C11: Reacting Masses
Applying the mole concept to balanced equations — calculating the masses of reactants needed and products formed, and understanding limiting and excess reactants.
Conservation of Mass
The law of conservation of mass states that no atoms are lost or made during a chemical reaction. The total mass of the reactants equals the total mass of the products.
This means the number of atoms of each element is the same on both sides of the equation. This is why chemical equations must always be balanced.
In some reactions mass may appear to change. If a gas is produced (e.g. thermal decomposition of metal carbonates), the gas escapes and the mass of the solid product appears lower. If measured in a closed system, mass is conserved.
When a substance burns in air, the mass of the products is greater than the mass of the original substance because oxygen from the air has been added. The total mass of reactants (substance + oxygen) still equals the total mass of products.
If a question asks why mass decreases in an open container, always mention the gas escaping. If it asks why mass increases, mention oxygen from the air reacting.
The Three-Step Method for Calculating Masses
To calculate the mass of a reactant or product from a balanced equation, use this three-step method:
Step 1: Write the balanced equation and identify the mole ratio.
Step 2: Calculate the moles of the substance you know the mass of.
Step 3: Use the mole ratio to find the moles (then mass) of the unknown substance.
Worked Example: Mass of Product from Mass of Reactant
Calculate the mass of magnesium oxide produced when 6 g of magnesium burns in oxygen.
Step 1: 2Mg + O2 → 2MgO (mole ratio Mg:MgO = 2:2 = 1:1)
Step 2: moles of Mg = mass / Ar = 6 / 24 = 0.25 mol
Step 3: From the ratio 1:1, moles of MgO = 0.25 mol
mass of MgO = moles × Mr = 0.25 × 40 = 10 g
Worked Example: Mass of Reactant Needed
What mass of calcium carbonate is needed to produce 11.2 g of calcium oxide?
Step 1: CaCO3 → CaO + CO2 (mole ratio CaCO3:CaO = 1:1)
Step 2: moles of CaO = mass / Mr = 11.2 / 56 = 0.2 mol
Step 3: From the ratio 1:1, moles of CaCO3 = 0.2 mol
mass of CaCO3 = 0.2 × 100 = 20 g
Worked Example: Non-1:1 Ratio
What mass of aluminium is needed to produce 102 g of aluminium oxide (Al2O3)?
Step 1: 4Al + 3O2 → 2Al2O3 (mole ratio Al:Al2O3 = 4:2 = 2:1)
Step 2: Mr of Al2O3 = (2 × 27) + (3 × 16) = 102
moles of Al2O3 = 102 / 102 = 1 mol
Step 3: From the ratio 2:1, moles of Al = 1 × 2 = 2 mol
mass of Al = 2 × 27 = 54 g
Always write out the balanced equation even if it is given — it helps you see the mole ratio clearly. Simplify ratios where possible (e.g. 4:2 becomes 2:1).
Calculating Masses from Balanced Equations — More Examples
Worked Example: Mass of Salt from Neutralisation
Calculate the mass of sodium chloride produced when 10 g of sodium hydroxide reacts with hydrochloric acid.
Step 1: NaOH + HCl → NaCl + H2O (mole ratio NaOH:NaCl = 1:1)
Step 2: Mr of NaOH = 23 + 16 + 1 = 40
moles of NaOH = 10 / 40 = 0.25 mol
Step 3: moles of NaCl = 0.25 mol
Mr of NaCl = 23 + 35.5 = 58.5
mass of NaCl = 0.25 × 58.5 = 14.625 g
Worked Example: Mass of Product from Iron Extraction
Calculate the mass of iron produced from 80 g of iron(III) oxide in the blast furnace.
Step 1: Fe2O3 + 3CO → 2Fe + 3CO2 (mole ratio Fe2O3:Fe = 1:2)
Step 2: Mr of Fe2O3 = (2 × 56) + (3 × 16) = 160
moles of Fe2O3 = 80 / 160 = 0.5 mol
Step 3: moles of Fe = 0.5 × 2 = 1 mol
mass of Fe = 1 × 56 = 56 g
Limiting and Excess Reactants
In most reactions, the reactants are not present in the exact ratio shown by the equation. One reactant will be completely used up first — this is the limiting reactant. The other reactant is in excess and some will be left over.
