, applying to atoms, molecules, ions and electrons

Mole = the amount of substance containing the same number of fundamental units as there are atoms in exactly 12.00 g of carbon-12

Molar mass

• Molar mass one mole of an element has a mass equal to its A_r in grams, so a mole of carbon is 12 g and a mole of water is 18 g

Particles

• Particles number of particles = moles × N_A
  What the symbols mean: N = number of particles · n = amount (mol) · N_A = Avogadro constant (6.02 × 10²³ mol⁻¹)

Exam note

• Exam note the value of the Avogadro constant need not be recalled

Section 5 of 32Particles in a mole

1 mole of	Atoms	Molecules	Mass / g
Na	6.02 × 1023	—	23.0
H2	1.204 × 1024	6.02 × 1023	2.0
NaCl	1.204 × 1024 (total)	6.02 × 1023 formula units	58.5

• Reasoning
  • H₂ has 2 atoms per molecule, and NaCl has one Na and one Cl per formula unit, so total atoms double

Section 6 of 32Moles, mass and molar mass

Equation

• Equation moles = mass ÷ molar mass, and mass = moles × molar mass

Example

• 0.250 mol of zinc (65.4 g mol^-1) has a mass of 16.4 g

Example

• 2.64 g of a compound with M_r 342.3 is 7.71 × 10^-3 mol

Formula triangle

• Formula triangle cover the quantity wanted to see the calculation

Section 7 of 32Moles, concentration and volume

Equation

• Equation moles = concentration × volume in dm³, with concentration in mol dm^-3

Use

• Use titration calculations, as in Required Practical 1

Example

• 80.0 g of NaOH (M_r 40.0) is 2 mol, and in 0.5 dm³ the concentration is 4 mol dm^-3

Units

• Units convert cm³ to dm³ by dividing by 1000

Section 8 of 32Kinetic theory and ideal gases

Assumptions of the kinetic theory

• Gas molecules move very fast and randomly.
• They have negligible volume.
• They do not attract or repel each other.
• Collisions are elastic, so no kinetic energy is lost.
• Temperature relates to the average kinetic energy.

Ideal gas = a gas that obeys these assumptions, whereas real gases fit them only approximately

Volume

• Volume it depends on pressure and temperature,
  • and at constant pressure volume is directly proportional to temperature because particles move further apart as they gain kinetic energy

Section 9 of 32Limitations of the ideal gas law

Conditions

• Conditions real gases deviate at very low temperature and high pressure, when molecules are close together

Attractive forces

• Attractive forces induced or permanent dipole forces exist between molecules
• They pull molecules away from the container wall, so the pressure is lower than expected.

Molecular volume

• Molecular volume it is not negligible, so the volume of the gas is smaller than expected for an ideal gas

Section 10 of 32The ideal gas equation

• Equation
  • PV = nRT
    What the symbols mean: p = pressure (Pa) · V = volume (m³) · n = amount of gas (mol) · R = 8.31 J mol⁻¹ K⁻¹ · T = temperature (K = °C + 273)

Symbol	Quantity	Unit
P	Pressure	Pa
V	Volume	m3
n	Amount of gas	mol
R	Gas constant (8.31)	J K-1 mol-1
T	Temperature	K

Conversions

• Conversions kPa to Pa by multiplying by 1000, °C to K by adding 273, and dm³ to m³ by dividing by 1000

Example

• 0.781 mol of oxygen at 220 kPa and 21 °C (294 K) has V = nRT ÷ P = 0.00867 m³ = 8.67 dm³

Section 11 of 32Finding molar mass of a gas from PV = nRT

1

Rearrange to n = PV ÷ RT.

2

Convert units: kPa to Pa, cm³ to m³ and °C to K by adding 273.

3

Calculate the moles of gas.

4

Use molar mass = mass ÷ moles.

Example

• 6.39 g of gas in 1000 cm³ at 300 kPa and 23 °C (296 K) is 0.12 mol, so the molar mass is 53.25 g mol^-1

Skill

• Skill rearrange the equation for any quantity, and take great care with units

Section 12 of 32Empirical and molecular formulae

Molecular formula = the formula showing the number and type of each atom in a molecule, such as C₂H₄O₂ for ethanoic acid

Empirical formula = the simplest whole-number ratio of elements in a molecule or formula unit, such as CH₂O for ethanoic acid

Pattern

• Pattern organic molecules often differ in empirical and molecular formula, simple inorganic molecules less often,
  • and ionic compounds always have the same

Source of data

• Source of data masses of each element, or percentage composition by mass

Section 13 of 32Calculating an empirical formula

1

Write the mass (or percentage) of each element.

2

Divide each by its A_r to give moles.

3

Divide all by the smallest value to find the ratio.

