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Chemistry

The Mole Concept

PDF
Matthew Williams
|May 15, 2026|10 min read
AvogadroConcentrationCSEC ChemistryMole ConceptPaper 01Paper 02Section AStoichiometry

The mole, molar mass, Avogadro's constant, relative formula mass, percentage composition, empirical and molecular formulae, gas volumes at rtp and stp, balanced equations, molar concentration, and volumetric analysis calculations.

Atoms and molecules are impossibly small — far too tiny to count or weigh individually. The mole is the bridge between the microscopic world of particles and the macroscopic quantities measured in the lab.

The Mole and Avogadro's Constant

A mole is the amount of substance that contains 6.02×10236.02 \times 10^{23}6.02×1023 particles. That number — 6.02×10236.02 \times 10^{23}6.02×1023 — is Avogadro's constant (NAN_ANA​). The particles can be atoms, molecules, ions, or electrons, depending on the substance.

One mole of any substance contains the same number of particles, regardless of what the substance is. One mole of carbon atoms, one mole of water molecules, and one mole of sodium ions all contain 6.02×10236.02 \times 10^{23}6.02×1023 particles.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in g mol⁻¹. Numerically, the molar mass of an element equals its relative atomic mass in grams. For a compound, the molar mass equals the sum of the relative atomic masses of all atoms in one formula unit.

SubstanceFormulaMolar mass
HydrogenH₂2 g mol⁻¹
CarbonC12 g mol⁻¹
WaterH₂O18 g mol⁻¹
Sodium chlorideNaCl58.5 g mol⁻¹
Calcium carbonateCaCO₃100 g mol⁻¹
Sulfuric acidH₂SO₄98 g mol⁻¹

Calculations Involving the Mole

The fundamental relationship between moles, mass, and molar mass is:

n=mMn = \frac{m}{M}n=Mm​

where nnn is the number of moles, mmm is the mass in grams, and MMM is the molar mass in g mol⁻¹.

Example

How many moles are in 44 g of carbon dioxide (CO₂)?

Molar mass of CO₂ = 12 + (2 × 16) = 44 g mol⁻¹

n=4444=1 moln = \frac{44}{44} = 1 \text{ mol}n=4444​=1 mol

Example

What is the mass of 0.25 mol of CaCO₃?

Molar mass of CaCO₃ = 40 + 12 + (3 × 16) = 100 g mol⁻¹

m=n×M=0.25×100=25 gm = n \times M = 0.25 \times 100 = 25 \text{ g}m=n×M=0.25×100=25 g

Relative Formula Mass

The relative formula mass (MrM_rMr​) of a compound is the sum of the relative atomic masses of all the atoms in one formula unit. For ionic compounds the term relative formula mass is used; for molecular compounds it is often called relative molecular mass. Neither is expressed in units — they are pure numbers relative to carbon-12.

Example

Find the relative formula mass of aluminium sulfate, Al₂(SO₄)₃.

Al: 2 × 27 = 54
S: 3 × 32 = 96
O: 12 × 16 = 192
MrM_rMr​ = 54 + 96 + 192 = 342

Percentage Composition by Mass

The percentage by mass of each element in a compound is:

% by mass=mass of element in one molemolar mass of compound×100\% \text{ by mass} = \frac{\text{mass of element in one mole}}{\text{molar mass of compound}} \times 100% by mass=molar mass of compoundmass of element in one mole​×100

Example

Find the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃.

Molar mass = 14 + 4 + 14 + 48 = 80 g mol⁻¹
Mass of N = 14 + 14 = 28 g

%N=2880×100=35%\% N = \frac{28}{80} \times 100 = 35\%%N=8028​×100=35%

Empirical and Molecular Formulae

The empirical formula gives the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual number of atoms in one molecule. The molecular formula is always a whole-number multiple of the empirical formula.

To find the empirical formula from percentage composition:

  1. Treat the percentages as masses in grams.
  2. Divide each mass by its relative atomic mass to find moles.
  3. Divide all mole values by the smallest to get the simplest ratio.
  4. Round to the nearest whole number (or multiply through if fractions appear).
Example

A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.

Element%÷ ArRatio
C4040/12 = 3.333.33/3.33 = 1
H6.76.7/1 = 6.76.7/3.33 = 2
O53.353.3/16 = 3.333.33/3.33 = 1

Empirical formula: CH₂O

To find the molecular formula, divide the given molar mass by the empirical formula mass and multiply the subscripts.

Avogadro's Law and Gas Volumes

Avogadro's Law states that equal volumes of all gases, measured at the same temperature and pressure, contain equal numbers of molecules.

This leads directly to the concept of molar volume. At the conditions used in CSEC:

ConditionsMolar volume
RTP (room temperature and pressure, 25 °C and 1 atm)24 dm³ mol⁻¹
STP (standard temperature and pressure, 0 °C and 1 atm)22.4 dm³ mol⁻¹

n=V24 (at RTP)n = \frac{V}{24} \text{ (at RTP)}n=24V​ (at RTP)

Example

What volume does 0.5 mol of oxygen gas occupy at RTP?

