Empirical and Molecular Formulas — AQA A-level Chemistry

SPEC 3.1.2.4 - AMOUNT OF SUBSTANCE

Empirical & Molecular Formula

Determining the simplest ratio and actual number of atoms in compounds from experimental data

What Formulas Tell Us

Chemical formulas describe the composition of compounds. In chemistry, we use two key types of formula that give us different information:

Key Distinction

Empirical Formula shows the simplest whole number ratio of atoms of each element in a compound.

Molecular Formula shows the actual number of atoms of each element in one molecule of the compound.

Learning Objectives

By the end of this lesson, you will be able to:

Define empirical and molecular formula and explain the relationship between them

Calculate empirical formula from mass or percentage composition data

Determine molecular formula using empirical formula and relative molecular mass

Empirical vs Molecular Formula

The empirical formula is the most reduced form of a molecular formula - you cannot simplify it further. The molecular formula is always a whole number multiple of the empirical formula.

MOLECULAR = n × EMPIRICAL

where n is a whole number (1, 2, 3, 4...)

Remember

For some compounds, the empirical and molecular formulas are the same! For example, water (H₂O) cannot be simplified further, so H₂O is both its molecular and empirical formula.

Why We Need Both Formulas

Different types of formula serve different purposes in chemistry:

Empirical FormulaMolecular Formula

Found from experimental analysis (combustion analysis, mass spectrometry)Needed to identify the actual compound

Shows ratio of atoms onlyShows exact number of atoms per molecule

Multiple compounds can share the same empirical formulaUnique to each molecular compound

Important

Different compounds can have the same empirical formula! For example, both glucose (C₆H₁₂O₆) and formaldehyde (CH₂O) have the empirical formula CH₂O. This is why we need the molecular formula to identify the actual compound.

The Relationship

Molecular Formula = n \times Empirical Formula

where

n

is a whole number (1, 2, 3, 4, ...)

Compound Types

Organic molecules often have different empirical and molecular formulae (e.g., glucose C₆H₁₂O₆ → CH₂O)

Simple inorganic molecules often have similar empirical and molecular formulae (e.g., H₂O, CO₂)

Ionic compounds always have similar empirical and molecular formulae (e.g., NaCl, MgO)

01Worked Example

Determine the empirical formula for each of the following molecular formulas:

(a) C₄H₁₀ (b) C₆H₆ (c) N₂O₄ (d) C₃H₆O₃

a)

C₄H₁₀ - Find the highest common factor (HCF) of 4 and 10 = 2

Divide both subscripts by 2: C₄÷₂H₁₀÷₂ = C₂H₅

b)

C₆H₆ - HCF of 6 and 6 = 6

Divide both by 6: C₆÷₆H₆÷₆ = CH

c)

N₂O₄ - HCF of 2 and 4 = 2

Divide both by 2: N₂÷₂O₄÷₂ = NO₂

d)

C₃H₆O₃ - HCF of 3, 6, and 3 = 3

Divide all by 3: C₃÷₃H₆÷₃O₃÷₃ = CH₂O

Calculating Empirical Formula

To find the empirical formula from experimental data, follow these four steps:

Memory Aid

"Mass - Moles - Minimise - Multiply" - The 4 M's of finding empirical formula!

Understanding Each Step

1Get the masses

If given percentages, treat them as masses by assuming 100g total. For example, 40% carbon becomes 40g of carbon.

2Convert mass to moles

Use

n = \frac{mass}{A _{r}}

for each element. This converts from grams to amount of substance.

3Divide by the smallest

Find the smallest mole value and divide ALL mole values by this number. This gives you the mole ratio.

4Convert to whole numbers

If the ratio contains decimals, multiply ALL values by an appropriate factor to get whole numbers.

Essential Formula

n = \frac{m}{M}

where n = moles, m = mass in grams, M = molar mass (Aᵣ for elements)

02Worked Example

A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

[Aᵣ: C = 12.0, H = 1.0, O = 16.0]

Empirical Formula: CH₂O

03Worked Example

Analysis of a compound shows it contains 2.4g of carbon and 0.4g of hydrogen. What is the empirical formula?

[Aᵣ: C = 12.0, H = 1.0]

1

Masses given: C = 2.4g, H = 0.4g

2

Convert to moles:

n(C) = 2.4 ÷ 12.0 = 0.20 mol

n(H) = 0.4 ÷ 1.0 = 0.40 mol

3

Divide by smallest (0.20):

C: 0.20 ÷ 0.20 = 1

H: 0.40 ÷ 0.20 = 2

4

Ratio: C : H = 1 : 2

Empirical Formula: CH₂

Empirical Formula from Combustion Data

When a hydrocarbon (or compound containing C, H, O) is completely combusted in oxygen, it produces CO₂ and H₂O. We can use the masses of these products to find the empirical formula.

