Reacting Masses

Summary: Calculate masses of reactants and products using balanced equations and mole ratios. The five-step method: balanced equation, moles of known, mole ratio, moles of unknown, mass of unknown. Includes limiting reactant calculations and percentage yield = (actual yield / theoretical yield) x 100%. Tags: igcse chemistry stoichiometry Created: 2026-07-14 Last Updated: 2026-07-16

The Principle of Reacting Masses

A balanced chemical equation tells you the mole ratio in which substances react and are produced. From the mole ratio, and knowing the Mr of each substance, the masses involved can be calculated.

The key insight: Equations work in MOLES, not grams. Mass must be converted to moles first, then the mole ratio from the equation used, then converted back to mass.

The Five-Step Method

For every reacting mass calculation, follow these steps:

If a mass is given, convert to moles. If a mass is asked for, moles are needed first.

Example 1: Basic Reacting Mass Calculate the mass of magnesium oxide (MgO) formed when 12 g of magnesium burns completely in oxygen. [Ar: Mg = 24, O = 16]

Step 1: Balanced equation
2Mg + O2 -> 2MgO

Step 2: Moles of known (Mg)
n(Mg) = m / Mr = 12 / 24 = 0.50 mol

Step 3: Mole ratio
From equation: 2 mol Mg produces 2 mol MgO
Ratio Mg : MgO = 2 : 2 = 1 : 1

Step 4: Moles of unknown (MgO)
n(MgO) = n(Mg) x (2/2) = 0.50 mol

Step 5: Mass of unknown
Mr(MgO) = 24 + 16 = 40
m(MgO) = n x Mr = 0.50 x 40 = 20 g

Example 2: Reacting Mass (Different Ratio) Calculate the mass of calcium oxide (CaO) produced when 50 g of calcium carbonate (CaCO3) is heated strongly (thermal decomposition). [Ar: Ca = 40, C = 12, O = 16]

Step 1: Balanced equation
CaCO3 -> CaO + CO2

Step 2: Moles of known (CaCO3)
Mr(CaCO3) = 40 + 12 + (3 x 16) = 100
n(CaCO3) = 50 / 100 = 0.50 mol

Step 3: Mole ratio
CaCO3 : CaO = 1 : 1

Step 4: Moles of unknown (CaO)
n(CaO) = 0.50 mol

Step 5: Mass of unknown
Mr(CaO) = 40 + 16 = 56
m(CaO) = 0.50 x 56 = 28 g

Example 3: Reacting Mass (Harder Ratio) Calculate the mass of iron produced when 80 g of iron(III) oxide (Fe2O3) is reduced by carbon monoxide. [Ar: Fe = 56, O = 16]

Step 1: Balanced equation
Fe2O3 + 3CO -> 2Fe + 3CO2

Step 2: Moles of known (Fe2O3)
Mr(Fe2O3) = (2 x 56) + (3 x 16) = 112 + 48 = 160
n(Fe2O3) = 80 / 160 = 0.50 mol

Step 3: Mole ratio
1 mol Fe2O3 produces 2 mol Fe
Ratio Fe2O3 : Fe = 1 : 2

Step 4: Moles of unknown (Fe)
n(Fe) = 0.50 x 2 = 1.0 mol

Step 5: Mass of unknown
m(Fe) = 1.0 x 56 = 56 g

Example 4: Reacting Mass (Finding Mass of Reactant Needed) What mass of hydrogen gas is needed to react completely with 42 g of nitrogen to produce ammonia? [Ar: N = 14, H = 1]

Step 1: Balanced equation
N2 + 3H2 -> 2NH3

Step 2: Moles of known (N2)
Mr(N2) = 2 x 14 = 28
n(N2) = 42 / 28 = 1.5 mol

Step 3: Mole ratio
1 mol N2 reacts with 3 mol H2
Ratio N2 : H2 = 1 : 3

Step 4: Moles of unknown (H2)
n(H2) = 1.5 x 3 = 4.5 mol

Step 5: Mass of unknown
Mr(H2) = 2 x 1 = 2
m(H2) = 4.5 x 2 = 9.0 g

Limiting Reactants

In many reactions, one reactant is present in excess while the other is used up completely. The reactant that runs out first is the limiting reactant — it determines the maximum amount of product that can be formed.

