Reacting Masses

This lesson shows you how to use balanced equations and moles to calculate the masses of reactants and products in a chemical reaction, as required by the Edexcel GCSE Combined Science specification (1SC0). This is one of the most important calculation skills in GCSE chemistry.

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The Principle

A balanced equation tells you the ratio of moles in which substances react and are produced. You can use this ratio, along with the moles formula, to convert between masses.

The Step-by-Step Method

flowchart LR
    A["Write balanced\nequation"] --> B["Calculate M_r\nof substances"]
    B --> C["Convert given\nmass → moles"]
    C --> D["Apply mole\nratio"]
    D --> E["Convert\nmoles → mass"]
    style A fill:#6366f1,color:#fff,stroke:#4f46e5
    style B fill:#8b5cf6,color:#fff,stroke:#7c3aed
    style C fill:#a855f7,color:#fff,stroke:#9333ea
    style D fill:#c084fc,color:#000,stroke:#a855f7
    style E fill:#e879f9,color:#000,stroke:#d946ef

1. Write the balanced equation.
2. Calculate the $M_r$Mr​ of the substances you need.
3. Convert the given mass to moles using $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​.
4. Use the ratio from the balanced equation to find moles of the desired substance.
5. Convert moles back to mass using $\text{mass} = \text{moles} \times M_r$mass=moles×Mr​.

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Worked Examples

Example 1 — Magnesium and Oxygen

Question: What mass of magnesium oxide is formed when 4.8 g of magnesium reacts completely with oxygen?

$2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$2Mg(s)+O2​(g)→2MgO(s)

Step 1: $M_r$Mr​ of Mg = 24, $M_r$Mr​ of MgO = 24 + 16 = 40

Step 2: Moles of Mg: $\text{moles of Mg} = \frac{4.8}{24} = 0.2 \text{ mol}$moles of Mg=244.8​=0.2 mol

Step 3: Use the equation ratio. The ratio of Mg : MgO = 2 : 2 = 1 : 1

$\text{moles of MgO} = 0.2 \text{ mol}$moles of MgO=0.2 mol

Step 4: Mass of MgO: $\text{mass} = 0.2 \times 40 = 8.0 \text{ g}$mass=0.2×40=8.0 g

Example 2 — Calcium Carbonate Decomposition

Question: What mass of calcium oxide is produced when 25 g of calcium carbonate is thermally decomposed?

$\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$CaCO3​(s)→CaO(s)+CO2​(g)

Step 1: $M_r$Mr​ of $\text{CaCO}_3$CaCO3​ = 100, $M_r$Mr​ of CaO = 56

Step 2: Moles of $\text{CaCO}_3$CaCO3​: $\text{moles} = \frac{25}{100} = 0.25 \text{ mol}$moles=10025​=0.25 mol

Step 3: Ratio of $\text{CaCO}_3$CaCO3​ : CaO = 1 : 1

$\text{moles of CaO} = 0.25 \text{ mol}$moles of CaO=0.25 mol

Step 4: Mass of CaO: $\text{mass} = 0.25 \times 56 = 14 \text{ g}$mass=0.25×56=14 g

Example 3 — Acid + Carbonate (Unequal Ratio)

Question: What mass of carbon dioxide is produced when 10.6 g of sodium carbonate reacts with excess hydrochloric acid?

$\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$Na2​CO3​(s)+2HCl(aq)→2NaCl(aq)+H2​O(l)+CO2​(g)

Step 1: $M_r$Mr​ of $\text{Na}_2\text{CO}_3$Na2​CO3​ = (2 × 23) + 12 + (3 × 16) = 106, $M_r$Mr​ of $\text{CO}_2$CO2​ = 44

Step 2: Moles of $\text{Na}_2\text{CO}_3$Na2​CO3​: $\text{moles} = \frac{10.6}{106} = 0.1 \text{ mol}$moles=10610.6​=0.1 mol

Step 3: Ratio of $\text{Na}_2\text{CO}_3$Na2​CO3​ : $\text{CO}_2$CO2​ = 1 : 1

$\text{moles of CO}_2 = 0.1 \text{ mol}$moles of CO2​=0.1 mol

Step 4: Mass of $\text{CO}_2$CO2​: $\text{mass} = 0.1 \times 44 = 4.4 \text{ g}$mass=0.1×44=4.4 g

Example 4 — Iron and Chlorine

Question: What mass of iron is needed to react with 14.2 g of chlorine gas?