Key Principle

All the carbon in the original compound ends up in CO₂

All the hydrogen in the original compound ends up in H₂O

Method

Step 1: Calculate moles of CO₂ and H₂O produced

Step 2: Find moles of C atoms (= moles of CO₂, since each CO₂ has 1 C)

Step 3: Find moles of H atoms (= 2 × moles of H₂O, since each H₂O has 2 H)

Step 4: Divide by smallest to get the ratio → empirical formula

Remember

Mᵣ(CO₂) = 12 + (2 × 16) = 44 | Mᵣ(H₂O) = (2 × 1) + 16 = 18

04Worked Example

A hydrocarbon sample weighing 2.20 g was completely combusted in oxygen, producing 3.30 g of CO₂ and 1.35 g of H₂O. Determine the empirical formula of the hydrocarbon.

[Mᵣ: CO₂ = 44.0, H₂O = 18.0]

1

Calculate moles of products:

n(CO₂) = 3.30 ÷ 44.0 = 0.075 mol

n(H₂O) = 1.35 ÷ 18.0 = 0.075 mol

2

Find moles of C atoms:

Each CO₂ contains 1 carbon atom

Moles of C = moles of CO₂ = 0.075 mol

3

Find moles of H atoms:

Each H₂O contains 2 hydrogen atoms

Moles of H = 2 × moles of H₂O = 2 × 0.075 = 0.150 mol

4

Find simplest ratio:

C : H = 0.075 : 0.150

Divide by smallest: 0.075 ÷ 0.075 : 0.150 ÷ 0.075 = 1 : 2

Empirical Formula: CH₂

From Empirical to Molecular Formula

To convert an empirical formula to a molecular formula, you need to know the relative molecular mass (Mᵣ) of the compound.

The Relationship

M_{r} (molecular) = n \times M_{r} (empirical)

Therefore:

n = \frac{M _{r} (molecular)}{M _{r} (empirical)}

Two-Step Process

Step 1: Calculate n = Mᵣ(molecular) ÷ Mᵣ(empirical)

Step 2: Multiply ALL subscripts in empirical formula by n

Calculating Empirical Formula Mass

To find n, you first need to calculate the Mᵣ of the empirical formula by adding up the Aᵣ values of all atoms:

M_{r} (empirical) = \sum (n \times A_{r})

for each element

Empirical FormulaCalculationMᵣ

CH₂O12 + (2 × 1) + 1630

CH₃12 + (3 × 1)15

NO₂14 + (2 × 16)46

CH₂Cl12 + (2 × 1) + 35.549.5

Exam Tip

Always show your calculation for Mᵣ of the empirical formula - this earns method marks even if your final answer is wrong!

05Worked Example

A compound has the empirical formula CH₂O and a relative molecular mass of 180. Determine its molecular formula.

[Aᵣ: C = 12.0, H = 1.0, O = 16.0]

1

Calculate Mᵣ of empirical formula (CH₂O):

Mᵣ = 12.0 + (2 × 1.0) + 16.0 = 30

2

Find n:

n = Mᵣ(molecular) ÷ Mᵣ(empirical)

n = 180 ÷ 30 = 6

3

Multiply empirical formula by n:

Molecular formula = 6 × CH₂O

= C₆H₁₂O₆

Molecular Formula: C₆H₁₂O₆ (glucose)

06Worked Example

A compound X contains 24.2% carbon, 4.1% hydrogen, and 71.7% chlorine by mass. The relative molecular mass of X is 99.0. Calculate the empirical formula and then the molecular formula of X.

[Aᵣ: C = 12.0, H = 1.0, Cl = 35.5]

Empirical formula: CH₂Cl

→

Mᵣ(empirical) = 12.0 + (2 × 1.0) + 35.5 = 49.5

→

n = 99.0 ÷ 49.5 = 2

→

Molecular formula = 2 × CH₂Cl = C₂H₄Cl₂

Molecular Formula: C₂H₄Cl₂ (1,2-dichloroethane)

Handling Non-Integer Ratios

Sometimes after dividing by the smallest, you get decimal values that are NOT close to whole numbers. In these cases, you need to multiply ALL values by an appropriate factor.

