Percentage Composition by Mass

Percentage composition tells you what fraction of a compound’s mass comes from a particular element. It appears on both Paper 2 and Paper 4, often as the first step before a question asks you to deduce an empirical formula.

The method

1. Calculate the Mr of the compound.
2. Find the total Ar contribution of the element you are interested in (Ar multiplied by the number of atoms of that element in the formula).
3. Divide by Mr.
4. Multiply by 100 to convert to a percentage.

The formula

Percentage by mass = (total Ar of the element in the formula / Mr of the compound) x 100

Where:

• Total Ar of the element = Ar of the element x number of atoms of that element in one formula unit
• Mr = relative formula mass of the entire compound

Worked examples

Example 1: Percentage of oxygen in water (H2O)

Question: Calculate the percentage by mass of oxygen in water. (Ar: H = 1, O = 16)

Working:

• Mr of H2O = (2 x 1) + 16 = 18
• Mass of O in formula = 16
• % O = (16 / 18) x 100 = 88.9%

Answer: 88.9%

Examiner note: The question asks for oxygen specifically. Do not calculate the hydrogen percentage unless asked.

Example 2: Percentage of carbon in methane (CH4)

Question: Calculate the percentage by mass of carbon in methane, CH4. (Ar: C = 12, H = 1)

Working:

• Mr of CH4 = 12 + (4 x 1) = 16
• Mass of C in formula = 12
• % C = (12 / 16) x 100 = 75.0%

Answer: 75.0%

Examiner note: This is a 2-mark question. One mark for Mr, one mark for the final answer. Show the Mr calculation even if it seems obvious.

Example 3: Percentage of nitrogen in ammonium nitrate (NH4NO3)

Question: Calculate the percentage by mass of nitrogen in ammonium nitrate, NH4NO3. (Ar: N = 14, H = 1, O = 16)

Working:

• Mr of NH4NO3 = 14 + (4 x 1) + 14 + (3 x 16) = 14 + 4 + 14 + 48 = 80
• Total mass of N in formula = 2 x 14 = 28 (there are two nitrogen atoms)
• % N = (28 / 80) x 100 = 35.0%

Answer: 35.0%

Examiner note: The formula contains two nitrogen atoms in different positions. Count both. Students who count only one nitrogen get 17.5%.

Example 4: Percentage of iron in iron(III) oxide (Fe2O3)

Question: Calculate the percentage by mass of iron in iron(III) oxide, Fe2O3. (Ar: Fe = 56, O = 16)

Working:

• Mr of Fe2O3 = (2 x 56) + (3 x 16) = 112 + 48 = 160
• Total mass of Fe in formula = 2 x 56 = 112
• % Fe = (112 / 160) x 100 = 70.0%

Answer: 70.0%

Examiner note: This type of question often links to extraction of metals. The percentage tells you the maximum mass of iron you could extract from a given mass of ore.

Example 5: Percentage of calcium in calcium hydroxide (Ca(OH)2)

Question: Calculate the percentage by mass of calcium in calcium hydroxide, Ca(OH)2. (Ar: Ca = 40, O = 16, H = 1)

Working:

• Mr of Ca(OH)2 = 40 + 2(16 + 1) = 40 + 34 = 74
• Mass of Ca in formula = 40
• % Ca = (40 / 74) x 100 = 54.1%

Answer: 54.1%

Examiner note: Expand the bracket first: Ca(OH)2 contains 1 Ca, 2 O, and 2 H. The bracket is the usual source of error.

Example 6: Which fertiliser has the highest percentage of nitrogen?

Question: Three fertilisers are ammonium sulfate (NH4)2SO4, urea CO(NH2)2, and ammonium nitrate NH4NO3. Which has the highest percentage of nitrogen by mass? (Ar: N = 14, H = 1, S = 32, O = 16, C = 12)

Working:

Ammonium sulfate (NH4)2SO4:

• Mr = (2 x 14) + (8 x 1) + 32 + (4 x 16) = 28 + 8 + 32 + 64 = 132
• % N = (28 / 132) x 100 = 21.2%

Urea CO(NH2)2:

• Mr = 12 + 16 + (2 x 14) + (4 x 1) = 12 + 16 + 28 + 4 = 60
• % N = (28 / 60) x 100 = 46.7%

Ammonium nitrate NH4NO3:

• Mr = 14 + 4 + 14 + 48 = 80
• % N = (28 / 80) x 100 = 35.0%

Answer: Urea has the highest percentage of nitrogen at 46.7%.

Examiner note: Comparison questions award marks for each correct Mr and percentage, plus one for the correct identification. Even if your values contain an error, you can still earn the comparison mark if your conclusion follows from your numbers.

Common mistakes

1. Counting atoms of the target element incorrectly. NH4NO3 has two nitrogen atoms, not one. Read the entire formula.
2. Bracket errors. (NH4)2SO4 has 2 nitrogen atoms and 8 hydrogen atoms. The subscript outside the bracket multiplies everything inside.
3. Dividing the wrong way round. It is element mass divided by compound Mr, not the other way round.
4. Rounding too early. Calculate Mr as an exact number before dividing. Rounding Mr introduces unnecessary error.
5. Forgetting to multiply by 100. The formula gives a decimal; the question asks for a percentage.

When this appears

Percentage composition questions appear directly on Paper 2 as multiple choice and on Paper 4 as structured calculation questions. They are closely linked to empirical formula calculations, where you reverse the process to go from percentages back to a formula. Understanding percentage composition also connects to the fertilisers subtopic and to comparing the efficiency of different compounds as sources of a particular element. The underlying theory is covered in the stoichiometry topic.