The Mole and Mole Calculations

Chemistry· 3.3· 25 min read

1. 1. Core Mole Definitions and Mass-Mole Conversions★★☆☆☆Extended only⏱ 7 min

📘 Definition

Mole

Unit: mol

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.

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}

📐 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. 2. Empirical and Molecular Formula Calculations★★★☆☆Extended only⏱ 6 min

📘 Definition

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

Write down the mass or percentage of each element in the compound
Divide each mass by the element's $A_{r}$ to get the number of moles of each element
Divide all mole values by the smallest mole value to get the atomic ratio
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
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. 3. Gas Volume and Solution Concentration Mole Calculations★★★☆☆Extended only⏱ 7 min

📘 Definition

Molar Gas Volume at r.t.p.

$V_{m} = 24 dm^{3} mol^{- 1}$

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.

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).

📐 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: $C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2}$ . $A_{r}$ 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. 4. Limiting Reactant and Yield Calculations★★★★☆Extended only⏱ 7 min

📘 Definition

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.

Calculate the number of moles of all reactants given in the question
Use the balanced equation mole ratio to find how much product each reactant can produce
The reactant that produces less product is the limiting reactant
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: $M g + 2 H C l \to M g C l_{2} + H_{2}$ . 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:
$V(H_2) = 0.025 \times 24 = 0.60 \text{ dm}^3$

5. Common Pitfalls

Wrong move:

Using relative atomic mass instead of relative formula mass for compounds in mole calculations

Why:

Relative atomic mass only applies to single elements, compounds have a combined mass that includes all atoms in their formula unit

Correct move:

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 move:

Forgetting to convert cm³ to dm³ when using molar gas volume or concentration equations

Why:

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 move:

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 move:

Rounding mole ratios to whole numbers too early when calculating empirical formula

Why:

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 move:

Keep 2-3 decimal places for intermediate ratio values, and multiply all ratios by the same integer to get whole numbers if needed

Wrong move:

Using the mole value of the excess reactant to calculate product yield

Why:

The excess reactant is not fully consumed, so its amount does not determine how much product is formed

Correct move:

Always identify the limiting reactant first, and use its mole value to calculate the maximum possible product yield

Wrong move:

Submitting empirical formula when the question asks for molecular formula, or vice versa

Why:

Exam questions explicitly specify which formula is required, and giving the wrong type loses marks even if calculations are correct

Correct move:

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

6. Quick Reference 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

7. Frequently Asked

What value of molar gas volume do I use for CIE IGCSE 0620 exams?

CIE specifies the molar gas volume at room temperature and pressure (r.t.p.) as 24 dm³ per mole (or 24000 cm³ per mole) for all 0620 questions, no other values are required.

When do I need to use Avogadro's constant in calculations?

You only use the Avogadro constant (6.0 × 10²³ mol⁻¹) if the question explicitly asks for the number of particles (atoms, molecules, ions). Most mole calculations use only mass, volume or concentration conversions that do not require this value.

How do I tell the difference between empirical and molecular formula questions?

Empirical formula questions ask for the simplest whole number ratio of atoms. If the question provides the relative molecular mass of the compound, you will need to calculate the molecular formula, which is an integer multiple of the empirical formula.

Going deeper

official_documentCIE IGCSE Chemistry 0620 2026-2028 SyllabusRefer to Section 3.3 for official mole calculation requirements

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.