Chemistry deals with atoms and molecules — far too small to count individually. The MOLE is the chemist's counting unit that bridges the particle world and the macroscopic (lab) world.
1 mole of any element contains Avogadro's number of atoms AND has a mass equal to the atomic mass from the periodic table expressed in grams (g·mol⁻¹). For example: 1 mole of carbon (C) has mass 12 g; 1 mole of iron (Fe) has mass 55.85 g; 1 mole of oxygen gas (O₂) has mass 32 g.
The molar mass (M) of a compound is the mass of 1 mole of that compound. It is calculated by adding together the molar masses of all atoms in the formula, multiplied by their subscripts. Units: g·mol⁻¹.
| Compound | Formula | Calculation | M (g·mol⁻¹) |
|---|---|---|---|
| Water | H₂O | 2(1.0) + 16.0 | 18.0 |
| Calcium carbonate | CaCO₃ | 40.1 + 12.0 + 3(16.0) | 100.1 |
| Sulfuric acid | H₂SO₄ | 2(1.0) + 32.1 + 4(16.0) | 98.1 |
| Sodium hydroxide | NaOH | 23.0 + 16.0 + 1.0 | 40.0 |
| Iron(III) oxide | Fe₂O₃ | 2(55.8) + 3(16.0) | 159.6 |
| Glucose | C₆H₁₂O₆ | 6(12.0) + 12(1.0) + 6(16.0) | 180.0 |
The three core mole equations link moles (n) to mass (m), number of particles (N), and volume of gas (V):
Worked Example 1 — How many moles in 54 g of water (H₂O)?
M(H₂O) = 18 g·mol⁻¹; n = m/M = 54/18 = 3 mol. Answer: 3 moles.
Worked Example 2 — How many molecules are in 3 mol of water?
N = n × Nₐ = 3 × 6.022 × 10²³ = 1.807 × 10²⁴ molecules.
Worked Example 3 — What volume does 2 mol of CO₂ occupy at STP?
V = n × Vₘ = 2 × 22.4 = 44.8 L.
Worked Example 4 — What is the concentration of a solution containing 0.5 mol NaOH in 250 mL water?
V = 250 mL = 0.250 L; c = n/V = 0.5/0.250 = 2 mol·L⁻¹.
Percentage composition tells us what fraction of a compound's mass comes from each element:
Example: % O in H₂O = (16.0 × 1)/18.0 × 100 = 88.9%. % H in H₂O = (1.0 × 2)/18.0 × 100 = 11.1%.
Example: % Ca in CaCO₃ = 40.1/100.1 × 100 = 40.1%. This explains why limestone (CaCO₃) is used to supply calcium in agriculture.
The EMPIRICAL FORMULA is the simplest whole-number ratio of atoms in a compound. The MOLECULAR FORMULA gives the actual number of atoms in one molecule (it is always a whole-number multiple of the empirical formula).
Example: A compound is 40.0% C, 6.7% H, 53.3% O by mass. Step 1: 40.0 g C, 6.7 g H, 53.3 g O. Step 2: moles C = 40.0/12 = 3.33; moles H = 6.7/1 = 6.7; moles O = 53.3/16 = 3.33. Step 3: ratio C:H:O = 3.33:6.7:3.33 → divide by 3.33 → 1:2:1. Empirical formula: CH₂O (glucose unit).
If M(actual) = 180 g·mol⁻¹ and M(empirical CH₂O) = 30 g·mol⁻¹, then n = 180/30 = 6. Molecular formula = 6 × CH₂O = C₆H₁₂O₆ (glucose).
A balanced chemical equation gives the MOLE RATIOS in which reactants combine and products form. These ratios are used directly in stoichiometry calculations.
Worked Example — Decomposition of calcium carbonate:
CaCO₃ → CaO + CO₂ Mole ratio: 1 : 1 : 1
"If 50 g of CaCO₃ decomposes completely, what mass of CO₂ is produced?"
Step 1: n(CaCO₃) = 50/100 = 0.5 mol
Step 2: From ratio, n(CO₂) = 0.5 mol (1:1 ratio)
Step 3: m(CO₂) = n × M = 0.5 × 44 = 22 g
Worked Example — Neutralisation:
NaOH + HCl → NaCl + H₂O Mole ratio: 1 : 1 : 1 : 1
"What mass of NaOH is needed to neutralise 3.65 g of HCl?"
n(HCl) = 3.65/36.5 = 0.100 mol; n(NaOH) = 0.100 mol (1:1); m(NaOH) = 0.100 × 40 = 4.0 g
In most reactions, one reactant runs out first — this is the LIMITING REAGENT. It determines the maximum amount of product that can form. The reagent that remains after the reaction is called the EXCESS reagent.
Example: 10 g Na₂CO₃ reacts with 10 g HCl. Equation: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
n(Na₂CO₃) = 10/106 = 0.0943 mol. Ratio requirement: each mole Na₂CO₃ needs 2 mol HCl → would need 0.1887 mol HCl.
n(HCl) = 10/36.5 = 0.274 mol. Divide by coefficient: HCl: 0.274/2 = 0.137; Na₂CO₃: 0.0943/1 = 0.0943.
Na₂CO₃ gives the smaller value → Na₂CO₃ is the limiting reagent. HCl is in excess.
Theoretical yield of CO₂: n = 0.0943 mol; m = 0.0943 × 44 = 4.15 g.
In practice, reactions rarely produce 100% of the theoretical yield due to: incomplete reactions, side reactions producing unwanted products, losses during separation/purification, or equilibrium limitations. A high % yield is desirable industrially for economic and environmental reasons.
Dilution: When a concentrated solution (stock) is diluted with water, the moles of solute remain constant. C₁V₁ = C₂V₂ where 1 = concentrated, 2 = diluted.
Example: What volume of 6 mol·L⁻¹ HCl is needed to make 250 mL of 0.5 mol·L⁻¹ HCl?
C₁V₁ = C₂V₂ → 6 × V₁ = 0.5 × 0.250 → V₁ = 0.125/6 = 0.0208 L = 20.8 mL.
Titration: a volumetric technique for finding the concentration of an unknown solution. A standard solution (known concentration) is added from a burette to a known volume of the unknown solution until the equivalence point is reached (indicated by a colour change of an indicator).
Full calculation: NaOH titrated with HCl. NaOH + HCl → NaCl + H₂O. At equivalence: n(NaOH) = n(HCl).
If 25.0 mL of NaOH is titrated with 22.4 mL of 0.100 mol·L⁻¹ HCl:
n(HCl) = c × V = 0.100 × 0.0224 = 0.00224 mol; n(NaOH) = 0.00224 mol; c(NaOH) = n/V = 0.00224/0.0250 = 0.0896 mol·L⁻¹.
Gas stoichiometry at non-STP conditions: use the ideal gas law PV = nRT where R = 8.314 J·mol⁻¹·K⁻¹ (or 8.314 Pa·m³·mol⁻¹·K⁻¹).
Example: What volume does 2 mol CO₂ occupy at 25°C and 150 kPa?
T = 298 K; P = 150 000 Pa; V = nRT/P = (2 × 8.314 × 298)/150 000 = 4955/150 000 = 0.0330 m³ = 33.0 L.
| Rough | Titre 1 | Titre 2 | Titre 3 | |
|---|---|---|---|---|
| Final reading (mL) | 24.60 | 23.45 | 23.40 | 23.42 |
| Initial reading (mL) | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume HCl used (mL) | 24.60 | 23.45 | 23.40 | 23.42 |