The mole and the Avogadro constant — Combined Science 0653 - Chemistry Revision Notes & Topical Questions

Section 1 of 26Reacting mass calculations

Reacting mass calculations

Use the information given to find the amount of the known substance in moles.

Use the ratio in the balanced equation to find the moles of the wanted substance.

Convert moles to grams using the relative atomic or formula mass.

• Molar ratio
  • the coefficients of the balanced equation give the ratio of moles.

Section 2 of 26Limiting reactant

Limiting reactant

• Limiting reactant the reactant used up first, which limits the amount of product.

Excess reactant

• Excess reactant the reactant left over when the reaction stops.

Product

• Product the amount of product is directly proportional to the amount of limiting reactant.

Section 3 of 26Finding the limiting reactant

1

Convert the mass of each reactant to moles by dividing by its molar mass.

2

Write the balanced equation and find the molar ratio.

3

Compare the moles of each reactant with the ratio required.

4

The reactant that runs out first is limiting; use it to find the maximum product.

Section 4 of 26Concentration of solutions

Solute

• = substance that dissolves.

Solvent

• = liquid in which the solute dissolves.

Solution

• = mixture of solute and solvent.

Concentration

• = amount of solute in a given volume of solution.
• The more solute in a given volume, the greater the concentration.

Section 5 of 26Concentration formulae

Quantity	Formula	Unit
Concentration (mass)	mass of solute ÷ volume in dm3	g/dm3
Concentration (moles)	moles ÷ volume in dm3	mol/dm3
Moles	concentration × volume in dm3	mol
g/dm3 to mol/dm3	divide by molar mass	mol/dm3
mol/dm3 to g/dm3	multiply by molar mass	g/dm3

• Volume units
  • divide cm³ by 1000 to get dm³.

Section 6 of 26More from the specification

More from the specification

Thermal decomposition to metal oxide

• Thermal decomposition to metal oxide heating copper or calcium carbonates produces solid metal oxide alongside gaseous carbon dioxide in quantitative stoichiometry.

Empirical Formula from Percentage Composition

• The empirical formula represents the simplest whole-number ratio of atoms of each element in a compound, calculated by dividing the percentage by mass of each element by its relative atomic mass (Ar).

Molecular Formula Determination

• The molecular formula specifies the actual number of atoms of each element in a molecule, found by dividing the relative molecular mass (Mr) of the compound by the empirical formula mass: Molecular formula = n × .
  • (empirical formula)

Avogadro Constant and Particle Quantification

• One mole of any substance contains exactly 6.02 × 10²³ constituent particles (atoms, molecules,
• or ions);
  • for example, one mole of water (H₂O) contains 6.02 × 10²³ molecules and 1.204 × 10²⁴ hydrogen atoms.

Water of Crystallisation and Percentage Yield

• Hydrated salts lose water molecules upon heating; stoichiometric calculations allow determination of the formula water content and experimental percentage yield.
  • (such as hydrated sodium carbonate, Na₂CO₃·xH2O)

Ammonium nitrate

• NH₄NO₃, a fertiliser with 35% nitrogen.

Composition

• What a substance or mixture is made of.

Molecule

• Two or more atoms held together by covalent bonds.

Purity

• Fraction of a sample that is the desired substance.

Empirical formula

• The simplest whole-number ratio of atoms or ions.

Molecular formula

• The number and type of atoms in one molecule.

Percentage yield

• Actual yield divided by theoretical yield, times 100.

Percentage composition

• The percentage by mass of each element in a compound.

Percentage purity

• The mass of pure substance divided by sample mass × 100.

Element

• A substance made of only one type of atom.

Hydrogen

• A gas that gives a squeaky pop with a lighted splint.

Compound

• Two or more elements chemically combined in a fixed ratio.

Oxygen

• A gas that relights a glowing splint.

Hydrated

• Containing water of crystallisation.

Section 7 of 26Titration and its purpose

Titration

• Titration analyses the concentration of a solution.

Acid-base titration

• Acid-base titration finds exactly how much alkali neutralises a quantity of acid, or the reverse.

