Why Do We Use Empirical Formulae?

Chemical experiments can often determine the relative quantities of elements in a compound.

For example, an experiment might tell us that a compound contains:

• 40.0% carbon
• 6.7% hydrogen
• 53.3% oxygen

From this information, we can determine the ratio of moles of atoms.

That ratio allows us to determine the empirical formula.

The general pathway is:

mass or percentage → moles → simplest mole ratio → empirical formula

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The Basic Method

Most empirical formula calculations follow four main steps.

Find the mass of each element

The question may provide:

• actual masses
• percentage composition
• experimental measurements

Convert each mass to moles

Use:

n = m/M

where:

• n = amount in moles
• m = mass in grams
• M = molar mass in g/mol

Find the simplest mole ratio

Divide every mole value by the smallest number of moles.

Convert the ratio to whole numbers

Use the resulting whole numbers as the subscripts in the empirical formula.

A useful memory aid is:

MASS → MOLES → DIVIDE → WHOLE NUMBERS → FORMULA

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Why We Convert Mass to Moles

An empirical formula represents the ratio of atoms, not the ratio of their masses.

Different atoms have different masses.

For example:

12 g carbon = 1 mol C

but:

1 g hydrogen ≈ 1 mol H

Therefore, a compound containing 12 g carbon and 1 g hydrogen contains approximately equal numbers of carbon and hydrogen atoms.

Its ratio is approximately:

C : H = 1 : 1

not:

12 : 1

This is why we must convert masses to moles before finding the formula.

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Empirical Formula from Mass Data

Suppose a compound contains:

• 24 g carbon
• 4 g hydrogen

Find the empirical formula.

Convert carbon to moles

n(C) = 24 / 12

n(C) = 2 mol

Convert hydrogen to moles

n(H) = 4 / 1

n(H) = 4 mol

Write the mole ratio

C : H = 2 : 4

Divide by the smallest value

Divide both by 2:

C : H = 1 : 2

Write the formula

CH₂

Answer

Empirical formula = CH₂

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Worked Example: Magnesium Oxide

A sample contains:

• 2.4 g Mg
• 1.6 g O

Use:

Ar(Mg) = 24

Ar(O) = 16

Calculate moles of Mg

n = 2.4 / 24

= 0.10 mol

Calculate moles of O

n = 1.6 / 16

= 0.10 mol

Find the ratio

Mg : O = 0.10 : 0.10

Divide by 0.10:

1 : 1

Empirical formula

MgO

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Worked Example: Iron Oxide

An iron oxide contains:

• 11.2 g Fe
• 4.8 g O

Use:

Ar(Fe) = 56

Ar(O) = 16

Iron

n(Fe) = 11.2 / 56

= 0.20 mol

Oxygen

n(O) = 4.8 / 16

= 0.30 mol

Ratio

Fe : O = 0.20 : 0.30

Divide by 0.20:

1 : 1.5

We cannot use 1.5 as a subscript in an empirical formula.

Multiply both numbers by 2:

2 : 3

Answer

Fe₂O₃

This is an important example because the first division does not always produce whole numbers.

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Dealing with Decimal Ratios

Sometimes dividing by the smallest mole value gives a ratio such as:

1 : 1.5

or:

1 : 1.33

or:

1 : 1.25

Do not simply round these values to the nearest whole number.

Instead, multiply all parts of the ratio by a suitable small integer.

Common patterns include:

For example:

1 : 1.5

multiply by 2:

2 : 3

Similarly:

1 : 1.33

multiply by 3:

3 : 4

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Do Not Round Too Quickly

Suppose your calculated ratio is:

1 : 1.49

This is probably intended to represent:

1 : 1.5

not:

1 : 1

Small differences often result from:

• experimental uncertainty
• rounded atomic masses
• measurement uncertainty

Use chemical reasoning when interpreting the numbers.

However, a value such as:

1.05

is reasonably close to:

1

The goal is to identify the most reasonable simple whole-number ratio.

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Empirical Formula from Percentage Composition

Percentage composition problems use the same method.

A useful shortcut is to imagine that you have:

100 g of the compound

Then:

percentage = mass in grams

For example:

40% carbon becomes:

40 g C

6.7% hydrogen becomes:

6.7 g H

53.3% oxygen becomes:

53.3 g O

This makes the calculation much easier.

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Worked Example: Percentage Composition

A compound contains:

• 40.0% C
• 6.7% H
• 53.3% O

Determine its empirical formula.

