Molecular Formulae

A molecular formula shows the actual number of atoms of each element in one molecule of a compound.

For example, glucose has the molecular formula:

C₆H₁₂O₆

This tells us that one molecule of glucose contains:

• 6 carbon atoms
• 12 hydrogen atoms
• 6 oxygen atoms

The molecular formula gives more information than the empirical formula because it shows the actual numbers of atoms rather than only their simplest ratio.

Molecular Formula vs. Empirical Formula

An empirical formula gives the simplest whole-number ratio of atoms.

A molecular formula gives the actual number of atoms in one molecule.

For glucose:

Molecular formula: C₆H₁₂O₆

Divide all subscripts by 6:

Empirical formula: CH₂O

For hydrogen peroxide:

Molecular formula: H₂O₂

Divide by 2:

Empirical formula: HO

For benzene:

Molecular formula: C₆H₆

Divide by 6:

Empirical formula: CH

The molecular formula is always a whole-number multiple of the empirical formula.

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Examples of Empirical and Molecular Formulae

Notice that several compounds can have the same empirical formula.

For example:

CH₂O

could represent a molecular formula such as:

CH₂O

C₂H₄O₂

C₃H₆O₃

C₆H₁₂O₆

Therefore, the empirical formula alone is not always enough to identify the molecular formula.

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The Relationship Between the Two Formulae

The relationship can be represented as:

Molecular formula = (empirical formula) × whole-number factor

For example:

Empirical formula:

CH₂O

Multiply all subscripts by 6:

C₆H₁₂O₆

Therefore:

CH₂O × 6 → C₆H₁₂O₆

The multiplier must be a positive whole number such as:

1, 2, 3, 4, 5, 6...

It cannot be:

1.5

or:

2.7

because molecules contain whole numbers of atoms.

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Empirical Formula Mass

To determine the molecular formula, we first calculate the empirical formula mass.

This is the total mass represented by the empirical formula.

For:

CH₂O

using:

• C = 12
• H = 1
• O = 16

Empirical formula mass:

12 + (2 × 1) + 16

= 30 g/mol

For:

CH

empirical formula mass:

12 + 1

= 13 g/mol

For:

NO₂

empirical formula mass:

14 + (2 × 16)

= 46 g/mol

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Finding the Whole-Number Factor

Once the empirical formula mass is known, compare it with the actual molar mass of the compound.

Calculate:

factor = molecular molar mass / empirical formula mass

The answer should be approximately a whole number.

Then multiply every subscript in the empirical formula by that factor.

The overall pathway is:

empirical formula → empirical formula mass → factor → molecular formula

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Worked Example: Glucose

A compound has:

Empirical formula = CH₂O

and:

Molar mass = 180 g/mol

Find its molecular formula.

Find the empirical formula mass

CH₂O

C:

1 × 12 = 12

H:

2 × 1 = 2

O:

1 × 16 = 16

Total:

30 g/mol

Find the factor

180 / 30 = 6

Multiply every subscript by 6

C:

1 × 6 = 6

H:

2 × 6 = 12

O:

1 × 6 = 6

Answer

C₆H₁₂O₆

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Worked Example: Benzene

A compound has:

Empirical formula = CH

and:

Molar mass = 78 g/mol

Empirical formula mass:

12 + 1 = 13 g/mol

Factor:

78 / 13 = 6

Multiply the empirical formula by 6:

C₆H₆

Answer

Molecular formula = C₆H₆

Worked Example: Hydrogen Peroxide

Empirical formula:

HO

Molar mass:

34 g/mol

Empirical formula mass:

1 + 16 = 17 g/mol

Factor:

34 / 17 = 2

Multiply:

H × 2 → H₂

O × 2 → O₂

Answer

H₂O₂

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Worked Example: Ethene

Empirical formula:

CH₂

Molar mass:

28 g/mol

Empirical formula mass:

12 + 2 = 14 g/mol

Factor:

28 / 14 = 2

Therefore:

CH₂ × 2 → C₂H₄

Answer

C₂H₄

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When the Empirical and Molecular Formulae Are the Same

Sometimes the factor is:

1

For example:

Empirical formula:

H₂O

Molar mass:

18 g/mol

Empirical formula mass:

(2 × 1) + 16

= 18 g/mol

Factor:

18 / 18 = 1

Therefore:

Molecular formula = H₂O

The empirical and molecular formulae are identical.

