# The Mole and Mole Calculations

> Chemistry · CIE IGCSE 0620 (2026-2028)
> Source: https://www.owlsprep.com/study/cie-0620-u3-the-mole-and-mole-calculations/

This Extended-only guide covers all CIE IGCSE 0620 3.3 mole calculation skills: formula derivation, mass/gas volume/concentration sums, and limiting reactant problems with exam-style worked examples.

**Prerequisites:** [CIE IGCSE 0620 Chemical Formulae and Balanced Equations](https://www.owlsprep.com/study/cie-0620-u3-chemical-formulae-and-equations/); [Relative Atomic and Formula Mass Calculations](https://www.owlsprep.com/study/cie-0620-u3-relative-mass-calculations/)

## Learning objectives

- Define the mole and Avogadro constant for CIE IGCSE exam use
- Calculate relative formula mass and convert between mass, moles and number of particles
- Derive empirical and molecular formulae from experimental composition data
- Perform mole calculations involving gas volume at r.t.p. and solution concentration
- Identify limiting reactants and calculate maximum product yield from reaction data

## 1. Core Mole Definitions and Mass-Mole Conversions

**Mole** — The amount of substance that contains the same number of particles (atoms, molecules, ions) as the number of atoms in 12 g of carbon-12. This fixed number is the Avogadro constant = 6.0 × 10²³ mol⁻¹.

*Notation:* Unit: mol

The core equation to convert between mass of a substance and moles uses $M_r$ (relative formula mass in g/mol, or $A_r$ for single elements), $m$ (mass in grams) and $n$ (number of moles):

$$n = \frac{m}{M_r}$$

> **tip**
>
> Use the formula triangle to rearrange the equation easily: cover the value you need to calculate to get the correct rearranged formula: $m = n \times M_r$ or $M_r = m / n$.

**Worked example:** Calculate the number of moles in 8.0 g of sodium hydroxide (NaOH). $A_r$ values: Na = 23, O = 16, H = 1.

1. Step 1: Calculate the relative formula mass of NaOH:

   $$M_r = 23 + 16 + 1 = 40$$
2. Step 2: Substitute values into the mole-mass equation:

   $$n = \frac{8.0}{40} = 0.20 \text{ mol}$$

## 2. Empirical and Molecular Formula Calculations

**Empirical vs Molecular Formula** — The empirical formula is the simplest whole number ratio of atoms of each element in a compound. The molecular formula is the actual number of each atom in one molecule of the compound, equal to (empirical formula) × n where n is a positive integer.

*Example:* Ethane: molecular formula = C₂H₆, empirical formula = CH₃, n = 2

1. Write down the mass or percentage of each element in the compound
2. Divide each mass by the element's $A_r$ to get the number of moles of each element
3. Divide all mole values by the smallest mole value to get the atomic ratio
4. Round to the nearest whole number to get the empirical formula; multiply all ratios by the same integer if you get values like 1.5 or 2.33 to get whole numbers
5. If given molecular $M_r$, calculate n = molecular $M_r$ / empirical $M_r$, then multiply empirical formula by n to get molecular formula

**Worked example:** A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180. Find its molecular formula. $A_r$ values: C=12, H=1, O=16.

1. Step 1: Assume 100 g of compound, so masses: C = 40.0 g, H = 6.7 g, O = 53.3 g
2. Step 2: Calculate moles of each element:

   $$\text{Moles C} = 40.0/12 = 3.33, \text{ Moles H} = 6.7/1 = 6.7, \text{ Moles O} = 53.3/16 = 3.33$$
3. Step 3: Divide all mole values by the smallest (3.33) to get ratio:

   $$\text{C:H:O} = 1:2:1$$
4. Step 4: Empirical formula = CH₂O, empirical $M_r$ = 12 + 2 + 16 = 30
5. Step 5: Calculate n and molecular formula:

   $$n = 180/30 = 6, \text{ Molecular formula} = C_6H_{12}O_6$$

## 3. Gas Volume and Solution Concentration Mole Calculations

**Molar Gas Volume at r.t.p.** — Volume occupied by 1 mole of any gas at room temperature and pressure (20°C, 1 atm). For CIE IGCSE exams this is always given as 24 dm³ (24000 cm³) per mole.

*Notation:* $V_m = 24 \text{ dm}^3 \text{ mol}^{-1}$

Two additional core equations for extended mole calculations: 1. Gas volume conversion: $n = V / 24$ (where V is gas volume in dm³ at r.t.p.), 2. Solution concentration: $n = c \times V$ (where c = concentration in mol/dm³, V = solution volume in dm³).

> **warning**
>
> Always check volume units: convert cm³ to dm³ by dividing by 1000 before using these equations, or use 24000 cm³/mol for gas volumes if you keep units in cm³.

**Worked example:** Calculate the volume of carbon dioxide gas produced at r.t.p. when 5.0 g of calcium carbonate reacts with excess hydrochloric acid: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$. $A_r$ values: Ca=40, C=12, O=16.

