Percentage Composition by Mass

This lesson teaches you how to calculate the percentage by mass of an element in a compound. This is a common calculation in GCSE Chemistry and is required by the Edexcel specification (1CH0).

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The Formula

To calculate the percentage by mass of an element in a compound:

$\text{percentage by mass} = \frac{A_r \times \text{number of atoms of that element}}{M_r \text{ of the compound}} \times 100$percentage by mass=Mr​ of the compoundAr​×number of atoms of that element​×100

Or more simply:

$\% = \frac{\text{total mass of the element in the formula}}{M_r \text{ of the compound}} \times 100$%=Mr​ of the compoundtotal mass of the element in the formula​×100

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Step-by-Step Method

1. Calculate Mr of the whole compound.
2. Find the total mass of the element you are interested in (Ar × number of atoms of that element in the formula).
3. Divide the total mass of the element by Mr.
4. Multiply by 100 to get the percentage.

Exam Tip: Always show your Mr calculation clearly. This earns marks and helps you avoid errors.

flowchart TD
    A[Compound formula] --> B["Calculate Mr of whole compound"]
    A --> C["For target element: Ar × number of atoms"]
    C --> D[Total mass of element in formula]
    B --> E["% by mass = (total mass of element ÷ Mr) × 100"]
    D --> E
    E --> F["Check: all element % values sum to 100%"]

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Worked Examples

Example 1: Percentage of Oxygen in Water (H₂O)

Step 1: Mr of H₂O = (2 × 1) + 16 = 18

Step 2: Total mass of oxygen = 1 × 16 = 16

Step 3: Percentage of O = (16 ÷ 18) × 100 = 88.9%

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Example 2: Percentage of Iron in Iron(III) Oxide (Fe₂O₃)

This is a classic exam question. Iron(III) oxide is the main component of rust.

Step 1: Mr of Fe₂O₃ = (2 × 56) + (3 × 16) = 112 + 48 = 160

Step 2: Total mass of iron = 2 × 56 = 112

Step 3: Percentage of Fe = (112 ÷ 160) × 100 = 70.0%

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Example 3: Percentage of Nitrogen in Ammonium Nitrate (NH₄NO₃)

Step 1: Mr of NH₄NO₃

• N: 2 atoms × 14 = 28
• H: 4 atoms × 1 = 4
• O: 3 atoms × 16 = 48
• Mr = 28 + 4 + 48 = 80

Step 2: Total mass of nitrogen = 2 × 14 = 28

Step 3: Percentage of N = (28 ÷ 80) × 100 = 35.0%

Exam Tip: In NH₄NO₃, there are two nitrogen atoms — one from the NH₄ part and one from the NO₃ part. Count carefully.

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Example 4: Percentage of Carbon in Ethanol (C₂H₅OH)

Step 1: Work out the formula carefully. C₂H₅OH is equivalent to C₂H₆O.

• C: 2 atoms × 12 = 24
• H: 6 atoms × 1 = 6
• O: 1 atom × 16 = 16
• Mr = 24 + 6 + 16 = 46

Step 2: Total mass of carbon = 2 × 12 = 24

Step 3: Percentage of C = (24 ÷ 46) × 100 = 52.2%

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Example 5: Percentage of Calcium in Calcium Hydroxide — Ca(OH)₂

Step 1: Mr of Ca(OH)₂

• Ca: 1 × 40 = 40
• O: 2 × 16 = 32
• H: 2 × 1 = 2
• Mr = 40 + 32 + 2 = 74

Step 2: Total mass of calcium = 40

Step 3: Percentage of Ca = (40 ÷ 74) × 100 = 54.1%

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Using Percentage Composition to Compare Fertilisers

Fertilisers contain essential elements like nitrogen (N), phosphorus (P) and potassium (K). Farmers and exam questions often compare fertilisers by their percentage of a key element.

Example 6: Comparing Nitrogen Fertilisers

Which fertiliser contains a higher percentage of nitrogen: ammonium nitrate (NH₄NO₃) or urea (CO(NH₂)₂)?

Ammonium nitrate:

• Mr = 80 (calculated above)
• % N = (28 ÷ 80) × 100 = 35.0%

Urea — CO(NH₂)₂:

• Mr: C = 12, O = 16, N = 2 × 14 = 28, H = 4 × 1 = 4
• Mr = 12 + 16 + 28 + 4 = 60
• Total mass of N = 28
• % N = (28 ÷ 60) × 100 = 46.7%

Urea has a higher percentage of nitrogen (46.7% compared to 35.0%), so it delivers more nitrogen per gram.

Exam Tip: Fertiliser comparison questions are common. Calculate the percentage of the key element in each compound and then compare.

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Reverse Calculations: Finding Mass from Percentage

You may also need to work backwards — given a percentage and a total mass, find the mass of a specific element.

Example 7

Question: A 50 g sample of limestone contains 92% calcium carbonate (CaCO₃) by mass. What mass of calcium does it contain?

Solution:

1. Mass of CaCO₃ in sample = 50 × 0.92 = 46 g
2. % of Ca in CaCO₃ = (40 ÷ 100) × 100 = 40%
3. Mass of Ca = 46 × 0.40 = 18.4 g

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Example 8

Question: What mass of iron can be obtained from 800 kg of iron(III) oxide (Fe₂O₃)?

Solution:

1. % of Fe in Fe₂O₃ = 70.0% (calculated earlier)
2. Mass of Fe = 800 × 0.70 = 560 kg

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Checking Your Answer

A useful check: the percentages of all elements in a compound must add up to 100%.

Example check for H₂O:

• % H = (2 ÷ 18) × 100 = 11.1%
• % O = (16 ÷ 18) × 100 = 88.9%
• Total = 11.1% + 88.9% = 100% ✓

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Common Mistakes

1. Forgetting to multiply Ar by the number of atoms — e.g. in Fe₂O₃, there are 2 iron atoms, so use 2 × 56 = 112, not just 56.
2. Using the wrong Mr — Always recalculate Mr; do not assume you remember it correctly.
3. Not counting atoms in brackets — In Ca(NO₃)₂, there are 2 nitrogen atoms and 6 oxygen atoms.
4. Confusing percentage composition with atom economy — Percentage composition is about elements in a compound; atom economy is about products in a reaction.

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Summary