Percentage Composition by Mass

Summary: Percentage composition by mass is the mass of each element in a compound expressed as a percentage of the compound’s total mass. It is calculated using the formula: % mass of element = (total mass of the element in one mole of compound / molar mass of compound) x 100. The percentages of all elements in a pure compound must sum to exactly 100%. Percentage composition can be determined both theoretically from a chemical formula and experimentally from combustion or decomposition data. This concept is fundamental to empirical formula determination, purity analysis, and industrial quality control in fields such as pharmaceuticals, mining, and materials science.

Tags: igcse chemistry stoichiometry percentage-composition moles quantitative-chemistry calculations

Created: 2025-09-12

Last Updated: 2026-01-20

Percentage composition by mass is the mass of each element present in a compound expressed as a percentage of the total mass of the compound. It is calculated using the formula % mass = (total mass of element in formula / relative molecular or formula mass) x 100, where the total mass of the element accounts for all atoms of that element in the formula unit. For any pure compound, the sum of the percentage compositions of all constituent elements must equal exactly 100%, a rule that serves as a useful check on calculation accuracy and is a direct consequence of the law of conservation of mass. Percentage composition can be determined theoretically from a known chemical formula, or experimentally by measuring the masses of products formed when a sample is decomposed or combusted, providing a bridge between theoretical stoichiometry and practical analytical chemistry. This concept is directly linked to empirical formula calculations — given percentage composition data, one can determine the simplest whole-number ratio of atoms in a compound — and is widely applied in industry for assessing ore purity, verifying pharmaceutical dosage, and ensuring the consistency of manufactured chemical products.

The Formula

The percentage by mass of an element in a compound is given by:

[\text{Percentage by mass} = \frac{\text{Total mass of the element in one formula unit}}{\text{Relative formula mass (M}_r\text{) of the compound}} \times 100]

Where:

• Total mass of the element = (number of atoms of that element in the formula) x (relative atomic mass, A_r, of the element)
• Relative formula mass (M_r) = the sum of the relative atomic masses of all atoms in the formula

Derivation: If one mole of the compound has mass M_r grams, and it contains n atoms of element X each with mass A_r grams, then the mass of X in one mole of the compound is n x A_r grams. The fraction of the total mass that is X is (n x A_r) / M_r, and multiplying by 100 gives the percentage.

Key Points:

• Always multiply the A_r by the number of atoms of that element in the formula.
• Use the relative formula mass for ionic compounds and the relative molecular mass for covalent compounds — the calculation method is identical.
• For hydrated salts (e.g. CuSO_4*5H_2O), include the mass of water molecules when calculating M_r.

Worked Examples

Example 1: Magnesium Oxide, MgO

Step 1: Determine the relative atomic masses.

• A_r of Mg = 24
• A_r of O = 16

Step 2: Calculate the relative formula mass (M_r).

• M_r of MgO = 24 + 16 = 40

Step 3: Apply the percentage composition formula.

• % Mg = (24 / 40) x 100 = 60.0%
• % O = (16 / 40) x 100 = 40.0%

Check: 60.0% + 40.0% = 100.0% (the sum-to-100% rule is satisfied)

Example 2: Calcium Carbonate, CaCO_3

Step 1: Determine the relative atomic masses.

• A_r of Ca = 40
• A_r of C = 12
• A_r of O = 16

Step 2: Calculate M_r.

• M_r = 40 + 12 + (3 x 16) = 40 + 12 + 48 = 100

Step 3: Apply the formula.

• % Ca = (40 / 100) x 100 = 40.0%
• % C = (12 / 100) x 100 = 12.0%
• % O = (48 / 100) x 100 = 48.0%

Check: 40.0% + 12.0% + 48.0% = 100.0%

Example 3: Ammonium Sulfate, (NH_4)_2SO_4

Step 1: Determine A_r values.

• A_r of N = 14
• A_r of H = 1
• A_r of S = 32
• A_r of O = 16

Step 2: Count the atoms in the formula.

• N: 2 x 1 = 2 atoms ⇒ mass = 2 x 14 = 28
• H: 2 x 4 = 8 atoms ⇒ mass = 8 x 1 = 8
• S: 1 atom ⇒ mass = 32
• O: 4 atoms ⇒ mass = 4 x 16 = 64

Step 3: Calculate M_r.