V=n×24=0.5×24=12 dm3V = n \times 24 = 0.5 \times 24 = 12 \text{ dm}^3V=n×24=0.5×24=12 dm3

Law of Conservation of Matter

The Law of Conservation of Matter states that matter is neither created nor destroyed in a chemical reaction. The total mass of reactants always equals the total mass of products. This is why chemical equations must be balanced — the same atoms must appear on both sides.

Balanced Equations and State Symbols

A balanced molecular equation shows the formulae of all reactants and products with state symbols, and has equal numbers of each type of atom on both sides.

State symbols:

  • (s) — solid
  • (l) — liquid
  • (g) — gas
  • (aq) — aqueous (dissolved in water)

Ionic equations show only the species that actually change in a reaction — spectator ions (ions present but unchanged) are omitted.

Example

Molecular equation:
HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}HCl(aq)+NaOH(aq)→NaCl(aq)+H2​O(l)

Full ionic equation:
H+(aq)+Cl−(aq)+Na+(aq)+OH−(aq)→Na+(aq)+Cl−(aq)+H2O(l)\text{H}^+\text{(aq)} + \text{Cl}^-\text{(aq)} + \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)} + \text{H}_2\text{O(l)}H+(aq)+Cl−(aq)+Na+(aq)+OH−(aq)→Na+(aq)+Cl−(aq)+H2​O(l)

Net ionic equation (spectators Na⁺ and Cl⁻ removed):
H+(aq)+OH−(aq)→H2O(l)\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)}H+(aq)+OH−(aq)→H2​O(l)

Applying the Mole Concept to Equations

The coefficients in a balanced equation are mole ratios — they tell you how many moles of each substance react or are produced. This allows calculation of reacting masses and volumes from any given starting quantity.

Method:

  1. Write the balanced equation.
  2. Identify the mole ratio between the substances of interest.
  3. Calculate moles of the given substance.
  4. Use the ratio to find moles of the unknown substance.
  5. Convert to mass (using m=nMm = nMm=nM) or volume (using V=n×24V = n \times 24V=n×24) as required.
Example

What mass of iron(III) oxide is produced when 5.6 g of iron reacts completely with oxygen?

4Fe(s)+3O2(g)→2Fe2O3(s)4\text{Fe(s)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{Fe}_2\text{O}_3\text{(s)}4Fe(s)+3O2​(g)→2Fe2​O3​(s)

Moles of Fe = 5.6 / 56 = 0.1 mol
Ratio Fe : Fe₂O₃ = 4 : 2 = 2 : 1
Moles of Fe₂O₃ = 0.1 / 2 = 0.05 mol
Molar mass of Fe₂O₃ = (2 × 56) + (3 × 16) = 160 g mol⁻¹
Mass of Fe₂O₃ = 0.05 × 160 = 8.0 g

Concentration and Standard Solutions

Molar concentration is the number of moles of solute dissolved per cubic decimetre (litre) of solution:

c=nVc = \frac{n}{V}c=Vn​

where ccc is in mol dm⁻³ and VVV is in dm³. Note: 1 dm³ = 1000 cm³, so always convert cm³ by dividing by 1000.

Mass concentration is the mass of solute dissolved per unit volume, expressed in g dm⁻³.

A standard solution is a solution whose concentration is accurately known. Standard solutions are prepared by dissolving an accurately weighed mass of solute in a known volume of solvent using a volumetric flask.

Exam Tip

When using n=cVn = cVn=cV, volume must be in dm³. A 25 cm³ pipette reading becomes 0.025 dm³. Forgetting this conversion is the single most common error in calculation questions.

Volumetric Analysis Calculations

In a titration, one solution of known concentration (the standard solution) reacts completely with another. The unknown concentration is calculated from the volume used and the mole ratio from the balanced equation.

General approach:

  1. Calculate moles of the known solution: n=cVn = cVn=cV
  2. Use the mole ratio from the balanced equation to find moles of the unknown.
  3. Calculate the concentration of the unknown: c=n/Vc = n/Vc=n/V
Example

25.0 cm³ of NaOH solution was neutralised by 20.0 cm³ of 0.10 mol dm⁻³ HCl.
Find the concentration of the NaOH.

HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}HCl(aq)+NaOH(aq)→NaCl(aq)+H2​O(l)

Moles HCl = 0.10 × 0.020 = 0.0020 mol
Ratio 1 : 1, so moles NaOH = 0.0020 mol
Concentration NaOH = 0.0020 / 0.025 = 0.080 mol dm⁻³

Example

15.0 cm³ of 0.60 mol dm⁻³ HNO₃ neutralised 25.0 cm³ of Na₂CO₃ solution.

2HNO3+Na2CO3→2NaNO3+H2O+CO22\text{HNO}_3 + \text{Na}_2\text{CO}_3 \rightarrow 2\text{NaNO}_3 + \text{H}_2\text{O} + \text{CO}_22HNO3​+Na2​CO3​→2NaNO3​+H2​O+CO2​

Moles HNO₃ = 0.60 × 0.015 = 0.0090 mol
Ratio HNO₃ : Na₂CO₃ = 2 : 1
Moles Na₂CO₃ = 0.0090 / 2 = 0.0045 mol
Concentration Na₂CO₃ = 0.0045 / 0.025 = 0.18 mol dm⁻³

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