Empirical and Molecular Formulae

Summary: Empirical formula is the simplest whole-number ratio of atoms in a compound. Molecular formula is the actual number of atoms of each element in a molecule. Empirical formula is calculated from mass or percentage composition data; molecular formula is found by comparing the Mr of the empirical formula to the actual Mr. Tags: igcse chemistry stoichiometry Created: 2026-07-14 Last Updated: 2026-07-16

Definitions

Empirical formula: The simplest whole-number ratio of atoms of each element in a compound.

Molecular formula: The actual number of atoms of each element in a molecule of the compound.

For many simple compounds (H2O, CO2, NaCl, HCl), the empirical formula and molecular formula are the same because the ratio cannot be simplified further.

Calculating Empirical Formula from Mass Data

Step-by-step method (using a table):

1. Write down the mass of each element present
2. Convert mass to moles: divide each mass by the Ar of that element (n = m / Ar)
3. Find the simplest whole-number ratio: divide all mole values by the smallest mole value
4. If the ratio is not close to whole numbers, multiply all values by an appropriate factor to get whole numbers (e.g., 1.5 x 2 = 3)

Set up a table with columns: Element | Mass (g) | Ar | Moles = mass/Ar | Ratio

Example 1: From Mass Data A compound contains 2.4 g of carbon and 0.60 g of hydrogen. Find its empirical formula. [Ar: C = 12, H = 1]

Empirical formula = CH3

Example 2: From Mass Data (Harder) 5.60 g of iron reacts with oxygen to form 8.00 g of an iron oxide. Find the empirical formula. [Ar: Fe = 56, O = 16]

Step 1: Mass of oxygen = 8.00 - 5.60 = 2.40 g

Ratio is 1 : 1.5 → multiply both by 2 → 2 : 3

Empirical formula = Fe2O3

Calculating Empirical Formula from Percentage Composition

The method is identical — just treat the percentages as masses, assuming 100 g of the compound.

Example 3: From Percentage Data A compound contains 40.0% carbon, 6.67% hydrogen, and 53.3% oxygen by mass. Find its empirical formula. [Ar: C = 12, H = 1, O = 16]

Empirical formula = CH2O

Calculating Empirical Formula from Combustion Data

In combustion analysis, a compound containing C, H, (and possibly O) is burned in excess oxygen. The masses of CO2 and H2O produced are measured.

Example 4: Combustion Analysis 0.60 g of an organic compound containing only C, H, and O is burned completely in excess oxygen. 0.88 g of CO2 and 0.36 g of H2O are produced. Find the empirical formula. [Ar: C = 12, H = 1, O = 16]

Step 1: Find mass of carbon (all C ends up in CO2)
Mr of CO2 = 44. 44 g CO2 contains 12 g C.
Mass of C in 0.88 g CO2 = (12/44) x 0.88 = 0.24 g

Step 2: Find mass of hydrogen (all H ends up in H2O)
Mr of H2O = 18. 18 g H2O contains 2 g H.
Mass of H in 0.36 g H2O = (2/18) x 0.36 = 0.040 g

Step 3: Find mass of oxygen (by subtraction)
Mass of O = 0.60 - (0.24 + 0.040) = 0.32 g

Empirical formula = CH2O

Molecular Formula from Empirical Formula

Once the empirical formula is known, the molecular formula can be found if the Mr of the compound is also known.

Step-by-step method:

1. Calculate the Mr of the empirical formula (the empirical formula mass)
2. Divide the actual Mr by the empirical formula mass to get a multiplication factor, n
3. Molecular formula = (empirical formula) x n

n = actual Mr / empirical formula mass
Molecular formula = (Empirical formula)n

Example 5: Finding Molecular Formula A compound has empirical formula CH2O and Mr = 180. Find its molecular formula.

Step 1: Mr of empirical formula CH2O = 12 + (2 x 1) + 16 = 30
Step 2: n = actual Mr / empirical Mr = 180 / 30 = 6
Step 3: Molecular formula = (CH2O)6 = C6H12O6

This is glucose.

