Relative Formula Mass

This lesson covers relative atomic mass (Ar) and relative formula mass (Mr) as required by the AQA GCSE Chemistry specification (4.3.1). These are essential skills for all quantitative chemistry calculations. You need to be able to read Ar values from the periodic table, calculate Mr for any compound, and understand what these values represent.

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Relative Atomic Mass (Ar)

The relative atomic mass (Ar) of an element is the average mass of one atom of that element compared to one-twelfth of the mass of a carbon-12 atom. It is a dimensionless quantity (it has no units) because it is a ratio.

Term	Symbol	Definition
Relative atomic mass	Ar	Average mass of one atom of an element relative to 1/12 of a carbon-12 atom
Mass number	A	Total number of protons and neutrons in the nucleus of a specific isotope

The Ar values are found on the periodic table. They are usually shown as the larger number associated with each element. Because elements often exist as a mixture of isotopes, the Ar is a weighted average of the masses of all the naturally occurring isotopes.

Key Ar Values to Know

Element	Symbol	Ar
Hydrogen	H	1
Carbon	C	12
Nitrogen	N	14
Oxygen	O	16
Sodium	Na	23
Magnesium	Mg	24
Aluminium	Al	27
Sulfur	S	32
Chlorine	Cl	35.5
Potassium	K	39
Calcium	Ca	40
Iron	Fe	56
Copper	Cu	63.5
Zinc	Zn	65

Exam Tip: You do not need to memorise Ar values — they are always given on the periodic table in the exam. However, knowing the common ones (H = 1, C = 12, O = 16, N = 14) will save you time.

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Calculating Relative Formula Mass (Mr)

The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of all the atoms shown in its chemical formula.

Mr = sum of all Ar values in the formula

Worked Example 1: Water (H2O)

• H2O contains 2 hydrogen atoms and 1 oxygen atom.
• Mr = (2 x Ar of H) + (1 x Ar of O)
• Mr = (2 x 1) + (1 x 16)
• Mr = 2 + 16
• Mr = 18

Worked Example 2: Calcium Carbonate (CaCO3)

• CaCO3 contains 1 calcium atom, 1 carbon atom, and 3 oxygen atoms.
• Mr = (1 x 40) + (1 x 12) + (3 x 16)
• Mr = 40 + 12 + 48
• Mr = 100

Worked Example 3: Magnesium Hydroxide (Mg(OH)2)

When a formula contains brackets, multiply everything inside the brackets by the subscript number outside.

• Mg(OH)2 contains 1 Mg, 2 O, and 2 H.
• Mr = (1 x 24) + (2 x 16) + (2 x 1)
• Mr = 24 + 32 + 2
• Mr = 58

Worked Example 4: Aluminium Sulfate (Al2(SO4)3)

• Al2(SO4)3 contains 2 Al, 3 S, and 12 O.
• Mr = (2 x 27) + (3 x 32) + (12 x 16)
• Mr = 54 + 96 + 192
• Mr = 342

Exam Tip: When dealing with brackets in formulae, be very careful with your counting. A common mistake is forgetting to multiply ALL atoms inside the brackets. For Al2(SO4)3, there are 3 x 4 = 12 oxygen atoms, not just 4.

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Step-by-Step Process for Mr Calculations

graph TD
    A["Write the chemical formula"] --> B["Identify all elements and count atoms"]
    B --> C["Look up Ar of each element"]
    C --> D["Multiply Ar by number of atoms"]
    D --> E["Add all values together"]
    E --> F["State the Mr value"]

    style A fill:#3498db,color:#fff
    style B fill:#2980b9,color:#fff
    style C fill:#e67e22,color:#fff
    style D fill:#d35400,color:#fff
    style E fill:#27ae60,color:#fff
    style F fill:#2ecc71,color:#fff

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Dealing with Formulae Containing Brackets

Brackets in chemical formulae indicate that a group of atoms is repeated. The subscript number outside the bracket tells you how many times that group appears.

Formula	Atoms Present	Mr Calculation	Mr
Ca(OH)2	1 Ca, 2 O, 2 H	40 + (2 x 16) + (2 x 1)	74
Mg(NO3)2	1 Mg, 2 N, 6 O	24 + (2 x 14) + (6 x 16)	148
Fe2(SO4)3	2 Fe, 3 S, 12 O	(2 x 56) + (3 x 32) + (12 x 16)	400
(NH4)2SO4	2 N, 8 H, 1 S, 4 O	(2 x 14) + (8 x 1) + 32 + (4 x 16)	132
Ca3(PO4)2	3 Ca, 2 P, 8 O	(3 x 40) + (2 x 31) + (8 x 16)	310

Exam Tip: Show your working clearly when calculating Mr. Even if you get the final answer wrong, you can still pick up method marks for correct Ar values and multiplication. Write out each element separately.

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Percentage Composition by Mass

You can use Mr to calculate the percentage by mass of an element in a compound:

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

Worked Example: Percentage of Oxygen in Water (H2O)

• Mr of H2O = (2 x 1) + (1 x 16) = 18
• Total mass of oxygen in formula = 1 x 16 = 16
• Percentage of oxygen = (16 / 18) x 100 = 88.9%

Worked Example: Percentage of Carbon in Carbon Dioxide (CO2)

• Mr of CO2 = 12 + (2 x 16) = 44
• Total mass of carbon in formula = 12
• Percentage of carbon = (12 / 44) x 100 = 27.3%

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

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

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