• M_r = 28 + 8 + 32 + 64 = 132

Step 4: Calculate percentages.

• % N = (28 / 132) x 100 = 21.2% (to 1 decimal place)
• % H = (8 / 132) x 100 = 6.1%
• % S = (32 / 132) x 100 = 24.2%
• % O = (64 / 132) x 100 = 48.5%

Check: 21.2% + 6.1% + 24.2% + 48.5% = 100.0%

Common mistake: Forgetting to multiply by the subscript outside the bracket. (NH_4)_2 means there are 2 N atoms and 8 H atoms in total.

Example 4: Hydrated Copper(II) Sulfate, CuSO_4*5H_2O

This example includes Water of Crystallisation.

Step 1: Determine A_r values.

• A_r of Cu = 63.5
• A_r of S = 32
• A_r of O = 16
• A_r of H = 1

Step 2: Count atoms and calculate mass contributions.

• Cu: 1 x 63.5 = 63.5
• S: 1 x 32 = 32
• O (in CuSO_4): 4 x 16 = 64
• H_2O (5 molecules): 5 x [(2 x 1) + 16] = 5 x 18 = 90

Step 3: Calculate M_r.

• M_r = 63.5 + 32 + 64 + 90 = 249.5

Step 4: Calculate percentages.

• % Cu = (63.5 / 249.5) x 100 = 25.5%
• % S = (32 / 249.5) x 100 = 12.8%
• % O (total, including water) = [(64 + 80) / 249.5] x 100 = (144 / 249.5) x 100 = 57.7%
• % H = (10 / 249.5) x 100 = 4.0%

Alternatively, percentage of water of crystallisation:

• % H_2O = (90 / 249.5) x 100 = 36.1%

Check: 25.5% + 12.8% + 57.7% + 4.0% = 100.0%

Exam tip: If asked for the percentage of water of crystallisation, calculate (mass of water / M_r of hydrated salt) x 100. This is a very common exam question.

Percentage Composition from Experimental Data

Percentage composition can also be determined experimentally. The method depends on the type of compound:

Worked Example (Experimental):

A 2.50 g sample of a hydrocarbon is completely burned in excess oxygen. The products are 7.69 g of CO_2 and 3.93 g of H_2O. Calculate the percentage composition by mass of carbon and hydrogen in the hydrocarbon.

Step 1: Find the mass of carbon in the CO_2.

• M_r of CO_2 = 12 + (2 x 16) = 44
• Mass fraction of C in CO_2 = 12/44
• Mass of C = (12/44) x 7.69 = 2.097 g

Step 2: Find the mass of hydrogen in the H_2O.

• M_r of H_2O = (2 x 1) + 16 = 18
• Mass fraction of H in H_2O = 2/18
• Mass of H = (2/18) x 3.93 = 0.437 g

Step 3: Calculate percentages.

• % C = (2.097 / 2.50) x 100 = 83.9%
• % H = (0.437 / 2.50) x 100 = 17.5%

Check: 83.9% + 17.5% = 101.4% (slight deviation due to rounding; acceptable within experimental error)

The Sum-to-100% Rule

For any pure compound, the sum of the percentage compositions of all elements must equal exactly 100%. This is a direct consequence of the Conservation of Mass: the whole is the sum of its parts.

Uses of the sum-to-100% rule:

Note on rounding: When percentages are rounded to 1 decimal place, the sum may be 99.9% or 100.1%. This is acceptable. Anything beyond +/- 0.2% indicates a calculation error.

Connection to Empirical Formula

Percentage composition data is the starting point for calculating Empirical and Molecular Formulae.

Procedure:

1. Assume 100 g of the compound (so each percentage becomes a mass in grams).
2. Convert each mass to moles by dividing by the A_r of the element (using Relative Atomic Mass).
3. Divide each mole value by the smallest number of moles obtained.
4. Round to the nearest whole number to get the simplest ratio.

Worked Example:

A compound is found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

Step 1: In 100 g, masses are C = 40.0 g, H = 6.7 g, O = 53.3 g.

