Relative Formula Mass

This lesson covers relative atomic mass (Ar) and relative formula mass (Mr) as required by the Edexcel GCSE Chemistry specification (1CH0), Topic 1: Key Concepts in Chemistry and Topic 9: Separate Chemistry 2. Understanding how to calculate Mr is essential because it underpins almost every quantitative calculation in chemistry.

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Relative Atomic Mass (Ar)

Every element on the periodic table has a relative atomic mass (Ar). This is the number shown at the top of each element's box on the periodic table (sometimes the bottom, depending on the version).

Relative atomic mass is defined as the average mass of the atoms of an element compared to 1/12 of the mass of a carbon-12 atom.

In practice, for GCSE, you simply read the Ar value from the periodic table:

Element	Symbol	Ar
Hydrogen	H	1
Carbon	C	12
Nitrogen	N	14
Oxygen	O	16
Sodium	Na	23
Magnesium	Mg	24
Chlorine	Cl	35.5
Calcium	Ca	40
Iron	Fe	56

Exam Tip: You will always be given a periodic table in the exam. You do not need to memorise Ar values. However, being familiar with common ones saves time.

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What Is 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 formula.

For example, water has the formula H₂O. This means it contains 2 hydrogen atoms and 1 oxygen atom.

Mr of H₂O = (2 × Ar of H) + (1 × Ar of O) = (2 × 1) + (1 × 16) = 18

The unit of Mr is no unit — it is a relative quantity (a ratio).

Exam Tip: Always show your working clearly when calculating Mr in exam questions. Write out which atoms you are adding and how many of each. This earns method marks even if you make an arithmetic slip.

flowchart TD
    A[Chemical formula] --> B[Identify each element and count atoms]
    B --> C{Brackets present?}
    C -- Yes --> D["Multiply atoms inside bracket<br/>by subscript outside"]
    C -- No --> E[Use atom counts directly]
    D --> F[Look up Ar for each element on periodic table]
    E --> F
    F --> G["For each element: Ar × number of atoms"]
    G --> H[Sum all contributions]
    H --> I["Mr = total (no unit)"]

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Worked Examples

Example 1: Sodium Chloride (NaCl)

Formula: NaCl

• 1 × Na = 1 × 23 = 23
• 1 × Cl = 1 × 35.5 = 35.5

Mr = 23 + 35.5 = 58.5

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Example 2: Carbon Dioxide (CO₂)

Formula: CO₂

• 1 × C = 1 × 12 = 12
• 2 × O = 2 × 16 = 32

Mr = 12 + 32 = 44

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Example 3: Calcium Carbonate (CaCO₃)

Formula: CaCO₃

• 1 × Ca = 1 × 40 = 40
• 1 × C = 1 × 12 = 12
• 3 × O = 3 × 16 = 48

Mr = 40 + 12 + 48 = 100

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Example 4: Sulfuric Acid (H₂SO₄)

Formula: H₂SO₄ (Ar values: H = 1, S = 32, O = 16)

• 2 × H = 2 × 1 = 2
• 1 × S = 1 × 32 = 32
• 4 × O = 4 × 16 = 64

Mr = 2 + 32 + 64 = 98

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Dealing with Brackets in Formulae

Brackets in a chemical formula mean that everything inside the brackets is multiplied by the subscript number outside.

Example 5: Magnesium Hydroxide — Mg(OH)₂

The formula Mg(OH)₂ means:

• 1 × Mg
• The (OH) group appears twice, so there are 2 × O and 2 × H

Calculation:

• 1 × Mg = 1 × 24 = 24
• 2 × O = 2 × 16 = 32
• 2 × H = 2 × 1 = 2

Mr = 24 + 32 + 2 = 58

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Example 6: Calcium Nitrate — Ca(NO₃)₂

The formula Ca(NO₃)₂ means:

• 1 × Ca
• The (NO₃) group appears twice, so there are 2 × N and 6 × O (2 × 3 = 6)

Calculation:

• 1 × Ca = 1 × 40 = 40
• 2 × N = 2 × 14 = 28
• 6 × O = 6 × 16 = 96

Mr = 40 + 28 + 96 = 164

Exam Tip: A very common mistake is forgetting to multiply all atoms inside the brackets. For Ca(NO₃)₂, there are 2 nitrogen atoms and 6 oxygen atoms, not just 2 oxygens.