The limiting reactant determines the maximum amount of product that can be formed. To identify the limiting reactant, calculate the moles of each reactant and compare to the mole ratio in the balanced equation.
An easy way to identify the limiting reactant: divide the moles of each reactant by its coefficient in the balanced equation. The reactant with the smaller result is the limiting reactant.
Worked Example: Identifying the Limiting Reactant
8 g of magnesium reacts with 8 g of oxygen. Which is the limiting reactant?
2Mg + O2 → 2MgO
moles of Mg = 8 / 24 = 0.333 mol (coefficient = 2, so 0.333/2 = 0.167)
moles of O2 = 8 / 32 = 0.25 mol (coefficient = 1, so 0.25/1 = 0.25)
Mg gives the smaller value, so magnesium is the limiting reactant.
The mass of MgO formed = 0.333 × 40 = 13.3 g
Worked Example: Limiting Reactant with Non-1:1 Ratio
2.7 g of aluminium reacts with 3.2 g of oxygen. Find the limiting reactant and the mass of Al2O3 produced.
4Al + 3O2 → 2Al2O3
moles of Al = 2.7 / 27 = 0.1 mol (divide by 4 → 0.025)
moles of O2 = 3.2 / 32 = 0.1 mol (divide by 3 → 0.0333)
Al gives the smaller value, so aluminium is the limiting reactant.
From ratio 4:2, moles of Al2O3 = 0.1 × (2/4) = 0.05 mol
Mr of Al2O3 = 102, mass = 0.05 × 102 = 5.1 g
When a question gives you masses of both reactants, it is almost certainly asking about limiting reactants. Always check which one limits before calculating the mass of product.
Practice Questions
Calculate the mass of water produced when 4 g of hydrogen reacts completely with oxygen. (2H2 + O2 → 2H2O)
moles of H2 = 4 / 2 = 2 mol
Mole ratio H2:H2O = 2:2 = 1:1
moles of H2O = 2 mol
mass = 2 × 18 = 36 g
What mass of carbon dioxide is produced when 25 g of calcium carbonate thermally decomposes? (CaCO3 → CaO + CO2) (Ca = 40, C = 12, O = 16)
Mr of CaCO3 = 40 + 12 + 48 = 100
moles of CaCO3 = 25 / 100 = 0.25 mol
Mole ratio 1:1, so moles of CO2 = 0.25 mol
Mr of CO2 = 12 + 32 = 44
mass of CO2 = 0.25 × 44 = 11 g
6 g of carbon is burned in 16 g of oxygen. Identify the limiting reactant. (C + O2 → CO2)
moles of C = 6 / 12 = 0.5 mol (coefficient 1 → 0.5)
moles of O2 = 16 / 32 = 0.5 mol (coefficient 1 → 0.5)
Both give the same value — neither is in excess. They are in the exact ratio required.
13.8 g of sodium (Na) reacts with 9.6 g of oxygen (O2). Find the limiting reactant and the mass of Na2O produced. (4Na + O2 → 2Na2O)
moles of Na = 13.8 / 23 = 0.6 mol (divide by 4 → 0.15)
moles of O2 = 9.6 / 32 = 0.3 mol (divide by 1 → 0.3)
Na gives smaller value — sodium is the limiting reactant.
moles of Na2O = 0.6 × (2/4) = 0.3 mol
Mr of Na2O = (2 × 23) + 16 = 62
mass = 0.3 × 62 = 18.6 g
Mass Changes in Reactions
When a gas is a reactant (e.g. burning), the product mass is greater than the starting mass because gas from the air has combined with the solid. The total mass of all reactants still equals the total mass of all products.
When a gas is produced in an open container, it escapes and the measured mass decreases. However, if the same reaction is carried out in a sealed container, the mass stays the same throughout.
| Situation | Apparent Mass Change | Explanation |
|---|---|---|
| Metal burned in open air | Mass increases | Oxygen from air combines with the metal to form the oxide |
| Metal carbonate heated in open crucible | Mass decreases | CO2 gas is produced and escapes to the atmosphere |
| Any reaction in a sealed container | No mass change | All products including gases are trapped — mass is conserved |
Never just say "mass is lost" or "mass is gained" in an exam. Always explain which gas is involved and which direction it moves (into or out of the reaction vessel).
