Percentage Composition

Percentage composition by mass tells us what percentage of a compound's total mass comes from each element.

A compound contains atoms of different elements, and those atoms do not all contribute equally to its mass. Percentage composition allows us to describe exactly how much of the total mass is contributed by each element.

The general equation is:

percentage by mass = (mass of element in formula / formula mass of compound) × 100

For an element X:

% X = (mass of X in formula / Mr of compound) × 100

The percentages of all elements in a pure compound should add to approximately:

100%

Small differences from 100% can occur because of rounding.

Why Percentage Composition Is Useful

A chemical formula tells us which elements are present and the ratio of their atoms.

Percentage composition gives us different information: how the mass of the compound is distributed among those elements.

Chemists use percentage composition to:

• analyze unknown substances
• compare compounds
• check the purity or identity of substances
• determine empirical formulae
• understand the composition of minerals and materials
• analyze fertilizers, foods, medicines, and industrial chemicals

For example, two compounds may both contain nitrogen, but one may contain a much larger percentage of nitrogen by mass.

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Atomic Ratios Are Not Mass Percentages

Consider water:

H₂O

There are:

• 2 hydrogen atoms
• 1 oxygen atom

By number of atoms:

2/3 of the atoms are hydrogen

However, this does not mean that water is 67% hydrogen by mass.

Hydrogen atoms are much lighter than oxygen atoms.

Using:

Ar(H) = 1

Ar(O) = 16

the mass contribution is:

Hydrogen:

2 × 1 = 2

Oxygen:

1 × 16 = 16

Total:

18

Therefore, most of the mass of water comes from oxygen.

This distinction between number of atoms and mass contribution is very important.

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Formula Mass

Before calculating percentage composition, determine the formula mass of the compound.

For:

H₂O

Hydrogen:

2 × 1 = 2

Oxygen:

1 × 16 = 16

Total:

Mr(H₂O) = 18

For:

CO₂

Carbon:

1 × 12 = 12

Oxygen:

2 × 16 = 32

Total:

Mr(CO₂) = 44

Percentage composition calculations use these mass contributions.

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Percentage Composition of Water

For:

H₂O

Formula mass:

18

Hydrogen

Mass contribution:

2 × 1 = 2

Percentage:

(2 / 18) × 100

= 11.1%

Oxygen

Mass contribution:

16

Percentage:

(16 / 18) × 100

= 88.9%

Answer

Water is approximately:

• 11.1% H
• 88.9% O

Check:

11.1 + 88.9 = 100.0%

Percentage Composition of Carbon Dioxide

Consider:

CO₂

Use:

Ar(C) = 12

Ar(O) = 16

Formula mass:

12 + (2 × 16)

= 44

Carbon

Carbon contributes:

12

Therefore:

% C = (12 / 44) × 100

= 27.3%

Oxygen

Oxygen contributes:

32

Therefore:

% O = (32 / 44) × 100

= 72.7%

Answer

CO₂ is approximately:

• 27.3% carbon
• 72.7% oxygen

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A Reliable Calculation Method

For most percentage composition problems, follow this sequence:

FORMULA → FORMULA MASS → ELEMENT MASS → DIVIDE → ×100

Find the formula mass

Add the mass contributions from every atom.

Find the mass contribution of the element

Multiply:

number of atoms × atomic mass

Divide

Divide the element's contribution by the total formula mass.

Multiply by 100

Convert the fraction into a percentage.

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Worked Example: Sodium Chloride

Calculate the percentage composition of:

NaCl

Use:

Ar(Na) = 23

Ar(Cl) = 35.5

Formula mass:

23 + 35.5 = 58.5

Sodium

% Na = (23 / 58.5) × 100

≈ 39.3%

Chlorine

% Cl = (35.5 / 58.5) × 100

≈ 60.7%

Answer

NaCl contains approximately:

• 39.3% Na
• 60.7% Cl

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Worked Example: Magnesium Oxide

Consider:

MgO

Use:

Ar(Mg) = 24

Ar(O) = 16

Formula mass:

24 + 16 = 40

Magnesium:

% Mg = (24 / 40) × 100

= 60%

Oxygen:

% O = (16 / 40) × 100

= 40%

Answer

MgO is:

• 60% Mg
• 40% O

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Compounds with More Than Two Elements

The same method works for compounds containing three or more elements.