4

Multiply to turn non-integer ratios into whole numbers, such as ×3 for 1 : 1.33.

5

Write the formula.

From masses

• 2.72 g C and 7.28 g O give 0.227 and 0.455 mol, ratio 1 : 2, so CO₂

From percentages

• 90.0% C and 10.0% H give 7.5 and 10.0 mol, ratio 1 : 1.33, multiplied by 3 to give C₃H₄

Section 14 of 32Calculating a molecular formula

1

Find the empirical formula.

2

Find the relative formula mass of the empirical formula.

3

Divide the M_r of the compound by that value.

4

Multiply every number in the empirical formula by the result.

Example

• 24.2% C, 4.1% H and 71.7% Cl give empirical formula CH₂Cl, with relative mass 49.5

Result

• Result an M_r of 99.0 gives a multiplier of 2, so the molecular formula is C₂H₄Cl₂

Section 15 of 32Balancing equations

Symbol equation

• Symbol equation it shows the number and type of each atom using symbols, and a word equation uses only words

Conservation

• Conservation atoms cannot be created or destroyed, so each side must have the same number of each atom

Rules

• Never change the formulae.
• Put balancing numbers in front of formulae.
• In combustion of organic compounds, balance carbon first, then hydrogen, then oxygen.

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.

4

Add state symbols.

State symbols

• (s) solid, (l) liquid, (g) gas and (aq) aqueous

Example

• 2Mg(s) + O₂(g) → 2MgO(s), showing 2 mol Mg reacts with 1 mol O₂ to form 2 mol MgO

Section 16 of 32Ionic equations

Dissociation

• Dissociation in aqueous solution ionic compounds split into ions

Spectator ions

• Spectator ions ions that take no part in the reaction

Ionic equation

• Ionic equation it shows only the ions or particles that react, with spectator ions cancelled

1

Write the full balanced equation with state symbols.

2

Split aqueous ionic compounds into ions.

3

Cancel the ions that appear unchanged on both sides.

4

Write what remains.

• Example
  • Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s) becomes Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with SO₄^2- as the spectator

Section 17 of 32Reacting masses

Purpose

• Purpose to find how much of each reactant exactly reacts, preventing waste

Stoichiometry

• Stoichiometry the balanced equation gives the ratio of moles of reactants and products

1

Write the balanced equation.

2

Find the molar masses.

3

Calculate moles of the given substance using n = mass ÷ M_r.

4

Use the equation ratio to find moles of the required substance.

5

Calculate its mass using moles × M_r.

• Example
  • 7.5 g of Mg (24.3) is 0.3086 mol, the ratio 2Mg : 2MgO is 1 : 1, so the mass of MgO (40.3) is 12.44 g

Section 18 of 32Molar gas volume

Value

• Value at room temperature and pressure, one mole of any gas occupies 24.0 dm³

Conditions

• Conditions room temperature is 20 °C and room pressure is 100 000 Pa

Alternative

• Alternative if RTP is not stated, use PV = nRT

Equations

• Equations volume (dm³) = moles × 24.0, and moles = volume (dm³) ÷ 24.0

Section 19 of 32Gas volume and moles calculations

• Equations
  • volume = moles × 24.0 dm³, and moles = volume ÷ 24.0
    What the symbols mean: V = volume of gas at r.t.p. (dm³) · n = amount (mol) · 24 dm³ mol⁻¹ = molar gas volume at room temperature and pressure

Gas	Moles	Volume / dm3
Hydrogen	3.0	72.0
Carbon dioxide	0.25	6.0
Oxygen	5.4	129.6
Ammonia	0.02	0.48

Gas	Volume / dm3	Moles
Methane	225.6	9.4
Carbon monoxide	7.2	0.30
Sulfur dioxide	960	40

Section 20 of 32Percentage yield

1Reasons for less than 100%

• Reasons for less than 100% other reactions occur at the same time, the reaction does not go to completion,
  • or reactants and products are lost to the atmosphere

2Equation

• Equation percentage yield = × 100
  • (actual yield ÷ theoretical yield)

3Actual yield

• Actual yield the moles or mass obtained experimentally

4Predicted (theoretical) yield

• Predicted (theoretical) yield the moles or mass calculated from the equation

1

Write the balanced equation.

2

Calculate moles of the limiting reactant.

3

Use the ratio to find the theoretical moles and mass of product.

4

Divide the actual mass by the theoretical mass and multiply by 100.