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Another Example

Empirical formula:

NO₂

Molar mass:

92 g/mol

Empirical formula mass:

14 + (2 × 16)

= 46 g/mol

Factor:

92 / 46 = 2

Multiply all subscripts by 2:

N₂O₄

Answer

Molecular formula = N₂O₄

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A Larger Molecular Formula

A compound has:

Empirical formula = C₂H₅

and:

Molar mass = 116 g/mol

Empirical formula mass:

(2 × 12) + (5 × 1)

= 24 + 5

= 29 g/mol

Factor:

116 / 29 = 4

Multiply:

C:

2 × 4 = 8

H:

5 × 4 = 20

Answer

C₈H₂₀

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Why the Factor Must Be a Whole Number

Suppose the empirical formula is:

CH₂

The possible molecular formulae include:

CH₂

C₂H₄

C₃H₆

C₄H₈

C₅H₁₀

and so on.

Notice that the empirical ratio:

C : H = 1 : 2

is preserved.

A molecular formula such as:

C₁.₅H₃

would make no physical sense because a molecule cannot contain half of a carbon atom.

Therefore, the multiplier must be a whole number.

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Experimental Information and Molecular Formulae

Often the empirical formula is not provided directly.

Instead, experimental information is given.

The problem may provide:

• percentage composition
• masses of elements
• combustion data
• experimental composition data

In these problems, first determine the empirical formula.

Then use the molar mass to determine the molecular formula.

The full pathway becomes:

experimental data → moles → simplest ratio → empirical formula → empirical formula mass → multiplier → molecular formula

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Worked Example from Percentage Composition

A compound contains:

• 40.0% C
• 6.7% H
• 53.3% O

Its molar mass is:

180 g/mol

Find its molecular formula.

Assume 100 g

This gives:

• C = 40.0 g
• H = 6.7 g
• O = 53.3 g

Convert to moles

C:

40.0 / 12 = 3.33 mol

H:

6.7 / 1 = 6.7 mol

O:

53.3 / 16 = 3.33 mol

Divide by the smallest

C : H : O

3.33 : 6.7 : 3.33

approximately:

1 : 2 : 1

Therefore:

Empirical formula = CH₂O

Calculate empirical formula mass

12 + 2 + 16 = 30 g/mol

Find the multiplier

180 / 30 = 6

Determine molecular formula

CH₂O × 6 → C₆H₁₂O₆

Answer

Molecular formula = C₆H₁₂O₆

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Worked Example from Mass Data

A compound contains:

• 4.8 g carbon
• 0.8 g hydrogen

Its molar mass is:

56 g/mol

Find its molecular formula.

Calculate moles

Carbon:

4.8 / 12 = 0.40 mol

Hydrogen:

0.8 / 1 = 0.80 mol

Find the simplest ratio

0.40 : 0.80

Divide by 0.40:

1 : 2

Empirical formula:

CH₂

Calculate empirical formula mass

12 + 2 = 14 g/mol

Calculate the factor

56 / 14 = 4

Multiply the subscripts

CH₂ × 4 → C₄H₈

Answer

C₄H₈

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Worked Example with Nitrogen and Oxygen

A compound contains:

• 30.4% N
• 69.6% O

Its molar mass is:

92 g/mol

Assume 100 g.

Nitrogen:

30.4 / 14 ≈ 2.17 mol

Oxygen:

69.6 / 16 = 4.35 mol

Divide by the smaller:

N : O ≈ 1 : 2

Empirical formula:

NO₂

Empirical formula mass:

14 + 32 = 46 g/mol

Factor:

92 / 46 = 2

Therefore:

Molecular formula = N₂O₄

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Checking Your Answer

After determining a molecular formula, calculate its molar mass.