Consider:

CaCO₃

Use:

• Ca = 40
• C = 12
• O = 16

Formula mass:

40 + 12 + (3 × 16)

= 40 + 12 + 48

= 100

This makes the percentages especially easy.

Calcium

(40 / 100) × 100 = 40%

Carbon

(12 / 100) × 100 = 12%

Oxygen

(48 / 100) × 100 = 48%

Answer

CaCO₃ contains:

• 40% Ca
• 12% C
• 48% O

Check:

40 + 12 + 48 = 100%

Worked Example: Ammonia

Calculate the percentage composition of:

NH₃

Use:

N = 14

H = 1

Formula mass:

14 + (3 × 1)

= 17

Nitrogen:

% N = (14 / 17) × 100

≈ 82.4%

Hydrogen:

% H = (3 / 17) × 100

≈ 17.6%

Answer

NH₃ contains approximately:

• 82.4% N
• 17.6% H

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Worked Example: Glucose

Consider:

C₆H₁₂O₆

Use:

• C = 12
• H = 1
• O = 16

Calculate each contribution

Carbon:

6 × 12 = 72

Hydrogen:

12 × 1 = 12

Oxygen:

6 × 16 = 96

Formula mass:

72 + 12 + 96 = 180

Carbon percentage

(72 / 180) × 100 = 40.0%

Hydrogen percentage

(12 / 180) × 100 ≈ 6.7%

Oxygen percentage

(96 / 180) × 100 ≈ 53.3%

Answer

Glucose contains approximately:

• 40.0% C
• 6.7% H
• 53.3% O

Understanding Mass Contribution

Consider glucose again:

C₆H₁₂O₆

It contains twice as many hydrogen atoms as carbon atoms:

12 H atoms compared with 6 C atoms

Yet hydrogen contributes much less to the total mass:

H contribution = 12

C contribution = 72

This happens because a carbon atom is approximately 12 times as massive as a hydrogen atom.

Therefore, percentage composition depends on both:

• the number of atoms
• the relative atomic mass of those atoms

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Worked Example: Sulfuric Acid

Calculate the percentage composition of:

H₂SO₄

Use:

• H = 1
• S = 32
• O = 16

Mass contributions:

H:

2 × 1 = 2

S:

1 × 32 = 32

O:

4 × 16 = 64

Formula mass:

2 + 32 + 64 = 98

Hydrogen

% H = (2 / 98) × 100

≈ 2.0%

Sulfur

% S = (32 / 98) × 100

≈ 32.7%

Oxygen

% O = (64 / 98) × 100

≈ 65.3%

Answer

H₂SO₄ contains approximately:

• 2.0% H
• 32.7% S
• 65.3% O

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Working with Parentheses

Chemical formulae containing parentheses require extra care.

Consider:

Ca(OH)₂

The subscript 2 applies to everything inside the parentheses.

Therefore, the compound contains:

• 1 Ca
• 2 O
• 2 H

Use:

• Ca = 40
• O = 16
• H = 1

Mass contributions:

Ca:

1 × 40 = 40

O:

2 × 16 = 32

H:

2 × 1 = 2

Formula mass:

40 + 32 + 2 = 74

Therefore:

% Ca = (40 / 74) × 100 ≈ 54.1%

% O = (32 / 74) × 100 ≈ 43.2%

% H = (2 / 74) × 100 ≈ 2.7%

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Another Example with Parentheses

Consider:

Al₂(SO₄)₃

Count the atoms carefully:

• Al = 2
• S = 3
• O = 12

Use:

• Al = 27
• S = 32
• O = 16

Mass contributions:

Al:

2 × 27 = 54

S:

3 × 32 = 96

O:

12 × 16 = 192

Formula mass:

54 + 96 + 192

= 342

Therefore:

% Al = (54 / 342) × 100 ≈ 15.8%

% S = (96 / 342) × 100 ≈ 28.1%

% O = (192 / 342) × 100 ≈ 56.1%

Check:

15.8 + 28.1 + 56.1 = 100.0%

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Finding the Mass of an Element in a Sample

Percentage composition can also be used in reverse.

