Quantitative Chemistry and Analysis

Empirical and Molecular Formulae – Revision Notes

Key Definitions and Terminology

**Empirical formula**: The simplest whole number ratio of atoms of each element present in a compound.
**Molecular formula**: The actual number of atoms of each element present in one molecule of a compound.
**Relative atomic mass (Aᵣ)**: The average mass of an atom of an element compared to ¹/₁₂ the mass of a carbon-12 atom.
**Relative molecular mass (Mᵣ)**: The sum of all the relative atomic masses of the atoms in a molecular formula.
**Relative formula mass**: The sum of all the relative atomic masses in a formula unit (used for ionic compounds).
**Mole (mol)**: The amount of substance that contains 6.02 × 10²³ particles (Avogadro's number); one mole of any substance has a mass equal to its Aᵣ or Mᵣ in grams.

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Main Concepts

1. Relationship Between Empirical and Molecular Formulae

The molecular formula is always a whole number multiple of the empirical formula.
**Molecular formula = (Empirical formula) × n**, where **n = Mᵣ of compound ÷ Mᵣ of empirical formula**.
If n = 1, the empirical and molecular formulae are the same (e.g., H₂O, CO₂).

2. How to Calculate an Empirical Formula

The standard method follows these steps:

Write down the mass (or percentage) of each element.
Divide each mass by the relative atomic mass (Aᵣ) of that element to find the number of moles.
Divide every mole value by the **smallest** mole value to get the simplest ratio.
If the ratio is not a whole number, multiply all values by the same factor to obtain whole numbers (e.g., multiply by 2 if you get 1.5, by 3 if you get 1.33).

3. How to Determine a Molecular Formula from an Empirical Formula

You need two pieces of information: the **empirical formula** and the **relative molecular mass (Mᵣ)** of the compound.
Calculate the Mᵣ of the empirical formula, then find n.
Multiply every subscript in the empirical formula by n.

4. Percentage Composition Data

When given percentage composition by mass, treat the percentages as if they are masses in grams (i.e., assume a 100 g sample).
This converts the problem directly into a mass-based empirical formula calculation.

5. Combustion Analysis and Empirical Formula

In combustion analysis, the masses of CO₂ and H₂O produced are used to determine the masses of carbon and hydrogen in the original compound.
Mass of C = mass of CO₂ × (12 ÷ 44)
Mass of H = mass of H₂O × (2 ÷ 18)
If oxygen is also present, its mass is found by subtraction: mass of O = total mass of compound − mass of C − mass of H.

6. Empirical Formulae of Ionic Compounds

Ionic compounds are always represented by their empirical formula because they form giant lattice structures, not discrete molecules (e.g., NaCl, MgO, CaCl₂ are already empirical formulae).

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Worked Examples

Worked Example 1: Finding the Empirical Formula from Masses

Question: A compound contains 2.4 g of carbon, 0.4 g of hydrogen and 3.2 g of oxygen. Find its empirical formula.

|------|-----------|--------------|------------|

| Mass (g) | 2.4 | 0.4 | 3.2 |

| ÷ Aᵣ | 2.4 ÷ 12 = 0.2 | 0.4 ÷ 1 = 0.4 | 3.2 ÷ 16 = 0.2 |

| ÷ smallest | 0.2 ÷ 0.2 = 1 | 0.4 ÷ 0.2 = 2 | 0.2 ÷ 0.2 = 1 |

Empirical formula = CH₂O

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Worked Example 2: Finding the Molecular Formula

Question: The empirical formula of a compound is CH₂O and its relative molecular mass is 180. Find the molecular formula.

Mᵣ of empirical formula CH₂O = 12 + (2 × 1) + 16 = **30**
n = 180 ÷ 30 = **6**
Molecular formula = C₍₁ₓ₆₎H₍₂ₓ₆₎O₍₁ₓ₆₎ = **C₆H₁₂O₆** (glucose)

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Worked Example 3: Using Percentage Composition

Question: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its Mᵣ is 60. Determine the empirical and molecular formulae.

| Step | C | H | O |

|------|---|---|---|

| Mass (assume 100 g) | 40.0 | 6.7 | 53.3 |

| ÷ Aᵣ | 40.0 ÷ 12 = 3.33 | 6.7 ÷ 1 = 6.7 | 53.3 ÷ 16 = 3.33 |

| ÷ smallest | 3.33 ÷ 3.33 = 1 | 6.7 ÷ 3.33 = 2 | 3.33 ÷ 3.33 = 1 |

Empirical formula = **CH₂O** (Mᵣ = 30)
n = 60 ÷ 30 = **2**
Molecular formula = **C₂H₄O₂** (ethanoic acid / acetic acid)

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Exam Technique Tips for Edexcel IGCSE

> Tip 1: Show every row of working clearly.

> Edexcel mark schemes award separate marks for (i) dividing masses by Aᵣ, (ii) dividing by the smallest value, and (iii) stating the correct formula. Even if your final answer is wrong, you can still pick up 2 out of 3 marks by showing the intermediate steps. Never skip straight to the answer.

> Tip 2: Watch for ratios that are not clean whole numbers.

> If dividing by the smallest gives values like 1 : 1.5 or 1 : 1.33, you must multiply through to get whole numbers (×2 gives 2 : 3; ×3 gives 3 : 4). A common Edexcel error is students leaving the ratio as decimals and losing the final mark. Always check: are all subscripts whole numbers?

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*End of revision notes — Empirical and Molecular Formulae*

Subtopics

Revision Method

Empirical and Molecular Formulae