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Percentage composition by mass

Percentage composition by mass: calculate the element contribution, divide by the compound's total relative mass, then multiply by 100.

  • SEC G3 Pure Chemistry 2027

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Percentage composition compares the mass of one element with the mass of the whole compound. It is a mass percentage, so counting atoms alone is not enough: different elements have different atomic masses.

Compare the element’s contribution with the whole

For an element X in a compound:

percentage by mass of X = (relative mass contribution of X)/(relative molecular or formula mass of the compound) × 100%

The numerator is the total number of X atoms in the formula multiplied by Aᵣ(X). The denominator includes every element. Both are on the same relative-mass scale, so their ratio is the same as the mass fraction in a sample of the pure compound.

For example, CO₂ contains one C atom and two O atoms. With Aᵣ(C) = 12 and Aᵣ(O) = 16, carbon contributes 12 and oxygen contributes 32 to the total relative molecular mass of 44. Oxygen is therefore about 72.7% of the mass, but two-thirds (about 66.7%) of the atom count. These are different comparisons.

Carbon dioxide: composition by massCarbon contributes 12 out of a total relative molecular mass of 44, giving 27.3% by mass. Oxygen contributes 32 out of 44, giving 72.7%.Carbon dioxide: composition by massElementPercentage by mass (%)
These mass percentages use Ar(C) = 12 and Ar(O) = 16. They differ from the atom-count percentages: carbon is one-third of the atoms and oxygen is two-thirds.
Data table
ElementCO₂
C27.3
O72.7

Count the requested element across the complete formula

In (NH₄)₂SO₄, the outside 2 multiplies both N and H: there are two N atoms and eight H atoms. It does not multiply the sulfate group. In NH₄NO₃, nitrogen occurs in two places, so there are two N atoms in total.

Part over whole
  1. Count each element and calculate the whole relative molecular or formula mass.
  2. Find the requested element’s relative mass contribution, including all of its atoms.
  3. Divide part by whole and multiply by 100. Label the answer with the element and %.

Check the numerator, denominator and rounding

  • Use only the requested element in the numerator, rather than a whole bracketed group.
  • Include every element in the denominator. Dividing by the requested element’s Aᵣ alone does not give its fraction of the compound.
  • Keep unrounded intermediate values and round the final percentage as requested, or to a suitable precision.
  • Each element’s percentage lies between 0% and 100%. Percentages for all elements add to 100%, apart from small rounding differences.

For a reminder of relative mass calculations, see Relative atomic, molecular and formula mass.

Worked examples

Modelled example 1

Carbon in carbon dioxide

Core

Problem

Find the percentage by mass of carbon in carbon dioxide, CO₂. Given Aᵣ(C) = 12, Aᵣ(O) = 16.

Study the worked solution

  1. Calculate the whole formula mass

    Method

    Add the mass contributions from one carbon and two oxygen atoms.

    Reason

    The denominator must represent the complete CO₂ formula.

    Working

    Mᵣ(CO₂) = 12 + 2(16) = 44.

  2. Identify the carbon contribution

    Method

    Use 12 as the mass due to carbon.

    Reason

    Each formula contains one carbon atom.

    Working

    Carbon contribution = 1(12) = 12.

  3. Convert the fraction to a percentage

    Method

    Divide the carbon contribution by the whole and multiply by 100.

    Reason

    Percentage by mass is part mass divided by total mass.

    Working

    %C = (12/44) × 100 = 27.3%.

Guided practice 2

Nitrogen in ammonium sulfate

About 7 min

Problem

Find the percentage by mass of nitrogen in ammonium sulfate, (NH₄)₂SO₄. Given Aᵣ(N) = 14, Aᵣ(H) = 1, Aᵣ(S) = 32, Aᵣ(O) = 16.

Apply the bracket multiplier

Hints

Hint 1: expand the bracket

The outer 2 multiplies both N and H: N₂H₈.

Hint 2: form part over whole

Use nitrogen mass 28 over total formula mass 132, then multiply by 100.

View solution step by step

  1. Expand and calculate formula mass

    Method

    Apply the outer 2 to N and H before adding all contributions.

    Reason

    The bracket means two complete NH₄ groups.

    Working

    Mᵣ = 2(14) + 8(1) + 32 + 4(16) = 132.

  2. Find nitrogen mass

    Method

    Multiply two nitrogen atoms by Aᵣ(N).

    Reason

    Only the nitrogen contribution belongs in the numerator.

    Working

    Mass of N = 2(14) = 28.

  3. Calculate percentage

    Method

    Divide part by whole and multiply by 100.

    Reason

    This converts the nitrogen mass fraction into a percentage.

    Working

    %N = (28/132) × 100 = 21.2%.