Titre

• Titre the average volume from the burette once concordant results are found.

Section 8 of 26Concentration, moles and volume

Formula

• Formula concentration = moles ÷ volume (dm³).
• Moles = concentration × volume (dm³).
• Volume (dm³) = moles ÷ concentration.

Unit conversion

• Unit conversion divide cm³ by 1000 to get dm³.

Moles from mass

• Moles from mass moles = mass ÷ Mr,
• then divide by the volume in dm³ for mol/dm³.
  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)

Section 9 of 26Finding an unknown concentration

1

Write the balanced equation.

2

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

3

Use the molar ratio to find the moles of the other reactant.

4

Divide those moles by the volume of the unknown solution in dm³.

Section 10 of 26Finding a volume to neutralise

1

Write the balanced equation.

2

Find the moles of the known solution.

3

Use the molar ratio to find the moles of the other solution.

4

Volume (dm³) = moles ÷ concentration; multiply by 1000 for cm³.

Section 11 of 26Using the mole ratio

1

1 ratio

• HCl + NaOH and HNO₃ + KOH react in a 1 : 1 ratio.

Other ratios

• Ba(OH)₂ + 2HNO₃ is 1 : 2 and Na₂CO₃ + 2HCl is 1 : 2, so moles are multiplied or divided by 2.

Section 12 of 26More from the specification

1Acid neutralisation with solid basic oxide

• Acid neutralisation with solid basic oxide reacting standardised sulfuric acid with copper(II) oxide allows volumetric determination of reacted moles and excess acid.

2The mole and Avogadro constant

• One mole of any chemical substance contains exactly 6.022 x 10²³ elementary entities (atoms, molecules,
• or ions), defined by the Avogadro constant N_A,
• allowing conversion between mass and amount in moles (n = m / M_r).

3Empirical formula from percentage composition

• The empirical formula is determined by converting the percentage mass of each element into moles, dividing by the smallest mole value to obtain the simplest whole-number ratio of atoms in the compound.

4Molecular formula and relative molecular mass

• The molecular formula gives the actual number of atoms of each element in a molecule, calculated by dividing the relative molecular mass (M_r) by the empirical formula mass.

5Percentage yield and water of crystallisation

• Calculating percentage yield ( * 100%) and the stoichiometric ratio of water in hydrated crystals (e.g. CuSO₄ . xH2O) utilizes precise mole-to-mass conversions.
  • (actual yield / theoretical yield)

6Ammonium nitrate

• NH₄NO₃, a fertiliser with 35% nitrogen.

7Hydrated

• Containing water of crystallisation.

8Composition

• What a substance or mixture is made of.

9Molecule

• Two or more atoms held together by covalent bonds.

10Solute

• A substance that dissolves in a solvent.

11Purity

• Fraction of a sample that is the desired substance.

12Empirical formula

• The simplest whole-number ratio of atoms or ions.

13Molecular formula

• The number and type of atoms in one molecule.

14Percentage yield

• Actual yield divided by theoretical yield, times 100.

15Element

• A substance made of only one type of atom.

16Hydrogen

• A gas that gives a squeaky pop with a lighted splint.

17Compound

• Two or more elements chemically combined in a fixed ratio.

18Oxygen

• A gas that relights a glowing splint.

Section 13 of 26Empirical formula calculation

1

Write each element and its mass or percentage.

2

Write the Ar of each element.

3

Calculate moles = mass ÷ Ar.

4

Divide all the moles by the smallest number.

5

Multiply to give whole numbers if needed.

6

Write the empirical formula.

Percentages

• Percentages if the percentages do not add to 100, find the missing element by subtraction.

Non-whole ratios

• Non-whole ratios a ratio such as 1.5 is multiplied by 2 along with the others, not rounded.

Section 14 of 26Molecular formula

• Molecular formula
  • gives the actual numbers of atoms and is a whole-number multiple of the empirical formula.
  • Methane CH₄, ethane C₂H₆, ethene C₂H₄ and benzene C₆H₆.

Molecular formula

Find the Mr of the empirical formula.

Divide the Mr of the compound by the Mr of the empirical formula.