Step 2: Moles of each:

• C: 40.0 / 12 = 3.33 mol
• H: 6.7 / 1 = 6.7 mol
• O: 53.3 / 16 = 3.33 mol

Step 3: Divide by smallest (3.33):

• C: 3.33 / 3.33 = 1
• H: 6.7 / 3.33 = 2.01
• O: 3.33 / 3.33 = 1

Step 4: Simplest ratio = 1 : 2 : 1

Empirical formula = CH_2O (this is methanal/formaldehyde)

Industrial and Laboratory Applications

1. Mining and Metallurgy Percentage composition is used to assess the grade of an ore. For example, iron ore (haematite, Fe_2O_3) can be analysed to determine the percentage of iron it contains, which determines whether extraction is economically viable. See Reacting Masses.

2. Pharmaceuticals Drug manufacturers must verify that each tablet or dose contains the correct percentage of the active ingredient. Percentage composition by mass is used in quality control assays to ensure batch consistency and regulatory compliance.

3. Fertilisers The percentage by mass of nitrogen, phosphorus, and potassium (NPK values) in fertilisers is a key specification. For example, ammonium nitrate (NH_4NO_3) is valued for its high nitrogen content, calculated using percentage composition. See Reacting Masses.

4. Purity Analysis By comparing the experimentally determined percentage composition with the theoretical value, chemists can assess the purity of a substance using Percentage Yield and Purity. A deviation indicates the presence of impurities.

5. Food and Nutrition Nutritional labelling relies on percentage composition analysis to report the mass of fats, proteins, carbohydrates, and other components per serving.

Key Points to Memorise

1. Formula: % mass of X = (total mass of X in formula / M_r) x 100
2. Total mass of element = number of atoms x A_r. Do not forget to multiply by the subscript.
3. Sum-to-100% rule: All percentages in a pure compound must add up to 100%.
4. Brackets: For formulas like Ca(OH)_2 or (NH_4)_2SO_4, the subscript outside the bracket multiplies everything inside.
5. Hydrated salts: Include the water molecules when calculating M_r and total mass contributions.
6. Experimental data: Mass of element = (fraction of element in product) x (mass of product).
7. Empirical formula: Convert % composition to masses (assume 100 g), then to moles, then to simplest ratio.
8. Rounding: Report percentages to 1 decimal place unless instructed otherwise; check that the sum is within 0.2 of 100%.

Related Notes

• Relative Atomic Mass
• Relative Molecular Mass
• Relative Masses and Moles
• Empirical and Molecular Formulae
• Reacting Masses
• Percentage Yield and Purity
• Water of Crystallisation
• Conservation of Mass
• IGCSE-Chem-Index

Sources

1. OpenStax Chemistry 2e (2023). Chapter 3: Composition of Substances and Solutions. OpenStax, Rice University. Available at: https://openstax.org/books/chemistry-2e/pages/3-2-determining-empirical-and-molecular-formulas
2. BBC Bitesize (2024). Higher Chemistry: Percentage Composition. Available at: https://www.bbc.co.uk/bitesize/guides/z84wfrd/revision/2
3. Cambridge IGCSE Chemistry Syllabus 0620 (2023-2025). Topic 3: Stoichiometry. Cambridge Assessment International Education.
4. CK-12 Foundation (2023). Percent Composition. CK-12 Chemistry for High School. Available at: https://flexbooks.ck12.org/cbook/ck-12-chemistry-flexbook-2.0/section/10.5/primary/lesson/percent-composition-chem/
5. ZNotes (2024). IGCSE Chemistry: Stoichiometry Revision Notes. Available at: https://znotes.org/caie/igcse/chemistry-0620/stoichiometry

Past Paper Sources

• Cambridge IGCSE Chemistry 0620/41/M/J/22 Q4 (percentage composition and empirical formula)
• Cambridge IGCSE Chemistry 0620/42/O/N/22 Q3 (water of crystallisation and % composition)
• Cambridge IGCSE Chemistry 0620/43/M/J/23 Q5 (experimental determination of % composition)
• Cambridge IGCSE Chemistry 0620/41/O/N/23 Q2 (percentage composition and reacting masses)

Common Misconceptions