How to identify the limiting reactant:

1. Calculate the moles of each reactant present
2. Compare the mole ratio from the balanced equation to the actual mole ratio
3. The reactant that gives the SMALLER amount of product is the limiting reactant

Example 5: Limiting Reactant 16 g of sulfur reacts with 24 g of oxygen to form SO2. Which reactant is in excess, and which is the limiting reactant? Calculate the mass of SO2 produced. [Ar: S = 32, O = 16]

Step 1: Balanced equation
S + O2 -> SO2

Step 2: Moles of each reactant
n(S) = 16 / 32 = 0.50 mol
n(O2) = 24 / 32 = 0.75 mol  (Mr of O2 = 2 x 16 = 32)

Step 3: Compare to mole ratio (1 : 1)
0.50 mol S needs 0.50 mol O2 to react completely
We have 0.75 mol O2, so O2 is in EXCESS
Sulfur is the LIMITING REACTANT (it will run out first)

Step 4: Moles of product (from limiting reactant)
0.50 mol S produces 0.50 mol SO2

Step 5: Mass of SO2
Mr(SO2) = 32 + 32 = 64
m(SO2) = 0.50 x 64 = 32 g

Example 6: Limiting Reactant (Trickier Ratio) 6.0 g of magnesium reacts with 4.0 g of oxygen. Which reactant is limiting? What mass of MgO is produced? [Ar: Mg = 24, O = 16]

Step 1: Balanced equation
2Mg + O2 -> 2MgO

Step 2: Moles of each reactant
n(Mg) = 6.0 / 24 = 0.25 mol
n(O2) = 4.0 / 32 = 0.125 mol  (Mr of O2 = 32)

Step 3: Identify limiting reactant using mole ratio (2 : 1)
For 0.25 mol Mg, O2 needed = 0.25 / 2 = 0.125 mol → EXACTLY enough!
For 0.125 mol O2, Mg needed = 0.125 x 2 = 0.25 mol → EXACTLY enough!
Neither is in excess — they are in exactly the stoichiometric ratio.

Moles of MgO from Mg: 0.25 x (2/2) = 0.25 mol
Moles of MgO from O2: 0.125 x (2/1) = 0.25 mol
Both give same result.

Step 4: Mass of MgO
Mr(MgO) = 24 + 16 = 40
m(MgO) = 0.25 x 40 = 10 g

Example 7: Standard Reacting Mass (Exam-style) Calculate the mass of CO2 produced when 20 g of CaCO3 is heated. [Ar: Ca = 40, C = 12, O = 16]

CaCO3 -> CaO + CO2
Mr(CaCO3) = 100
n(CaCO3) = 20 / 100 = 0.20 mol
Mole ratio CaCO3 : CO2 = 1 : 1
n(CO2) = 0.20 mol
Mr(CO2) = 44
m(CO2) = 0.20 x 44 = 8.8 g

Example 8: Limiting Reactant + Yield 2.7 g of aluminium reacts with 8.0 g of bromine (Br2). (a) Identify the limiting reactant. (b) Calculate the theoretical yield of AlBr3. (c) If 6.5 g of AlBr3 was obtained, calculate the percentage yield. [Ar: Al = 27, Br = 80]

(a) Balanced equation: 2Al + 3Br2 -> 2AlBr3

n(Al) = 2.7 / 27 = 0.10 mol
n(Br2) = 8.0 / 160 = 0.050 mol  (Mr of Br2 = 2 x 80 = 160)

From equation: 2 mol Al reacts with 3 mol Br2
0.10 mol Al needs (3/2) x 0.10 = 0.15 mol Br2 — but only 0.050 mol is available
Therefore Br2 is the LIMITING REACTANT.