If you get...Multiply all byExample

0.5 or X.521 : 1.5 becomes 2 : 3

0.33 or 0.6731 : 1.33 becomes 3 : 4

0.25 or 0.7541 : 1.25 becomes 4 : 5

0.2 or 0.4 or 0.6 or 0.851 : 1.4 becomes 5 : 7

Common Trap

Values like 0.99, 1.01, or 2.02 should be rounded to the nearest whole number (1, 1, 2). These small deviations are due to experimental error or rounding. Only multiply when you have clear fractions like 0.5, 0.33, 0.67, etc.

Recognition Tip

0.33 ≈ ⅓, 0.67 ≈ ⅔, 0.25 = ¼, 0.75 = ¾

Recognise these fractions and multiply by the denominator!

07Worked Example

A compound contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass. Determine its empirical formula.

[Aᵣ: C = 12.0, H = 1.0, O = 16.0]

1

Assume 100g: C = 52.2g, H = 13.0g, O = 34.8g

2

Convert to moles:

n(C) = 52.2 ÷ 12.0 = 4.35 mol

n(H) = 13.0 ÷ 1.0 = 13.0 mol

n(O) = 34.8 ÷ 16.0 = 2.175 mol

3

Divide by smallest (2.175):

C: 4.35 ÷ 2.175 = 2.0 H: 13.0 ÷ 2.175 = 5.98 ≈ 6 O: 2.175 ÷ 2.175 = 1

4

Ratio is already whole numbers: C : H : O = 2 : 6 : 1

Empirical Formula: C₂H₆O (ethanol)

Exam Tips

AQA Command Words

"Determine" - Use data to calculate the formula (show all working)

"Calculate" - A numerical answer is required (show method)

"Deduce" - Reach a conclusion from information given

Common Mistakes to Avoid

Forgetting to divide by the SMALLEST mole value (not the largest!)

Rounding too early - keep at least 3 significant figures during calculations

Confusing Aᵣ (relative atomic mass) with mass in grams

Forgetting to multiply ALL subscripts when finding molecular formula

Not recognising common fractions (0.5, 0.33, 0.67, etc.)

Checking Your Answer

Calculate the percentage composition from your empirical formula - it should match the original data (within rounding error).

Practice

Question 1: A compound contains 75.0% carbon and 25.0% hydrogen by mass. Determine its empirical formula.

[Aᵣ: C = 12.0, H = 1.0]

1

Assume 100g: C = 75.0g, H = 25.0g

2

n(C) = 75.0 ÷ 12.0 = 6.25 mol

n(H) = 25.0 ÷ 1.0 = 25.0 mol

3

Divide by smallest (6.25):

C: 6.25 ÷ 6.25 = 1 H: 25.0 ÷ 6.25 = 4

Empirical Formula: CH₄

Practice

Question 2: A compound has the empirical formula NO₂ and a relative molecular mass of 92. Determine its molecular formula.

[Aᵣ: N = 14.0, O = 16.0]

1

Calculate Mᵣ of NO₂:

Mᵣ = 14.0 + (2 × 16.0) = 46

2

Find n:

n = 92 ÷ 46 = 2

3

Molecular formula = 2 × NO₂ = N₂O₄

Molecular Formula: N₂O₄ (dinitrogen tetroxide)

Practice

Question 3: A compound contains 26.7% carbon, 2.2% hydrogen, and 71.1% oxygen by mass. Its Mᵣ is 90. Determine its molecular formula.

[Aᵣ: C = 12.0, H = 1.0, O = 16.0]

1

Assume 100g: n(C) = 26.7 ÷ 12.0 = 2.225 mol

n(H) = 2.2 ÷ 1.0 = 2.2 mol n(O) = 71.1 ÷ 16.0 = 4.44 mol

2

Divide by smallest (2.2):

C: 2.225 ÷ 2.2 = 1.01 ≈ 1 H: 2.2 ÷ 2.2 = 1 O: 4.44 ÷ 2.2 = 2.02 ≈ 2

Empirical formula = CHO₂

3

Mᵣ of CHO₂ = 12 + 1 + 32 = 45

n = 90 ÷ 45 = 2

4

Molecular formula = 2 × CHO₂ = C₂H₂O₄

Molecular Formula: C₂H₂O₄ (oxalic acid)

Summary

Key Definitions

TermDefinition

Empirical FormulaSimplest whole number ratio of atoms of each element in a compound

Molecular FormulaActual number of atoms of each element in one molecule

RelationshipMolecular = n × Empirical (where n is a whole number)

The 4-Step Method

Mass - Get masses (assume 100g if given percentages)

Moles - Convert each mass to moles using n = m ÷ Aᵣ

Minimise - Divide all mole values by the smallest

Multiply - If needed, multiply to get whole numbers

n = \frac{M _{r} (molecular)}{M _{r} (empirical)}