Assume:

100 g compound

Therefore:

• C = 40.0 g
• H = 6.7 g
• O = 53.3 g

Convert carbon to moles

n(C) = 40.0 / 12.0

= 3.33 mol

Convert hydrogen to moles

n(H) = 6.7 / 1.0

= 6.7 mol

Convert oxygen to moles

n(O) = 53.3 / 16.0

= 3.33 mol

Find the ratio

C : H : O

3.33 : 6.7 : 3.33

Divide everything by 3.33:

1 : 2.01 : 1

Approximately:

1 : 2 : 1

Answer

CH₂O

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Another Percentage Example

A compound contains:

• 52.2% C
• 13.0% H
• 34.8% O

Assume a 100 g sample.

Therefore:

• C = 52.2 g
• H = 13.0 g
• O = 34.8 g

Convert to moles

Carbon:

52.2 / 12 = 4.35 mol

Hydrogen:

13.0 / 1 = 13.0 mol

Oxygen:

34.8 / 16 = 2.175 mol

Divide by the smallest

Smallest = 2.175

Carbon:

4.35 / 2.175 = 2

Hydrogen:

13.0 / 2.175 ≈ 6

Oxygen:

2.175 / 2.175 = 1

Ratio:

2 : 6 : 1

Answer

C₂H₆O

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A Three-Element Example

A compound contains:

• 27.3% C
• 72.7% O

Assume:

100 g

Therefore:

27.3 g C

72.7 g O

Calculate moles:

C = 27.3 / 12 = 2.275 mol

O = 72.7 / 16 = 4.544 mol

Divide by the smallest:

C = 2.275 / 2.275 = 1

O = 4.544 / 2.275 ≈ 2

Answer

CO₂

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Experimental Determination of Empirical Formulae

Empirical formulae can be determined experimentally by measuring how much of each element combines.

A classic example is the formation of magnesium oxide.

Magnesium reacts with oxygen:

magnesium + oxygen → magnesium oxide

Symbolically:

2Mg + O₂ → 2MgO

A student can measure:

1. the mass of the empty crucible
2. the mass of the crucible + magnesium
3. the mass of the crucible + magnesium oxide after heating

From these measurements, the masses of magnesium and oxygen can be determined.

Finding Masses from Experimental Data

Suppose:

mass of crucible = 25.60 g

mass of crucible + Mg = 28.00 g

mass of crucible + MgO = 29.60 g

First find the mass of magnesium:

28.00 − 25.60 = 2.40 g Mg

Then find the mass of magnesium oxide:

29.60 − 25.60 = 4.00 g MgO

The oxygen mass is:

4.00 − 2.40 = 1.60 g O

Now calculate moles.

Mg:

2.40 / 24 = 0.100 mol

O:

1.60 / 16 = 0.100 mol

Ratio:

1 : 1

Answer

MgO

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Why Heat to Constant Mass?

In experiments such as the magnesium oxide investigation, the sample may be:

1. heated
2. cooled
3. weighed
4. reheated
5. cooled
6. weighed again

This continues until the mass no longer changes significantly.

This is called heating to constant mass.

A constant mass provides evidence that the reaction is complete.

If heating causes the mass to increase again, more oxygen may still be reacting with the magnesium.

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Experimental Errors

Experimental empirical formulae are not always perfect.

Suppose the true ratio should be:

1 : 1

but the experiment produces:

1 : 0.92

This may result from experimental error.

Possible sources include:

• incomplete reaction
• loss of product
• contamination
• inaccurate measurements
• side reactions
• material escaping during heating

Scientists must evaluate whether their calculated ratio is reasonably consistent with a simple whole-number ratio.

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Example: Finding an Empirical Formula from Combustion Data

A compound contains only carbon and hydrogen.

A sample contains:

• 3.6 g C
• 0.6 g H

Calculate moles.

Carbon:

3.6 / 12 = 0.30 mol

Hydrogen:

0.6 / 1 = 0.60 mol

Ratio:

0.30 : 0.60

Divide by 0.30:

1 : 2

Answer

CH₂

This tells us the simplest ratio but not necessarily the actual molecular formula.

The compound could potentially have a molecular formula such as:

C₂H₄

C₃H₆

C₄H₈

and so on.

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Empirical Formula Does Not Identify a Compound Uniquely

Consider:

CH₂O

This is an empirical formula.

Possible molecular formulae include:

CH₂O

C₂H₄O₂

C₃H₆O₃

C₆H₁₂O₆

All of these reduce to:

CH₂O

Therefore, knowing the empirical formula alone is not always enough to identify a compound.

Additional information, such as the molar mass, may be required.