Common misconception 3

Repair a missing subscript

Find and correct the mistake

Learner response

A student writes %O in CO₂ = (16/44) × 100. Explain what is wrong and correct it.

Locate the first incorrect quantity

Error

View solution step by step

  1. Locate the missing multiplier

    Method

    Replace 16 with 2(16) in the numerator.

    Reason

    The formula CO₂ contains two oxygen atoms.

    Working

    Oxygen contribution = 2(16) = 32.

  2. Correct the percentage

    Method

    Use the full oxygen contribution over the unchanged total mass.

    Reason

    Mᵣ(CO₂) = 44 was already correct.

    Working

    %O = (32/44) × 100 = 72.7%.

Examiner practice 4

Compare nitrogen percentages in fertilisers

6 marks

Examination question

Which fertiliser is richer in nitrogen by mass: ammonium nitrate, NH₄NO₃, or urea, CO(NH₂)₂? Given Aᵣ(N) = 14, Aᵣ(H) = 1, Aᵣ(C) = 12, Aᵣ(O) = 16. [6 marks]

Calculate both percentages before comparing

View solution step by step

  1. Calculate ammonium nitrate formula mass

    1 mark

    Method

    Sum all atom contributions in NH₄NO₃.

    Reason

    The complete formula mass is the denominator for nitrogen percentage.

    Working

    Mᵣ(NH₄NO₃) = 14 + 4(1) + 14 + 3(16) = 80.

  2. Calculate ammonium nitrate nitrogen percentage

    1 mark

    Method

    Use the mass of both nitrogen atoms over formula mass.

    Reason

    NH₄NO₃ contains two nitrogen atoms in total.

    Working

    %N = [2(14)/80] × 100 = 35.0%.

  3. Calculate urea formula mass

    1 mark

    Method

    Apply the outer 2 to the complete NH₂ group.

    Reason

    The bracket multiplier counts two nitrogen atoms and four hydrogen atoms.

    Working

    Mᵣ(CO(NH₂)₂) = 12 + 16 + 2[14 + 2(1)] = 60.

  4. Calculate urea nitrogen percentage

    1 mark

    Method

    Divide the two-nitrogen contribution by 60.

    Reason

    Percentage compares nitrogen mass with the whole urea formula mass.

    Working

    %N = [2(14)/60] × 100 = 46.7%.

  5. Compare like quantities

    2 marks

    Method

    Compare the two nitrogen percentages and name the richer fertiliser.

    Reason

    Both values describe nitrogen mass per 100 g, so they are directly comparable.

    Working

    46.7% > 35.0%, so urea is richer in nitrogen by mass.

Challenge 5

Element percentage in an unfamiliar formula

Minimal support

Formula transfer

An oxide has formula X₂O₃, where Aᵣ(X) = 27. Calculate the percentage by mass of oxygen. Use Aᵣ(O) = 16.

Find part and whole from the formula

Hints

Hint 1: calculate the whole

Use two X atoms and three oxygen atoms in the denominator.

Hint 2: calculate the requested part

Only the three oxygen atoms belong in the numerator.

View solution step by step

  1. Calculate the whole formula mass

    Method

    Add the two X and three O contributions.

    Reason

    The denominator represents the complete X₂O₃ formula.

    Working

    Relative formula mass = 2(27) + 3(16) = 102.

  2. Calculate oxygen percentage

    Method

    Divide the oxygen contribution by the whole and multiply by 100.

    Reason

    Three oxygen atoms contribute 3(16) = 48.

    Working

    %O = (48/102) × 100 = 47.1%.

Try it yourself

Mind stretcher 1: Find the remaining element’s percentageExtension

Question: A compound contains only carbon, hydrogen and oxygen. It has 40.0% carbon and 6.67% hydrogen by mass. What percentage is oxygen? Explain why your method is valid.

Show answer

Percentages of all elements in a pure compound must add to 100%.

%O = 100 - 40.0 - 6.67; = 53.33% before rounding; ≈ 53.3% to one decimal place

Mind stretcher 2: Use a percentage to distinguish two candidatesExtension

Question: A pure carbonate contains 40.0% of its metal by mass. Which candidate fits: CaCO₃ or MgCO₃? Use Aᵣ(Ca) = 40, Aᵣ(Mg) = 24, Aᵣ(C) = 12 and Aᵣ(O) = 16.

Show answer

For CaCO₃, the relative formula mass is 40 + 12 + 3(16) = 100. Its calcium percentage is (40/100) × 100 = 40.0%.

For MgCO₃, the relative formula mass is 24 + 12 + 3(16) = 84. Its magnesium percentage is (24/84) × 100 ≈ 28.6%.

Of the two supplied candidates, CaCO₃ fits the data. A matching percentage alone does not identify a compound among every possible substance.

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