Suppose a compound is:

40% calcium by mass

and you have:

250 g of the compound

Mass of calcium:

mass = percentage × total mass

Convert the percentage to a decimal:

40% = 0.40

Then:

mass Ca = 0.40 × 250

= 100 g

Therefore, the sample contains:

100 g Ca

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Worked Example: Oxygen in Calcium Carbonate

CaCO₃ is:

48% oxygen by mass

How much oxygen is present in:

500 g CaCO₃?

Use:

mass O = 0.48 × 500

= 240 g

Answer

240 g oxygen

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Worked Example: Nitrogen in Ammonia

NH₃ contains approximately:

82.4% nitrogen

How much nitrogen is present in:

50.0 g NH₃?

mass N = 0.824 × 50.0

= 41.2 g

Answer

41.2 g nitrogen

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Finding the Total Mass

We can also work backwards.

Suppose a fertilizer sample contains:

28 g nitrogen

and is:

35% nitrogen by mass

Let the total mass be m.

0.35m = 28

Therefore:

m = 28 / 0.35

= 80 g

Answer

The total sample mass is:

80 g

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Comparing Compounds

Percentage composition is particularly useful when comparing compounds containing the same element.

Consider:

CO

and:

CO₂

For CO:

Formula mass:

12 + 16 = 28

Carbon percentage:

(12 / 28) × 100 ≈ 42.9%

For CO₂:

Formula mass:

12 + 32 = 44

Carbon percentage:

(12 / 44) × 100 ≈ 27.3%

Therefore:

CO has a greater percentage of carbon by mass than CO₂.

Even though both compounds contain carbon and oxygen, their compositions are different.

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Comparing Nitrogen Compounds

Suppose we want to compare the percentage of nitrogen in:

NH₃

and:

NH₄NO₃

For NH₃:

Formula mass:

17

Nitrogen contribution:

14

% N ≈ 82.4%

For NH₄NO₃:

There are:

• 2 N
• 4 H
• 3 O

Formula mass:

(2 × 14) + (4 × 1) + (3 × 16)

= 28 + 4 + 48

= 80

Nitrogen contribution:

28

% N = (28 / 80) × 100

= 35%

Therefore, NH₃ contains a much higher percentage of nitrogen by mass.

This type of comparison is important in areas such as fertilizer chemistry.

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Percentage Composition and Empirical Formulae

Percentage composition can be used to determine an empirical formula.

Suppose a compound contains:

• 40.0% C
• 6.7% H
• 53.3% O

Assume a:

100 g sample

Then:

• C = 40.0 g
• H = 6.7 g
• O = 53.3 g

Convert each mass 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:

1 : 2 : 1

Therefore:

Empirical formula = CH₂O

This demonstrates an important connection:

percentage composition → mass → moles → empirical formula

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Percentage Composition and Molecular Formulae

Percentage composition can also contribute to determining a molecular formula.

For example, suppose experimental analysis gives:

• 40.0% C
• 6.7% H
• 53.3% O

First determine:

Empirical formula = CH₂O

Empirical formula mass:

30 g/mol

If the experimentally measured molecular molar mass is:

180 g/mol

then:

180 / 30 = 6

Therefore:

Molecular formula = C₆H₁₂O₆

So the complete pathway is:

percentage composition → empirical formula → molecular formula

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Percentage Composition from Experimental Masses

Sometimes percentage composition is determined directly from experimental measurements.

Suppose a 20.0 g sample contains:

• 8.0 g element A
• 12.0 g element B

Percentage A:

(8.0 / 20.0) × 100

= 40%

Percentage B:

(12.0 / 20.0) × 100

= 60%

The same principle applies regardless of the sample size.

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Worked Experimental Example

A compound contains:

• 5.4 g aluminium
• 4.8 g oxygen

Total mass:

5.4 + 4.8 = 10.2 g

Aluminium:

% Al = (5.4 / 10.2) × 100

≈ 52.9%

Oxygen:

% O = (4.8 / 10.2) × 100

≈ 47.1%

Check:

52.9 + 47.1 = 100.0%

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Conservation of Mass

Percentage composition is closely connected to the law of conservation of mass.

If a pure compound contains only elements A and B:

% A + % B = 100%

If it contains A, B, and C:

% A + % B + % C = 100%

Therefore, if one percentage is missing, it can often be found by subtraction.