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From Empirical Formula to Molecular Formula

Suppose the empirical formula is:

CH₂O

First calculate the empirical formula mass:

C:

1 × 12 = 12

H:

2 × 1 = 2

O:

1 × 16 = 16

Total:

12 + 2 + 16 = 30 g/mol

Suppose the actual molecular molar mass is:

180 g/mol

Calculate:

180 / 30 = 6

Therefore, multiply every subscript in CH₂O by 6:

C₆H₁₂O₆

Molecular formula

C₆H₁₂O₆

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Another Molecular Formula Example

The empirical formula of a compound is:

CH

Empirical formula mass:

12 + 1 = 13 g/mol

The molecular molar mass is:

78 g/mol

Calculate:

78 / 13 = 6

Multiply the empirical formula by 6:

C₆H₆

Answer

Molecular formula = C₆H₆

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When Empirical and Molecular Formulae Are the Same

Suppose the empirical formula is:

H₂O

Its empirical formula mass is:

18 g/mol

If the molecular molar mass is also:

18 g/mol

then:

18 / 18 = 1

Therefore, the molecular formula is simply:

H₂O

No multiplication is needed.

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Percentage Composition from an Empirical Formula

We can also work in the opposite direction.

Suppose a compound has empirical formula:

CH₂O

Formula mass:

12 + 2 + 16 = 30

Carbon percentage:

(12 / 30) × 100 = 40.0%

Hydrogen percentage:

(2 / 30) × 100 ≈ 6.7%

Oxygen percentage:

(16 / 30) × 100 ≈ 53.3%

This matches the composition used in our earlier example.

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Worked Example with a Decimal Ratio

A compound contains:

• 1.2 g C
• 0.30 g H

Calculate moles:

C:

1.2 / 12 = 0.10 mol

H:

0.30 / 1 = 0.30 mol

Ratio:

0.10 : 0.30

Divide by 0.10:

1 : 3

Answer

CH₃

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Worked Example Requiring Multiplication

A compound contains:

• 5.6 g Fe
• 2.4 g O

Calculate moles:

Fe:

5.6 / 56 = 0.10 mol

O:

2.4 / 16 = 0.15 mol

Divide by 0.10:

1 : 1.5

Multiply both by 2:

2 : 3

Answer

Fe₂O₃

Never round:

1 : 1.5

to:

1 : 2

That would incorrectly produce FeO₂.

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Worked Example with a 1.33 Ratio

Suppose a calculation gives:

C : H = 1 : 1.33

Recognize:

1.33 ≈ 4/3

Multiply the entire ratio by 3:

3 : 4

Therefore:

C₃H₄

The important step is multiplying every value in the ratio, not just the decimal value.

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Percentage Data Must Total Approximately 100%

If all elements in the compound are listed, their percentages should add to approximately:

100%

For example:

40.0 + 6.7 + 53.3 = 100.0%

Small differences may occur because percentages have been rounded.

If the total is very different from 100%, check whether:

• an element is missing
• the data were copied incorrectly
• the question provides only partial composition information

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Finding a Missing Percentage

Suppose a compound contains:

• 48.0% carbon
• 8.0% hydrogen
• the remainder oxygen

Find the oxygen percentage:

100 − 48.0 − 8.0

= 44.0% oxygen

You can then proceed with the usual empirical formula calculation.

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Worked Example: Missing Percentage

A compound contains:

• 40.0% sulfur
• 60.0% oxygen

Use:

Ar(S) = 32

Ar(O) = 16

Assume 100 g.

S:

40.0 / 32 = 1.25 mol

O:

60.0 / 16 = 3.75 mol

Divide by 1.25:

S : O = 1 : 3

Answer

SO₃

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Experimental Example: Metal Oxide

A metal oxide is produced by reacting:

5.40 g aluminium

with oxygen.

The final oxide has a mass of:

10.20 g

Find the empirical formula.

Find oxygen mass

10.20 − 5.40 = 4.80 g O

Calculate aluminium moles

Use:

Ar(Al) = 27

n(Al) = 5.40 / 27

= 0.200 mol

Calculate oxygen moles

n(O) = 4.80 / 16

= 0.300 mol

Divide by the smallest

Al : O = 0.200 : 0.300

Divide by 0.200:

1 : 1.5

Multiply by 2:

2 : 3

Answer

Al₂O₃

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Experimental Example: Copper Oxide

A sample contains:

6.35 g Cu

and reacts with:

1.60 g O

Use:

Ar(Cu) = 63.5

Ar(O) = 16.0

Copper:

6.35 / 63.5 = 0.100 mol

Oxygen:

1.60 / 16.0 = 0.100 mol

Ratio:

1 : 1

Answer

CuO

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Empirical Formulae and Ionic Compounds

For ionic compounds, the formula already represents the simplest whole-number ratio of ions.