• Example
  • 6.54 g of zinc is 0.10 mol, giving a theoretical 6.35 g of copper, and 4.80 g obtained is 75.6%

Section 21 of 32Atom economy

Definition

• Definition it shows how much of the mass of the reactants becomes the desired product, with the rest wasted

Equation

• Equation percentage atom economy = (M_r of desired product ÷ sum of M_r of all reactants) × 100

Source

• Source it is found directly from the balanced equation

Addition reactions

• Addition reactions they have 100% atom economy because all atoms form the desired product, as does any reaction with one product

Mass version

• Mass version divide the mass of desired product by the total mass of reactants

Importance

• Importance high atom economy means fewer resources used and less waste, which supports sustainable development

Section 22 of 32Atom economy comparisons and calculation

Atom economy comparisons and calculation

Making ethanol

• Hydration of ethene, C₂H₄ + H₂O → C₂H₅OH, has 100% atom economy.
• Substitution of bromoethane, C₂H₅Br + NaOH → C₂H₅OH + NaBr, wastes NaBr.

Blast furnace, Fe₂O₃ + 3CO → 2Fe + 3CO₂

• Desired product 2 × 55.8, reactants 159.6 + (3 × 28.0).
• Atom economy is 45.8%, and the M_r of CO₂ is not needed.

Section 23 of 32Concentration and volumes of solutions

Concentration

• Concentration the amount of solute dissolved in a solvent to make 1 dm³ of solution

Equation

• Equation concentration (mol dm^-3) = moles of solute ÷ volume of solution (dm³)

Concentrated and dilute

• Concentrated and dilute a concentrated solution has a high concentration and a dilute solution a low one

Points to remember

• Convert mass in grams to moles.
• Convert cm³ to dm³.

Mass of solute

• Mass of solute moles = concentration × volume (dm³), then mass = moles × molar mass

Section 24 of 32Reaction calculations in solution

1

Write the balanced equation.

2

Calculate moles of the substance with known mass or concentration and volume.

3

Use the equation ratio to find moles of the other substance.

4

Calculate volume = moles ÷ concentration, or concentration = moles ÷ volume.

Volume example

• 2.5 g CaCO₃ (100) is 0.025 mol and needs 0.05 mol HCl for CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, so 0.05 dm³ of 1.0 mol dm^-3 acid

Concentration example

• 25.0 cm³ of 0.050 mol dm^-3 Na₂CO₃ is 0.00125 mol and needs 0.00250 mol HCl, so 20.0 cm³ of acid is 0.125 mol dm^-3

Section 25 of 32Required Practical 1 Part A: making a volumetric solution

Required Practical 1 Part A: making a volumetric solution

Aim

• Aim to prepare 250 cm³ of 0.100 mol dm^-3 sodium hydrogensulfate, NaHSO₄ (M_r 120.1)

Apparatus

• Apparatus weighing bottle, 250 cm³ volumetric flask, solid NaHSO₄, filter funnel, spatula, deionised water, 250 cm³ beaker, glass rod and a balance reading to 2 or 3 decimal places

Mass needed

• 0.100 mol dm^-3 is 12.01 g per dm³, so 250 cm³ needs 3.0025 g (about 3.02 g)

1

Calculate the mass of solid needed.

2

Weigh out about that mass into the weighing bottle and record the exact mass.

3

Transfer the solid to the beaker.

4

Rinse the weighing bottle with deionised water and add the washings to the beaker.

5

Add about 100 cm³ of deionised water and stir until all the solid dissolves.

6

Transfer the solution to the volumetric flask.

7

Rinse the beaker and glass rod and add the washings to the flask.

8

Make up to the graduation mark.

9

Stopper and invert to mix.

10

Calculate the exact concentration from the exact mass.

Required Practical 1 Part A: making a volumetric solution

Source of error

• Source of error the most common error is loss of solid when transferring it between vessels

Result

• Result the only measurement is the mass, recorded to the balance’s number of decimal places

Section 26 of 32Exact concentration of the standard solution

Exact concentration of the standard solution

Record the exact mass of solid weighed.

Calculate moles = mass ÷ M_r.

Divide by the volume in dm³.

• Example
  • 3.02 g of NaHSO₄ (M_r 120.1) is 0.0251 mol, and in 0.250 dm³ the concentration is 0.1004 mol dm^-3

Section 27 of 32Mass of a hydrated solute

Principle

• Principle use the M_r of the hydrated formula, including the water of crystallisation

Method

• Method moles = concentration × volume (dm³), then mass = moles × M_r

Example

• 250 cm³ of 0.050 mol dm^-3 NaHSO₄·H₂O needs 0.0125 mol, and with M_r 138.1 the mass is 1.73 g to 3 significant figures

Section 28 of 32Required Practical 1 Part B: titration set-up

Required Practical 1 Part B: titration set-up

Aim

• Aim to find the concentration of sodium hydroxide solution using a standard sodium hydrogensulfate solution

Apparatus

• Apparatus burette, stand and clamp, 25 cm³ volumetric pipette with filler, 250 cm³ conical flasks, funnel, wash bottle, phenolphthalein, 150 cm³ standard NaHSO₄ solution and 150 cm³ NaOH of unknown concentration

Section 29 of 32Required Practical 1 Part B: method

1

Rinse the burette with the standard solution, then fill it.