1. Step 1: Calculate moles of CaCO₃:

   $$M_r(CaCO_3) = 40 + 12 + 48 = 100, n(CaCO_3) = 5.0/100 = 0.050 \text{ mol}$$
2. Step 2: Use mole ratio from balanced equation: 1 mol CaCO₃ produces 1 mol CO₂, so n(CO₂) = 0.050 mol
3. Step 3: Calculate volume of CO₂ at r.t.p.:

   $$V = 0.050 \times 24 = 1.2 \text{ dm}^3$$

## 4. Limiting Reactant and Yield Calculations

**Limiting Reactant** — The reactant that is completely used up first in a reaction, which limits the maximum amount of product that can be formed. All other reactants are present in excess.

*Example:* 2 mol H₂ reacts with 1 mol O₂ to make H₂O: O₂ is limiting because 2 mol H₂ requires 1 mol O₂ to react fully.

1. Calculate the number of moles of all reactants given in the question
2. Use the balanced equation mole ratio to find how much product each reactant can produce
3. The reactant that produces less product is the limiting reactant
4. Use the moles of the limiting reactant to calculate the maximum yield of product

**Worked example:** 1.2 g of magnesium is added to 500 cm³ of 0.10 mol/dm³ hydrochloric acid: $Mg + 2HCl \rightarrow MgCl_2 + H_2$. Find the limiting reactant and maximum volume of H₂ produced at r.t.p. $A_r$: Mg=24.

1. Step 1: Calculate moles of Mg:

   $$n(Mg) = 1.2/24 = 0.050 \text{ mol}$$
2. Step 2: Convert solution volume to dm³ (500 cm³ = 0.5 dm³) and calculate moles of HCl:

   $$n(HCl) = 0.10 \times 0.5 = 0.050 \text{ mol}$$
3. Step 3: Use mole ratio 1 Mg : 2 HCl: 0.05 mol Mg needs 0.10 mol HCl to react fully, but only 0.05 mol HCl is available. HCl is the limiting reactant.
4. Step 4: Mole ratio HCl : H₂ = 2:1, so n(H₂) = 0.050 / 2 = 0.025 mol
5. Step 5: Calculate H₂ volume:

   $$V(H_2) = 0.025 \times 24 = 0.60 \text{ dm}^3$$

## Common pitfalls

- **Wrong:** Using relative atomic mass instead of relative formula mass for compounds in mole calculations
  - Why it fails: Relative atomic mass only applies to single elements, compounds have a combined mass that includes all atoms in their formula unit
  - Correct: Always calculate $M_r$ for all compounds by adding the $A_r$ values of every atom in their formula before substituting into mole equations
- **Wrong:** Forgetting to convert cm³ to dm³ when using molar gas volume or concentration equations
  - Why it fails: Standard molar gas volume is 24 dm³ per mole and concentration units are mol/dm³, so using cm³ directly gives results off by a factor of 1000
  - Correct: Divide volume in cm³ by 1000 to get dm³, or use 24000 cm³/mol for gas volume calculations if you keep units in cm³
- **Wrong:** Rounding mole ratios to whole numbers too early when calculating empirical formula
  - Why it fails: Rounding intermediate values can lead to incorrect ratios, especially if values are 1.5, 2.33 or 2.5 that need scaling to whole numbers
  - Correct: Keep 2-3 decimal places for intermediate ratio values, and multiply all ratios by the same integer to get whole numbers if needed
- **Wrong:** Using the mole value of the excess reactant to calculate product yield
  - Why it fails: The excess reactant is not fully consumed, so its amount does not determine how much product is formed
  - Correct: Always identify the limiting reactant first, and use its mole value to calculate the maximum possible product yield
- **Wrong:** Submitting empirical formula when the question asks for molecular formula, or vice versa
  - Why it fails: Exam questions explicitly specify which formula is required, and giving the wrong type loses marks even if calculations are correct
  - Correct: Check the question carefully: if molecular $M_r$ is provided, calculate molecular formula as a multiple of the empirical formula; if no $M_r$ is given, submit the empirical formula

## Cheatsheet

| Calculation Type | Formula | Units to Use | Key Exam Tip |
| --- | --- | --- | --- |
| Moles from mass | $n = m / M_r$ | m = g, $M_r$ = g/mol, n = mol | Use $A_r$ for elements, $M_r$ for compounds |
| Gas volume at r.t.p. | $V = n \times 24$ | V = dm³, n = mol | Use 24000 if V is in cm³ |
| Moles from solution | $n = c \times V$ | c = mol/dm³, V = dm³ | Divide cm³ by 1000 to get dm³ |
| Empirical formula | Mass/% → divide by $A_r$ → divide by smallest value → whole number ratio |  | Do not round intermediate ratios early |
| Limiting reactant | Calculate moles of all reactants → compare to equation ratio |  | Only use limiting reactant moles for product calculations |

## What's next

Now that you have mastered core mole calculation skills for CIE IGCSE Chemistry 0620 Extended, you can apply these to more complex stoichiometry problems including percentage yield and purity calculations, which appear frequently in both Paper 2 and Paper 4 extended exams. These foundational skills are also required for all quantitative chemistry topics in the syllabus, including electrolysis calculations and energy change sums. Be sure to practice a wide range of exam-style structured questions to avoid common unit and ratio mistakes, and always show your full working for all calculation questions to earn partial marks even if your final answer is incorrect.

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