For example:

NaCl

represents:

Na⁺ : Cl⁻ = 1 : 1

and:

CaCl₂

represents:

Ca²⁺ : Cl⁻ = 1 : 2

We do not normally speak of a molecular formula for ionic compounds because they form extended ionic lattices rather than discrete molecules.

Empirical Formulae and Experimental Evidence

Empirical formula calculations illustrate an important feature of chemistry:

We can use macroscopic measurements such as mass to determine information about particles that are far too small to observe individually.

A balance might tell us that two elements combined in certain masses.

Atomic masses allow us to convert those measurements to moles.

Mole ratios then reveal the relative numbers of atoms.

So:

measured mass → moles → atomic ratio → chemical formula

This is an excellent example of how experimental evidence can reveal microscopic chemical structure.

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A Reliable Problem-Solving Method

For almost every empirical formula problem, follow this sequence:

Write the elements

Keep the elements in the same order throughout the calculation.

Write the masses

If percentages are given, assume a 100 g sample.

Divide by atomic mass

Calculate:

moles = mass / molar mass

Divide by the smallest mole value

This produces the simplest relative ratio.

Examine the decimals

If necessary:

• ×2 for approximately 0.5
• ×3 for approximately 0.33 or 0.67
• ×4 for approximately 0.25 or 0.75

Write the empirical formula

Use the whole-number ratios as subscripts.

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A Calculation Table Can Help

For example:

Therefore:

CH₂O

This format can make multi-element calculations easier to organize.

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Common Mistakes

Using the Mass Ratio as the Formula

If a compound contains:

24 g C

and:

4 g H

do not use:

24 : 4 = 6 : 1

The empirical formula is based on the mole ratio, not the mass ratio.

Always convert:

mass → moles

first.

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Multiplying Instead of Dividing by Atomic Mass

Use:

n = m/M

not:

n = m × M

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Forgetting the 100 g Assumption

For percentage composition problems, assuming:

100 g compound

allows percentages to be treated directly as grams.

For example:

32% oxygen → 32 g oxygen

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Rounding 1.5 to 2

A ratio of:

1 : 1.5

should not be rounded to:

1 : 2

Instead multiply by 2:

2 : 3

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Rounding Too Early

Keep several decimal places during intermediate calculations.

Early rounding can turn a recognizable ratio into an incorrect one.

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Forgetting to Divide Every Value

If the mole amounts are:

0.20 : 0.40 : 0.60

divide all values by 0.20:

1 : 2 : 3

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Forgetting to Multiply Every Ratio

If the result is:

1 : 1.5 : 1

multiply every number by 2:

2 : 3 : 2

not:

1 : 3 : 1

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Assuming Empirical Formula Means Molecular Formula

An empirical formula of:

CH₂O

does not automatically mean the molecule is CH₂O.

Its molecular formula could be:

C₂H₄O₂

C₃H₆O₃

C₆H₁₂O₆

or another whole-number multiple.

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Using the Wrong Experimental Mass

In a metal oxide experiment:

mass of oxygen = mass of oxide − mass of metal

Do not use the total oxide mass as the oxygen mass.

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Key Terms

Empirical formula — The chemical formula showing the simplest whole-number ratio of atoms of each element in a compound.

Molecular formula — The formula showing the actual number of atoms of each element in one molecule.

Percentage composition — The percentage by mass contributed by each element in a compound.

Mole ratio — The relative numbers of moles of substances or elements.

Relative atomic mass (Ar) — The average mass of an atom relative to 1/12 of the mass of carbon-12.

Molar mass — The mass of one mole of a substance, usually expressed in g/mol.

Formula mass — The sum of the relative atomic masses represented by a chemical formula.

Whole-number ratio — A ratio expressed using integers such as 1:2, 2:3, or 3:4.

Constant mass — A mass that remains unchanged after repeated heating, cooling, and weighing, suggesting a reaction or drying process is complete.

Percentage by mass — The fraction of a compound's total mass contributed by a particular element, expressed as a percentage.

Experimental data — Measurements obtained during an investigation and used to determine quantities such as masses and mole ratios.