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Finding a Missing Percentage

A compound contains:

• 32% sulfur
• 64% oxygen
• the remainder hydrogen

Hydrogen percentage:

100 − 32 − 64

= 4%

Therefore:

4% hydrogen

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Another Missing Percentage Example

A compound contains:

• 54.5% carbon
• 9.1% hydrogen
• the remainder oxygen

Oxygen:

100 − 54.5 − 9.1

= 36.4%

Therefore:

36.4% oxygen

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Real-World Applications

Percentage composition is widely used outside classroom calculations.

It can help describe:

• the amount of nitrogen in fertilizers
• the mineral composition of ores
• the concentration of nutrients in foods
• the composition of pharmaceuticals
• the percentage of metals in alloys
• the elemental composition of fuels
• the composition of environmental samples

Percentage Composition and Fertilizers

Plants require nitrogen for growth.

Different nitrogen-containing compounds contain different percentages of nitrogen.

For example:

NH₃ ≈ 82.4% N

NH₄NO₃ = 35.0% N

This does not automatically mean that ammonia is always the better fertilizer. Other factors matter, including:

• chemical form
• handling
• safety
• storage
• how nitrogen becomes available to plants

Percentage composition provides useful quantitative information, but it is only one factor in real-world decisions.

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Percentage Composition and Purity

Suppose a pure compound should contain:

40.0% element X

but laboratory analysis of a sample gives:

32.5% element X

This difference could indicate:

• impurities
• experimental error
• incorrect identification of the substance
• contamination

Chemists can therefore compare experimental composition with theoretical composition when analyzing substances.

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Percentage Composition Is Independent of Sample Size

A pure compound has the same percentage composition regardless of the amount of sample.

If water is:

11.1% H

then:

• 18 g water is 11.1% H
• 100 g water is 11.1% H
• 1 kg water is 11.1% H

The amount changes, but the proportion does not.

This happens because a pure compound always has its elements combined in a fixed ratio.

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A Useful Calculation Table

For a compound such as:

C₂H₆O

you can organize the calculation like this:

Then:

% C = (24 / 46) × 100 ≈ 52.2%

% H = (6 / 46) × 100 ≈ 13.0%

% O = (16 / 46) × 100 ≈ 34.8%

Check:

52.2 + 13.0 + 34.8 = 100.0%

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Multi-Step Example

Calculate the percentage of oxygen in:

Al₂(SO₄)₃

Count oxygen atoms

There are:

3 × 4 = 12 oxygen atoms

Find oxygen's mass contribution

12 × 16 = 192

Calculate formula mass

Al:

2 × 27 = 54

S:

3 × 32 = 96

O:

12 × 16 = 192

Total:

342

Calculate oxygen percentage

% O = (192 / 342) × 100

≈ 56.1%

Answer

Al₂(SO₄)₃ is approximately 56.1% oxygen by mass.

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Common Mistakes

Using the Number of Atoms Instead of Their Masses

In H₂O, 2 of the 3 atoms are hydrogen.

This does not mean water is:

66.7% hydrogen by mass

You must account for atomic masses.

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Forgetting Subscripts

For CO₂, oxygen contributes:

2 × 16 = 32

not:

16

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Ignoring Parentheses

In:

Ca(OH)₂

there are:

• 2 oxygen atoms
• 2 hydrogen atoms

The subscript outside the parentheses applies to the entire group.

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Dividing by the Element Mass Instead of Total Mass

Use:

element contribution / total formula mass

not the reverse.

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Forgetting to Multiply by 100

For example:

12 / 44 = 0.273

This is the decimal fraction.

The percentage is:

0.273 × 100 = 27.3%

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Using the Wrong Formula Mass

Every atom in the chemical formula must be included when calculating the total formula mass.

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Rounding Too Early

Keep extra decimal places during calculations and round the final answer.

Early rounding can cause the final percentages to differ unnecessarily from 100%.

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Expecting Exactly 100.0% After Rounding

For some calculations, individually rounded percentages might total:

99.9%

or:

100.1%

This is normally just a rounding effect.

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Confusing Percentage Composition with Concentration

Percentage composition describes the fraction of a compound's mass contributed by an element.

It is different from solution concentration such as:

mol/L

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Key Terms

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

Percentage by mass — The mass of a component expressed as a percentage of the total mass.