Multiply every subscript in the empirical formula by that number.

Section 15 of 26Hydrated salts

Hydrated salt

• Hydrated salt a crystallised salt that contains water of crystallisation, such as CuSO₄·5H₂O.

Experiment

• Experiment heat a weighed sample until constant mass and reweigh to find the mass of water lost.

1

Find the mass of anhydrous salt and the mass of water.

2

Calculate moles of each = mass ÷ Mr.

3

Divide both by the smaller number of moles.

4

Write the ratio 1 salt : x water.

Section 16 of 26Percentage yield

Yield

• Yield the amount of product obtained from a reaction.

Percentage yield

• = actual yield ÷ theoretical yield × 100.
• Actual yield is the mass obtained.
• Theoretical yield is the maximum mass from the balanced equation.

Importance

• Importance a high percentage yield increases profit and reduces waste.

Section 17 of 26Why yield is below 100%

Reversible reaction

• Reversible reaction does not go to completion.

Losses

• Losses product is lost during separation, filtering, distillation and transfer.

Side reactions

• Side reactions reactants may react with air or impurities.

Left behind

• Left behind some reactant remains in the equipment.

Section 18 of 26More from the specification

More from the specification

Molar Mass and Amount of Substance

• The number of moles of a substance is calculated by dividing mass by molar mass (n = m / M),
• where molar mass is calculated from relative atomic masses (Ar).

Solution Concentration Formula

• Concentration of a solution is defined as the amount of solute per unit volume: c = n / V in mol dm⁻³, or mass per volume in g dm⁻³.

Empirical Formula of Metal Oxides

• When magnesium burns in oxygen to form magnesium oxide, reacting masses determine the simplest whole-number ratio of magnesium to oxygen atoms.
  • (2Mg + O₂ → 2MgO)

Molar Ratio in Stoichiometry

• In chemical reactions such as sodium reacting with water or acid, balanced stoichiometric coefficients provide the exact molar ratio between reacting species.

Purity

• Fraction of a sample that is the desired substance.

Molar volume

• Volume of one mole of gas24 dm³ at RTP.

Molar ratio

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

Percentage purity

• The mass of pure substance divided by sample mass × 100.

Ammonium nitrate

• NH₄NO₃, a fertiliser with 35% nitrogen.

Volume

• The space taken up by an object.

Concentration

• Moles ÷ volume in dm³, or g/dm³.

Substance

• A material with only one type of particle.

Oxygen

• A gas that relights a glowing splint.

Composition

• What a substance or mixture is made of.

Solute

• A substance that dissolves in a solvent.

Section 19 of 26Percentage composition by mass

Percentage composition by mass

Formula

• Formula percentage of an element = total Ar of that element in the formula ÷ Mr of the compound × 100.

Check

• Check the percentages of all elements in a compound add up to 100%.
• Water is 11.1% hydrogen and 88.9% oxygen.

Total mass of the element

• Total mass of the element use all the atoms of that element in the formula.
• Fe₂O₃ is 70% iron because 112 ÷ 160.
• NH₄NO₃ contains two nitrogen atoms, so nitrogen is 35%.

Section 20 of 26Method for percentage composition

1

Write the formula and the Ar of each element.

2

Calculate the total mass of the element wanted.

3

Calculate the Mr of the compound.

4

Divide the element total by the Mr and multiply by 100.

Section 21 of 26Purity

Pure substance

• Pure substance has nothing else mixed with it.

Contamination

• Contamination products may contain unreacted reactants, catalysts and other impurities.

Percentage purity

• = mass of pure substance ÷ total mass of sample × 100.

Section 22 of 26Checking purity

Melting point

• Melting point a pure solid melts sharply at one temperature.

Chromatography

• Chromatography a pure substance gives a single spot.

Methods of purifying

• Methods of purifying crystallisation, filtration, distillation and chromatography.

Section 23 of 26Using purity in calculations

Impure sample

• Impure sample only the pure fraction takes part, so 80% pure means 80% of the mass reacts.

Method

• Method find the mass of pure reactant first, then convert to moles and use the equation ratio.

