Mole Calculations

Learning Objectives (OCR Spec 2.1.3)

• Use n = m/M in reacting-mass problems
• Use c = n/V for solution concentrations
• Perform stoichiometric calculations using balanced equations
• Identify and use limiting reagents
• Calculate volumes and concentrations of acids and alkalis in neutralisation reactions
• Calculate theoretical yields and compare with actual yields
• Perform multi-step stoichiometry involving gases, solutions and solids

The Three Core Mole Equations

You must be fluent with these three relationships:

$n = \frac{m}{M} \qquad c = \frac{n}{V} \qquad N = nL$n=Mm​c=Vn​N=nL

Where:

• m = mass (g)
• M = molar mass (g mol⁻¹)
• c = concentration (mol dm⁻³)
• V = volume (dm³) — note: dm³, not cm³
• N = number of particles
• L = Avogadro constant

Unit Conversion for Volumes

• 1 dm³ = 1000 cm³ = 1 L = 1000 mL
• Convert cm³ to dm³ by dividing by 1000
• Convert dm³ to cm³ by multiplying by 1000

Tip: Burettes read in cm³. Concentrations are given in mol dm⁻³. Always convert burette readings to dm³ before substituting into c = n/V.

Concentration

Concentration is the amount of solute per unit volume of solution.

$c = \frac{n}{V}$c=Vn​

Worked Example 1: Concentration from Mass

A 250 cm³ solution of NaOH contains 10.0 g of NaOH. Calculate the concentration in mol dm⁻³.

• M(NaOH) = 40.0 g mol⁻¹
• n = m/M = 10.0 / 40.0 = 0.250 mol
• V = 250 / 1000 = 0.250 dm³
• c = n/V = 0.250 / 0.250 = 1.00 mol dm⁻³

Worked Example 2: Mass from Concentration

What mass of potassium hydroxide (KOH) is needed to prepare 500 cm³ of a 0.200 mol dm⁻³ solution?

• V = 0.500 dm³
• n = cV = 0.200 × 0.500 = 0.100 mol
• M(KOH) = 56.1 g mol⁻¹
• m = n × M = 0.100 × 56.1 = 5.61 g

Worked Example 3: Converting Between g dm⁻³ and mol dm⁻³

A solution of Na₂CO₃ has concentration 21.2 g dm⁻³. Express this in mol dm⁻³.

• M(Na₂CO₃) = 2(23.0) + 12.0 + 3(16.0) = 106.0 g mol⁻¹
• c = 21.2 / 106.0 = 0.200 mol dm⁻³

Worked Example 4: Dilution

A student takes 25.0 cm³ of 2.00 mol dm⁻³ HCl and dilutes it to 250 cm³ in a volumetric flask. Calculate the new concentration.

• n(HCl) = cV = 2.00 × 0.0250 = 0.0500 mol (moles conserved during dilution)
• New c = n/V = 0.0500 / 0.250 = 0.200 mol dm⁻³

A quicker dilution formula: c₁V₁ = c₂V₂ → 2.00 × 25.0 = c₂ × 250 → c₂ = 0.200 mol dm⁻³ ✓

Reacting Masses (Stoichiometry)

The stoichiometry of a balanced equation tells you the ratio in which moles of reactants and products are related. Follow this general strategy:

1. Write the balanced equation
2. Convert the known mass (or volume) to moles using n = m/M
3. Use the stoichiometric ratio to find moles of the target substance
4. Convert moles back to mass (or volume) as required

flowchart LR
    A[Mass of A g] -- divide by M_A --> B[Moles of A]
    B -- ratio from balanced equation --> C[Moles of B]
    C -- multiply by M_B --> D[Mass of B g]

Worked Example 5: Reacting Mass

Calculate the mass of calcium oxide produced when 25.0 g of calcium carbonate is fully decomposed:

CaCO₃(s) → CaO(s) + CO₂(g)

• M(CaCO₃) = 100.1; M(CaO) = 56.1
• n(CaCO₃) = 25.0 / 100.1 = 0.2498 mol
• Ratio 1:1, so n(CaO) = 0.2498 mol
• m(CaO) = 0.2498 × 56.1 = 14.0 g

Worked Example 6: With a Non-1:1 Ratio

What mass of iron is produced when 80.0 g of iron(III) oxide reacts with excess carbon monoxide?

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

• M(Fe₂O₃) = 159.6; M(Fe) = 55.8
• n(Fe₂O₃) = 80.0 / 159.6 = 0.5013 mol
• Ratio 1 Fe₂O₃ : 2 Fe → n(Fe) = 2 × 0.5013 = 1.003 mol
• m(Fe) = 1.003 × 55.8 = 55.9 g

Worked Example 7: Multi-Step Stoichiometry

Calculate the mass of CaCO₃ required to neutralise 5.00 g of H₂SO₄:

CaCO₃(s) + H₂SO₄(aq) → CaSO₄(s) + H₂O(l) + CO₂(g)

• n(H₂SO₄) = 5.00 / 98.1 = 0.05097 mol
• Ratio 1:1 → n(CaCO₃) = 0.05097 mol
• m(CaCO₃) = 0.05097 × 100.1 = 5.10 g

Limiting Reagents

When two reactants are given in amounts that are not in the stoichiometric ratio, one of them runs out first and determines how much product can form. This is the limiting reagent. The other is said to be in excess.

Method for Identifying the Limiting Reagent

1. Calculate moles of each reactant
2. Divide each by its coefficient in the balanced equation
3. The smallest result corresponds to the limiting reagent
4. Use moles of limiting reagent (with its ratio) to calculate the product

Worked Example 8: Limiting Reagent

When 6.5 g of zinc reacts with 30.0 cm³ of 2.00 mol dm⁻³ HCl:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

• n(Zn) = 6.5 / 65.4 = 0.0994 mol
• n(HCl) = c × V = 2.00 × 0.0300 = 0.0600 mol
• Divide by coefficients: Zn → 0.0994 / 1 = 0.0994; HCl → 0.0600 / 2 = 0.0300
• HCl is smaller → HCl is the limiting reagent

Moles of H₂ formed = 0.0600 / 2 = 0.0300 mol Mass of H₂ = 0.0300 × 2.0 = 0.0600 g (Zn left over = 0.0994 − 0.0300 = 0.0694 mol)

Worked Example 9: Limiting Reagent with Two Solids