Relative atomic mass (Ar) — The average relative mass of atoms of an element.

Relative formula mass (Mr) — The sum of the relative atomic masses represented by a chemical formula.

Mass contribution — The amount an element contributes to the total formula mass.

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

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.

Percentage composition data — Information showing the percentage by mass of each element in a compound.

Purity — A measure of how much of a sample consists of the intended substance rather than impurities.

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Key Takeaways

• Percentage composition describes how the mass of a compound is divided among its elements.
• The central equation is:

% element = (mass contribution of element / total formula mass) × 100

• Atomic ratios and mass percentages are not the same.
• Heavier atoms can contribute a large percentage of the mass even when relatively few are present.
• Subscripts must be included when calculating mass contributions.
• Parentheses must be interpreted carefully.
• The percentages of all elements should total approximately 100%.
• Percentage composition is independent of sample size for a pure compound.
• Percentage data can be used to compare compounds.
• Percentage composition can be determined from either a chemical formula or experimental mass measurements.
• If the percentage and total sample mass are known:

mass of element = percentage as decimal × total mass

• Percentage composition can be used to determine empirical formulae.
• When combined with molar mass, composition data can help determine molecular formulae.
• Percentage composition has practical applications in materials, fertilizers, foods, pharmaceuticals, and chemical analysis.

The main calculation pathway is:

FORMULA → FORMULA MASS → ELEMENT MASS → DIVIDE → ×100

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Check Your Understanding

Basic Concepts

1. Define percentage composition by mass.

2. Explain why H₂O is not 66.7% hydrogen by mass even though two of its three atoms are hydrogen.

3. What should the percentages of all elements in a pure compound add up to?

4. Explain why percentage composition remains the same when the sample size changes.

5. State the general equation used to calculate percentage composition.

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Basic Calculations

Calculate the percentage composition of each compound.

6. H₂O

7. CO₂

8. MgO

9. NaCl

10. NH₃

11. CH₄

12. SO₂

13. HCl

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Multi-Element Compounds

Calculate the percentage by mass of each element.

14. CaCO₃

15. H₂SO₄

16. C₆H₁₂O₆

17. C₂H₆O

18. Na₂CO₃

19. HNO₃

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Parentheses

Calculate the percentage composition.

20. Ca(OH)₂

21. Mg(NO₃)₂

22. Al₂(SO₄)₃

23. (NH₄)₂SO₄

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Finding Element Mass

24. CaCO₃ contains 40% calcium. How much calcium is present in 250 g CaCO₃?

25. CaCO₃ contains 48% oxygen. How much oxygen is present in 500 g CaCO₃?

26. NH₃ is approximately 82.4% nitrogen. How much nitrogen is present in 75.0 g NH₃?

27. A compound is 25% element X by mass. How much X is present in a 360 g sample?

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Finding Total Mass

28. A sample contains 20 g of element X and is 25% X by mass. Calculate the total mass.

29. A fertilizer contains 14 g nitrogen and is 35% nitrogen by mass. Calculate the total mass of fertilizer.

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Comparing Compounds

30. Calculate the percentage of carbon in CO and CO₂. Which contains the greater percentage of carbon?

31. Calculate the percentage of nitrogen in NH₃ and NH₄NO₃. Which has the greater nitrogen percentage?

32. Compare the percentage of oxygen in CO₂ and SO₂.

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Experimental Data

33. A 20.0 g compound contains 8.0 g element A and 12.0 g element B. Calculate the percentage composition.

34. A 15.0 g oxide contains 9.0 g metal and 6.0 g oxygen. Calculate the percentage composition.

35. A compound contains 5.4 g Al and 4.8 g O. Calculate the percentage of each element.

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Analysis and Application

36. A compound contains 32% sulfur and 64% oxygen. The remainder is hydrogen. Determine the percentage of hydrogen.

37. A student calculates the percentage composition of a compound as 25.3%, 35.6%, and 39.0%. The total is 99.9%. Explain why this does not necessarily indicate an error.

38. Explain how percentage composition data can be used to determine an empirical formula.

39. A laboratory determines that an unknown sample has a noticeably different percentage composition from the expected pure compound. Give two possible explanations.

40. Describe the complete procedure for calculating the percentage by mass of every element in a compound when only its chemical formula is given.