(b) From Br2: 3 mol Br2 -> 2 mol AlBr3
n(AlBr3) = 0.050 x (2/3) = 0.0333 mol
Mr(AlBr3) = 27 + (3 x 80) = 267
Theoretical yield = 0.0333 x 267 = 8.9 g

(c) Percentage yield = (6.5 / 8.9) x 100% = 73%

Percentage Yield

In practice, reactions rarely give 100% of the theoretical amount of product. The percentage yield measures how much product was actually obtained compared to the theoretical maximum.

Percentage yield = (actual yield / theoretical yield) x 100%

Reasons for yields being less than 100%:

• Reaction may be reversible (does not go to completion)
• Side reactions produce other products
• Product is lost during purification (filtration, crystallisation, distillation)
• Reactants may not be pure
• Some product may be left on apparatus (transfer losses)

Example: Percentage Yield In the thermal decomposition of 50 g of CaCO3, the theoretical yield of CaO is 28 g. The actual yield obtained was 22.4 g of CaO. Calculate the percentage yield.

Percentage yield = (22.4 / 28.0) x 100% = 80%

Combined example: When 80 g of Fe2O3 is reduced, the theoretical yield of iron is 56 g. The actual yield was 44.8 g.

Percentage yield = (44.8 / 56.0) x 100% = 80%

Key Points

• Balanced equations give MOLE ratios, not mass ratios
• The five-step method: Equation → Moles of known → Mole ratio → Moles of unknown → Mass of unknown
• Limiting reactant = the reactant that runs out first; determines the maximum amount of product
• Excess reactant = the reactant that is left over after the reaction stops
• Percentage yield = (actual yield / theoretical yield) x 100%
• Yields are rarely 100% due to reversible reactions, side reactions, and product loss during purification
• Always show the mole ratio step clearly using the coefficients from the balanced equation

Key Concepts from Past Papers

• Limiting reactant: the reactant that is completely used up in a reaction and determines the amount of product formed
• Excess reactant: the reactant that is not completely used up; some remains after the reaction
• Theoretical yield: the maximum mass of product that can be formed, calculated from the balanced equation
• Actual yield: the mass of product actually obtained from the reaction
• Percentage yield: (actual yield / theoretical yield) x 100%
• Convert mass to moles using n = m / Mr
• Use the mole ratio from the balanced equation
• Yield less than 100% because of reversible reaction / side reactions / product lost during purification

Keywords from Past Papers

molecules, particles, concentration, movement, rate, reaction, lower, spread, koh, move, collide, mix, volume, amount, increases

Related Notes

• Relative Masses and Moles
• Empirical and Molecular Formulae
• Concentration and Gas Volumes

Sources

• OpenStax Chemistry 2e — [Chapter 4: Stoichiometry of Chemical Reactions], Rice University (free, CC BY 4.0)
• BBC Bitesize GCSE Chemistry — [Reacting Mass Calculations], BBC (free educational resource)
• Cambridge IGCSE Chemistry 0620 — Syllabus Section 3: Stoichiometry (Reacting Masses), Cambridge Assessment International Education
• CK-12 Chemistry for High School — [Chapter 12: Stoichiometry], CK-12 Foundation (free, CC BY-NC 3.0)

Past Paper Sources

• 0620/31 May/June 2015: Q22(b)(iv) (0m), Q44(a)(iii) (0m), Q55(c)(i) (0m) (+3 more)
• 0620/32 Feb/March 2017: Q33(b)(v) (2m)
• 0620/32 May/June 2018: Q11(b)(i) (1m)
• 0620/32 May/June 2020: Q66(b)(i) (1m)
• 0620/33 May/June 2016: Q66(d)(i) (1m)
• 0620/33 May/June 2017: Q66(b)(iv) (1m)
• 0620/33 May/June 2021: Q22(d)(ii) (3m)
• 0620/33 May/June 2022: Q33(b)(ii) (1m)
• 0620/33 May/June 2023: Q88(c)(v) (0m)
• 0620/33 May/June 2024: Q66(c)(ii) (1m)
• 0620/33 October/November 2015: Q77(b)(i) (2m), Q77(b)(ii) (0m)
• 0620/33 October/November 2018: Q22(a)(i) (2m)

Common Misconceptions