For example, suppose your answer is:

C₄H₈

Calculate:

(4 × 12) + (8 × 1)

= 48 + 8

= 56 g/mol

If the question gave:

molar mass = 56 g/mol

the answer is consistent.

This is a useful way to catch calculation mistakes.

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Experimental Molar Mass

The molar mass used in molecular formula calculations may itself have been determined experimentally.

Depending on the substance, chemists can estimate or determine molar mass using measurements involving:

• gas properties
• mass spectrometry
• solution properties
• other analytical techniques

Once the molar mass and elemental composition are known, a molecular formula can often be determined.

Molecular Formula and Structure

A molecular formula tells us how many atoms are present, but it does not necessarily tell us how those atoms are arranged.

For example:

C₂H₆O

could represent different structures.

Two examples are:

ethanol

and:

dimethyl ether

Both have the same molecular formula:

C₂H₆O

but their atoms are connected differently.

These substances are called structural isomers.

Molecular Formula vs. Structural Formula

Consider ethanol.

Molecular formula:

C₂H₆O

A condensed structural representation is:

CH₃CH₂OH

The molecular formula tells us the total numbers of each atom.

The structural formula provides additional information about how the atoms are connected.

Therefore:

molecular formula → composition

structural formula → arrangement

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Molecular Formulae of Ionic Compounds

The term molecular formula is mainly used for substances that exist as discrete molecules.

For ionic compounds such as:

NaCl

or:

MgCl₂

there are no individual NaCl or MgCl₂ molecules in the crystal.

Instead, the formula gives the simplest ratio of ions in an ionic lattice.

For this reason, the term formula unit is generally used for ionic compounds.

Molecular Formula from Experimental Data: Full Process

Suppose a compound contains:

• 85.7% carbon
• 14.3% hydrogen

and has a molar mass of:

56 g/mol

Assume 100 g

C:

85.7 g

H:

14.3 g

Convert to moles

C:

85.7 / 12 ≈ 7.14 mol

H:

14.3 / 1 = 14.3 mol

Divide by the smallest

C:

7.14 / 7.14 = 1

H:

14.3 / 7.14 ≈ 2

Empirical formula:

CH₂

Empirical formula mass

12 + 2 = 14 g/mol

Find the factor

56 / 14 = 4

Molecular formula

C₄H₈

Check

(4 × 12) + (8 × 1)

= 48 + 8

= 56 g/mol

Correct.

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Worked Example Requiring a More Complex Empirical Formula

A compound has the empirical formula:

C₂H₅O

and a molar mass of:

90 g/mol

Empirical formula mass:

(2 × 12) + (5 × 1) + 16

= 24 + 5 + 16

= 45 g/mol

Factor:

90 / 45 = 2

Multiply every subscript by 2:

C₄H₁₀O₂

Answer

C₄H₁₀O₂

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Using Relative Molecular Mass Instead

Sometimes a problem gives relative molecular mass (Mr) instead of molar mass.

The calculation method is essentially the same.

For example:

Empirical formula:

CH₂

Empirical formula mass:

14

Given:

Mr = 42

Factor:

42 / 14 = 3

Therefore:

Molecular formula = C₃H₆

The numerical values of Mr and molar mass correspond, although Mr itself is dimensionless while molar mass is normally expressed in g/mol.

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Interpreting a Non-Whole-Number Factor

Suppose your calculation gives:

factor = 2.01

This is almost certainly intended to represent:

2

Small differences can occur because:

• experimental measurements have uncertainty
• percentages have been rounded
• atomic masses have been rounded
• molar mass measurements have uncertainty

However, suppose you obtain:

factor = 2.6

That is not close to a whole number.

You should check:

• the empirical formula
• atomic masses
• arithmetic
• molar mass
• experimental data

Do not automatically round a clearly unreasonable result.

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Why the Molecular Molar Mass Cannot Be Smaller

Suppose:

Empirical formula = CH₂

Empirical formula mass:

14 g/mol

The molecular molar mass cannot be:

10 g/mol

because the molecular formula must contain at least one complete empirical formula unit.

Therefore:

molecular molar mass ≥ empirical formula mass

and it should be a whole-number multiple of the empirical formula mass.

