Empirical & Molecular Formula

Empirical and molecular formula: find simplest whole-number ratios from mass/% data, then use Mr to get the molecular formula.

SEC G3 Pure Chemistry 2027

On this page

Learning objectives

calculate empirical and molecular formulae from relevant data

Chemical formula questions usually test two ideas: the simplest ratio of atoms (empirical formula) and the actual numbers of atoms in a molecule (molecular formula).

1. Definition

A. Empirical Formula

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound.

B. Molecular Formula

The molecular formula is the actual number of atoms of each element in one molecule of a substance.

C. Empirical vs Molecular (What Examiners Want)

Idea	Empirical formula	Molecular formula
Meaning	Simplest ratio of atoms	Actual numbers of atoms in a molecule
How you get it	Convert masses/% to moles, then ratio	Find empirical first, then use Mᵣ to scale up
Applies to	Ionic + covalent compounds	Covalent substances (molecules)

2. Key Ideas

A. Empirical Formula (Table Method)

To find an empirical formula from masses or percentages:

Convert each element’s mass/% into moles by dividing by its relative atomic mass (Aᵣ).
Divide every mole value by the smallest mole value to get a ratio.
Multiply if needed to get whole numbers.

B. Molecular Formula (Scaling Up)

Find the empirical formula.
Find the empirical formula mass (sometimes called formula mass).
Use: n = Mᵣ/(Empirical formula mass)
Multiply every subscript in the empirical formula by n.

C. Ratios That Are Not Whole Numbers

If you get ratios like 1 : 1.5 : 1, multiply all ratios by the same number to remove the decimal (e.g., multiply by 2 to get 2 : 3 : 2).

3. Detailed Explanations

If given percentages, assume 100 g so % becomes mass in g.
Convert mass → moles using moles = mass/Aᵣ.
Divide all moles by the smallest to get a ratio; multiply to remove decimals (e.g., 1.5 → ×2).
Molecular formula: n = Mᵣ/(empirical formula mass), then multiply subscripts by n.

A. Why You Must Use Moles

Masses are misleading because atoms have different masses. Ratio questions are about numbers of atoms, so you must convert to moles first.

B. Common Multipliers (When You Get Decimals)

Decimal part (approx.)	Multiply all ratios by
0.5	2
0.33	3
0.25	4
0.67	3

C. Ionic vs Covalent (Exam-Safe)

Ionic compounds do not exist as molecules, so their written formula is a ratio (empirical formula).
Covalent substances can have a molecular formula different from the empirical formula (e.g., n > 1).

If relative atomic mass (Aᵣ) and relative molecular mass (Mᵣ) feel uncertain, revise them here: Atomic & Molecular Mass

4. Common Mistakes

A. Empirical Formula Mistakes

Dividing by Mᵣ instead of Aᵣ (wrong quantity).
Rounding intermediate mole values too early. Keep guard digits until the final ratio.
Forgetting to divide by the smallest mole value.
Ignoring a decimal ratio instead of multiplying (1 : 1.5 is not “close enough”).

B. Molecular Formula Mistakes

Using Mᵣ before finding the empirical formula first.
Getting n not a whole number and “forcing” it to be (usually your empirical formula mass is wrong).

5. Exam Tips

For percentage composition, assume you have 100 g of the compound. Then “% by mass” becomes a mass in grams immediately.

Always show a clear mole calculation line: “moles = mass ÷ Aᵣ”.
State the final answer as a formula, not just a ratio (e.g., “ratio 1:2:1, so empirical formula is CH₂O”).

6. Worked Examples

Problem

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

Aᵣ:

C = 12,

H = 1,

O = 16.

Study the worked solution

Convert percentages to masses
Method
Assume a 100 g sample.
Reason
Each percentage then has the same numerical value as its mass in grams.
Working
m(C) = 40.0 g, m(H) = 6.7 g, m(O) = 53.3 g.
Convert each mass to moles
Method
Divide each element’s mass by its Aᵣ.
Reason
Formula subscripts represent amount ratios, not mass ratios.
Working
n(C) = 40.0/12 = 3.33; n(H) = 6.7/1 = 6.7; n(O) = 53.3/16 = 3.33.
Find the simplest whole-number ratio
Method
Divide all amounts by the smallest value, 3.33.
Reason
This preserves the mole ratio while scaling its smallest term to one.
Working
C:H:O = 1:2:1.
Write the empirical formula
Method
Use the ratio as subscripts.
Reason
A subscript of one is omitted.
Working
Empirical formula = CH₂O.

Problem

2.40 g of magnesium combines with 1.60 g of oxygen to form an oxide. Find the empirical formula. Use

Aᵣ:

Mg = 24,

O = 16.

Compare amounts rather than masses

Moles of Mg

Moles of O

Empirical formula

MgOMg2OMgO2

Hints

Hint 1: convert both masses

Calculate

2.40/24 and

1.60/16.

Hint 2: simplify together

Divide both mole values by 0.100.

View solution step by step

Calculate element amounts
Method
Divide each mass by the corresponding Aᵣ.
Reason
Atoms combine in mole ratios.
Working
n(Mg) = 2.40/24 = 0.100 mol; n(O) = 1.60/16 = 0.100 mol.
Simplify the ratio
Method
Divide both amounts by 0.100.
Reason
This gives the smallest whole-number ratio.
Working
Mg:O = 1:1.
Write the formula
Working
Empirical formula = MgO.