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Key Takeaways

• An empirical formula gives the simplest whole-number ratio of atoms in a compound.
• A molecular formula gives the actual number of atoms in a molecule.
• The empirical and molecular formulae may be the same, but they do not have to be.
• Empirical formulae are based on mole ratios, not mass ratios.
• The main calculation pathway is:

MASS → MOLES → DIVIDE BY SMALLEST → WHOLE-NUMBER RATIO → FORMULA

• Convert mass to moles using:

n = m/M

• For percentage composition problems, assuming a 100 g sample allows percentages to be treated as masses in grams.
• Divide all mole values by the smallest mole value.
• Ratios such as 1.5, 1.33, and 1.25 should usually be converted to whole numbers by multiplying the entire ratio.
• Do not round fractional ratios too aggressively.
• Experimental measurements can be used to determine empirical formulae.
• For an oxide experiment:

mass of oxygen = mass of oxide − mass of original element

• Experimental uncertainty may produce ratios that are close to, but not exactly, whole numbers.
• More than one molecular compound can have the same empirical formula.
• Molecular molar mass can be used with an empirical formula to determine a molecular formula.

The central strategy is:

MASS → MOLES → RATIO → FORMULA

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Check Your Understanding

Basic Concepts

1. Define an empirical formula.

2. Explain the difference between an empirical formula and a molecular formula.

3. Determine the empirical formula corresponding to C₂H₄.

4. Determine the empirical formula corresponding to C₆H₁₂O₆.

5. Determine the empirical formula corresponding to H₂O₂.

6. Why is the empirical formula of H₂O still H₂O?

7. Explain why empirical formula calculations use mole ratios rather than mass ratios.

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Empirical Formula from Mass

8. A compound contains 12 g C and 2 g H. Determine its empirical formula.

9. A compound contains 2.4 g Mg and 1.6 g O. Determine its empirical formula.

10. A compound contains 5.6 g Fe and 2.4 g O. Determine its empirical formula.

11. A compound contains 2.7 g Al and 2.4 g O. Determine its empirical formula.

12. A compound contains 6.35 g Cu and 1.60 g O. Determine its empirical formula.

13. A compound contains 3.1 g P and 4.0 g O. Determine its empirical formula. Use Ar(P) = 31.

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Percentage Composition

14. A compound contains 40.0% C, 6.7% H, and 53.3% O. Determine its empirical formula.

15. A compound contains 27.3% C and 72.7% O. Determine its empirical formula.

16. A compound contains 40.0% S and 60.0% O. Determine its empirical formula.

17. A compound contains 43.7% P and 56.3% O. Determine its empirical formula.

18. Explain why it is convenient to assume a 100 g sample when percentage composition is given.

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Decimal Ratios

19. A mole ratio is calculated as 1 : 1.5. What whole-number ratio should be used?

20. A ratio is 1 : 1.33. What whole-number ratio is most likely?

21. A ratio is 1 : 1.25. What should you multiply the entire ratio by?

22. Explain why a ratio of 1 : 1.5 should not simply be rounded to 1 : 2.

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Experimental Data

23. A student heats 2.40 g Mg and produces 4.00 g MgO. Calculate the mass of oxygen that reacted.

24. Use the data from Question 23 to determine the empirical formula.

25. A student reacts 5.40 g Al with oxygen and obtains 10.20 g aluminium oxide. Determine the mass of oxygen that reacted.

26. Use the data from Question 25 to determine the empirical formula.

27. Explain why a sample may be heated repeatedly until constant mass is reached.

28. Give two experimental errors that could affect an empirical formula calculation.

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Molecular and Empirical Formulae

29. The empirical formula is CH₂O and the molecular molar mass is 180 g/mol. Determine the molecular formula.

30. The empirical formula is CH and the molecular molar mass is 78 g/mol. Determine the molecular formula.

31. The empirical formula is NO₂ and the molecular molar mass is 92 g/mol. Determine the molecular formula.

32. The empirical formula is CH₂ and the molecular molar mass is 56 g/mol. Determine the molecular formula.

33. Explain why two different compounds can have the same empirical formula.

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Analysis and Application

34. A student obtains mole values of 0.20 mol C, 0.40 mol H, and 0.20 mol O. Determine the empirical formula.

35. A student obtains a ratio of 1.00 : 1.49. Explain how this ratio should be interpreted.

36. A student uses the masses of two elements directly as subscripts in a chemical formula. Explain why this method is incorrect.

37. A compound contains 48.0% C, 8.0% H, and the remainder O. Calculate the percentage of oxygen and then determine the empirical formula.

38. An oxide is formed from 7.00 g of a metal. The final oxide mass is 9.00 g. Explain the steps you would use to determine the empirical formula if the atomic mass of the metal were known.

39. A student obtains an experimental mole ratio of Mg : O = 1.00 : 0.91 instead of exactly 1 : 1. Give possible experimental reasons for the difference and state the most reasonable empirical formula.

40. Explain how measurements made with a laboratory balance can ultimately be used to determine the simplest ratio of atoms in a chemical compound.