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Using Experimental Evidence

Consider a compound with:

Empirical formula = CH

Suppose experimental analysis gives:

molar mass ≈ 52 g/mol

Empirical formula mass:

13 g/mol

Factor:

52 / 13 = 4

Therefore:

molecular formula = C₄H₄

Experimental molar mass has allowed us to distinguish C₄H₄ from other possibilities such as:

CH

C₂H₂

C₃H₃

C₅H₅

and so on.

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A Reliable Problem-Solving Method

For a problem where the empirical formula is already known:

EMPIRICAL FORMULA → FORMULA MASS → DIVIDE → MULTIPLY SUBSCRIPTS

Find the empirical formula mass

Add the atomic masses represented in the empirical formula.

Divide the molecular molar mass by the empirical formula mass

This gives the whole-number factor.

Multiply every subscript

Multiply all subscripts in the empirical formula by the factor.

Check the molecular molar mass

Calculate the mass of your final formula to make sure it matches the value given.

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For Experimental Composition Problems

When percentages or masses are given, use the longer pathway:

MASS/PERCENTAGE → MOLES → SIMPLEST RATIO → EMPIRICAL FORMULA → EMPIRICAL FORMULA MASS → FACTOR → MOLECULAR FORMULA

This combines empirical formula calculations with molar mass calculations.

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Common Mistakes

Confusing Empirical and Molecular Formulae

CH₂O

may be an empirical formula.

C₆H₁₂O₆

may be the corresponding molecular formula.

They provide different information.

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Dividing the Wrong Way

Use:

molecular molar mass / empirical formula mass

For example:

180 / 30 = 6

not:

30 / 180

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Multiplying Only One Subscript

If:

empirical formula = CH₂O

and:

factor = 3

multiply every subscript:

C₃H₆O₃

not:

C₃H₂O

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Forgetting an Implied Subscript of 1

In:

CH₂O

C has an implied subscript of 1.

O also has an implied subscript of 1.

If the multiplier is 4:

C₄H₈O₄

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Simplifying the Final Molecular Formula

Suppose you determine:

C₆H₁₂O₆

Do not simplify it back to:

CH₂O

The goal is to find the actual molecular formula.

Simplifying would return you to the empirical formula.

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Assuming the Empirical Formula Is Always Different

Sometimes:

empirical formula = molecular formula

This occurs when the factor is:

1

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Ignoring Units and Meaning

A molar mass such as:

180 g/mol

represents the mass of one mole of molecules.

The empirical formula mass represents the mass corresponding to one empirical formula unit.

Their ratio tells us how many empirical formula units are contained in one molecular formula.

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Rounding a Bad Result

If the calculated factor is:

2.01

use:

2

If it is:

2.63

do not simply round to 3 without investigating the calculation.

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Key Terms

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

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

Empirical formula mass — The total relative mass represented by one empirical formula unit.

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

Relative molecular mass (Mr) — The relative mass of a molecule compared with 1/12 of the mass of a carbon-12 atom.

Whole-number factor — The integer by which the empirical formula subscripts must be multiplied to obtain the molecular formula.

Subscript — A small number in a chemical formula indicating the number of atoms of an element.

Percentage composition — The percentage of a compound's mass contributed by each element.

Structural formula — A representation showing how atoms are connected within a molecule.

Structural isomer — A compound with the same molecular formula as another compound but a different arrangement of atoms.

Formula unit — The simplest ratio of ions represented in an ionic compound.

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Key Takeaways

• A molecular formula shows the actual number of atoms of each element in one molecule.
• An empirical formula shows the simplest whole-number ratio.
• A molecular formula is a whole-number multiple of its empirical formula.
• The empirical and molecular formulae can sometimes be identical.
• Calculate the empirical formula mass by adding the atomic masses represented in the empirical formula.
• Determine the multiplication factor by comparing the molecular molar mass with the empirical formula mass.
• The factor should be a positive whole number or very close to one.
• Multiply every subscript in the empirical formula by this factor.
• Do not simplify the molecular formula after determining it.
• Experimental composition data may first need to be used to determine the empirical formula.
• The complete calculation pathway is:

experimental data → moles → empirical formula → empirical formula mass → factor → molecular formula

• A molecular formula gives composition but does not necessarily reveal how the atoms are arranged.
• Different compounds can have the same molecular formula but different structures.
• Molecular formula terminology is mainly used for substances consisting of discrete molecules.
• Checking the calculated molar mass of the final formula is an effective way to verify your answer.