Importance

• Importance medicines and food products need very high purity.

Section 24 of 26More from the specification

1Gravimetric yield of magnesium oxide

• Gravimetric yield of magnesium oxide comparing experimental product weight against theoretical mass yields the percentage conversion to pure magnesium oxide.

2Molar Concentration and Solution Volumes

• The concentration of a solute in an aqueous solution is the amount of substance (in moles) dissolved per unit volume: Concentration (mol dm^-3 = amount (mol) / volume (dm³.

3Molar Mass and Stoichiometric Calculations

• The molar mass (M) of a substance is calculated by summing the relative atomic masses (Ar) of its constituent atoms;
  • multiplying the amount in moles by molar mass gives the physical mass in grams (mass = n × M).

4Empirical vs. Molecular Formula Determination

• The empirical formula expresses the simplest whole-number ratio of atoms in a compound derived from percentage composition,
• whereas the molecular formula gives the actual number of atoms in a molecular unit.

5Percentage Yield and Limiting Reactants

• In chemical preparations, percentage yield is determined by: (actual experimental yield / theoretical yield) × 100%, reflecting product loss during synthesis and purification.
  • (such as precipitating insoluble sodium salts)

6Molecule

• Two or more atoms held together by covalent bonds.

7Solute

• A substance that dissolves in a solvent.

8Empirical formula

• The simplest whole-number ratio of atoms or ions.

9Molecular formula

• The number and type of atoms in one molecule.

10Percentage yield

• Actual yield divided by theoretical yield, times 100.

11Molar volume

• Volume of one mole of gas24 dm³ at RTP.

12Molar ratio

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

13Ammonium nitrate

• NH₄NO₃, a fertiliser with 35% nitrogen.

14Volume

• The space taken up by an object.

15Concentration

• Moles ÷ volume in dm³, or g/dm³.

16Hydrated

• Containing water of crystallisation.

Section 25 of 26Key terms

Term	Meaning
Limiting reactant	Reactant that is used up first.
Excess reactant	Reactant left over.
Molar ratio	Ratio of moles in the balanced equation.
Solute	Substance that dissolves.
Solvent	Liquid that dissolves a solute.
Concentration	Amount of solute per dm^{3} of solution.

Term	Meaning
Titration	Method of finding the volume of one solution that reacts with another.
Titre	Volume added from the burette.
Concordant results	Titres that agree within 0.10 cm^{3}.
Concentration	Moles of solute per dm^{3}.
Molar ratio	Ratio of moles in the balanced equation.
Neutralisation	Reaction of an acid with an alkali.

Term	Meaning
Empirical formula	Simplest whole-number ratio of atoms.
Molecular formula	Actual numbers of atoms in a molecule.
Hydrated salt	Salt containing water of crystallisation.
Anhydrous	Without water of crystallisation.
Actual yield	Mass of product obtained.
Theoretical yield	Maximum mass of product expected from the equation.

Term	Meaning
Percentage composition	Percentage by mass of an element in a compound.
Purity	Fraction of a sample that is the desired substance.
Impurity	Unwanted substance in a product.
Pure substance	Substance with nothing else mixed in.
Sharp melting point	Melting at a single fixed temperature.

Section 26 of 26Exam technique

1

Forgetting the molar ratio from the balanced equation.

2

Not converting cm^{3} to dm^{3}.

3

Using the excess reactant to find the amount of product.

4

Mixing up g/dm^{3} and mol/dm^{3}.

1

Forgetting to convert cm^{3} to dm^{3}.

2

Assuming a 1 : 1 ratio when the equation shows another.

3

Using an unbalanced equation.

4

Calculating the substance asked about with the wrong solution's volume.

1

Rounding a ratio such as 1.5 to 2 instead of multiplying every ratio.

2

Dividing theoretical yield by actual yield.

3

Using the empirical Mr instead of the compound Mr when finding the multiple.

4

Giving a percentage yield above 100%.

1

Using one nitrogen atom instead of both in ammonium nitrate.

2

Dividing by the mass of the element instead of the compound.

3

Saying a pure substance melts over a range.

4

Forgetting to multiply by 100.