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.
A mole is the amount of substance that contains particles. That number — — is Avogadro's constant (). 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 particles.
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.
| Substance | Formula | Molar mass |
|---|---|---|
| Hydrogen | H₂ | 2 g mol⁻¹ |
| Carbon | C | 12 g mol⁻¹ |
| Water | H₂O | 18 g mol⁻¹ |
| Sodium chloride | NaCl | 58.5 g mol⁻¹ |
| Calcium carbonate | CaCO₃ | 100 g mol⁻¹ |
| Sulfuric acid | H₂SO₄ | 98 g mol⁻¹ |
The fundamental relationship between moles, mass, and molar mass is:
where is the number of moles, is the mass in grams, and is the molar mass in g mol⁻¹.
How many moles are in 44 g of carbon dioxide (CO₂)?
Molar mass of CO₂ = 12 + (2 × 16) = 44 g mol⁻¹
What is the mass of 0.25 mol of CaCO₃?
Molar mass of CaCO₃ = 40 + 12 + (3 × 16) = 100 g mol⁻¹
The relative formula mass () 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.
Find the relative formula mass of aluminium sulfate, Al₂(SO₄)₃.
Al: 2 × 27 = 54
S: 3 × 32 = 96
O: 12 × 16 = 192
= 54 + 96 + 192 = 342
The percentage by mass of each element in a compound is:
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
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:
A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.
| Element | % | ÷ Ar | Ratio |
|---|---|---|---|
| C | 40 | 40/12 = 3.33 | 3.33/3.33 = 1 |
| H | 6.7 | 6.7/1 = 6.7 | 6.7/3.33 = 2 |
| O | 53.3 | 53.3/16 = 3.33 | 3.33/3.33 = 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 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:
| Conditions | Molar 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 |
What volume does 0.5 mol of oxygen gas occupy at RTP?
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.
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:
Ionic equations show only the species that actually change in a reaction — spectator ions (ions present but unchanged) are omitted.
Molecular equation:
Full ionic equation:
Net ionic equation (spectators Na⁺ and Cl⁻ removed):
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:
What mass of iron(III) oxide is produced when 5.6 g of iron reacts completely with oxygen?
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
Molar concentration is the number of moles of solute dissolved per cubic decimetre (litre) of solution:
where is in mol dm⁻³ and 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.
When using , 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.
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:
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.
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
15.0 cm³ of 0.60 mol dm⁻³ HNO₃ neutralised 25.0 cm³ of Na₂CO₃ solution.
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