Example 6: From Percentage to Molecular Formula A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its Mr is 56. Find its molecular formula. [Ar: C = 12, H = 1]

Step 1: Find empirical formula
Assume 100 g: 85.7 g C, 14.3 g H

Element | Mass | Ar | Moles     | Ratio
C       | 85.7 | 12 | 85.7/12 = 7.14 | 7.14/7.14 = 1
H       | 14.3 | 1  | 14.3/1 = 14.3  | 14.3/7.14 = 2

Empirical formula = CH2

Step 2: Find molecular formula
Mr of CH2 = 12 + 2 = 14
n = 56 / 14 = 4
Molecular formula = (CH2)4 = C4H8

Further examples:

Example 7: A compound contains 5.85 g of potassium, 2.10 g of nitrogen, and 7.20 g of oxygen. Find its empirical formula. [Ar: K = 39, N = 14, O = 16]

Empirical formula = KNO3 (potassium nitrate)

Example 8: A compound has empirical formula NO2 and Mr = 92. What is its molecular formula? [Ar: N = 14, O = 16]

Mr of NO2 = 14 + 32 = 46
n = 92 / 46 = 2
Molecular formula = (NO2)2 = N2O4

Key Points

• Empirical formula = simplest whole-number ratio of atoms
• Molecular formula = actual number of atoms in one molecule
• Method: mass (or %) → moles (divide by Ar) → ratio (divide by smallest) → empirical formula
• For percentage data: assume 100 g total, so % become gram values
• Molecular formula = (empirical formula) x n, where n = actual Mr / empirical formula mass
• Common ratios that need multiplying: 1:1.5 → 2:3; 1:1.33 → 3:4; 1:1.67 → 3:5; 0.5:1 → 1:2

Key Concepts from Past Papers

• Empirical formula: the simplest whole-number ratio of atoms in a compound
• Molecular formula: the actual number of atoms of each element in a molecule
• Divide mass (or %) by Ar to find moles; divide by the smallest to find the ratio
• Molecular formula = (empirical formula) x n, where n = Mr / empirical formula mass

Keywords from Past Papers

molecules, particles, concentration, movement, rate, reaction, lower, spread, koh, move, collide, mix, volume, amount, increases

Related Notes

• Relative Masses and Moles
• Reacting Masses
• Concentration and Gas Volumes

Sources

• OpenStax Chemistry 2e — [Chapter 3: Composition of Substances and Solutions (Empirical and Molecular Formulas)], Rice University (free, CC BY 4.0)
• BBC Bitesize GCSE Chemistry — [Empirical Formulae], BBC (free educational resource)
• Cambridge IGCSE Chemistry 0620 — Syllabus Section 3: Stoichiometry (Empirical and Molecular Formulae), Cambridge Assessment International Education
• CK-12 Chemistry for High School — [Chapter 10: Empirical and Molecular Formulas], CK-12 Foundation (free, CC BY-NC 3.0)

Past Paper Sources

• 0620/31 May/June 2015: Q22(b)(iv) (0m), Q44(a)(iii) (0m), Q55(c)(i) (0m) (+3 more)
• 0620/32 Feb/March 2017: Q33(b)(v) (2m)
• 0620/32 May/June 2018: Q11(b)(i) (1m)
• 0620/32 May/June 2020: Q66(b)(i) (1m)
• 0620/33 May/June 2016: Q66(d)(i) (1m)
• 0620/33 May/June 2017: Q66(b)(iv) (1m)
• 0620/33 May/June 2021: Q22(d)(ii) (3m)
• 0620/33 May/June 2022: Q33(b)(ii) (1m)
• 0620/33 May/June 2023: Q88(c)(v) (0m)
• 0620/33 May/June 2024: Q66(c)(ii) (1m)
• 0620/33 October/November 2015: Q77(b)(i) (2m), Q77(b)(ii) (0m)
• 0620/33 October/November 2018: Q22(a)(i) (2m)

Common Misconceptions