Conservation of Mass

This lesson covers the law of conservation of mass as required by the Edexcel GCSE Combined Science specification (1SC0). You will learn why mass is conserved in chemical reactions, how balanced equations demonstrate this, and how to explain apparent changes in mass using practical evidence.

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The Law of Conservation of Mass

No atoms are created or destroyed during a chemical reaction.

This fundamental law, first clearly stated by Antoine Lavoisier in the 18th century, means:

$\text{Total mass of reactants} = \text{Total mass of products}$Total mass of reactants=Total mass of products

The atoms present at the start are simply rearranged to form new substances — they do not appear from nowhere or vanish.

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Why Is Mass Conserved?

In a chemical reaction:

1. Bonds between atoms in the reactants are broken.
2. New bonds are formed to make the products.
3. The same atoms are present before and after — just arranged differently.

Since no atoms are gained or lost, the total mass cannot change.

Demonstration With a Balanced Equation

Consider the reaction between magnesium and oxygen:

$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO

	Reactant Side	Product Side
Mg atoms	2	2
O atoms	2	2
Total $M_r$Mr​	$(2 \times 24) + 32 = 80$(2×24)+32=80	$2 \times (24 + 16) = 80$2×(24+16)=80

The mass on both sides is 80 — mass is conserved.

Exam Tip: If a question asks you to "explain conservation of mass", always state: (1) no atoms are created or destroyed, and (2) atoms are rearranged — the same number and type of atoms are on each side of the equation.

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Balanced Equations Prove Conservation

A correctly balanced equation has the same number of each type of atom on both sides. This is a direct consequence of conservation of mass.

Worked Example

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

Check atom count:

Atom	Left	Right
Ca	1	1
C	1	1
O	3	1 + 2 = 3

Atoms balance — mass is conserved.

Check mass:

• Reactant: $M_r$Mr​ of $\text{CaCO}_3$CaCO3​ = 40 + 12 + 48 = 100
• Products: $M_r$Mr​ of CaO + $M_r$Mr​ of $\text{CO}_2$CO2​ = (40 + 16) + (12 + 32) = 56 + 44 = 100

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Apparent Changes in Mass

Sometimes when you carry out a reaction on a balance, the mass seems to change. This does NOT mean conservation of mass is wrong — it means a gas has entered or left the reaction vessel.

Case 1: Mass Appears to Decrease

Example: Heating calcium carbonate in an open container.

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

• The $\text{CO}_2$CO2​ gas escapes into the atmosphere.
• The mass reading on the balance decreases.
• But the total mass of CaO + $\text{CO}_2$CO2​ still equals the mass of $\text{CaCO}_3$CaCO3​ — you just cannot weigh the gas that left.

Case 2: Mass Appears to Increase

Example: Burning magnesium in an open crucible.

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

• Oxygen from the air reacts with the magnesium.
• The product (MgO) contains the original Mg plus extra oxygen atoms.
• The mass reading on the balance increases because atoms from the air have been incorporated.

Case 3: No Change in Mass

If the reaction takes place in a sealed container (closed system), the mass before and after will be exactly the same, because no gas can enter or leave.

flowchart TD
    A[Chemical reaction on a balance] --> B{Open or closed system?}
    B -- Closed sealed --> C[No atoms enter or leave]
    C --> D[Balance reading unchanged]
    B -- Open --> E{Gas reactant or gas product?}
    E -- Gas product escapes --> F["Apparent mass decrease<br/>e.g. CaCO3 to CaO + CO2"]
    E -- Gas reactant from air absorbed --> G["Apparent mass increase<br/>e.g. Mg + O2 to MgO"]
    F --> H["Total mass of all substances<br/>including the gas is still conserved"]
    G --> H
    D --> H

Exam Tip: Questions about apparent mass changes in open containers come up frequently. Always explain by identifying which gas has entered or escaped.

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Practical Evidence

Experiment 1: Precipitation Reaction in a Sealed Flask

Mix lead nitrate solution with potassium iodide solution in a sealed flask on a balance:

$\text{Pb(NO}_3)_2(aq) + 2\text{KI}(aq) \rightarrow \text{PbI}_2(s) + 2\text{KNO}_3(aq)$Pb(NO3​)2​(aq)+2KI(aq)→PbI2​(s)+2KNO3​(aq)

• A yellow precipitate of lead iodide forms.
• The mass on the balance does not change — it is a closed system.

Experiment 2: Acid + Carbonate in an Open Flask

Add hydrochloric acid to calcium carbonate in an open flask:

$\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$CaCO3​(s)+2HCl(aq)→CaCl2​(aq)+H2​O(l)+CO2​(g)

• Fizzing occurs as $\text{CO}_2$CO2​ is produced.
• The mass decreases as $\text{CO}_2$CO2​ escapes.
• If done in a sealed flask with a cotton wool plug (to prevent shattering), less mass is lost, but some $\text{CO}_2$CO2​ still escapes through the plug.

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Calculating Mass Changes

If you know the mass of reactants and one product, you can calculate the mass of the other product using conservation of mass.

Worked Example

10.0 g of calcium carbonate is heated. 5.6 g of calcium oxide remains. What mass of carbon dioxide was produced?

$\text{mass of CO}_2 = \text{mass of CaCO}_3 - \text{mass of CaO} = 10.0 - 5.6 = 4.4 \text{ g}$mass of CO2​=mass of CaCO3​−mass of CaO=10.0−5.6=4.4 g

Worked Example — Mass Increase

4.8 g of magnesium is burned in excess oxygen. 8.0 g of magnesium oxide is formed. What mass of oxygen reacted?

$\text{mass of O}_2 = \text{mass of MgO} - \text{mass of Mg} = 8.0 - 4.8 = 3.2 \text{ g}$mass of O2​=mass of MgO−mass of Mg=8.0−4.8=3.2 g

Exam Tip: These "mass arithmetic" questions are straightforward — just use total reactants = total products. Always check that your answer makes physical sense.

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Summary

• The law of conservation of mass states that no atoms are created or destroyed in a chemical reaction.
• Total mass of reactants = total mass of products.
• Balanced equations show equal numbers of each type of atom on both sides.
• In open systems, mass may appear to change because gases enter or leave.
• In closed systems, the measured mass stays the same.
• You can use conservation of mass to calculate unknown masses in a reaction.

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Extended Worked Examples

These examples show you how to use conservation of mass as a tool for answering GCSE questions — not just a slogan. Lay every calculation out as an equation, a substitution and an answer with units.

Example 4 — Combustion of Magnesium With Full Atom Check

Question: 2.40 g of magnesium is burned completely in oxygen. What mass of magnesium oxide is formed? Use $A_r$Ar​(Mg) = 24, $A_r$Ar​(O) = 16.

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

Step 1 — moles of Mg. $n = 2.40 / 24 = 0.100 \text{ mol}$n=2.40/24=0.100 mol.

Step 2 — mole ratio Mg : MgO = 2 : 2 = 1 : 1. So $n(\text{MgO}) = 0.100 \text{ mol}$n(MgO)=0.100 mol.