The central strategy is:

EMPIRICAL FORMULA → FORMULA MASS → FACTOR → MOLECULAR FORMULA

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Check Your Understanding

Basic Concepts

1. Define a molecular formula.

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

3. What information does C₆H₁₂O₆ provide about one molecule of glucose?

4. Determine the empirical formula of C₄H₈.

5. Determine the empirical formula of C₆H₆.

6. Give an example in which the empirical and molecular formulae are identical.

7. Explain why the molecular formula must be a whole-number multiple of the empirical formula.

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Empirical Formula Mass

Calculate the empirical formula mass of each.

8. CH₂

9. CH₂O

10. HO

11. NO₂

12. C₂H₅O

13. CH₃O

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Determine the Molecular Formula

14. Empirical formula = CH₂; molar mass = 28 g/mol.

15. Empirical formula = CH₂; molar mass = 42 g/mol.

16. Empirical formula = CH₂; molar mass = 56 g/mol.

17. Empirical formula = CH; molar mass = 78 g/mol.

18. Empirical formula = HO; molar mass = 34 g/mol.

19. Empirical formula = NO₂; molar mass = 92 g/mol.

20. Empirical formula = CH₂O; molar mass = 180 g/mol.

21. Empirical formula = C₂H₅O; molar mass = 90 g/mol.

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Full Experimental Problems

22. A compound contains 85.7% C and 14.3% H. Its molar mass is 28 g/mol. Determine its empirical and molecular formulae.

23. A compound contains 85.7% C and 14.3% H. Its molar mass is 56 g/mol. Determine its empirical and molecular formulae.

24. A compound contains 40.0% C, 6.7% H, and 53.3% O. Its molar mass is 180 g/mol. Determine its empirical and molecular formulae.

25. A compound contains 30.4% N and 69.6% O. Its molar mass is 92 g/mol. Determine its empirical and molecular formulae.

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Mass Data

26. A compound contains 2.4 g C and 0.4 g H. Its molar mass is 42 g/mol. Determine its empirical and molecular formulae.

27. A compound contains 4.8 g C and 0.8 g H. Its molar mass is 56 g/mol. Determine its empirical and molecular formulae.

28. A compound contains 3.6 g C and 0.6 g H. Its molar mass is 84 g/mol. Determine its empirical and molecular formulae.

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Analysis

29. A compound has empirical formula CH₂O and molecular formula C₃H₆O₃. What is the whole-number factor?

30. A compound has empirical formula NO₂ and molecular formula N₂O₄. Determine the factor.

31. A compound has empirical formula CH and a molar mass of 26 g/mol. Determine its molecular formula.

32. A student calculates an empirical formula mass of 30 g/mol and a molecular molar mass of 150 g/mol. Determine the multiplier.

33. The empirical formula is CH₂ and the molar mass is 70 g/mol. Determine the molecular formula.

34. A student calculates a multiplier of 3.02. What whole-number factor should probably be used? Explain.

35. A student calculates a multiplier of 2.67. Explain why simply rounding it to 3 may not be appropriate.

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Error Analysis and Reasoning

36. A student determines that the empirical formula is CH₂O and the factor is 4, but writes C₄H₂O. Identify the error and give the correct molecular formula.

37. A student finds C₆H₁₂O₆ and then simplifies it to CH₂O. Explain why this is incorrect when the question asks for the molecular formula.

38. A compound has empirical formula CH₂ and molar mass 10 g/mol. Explain why these data appear inconsistent.

39. Explain how percentage composition and molar mass can be combined to determine a molecular formula.

40. A scientist determines the percentage composition and molar mass of an unknown molecular compound experimentally. Describe the complete sequence of calculations needed to determine its molecular formula.