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Example 7: Aluminium Sulfate — Al₂(SO₄)₃

The formula Al₂(SO₄)₃ means:

• 2 × Al
• The (SO₄) group appears three times, so there are 3 × S and 12 × O (3 × 4 = 12)

Calculation (Ar values: Al = 27, S = 32, O = 16):

• 2 × Al = 2 × 27 = 54
• 3 × S = 3 × 32 = 96
• 12 × O = 12 × 16 = 192

Mr = 54 + 96 + 192 = 342

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Using Mr in Further Calculations

Once you can calculate Mr, you can use it in many other calculations throughout the course:

Calculation	Formula
Number of moles	moles = mass ÷ Mr
Mass of a substance	mass = moles × Mr
Percentage composition	% = (Ar × number of atoms ÷ Mr) × 100
Concentration in mol/dm³	mol/dm³ = (g/dm³) ÷ Mr

These will all be covered in detail in later lessons.

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Common Mistakes

1. Forgetting the subscript — In H₂O, there are 2 hydrogens, not 1.
2. Ignoring brackets — In Mg(OH)₂, multiply both O and H by 2.
3. Using the wrong Ar — Always double-check the periodic table.
4. Rounding Ar values — Use the values given in the exam. Chlorine is 35.5, not 35 or 36.

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Summary

• Relative atomic mass (Ar) is the average mass of an atom of an element relative to carbon-12. Read it from the periodic table.
• Relative formula mass (Mr) is the sum of all Ar values in a formula.
• To handle brackets, multiply every atom inside the bracket by the subscript outside.
• Mr has no unit — it is a dimensionless ratio.
• Always show your working clearly in calculations.
• Mr is the foundation for moles, reacting masses, concentration and many other calculations in chemistry.

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Extended Worked Examples

These worked examples take you from a bare formula through to a full quantitative answer. Notice how we always show the unit at each step, state the ratio clearly and check the answer is sensible.

Worked Example A — Mr with brackets and waters of crystallisation

Calculate the relative formula mass Mr of hydrated magnesium sulfate, MgSO₄·7H₂O.

Step 1 — list atoms:

• 1 × Mg
• 1 × S
• 4 × O (in SO₄)
• 7 × H₂O = 14 × H and 7 × O

Step 2 — multiply by Ar:

• Mg: 1 × 24 = 24
• S: 1 × 32 = 32
• O (sulfate): 4 × 16 = 64
• H (water): 14 × 1 = 14
• O (water): 7 × 16 = 112

Step 3 — add: Mr = 24 + 32 + 64 + 14 + 112 = 246

Check: A hydrated salt should have a larger Mr than the anhydrous salt (MgSO₄ = 120). 246 is bigger, so this is sensible.

Worked Example B — Mr of ammonium sulfate, (NH₄)₂SO₄

Step 1 — expand the brackets. (NH₄)₂ means 2 × N and 8 × H. Step 2 — total atoms: 2 × N, 8 × H, 1 × S, 4 × O. Step 3 — multiply and add:

• N: 2 × 14 = 28
• H: 8 × 1 = 8
• S: 1 × 32 = 32
• O: 4 × 16 = 64
• Mr = 28 + 8 + 32 + 64 = 132

Worked Example C — Using Mr to convert mass to moles

A student weighs out 5.85 g of sodium chloride, NaCl. How many moles are present?

Step 1 — Mr of NaCl = 23 + 35.5 = 58.5 Step 2 — moles = mass ÷ Mr = 5.85 ÷ 58.5 = 0.100 mol

Unit check: g ÷ (g/mol) = mol. Always keep the unit on display.

Worked Example D — Mr in a balanced equation (foundation for later topics)

For the reaction 2Mg + O₂ → 2MgO, confirm that mass is conserved by comparing total Mr on each side (treating the coefficient as a multiplier):

• LHS: 2 × Ar(Mg) + Mr(O₂) = 2 × 24 + 32 = 80
• RHS: 2 × Mr(MgO) = 2 × (24 + 16) = 80

Totals match — conservation confirmed.