2

Rinse a conical flask with deionised water.

3

Rinse the pipette with the sodium hydroxide solution.

4

Transfer exactly 25.0 cm³ of sodium hydroxide into the flask.

5

Add two to three drops of phenolphthalein.

6

Construct a results table and record the initial burette reading to the appropriate precision.

7

Add the standard solution slowly with swirling, and dropwise near the end-point where the indicator changes colour.

8

Record the final reading and calculate the volume used.

9

Repeat until two results are concordant, within 0.1 cm³, doing at least three titrations.

10

Calculate the mean volume and then the concentration of sodium hydroxide, showing working.

• Practical points
  • A rough titration first shows the approximate end-point, so the burette can be run to within 2–3 cm³ and then added dropwise.
  • Use only a few drops of indicator, because indicators are weak acids and can affect results.

Section 30 of 32Required Practical 1 Part B: results and calculation

• Table
  • columns for rough, run 1, run 2 and run 3 with initial reading, final reading and titre volume

1

Identify the concordant results, within 0.1 cm³.

2

Calculate the mean titre.

3

Calculate moles of the standard solution using concentration × volume (dm³).

4

Use the equation ratio to find moles of the unknown, as NaHSO₄ + NaOH → Na₂SO₄ + H₂O is 1 : 1.

5

Divide by the volume in dm³ to get the concentration.

Required Practical 1 Part B: results and calculation

Example

• Example titres 25.10 and 25.00 cm³ give a mean of 25.05 cm³,
• so 0.002505 mol of 0.100 mol dm^-3 acid reacts with 25.0 cm³ of alkali,
• giving 0.1002 mol dm^-3 NaOH

Careful

• Careful convert cm³ to dm³ before the final step

Section 31 of 32Key terms

Term	Meaning
Relative atomic mass	The average mass of an element’s atoms compared with one twelfth of a carbon-12 atom.
Relative molecular mass	The weighted average mass of a molecule compared with one twelfth of a carbon-12 atom.
Mole	The amount of substance containing 6.02 × 1023 particles.
Avogadro constant	The number of particles in one mole.
Ideal gas	A gas that obeys the kinetic theory assumptions exactly.
Gas constant	R, 8.31 J K-1 mol-1.

Term	Meaning
Empirical formula	The simplest whole-number ratio of atoms of each element in a compound.
Molecular formula	The actual number of atoms of each element in a molecule.
Spectator ion	An ion that does not take part in a reaction.
Ionic equation	An equation showing only the ions or particles that react.
Stoichiometry	The ratio of moles of substances in a balanced equation.
Molar gas volume	24.0 dm3 per mole at room temperature and pressure.

Term	Meaning
Percentage yield	Actual yield divided by theoretical yield, times 100.
Atom economy	The proportion of the mass of reactants that becomes the desired product.
Volumetric solution	A solution of accurately known concentration made in a volumetric flask.
Concentration	The amount of solute per dm3 of solution.
Theoretical yield	The maximum product calculated from the balanced equation.

Term	Meaning
Standard solution	A solution of accurately known concentration.
Titre	The volume of solution added from the burette to reach the end-point.
Concordant results	Titres within 0.1 cm3 of each other.
End-point	The point at which the indicator changes colour.
Hydrated salt	A salt containing water of crystallisation in its formula.

Section 32 of 32Exam technique

1

Using the Avogadro constant as grams per mole — it is per mole, and the mass of a mole is the molar mass.

2

Using cm³ in moles = concentration × volume — convert to dm³.

3

Using kPa or dm³ in PV = nRT — use Pa and m³.

4

Saying A_r has units — it is a ratio.

1

Using cm³ directly in PV = nRT — convert to m³.

2

Forgetting to convert °C to K — add 273.

3

Changing a subscript to balance an equation — only change the coefficients.

4

Leaving a non-integer ratio in an empirical formula — multiply to get whole numbers.

1

Using the molar volume 24.0 when RTP is not stated — use PV = nRT.

2

Calculating atom economy using the mass of a non-desired product — only the desired product goes on top.

3

Using actual yield on the bottom of percentage yield — theoretical yield goes on the bottom.

4

Not rinsing the beaker and weighing bottle — loss of solid makes the concentration inaccurate.

1

Using the rough titre in the mean — use only concordant results.

2

Using the unhydrated M_r for a hydrated solid — include the water of crystallisation.

3

Adding too much indicator — it is a weak acid and affects the result.

4

Forgetting the mole ratio from the equation — the 1 : 1 here may differ elsewhere.