Learner response

A compound has 0.020 mol of X and 0.030 mol of Y. A student obtains

1:1.5, rounds it to

1:2 and writes

XY₂. Explain the error and find the empirical formula.

View solution step by step

Form the ratio
Method
Divide both amounts by the smallest, 0.020 mol.
Reason
Formula subscripts must preserve the measured amount ratio.
Working
X:Y = (0.020/0.020):(0.030/0.020) = 1:1.5.
Remove the half without rounding
Method
Multiply every ratio term by two.
Reason
Scaling both terms preserves the ratio; rounding only one changes it.
Working
1:1.5 → 2:3.
Correct the formula
Working
Empirical formula = X₂Y₃.

Examination question

The empirical formula of a compound is

CH₂. Its

Mᵣ is 56. Find its molecular formula. [3 marks]

Show the empirical-mass multiplier

Write the calculation and formula

View solution step by step

Find empirical formula mass
1 mark
Method
Add the mass represented by CH₂.
Reason
The molecular Mᵣ must be compared with one empirical unit.
Working
Empirical formula mass = 12 + 2(1) = 14.
Find the whole-number multiplier
1 mark
Method
Divide molecular Mᵣ by empirical formula mass.
Reason
A molecular formula contains a whole number of empirical units.
Working
n = 56/14 = 4.
Scale every subscript
1 mark
Method
Multiply both empirical subscripts by four.
Reason
The complete empirical unit repeats four times.
Working
Molecular formula = C₄H₈.

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Self-mark empirical formula mass, integer multiplier and scaled formula.

Mark points

1 mark: Calculates empirical formula mass 14.
1 mark: Calculates 56/14 = 4.
1 mark: Obtains C4H8 by scaling every subscript.

Combined-data transfer

A compound contains 92.3% carbon and 7.7% hydrogen by mass. Its

Mᵣ is 78. Find its molecular formula. Use

Aᵣ:

C = 12,

H = 1.

Hints

Hint 1: first find the empirical unit

Assume 100 g and compare

92.3/12 with

7.7/1.

Hint 2: then use Mr

Once the empirical formula is

CH, compare its mass 13 with 78.

View solution step by step

Convert composition to amounts
Method
Assume 100 g and divide each mass by Aᵣ.
Reason
Percentage data become gram masses in a 100 g sample.
Working
n(C) = 92.3/12 = 7.69; n(H) = 7.7/1 = 7.7.
Find empirical formula
Method
Divide by the smaller amount.
Reason
The values are equal within the precision of the data.
Working
C:H = 1:1.00, so empirical formula = CH.
Find molecular multiplier
Method
Divide the molecular Mᵣ by empirical formula mass.
Reason
The empirical unit CH has mass 13.
Working
n = 78/13 = 6.
Write molecular formula
Method
Multiply every empirical subscript by six.
Reason
The molecule contains six empirical units.
Working
Molecular formula = C₆H₆.

Next step

Try Mind Stretcher 1 independently.

7. Mind Stretchers

Mind stretcher 1: Use a mass change to find an empirical formulaExtension

5.60 g of iron reacts with oxygen to form 8.00 g of an oxide. Find the empirical formula of the oxide. Use Aᵣ(Fe) = 56 and Aᵣ(O) = 16.

Show Answer

First find the oxygen mass from the increase:

m(O) = 8.00-5.60 = 2.40 g

Then convert both element masses to moles:

n(Fe) = 5.60/56 = 0.100; n(O) = 2.40/16 = 0.150

The ratio is 0.100:0.150 = 1:1.5. Multiply both terms by 2 to obtain 2:3.

Final: the empirical formula is Fe₂O₃.

Mind stretcher 2: Combustion Data (Find Empirical, Then Molecular)Extension

0.260 g of a hydrocarbon burns completely to form 0.88 g of CO₂ and 0.18 g of H₂O. The hydrocarbon has Mᵣ = 78. Find its molecular formula. (Use Aᵣ: C = 12, H = 1, O = 16.)

Show Answer

From products:

n(CO₂) = 0.88/44 = 0.0200 ⇒ n(C) = 0.0200; n(H₂O) = 0.18/18 = 0.0100 ⇒ n(H) = 2 × 0.0100 = 0.0200

Ratio C:H = 0.0200 : 0.0200 = 1 : 1, so empirical formula is CH.

Empirical formula mass = 13, so n = 78/13 = 6.

Molecular formula: C₆H₆.

8. Quiz

Ready to check your understanding? Try the interactive quiz, then review any questions you missed.

Go to quiz

Continue where you stopped

Learning objectives

1. Definition

A. Empirical Formula

B. Molecular Formula

C. Empirical vs Molecular (What Examiners Want)

2. Key Ideas

A. Empirical Formula (Table Method)

B. Molecular Formula (Scaling Up)

C. Ratios That Are Not Whole Numbers

3. Detailed Explanations

A. Why You Must Use Moles

B. Common Multipliers (When You Get Decimals)

C. Ionic vs Covalent (Exam-Safe)

4. Common Mistakes

A. Empirical Formula Mistakes

B. Molecular Formula Mistakes

5. Exam Tips

6. Worked Examples

7. Mind Stretchers

Mind stretcher 1: Use a mass change to find an empirical formulaExtension

Mind stretcher 2: Combustion Data (Find Empirical, Then Molecular)Extension

8. Quiz