Worked Example E — Concentration in g/dm³ with Mr

A solution contains 10.6 g of sodium carbonate (Na₂CO₃) in 250 cm³ of water. Calculate the concentration in g/dm³ and in mol/dm³ (Higher).

Step 1 — volume in dm³: 250 ÷ 1000 = 0.250 dm³ Step 2 — concentration in g/dm³ = 10.6 ÷ 0.250 = 42.4 g/dm³ Step 3 — Mr of Na₂CO₃ = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106 Step 4 — moles = 10.6 ÷ 106 = 0.100 mol Step 5 — concentration in mol/dm³ = 0.100 ÷ 0.250 = 0.400 mol/dm³

Worked Example F — Limiting reactant check using Mr (preview)

1.2 g of Mg is added to 50 cm³ of 1.0 mol/dm³ HCl. Which is the limiting reactant?

• Moles of Mg = 1.2 ÷ 24 = 0.050 mol
• Moles of HCl = 0.050 × 1.0 = 0.050 mol
• Equation: Mg + 2HCl → MgCl₂ + H₂. Ratio is 1:2.
• Mg needs 2 × 0.050 = 0.100 mol HCl but only 0.050 mol available.
• Therefore HCl is limiting; Mg is in excess.

This preview shows why Mr is the gateway skill for moles, limiting reactant and yield questions later in the course.

Common Pitfalls in Mr Calculations

Examiner reports from recent Edexcel papers flag these errors repeatedly:

1. Forgetting a bracket. For Al₂(SO₄)₃, students often forget the subscript 3 applies to everything in the bracket. The correct atom count is 2 Al, 3 S, 12 O and Mr = (2 × 27) + (3 × 32) + (12 × 16) = 54 + 96 + 192 = 342. Missing the 3 outside the bracket gives the wrong answer of 150.
2. Mis-reading Ar. Chlorine is 35.5, not 35 or 36. Copper is 63.5 or 64 depending on the periodic table provided — always check the data booklet.
3. Writing an incorrect unit. Mr has no unit. Writing "42 g" or "42 g/mol" is wrong at GCSE. Molar mass (in g/mol) is numerically equal to Mr but is a separate quantity.
4. Not multiplying by the coefficient. When the balanced equation has a 2 in front of a formula, that coefficient counts for moles and masses but does not alter Mr of the substance itself.

Worked Example G — Mr for an ionic compound with variable oxidation state

Calculate Mr of iron(III) sulfate, Fe₂(SO₄)₃.

• 2 Fe: 2 × 56 = 112
• 3 S: 3 × 32 = 96
• 12 O: 12 × 16 = 192
• Mr = 112 + 96 + 192 = 400

Sanity check: A large polyatomic compound should have a three-digit Mr. 400 is reasonable for iron(III) sulfate.

Worked Example H — Reverse calculation

A compound with formula CaX₂ has Mr = 111. Identify element X.

• Mr contribution of Ca = 40
• Mr contribution of 2X = 111 − 40 = 71
• Ar of X = 71 ÷ 2 = 35.5 → X is chlorine, compound is CaCl₂.

Answering at different grade levels

Grade band	Typical response
Grades 1–3	States that Mr is "adding up the numbers" but forgets brackets; may write a unit such as g.
Grades 4–5	Correctly uses relative formula mass Mr, handles simple formulae (NaCl, CO₂), states Mr has no unit.
Grades 6–7	Handles brackets and hydrated salts; links Mr to mole using moles = mass ÷ Mr; uses concentration mol/dm³ confidently.
Grades 8–9	Integrates Mr into multi-step calculations for limiting reactant, percentage yield and atom economy; explains why Mr is dimensionless; checks answers by conservation of mass.

Edexcel alignment: This content is aligned with Edexcel GCSE Chemistry (1CH0) specification Topic 3 Chemical changes — specifically 3.1–3.4 quantitative chemistry; Topic 4 Extracting metals and equilibria also applies for yield/atom economy. Assessed on Paper 1.