$2\text{Fe}(s) + 3\text{Cl}_2(g) \rightarrow 2\text{FeCl}_3(s)$2Fe(s)+3Cl2​(g)→2FeCl3​(s)

Step 1: $M_r$Mr​ of $\text{Cl}_2$Cl2​ = 71, $M_r$Mr​ of Fe = 56

Step 2: Moles of $\text{Cl}_2$Cl2​: $\text{moles} = \frac{14.2}{71} = 0.2 \text{ mol}$moles=7114.2​=0.2 mol

Step 3: Ratio of Fe : $\text{Cl}_2$Cl2​ = 2 : 3

$\text{moles of Fe} = \frac{2}{3} \times 0.2 = 0.1\overline{3} \text{ mol}$moles of Fe=32​×0.2=0.13 mol

Step 4: Mass of Fe: $\text{mass} = 0.1\overline{3} \times 56 = 7.47 \text{ g (3 s.f.)}$mass=0.13×56=7.47 g (3 s.f.)

Exam Tip: The most common mistake is forgetting to use the mole ratio from the balanced equation. Always identify the ratio first before calculating.

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Reacting Masses Without Calculating Moles

For simple cases, you can use a ratio method directly with masses. This is sometimes called the "direct proportion method".

Example — Quick Ratio Method

$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$CaCO3​→CaO+CO2​

From the equation: 100 g of $\text{CaCO}_3$CaCO3​ produces 56 g of CaO.

If you start with 50 g: $\text{mass of CaO} = \frac{56}{100} \times 50 = 28 \text{ g}$mass of CaO=10056​×50=28 g

Exam Tip: Both the moles method and the ratio method give the same answer. The moles method is more versatile for complex questions, but the ratio method is quicker for 1:1 reactions. Use whichever you are most comfortable with.

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Setting Out Your Working for Maximum Marks

The mark scheme usually awards marks for:

1. Correctly calculating $M_r$Mr​ values (1 mark)
2. Correctly calculating moles of the given substance (1 mark)
3. Using the correct mole ratio (1 mark)
4. Converting back to mass with the correct answer (1 mark)

Always show each step clearly — even if you make an arithmetic mistake, you can still earn method marks.

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Common Mistakes to Avoid

Mistake	Correction
Ignoring the mole ratio	Always read the ratio from the balanced equation
Using the wrong $M_r$Mr​	Double-check which substance you are converting
Not balancing the equation first	An unbalanced equation gives wrong ratios
Rounding too early	Keep full calculator values until the final answer

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Summary

• A balanced equation shows the mole ratio of reactants and products.
• To find reacting masses: mass → moles → use ratio → moles → mass.
• Always write the balanced equation first.
• Show every step of working for full marks.
• Keep calculations precise — do not round intermediate values.

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Extended Worked Examples

Use the mass → moles → ratio → moles → mass framework for every problem below. Examiners award method marks for each stage, so even if an arithmetic slip creeps in you can still earn most of the marks.

Example 5 — Hydrogen From Zinc and Dilute Sulfuric Acid

Question: What mass of hydrogen is produced when 6.50 g of zinc reacts completely with excess dilute sulfuric acid?

$\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{H}_2(g)$Zn(s)+H2​SO4​(aq)→ZnSO4​(aq)+H2​(g)

Step 1 — $M_r$Mr​: $A_r$Ar​(Zn) = 65; $M_r$Mr​($\text{H}_2$H2​) = 2.

Step 2 — moles of Zn: $n = 6.50 / 65 = 0.100 \text{ mol}$n=6.50/65=0.100 mol.

Step 3 — ratio: Zn : $\text{H}_2$H2​ = 1 : 1, so $n(\text{H}_2) = 0.100 \text{ mol}$n(H2​)=0.100 mol.

Step 4 — mass: $m = 0.100 \times 2 = \mathbf{0.20 \text{ g}}$m=0.100×2=0.20 g.

Example 6 — Mass of Reactant Needed

Question: What mass of aluminium reacts with 48.0 g of oxygen to form aluminium oxide?

$4\text{Al}(s) + 3\text{O}_2(g) \rightarrow 2\text{Al}_2\text{O}_3(s)$4Al(s)+3O2​(g)→2Al2​O3​(s)