Maths Skills
The Three-Step Method for Reacting Masses
The three-step method is essential: (1) write the balanced equation and identify the mole ratio, (2) calculate moles of the known substance using moles = mass / Mr, (3) use the ratio to find moles and then mass of the unknown substance. Always show each step clearly to gain full marks.
Conservation of mass means the total mass of reactants equals the total mass of products. Use this to check your answers — if masses do not balance, you have made an error in your mole ratio or Mr calculation.
Identifying Limiting Reactants
For limiting reactant calculations, divide moles of each reactant by its coefficient in the balanced equation. The reactant giving the smallest value is the limiting reactant. Always use the limiting reactant to calculate the mass of product — the excess reactant will have some left over.
When converting between moles and mass, always show your working: moles = mass / Mr, then mass = moles × Mr. Do not try to skip steps, as this is where errors commonly occur. A common mistake is forgetting to simplify the mole ratio before using it (e.g. 4:2 becomes 2:1).
For non-1:1 ratios, take extra care: if the ratio of A:B is 2:3 and you have 0.4 mol of A, then moles of B = 0.4 × (3/2) = 0.6 mol. Always multiply by the ratio factor, never divide.
Common Misconceptions
Limiting Reactants
Wrong: The reactant with the smaller mass is always the limiting reactant Correct: The limiting reactant depends on moles and the mole ratio, not mass alone. A reactant with a small mass but low Mr could have more moles than a heavier reactant with a high Mr. For example, 2 g of H2 (1 mol) has more moles than 32 g of O2 (1 mol), but in the reaction 2H2 + O2, 2 g of H2 would still limit because the ratio requires 2 mol H2 per 1 mol O2
Percentage Yield
Wrong: 100% yield is always possible if you carry out the reaction carefully enough Correct: 100% yield is never achieved in practice because of side reactions, product losses during transfer, reversible reactions, and the difficulty of collecting all the product. Even with perfect technique, some product is always lost
Wrong: Mass can be lost in a closed system Correct: In a closed system, mass is always conserved. Apparent mass changes only happen in open systems where gases enter or escape
6-Mark Extended Question
Calculating Mass of Product
Explain how to calculate the mass of product formed in a reaction. Describe the three-step method.
Step 1: Write the balanced symbol equation for the reaction and identify the mole ratio between the known reactant and the desired product. For example, in 2Mg + O2 → 2MgO, the mole ratio of Mg:MgO is 2:2, which simplifies to 1:1. [2 marks]
Step 2: Calculate the number of moles of the known substance using the equation moles = mass / Mr. For example, if 6 g of magnesium is used, moles of Mg = 6 / 24 = 0.25 mol. You must calculate the Mr carefully from the periodic table. [2 marks]
Step 3: Use the mole ratio from the balanced equation to find the moles of the product, then convert to mass using mass = moles × Mr. From the 1:1 ratio, moles of MgO = 0.25 mol. Mr of MgO = 24 + 16 = 40, so mass of MgO = 0.25 × 40 = 10 g. Always check your answer makes sense — the mass of product should be consistent with the mass of reactant used. [2 marks]
AO3: Analyse and Evaluate
Identifying the Limiting Reactant
A student carries out the reaction 2Mg + O2 → 2MgO using different masses of magnesium. In experiment 1, they use 2.4 g of Mg with 3.2 g of O2. In experiment 2, they use 4.8 g of Mg with 3.2 g of O2. Identify the limiting reactant in each experiment and calculate the expected mass of MgO. (Mr: Mg = 24, O2 = 32, MgO = 40)
Experiment 1: moles of Mg = 2.4 / 24 = 0.1 mol (divide by coefficient 2 = 0.05). Moles of O2 = 3.2 / 32 = 0.1 mol (divide by coefficient 1 = 0.1). Mg gives the smaller value, so magnesium is the limiting reactant. Moles of MgO = 0.1 mol (from 1:1 ratio). Mass of MgO = 0.1 × 40 = 4.0 g.
Experiment 2: moles of Mg = 4.8 / 24 = 0.2 mol (divide by coefficient 2 = 0.1). Moles of O2 = 3.2 / 32 = 0.1 mol (divide by coefficient 1 = 0.1). Both give the same value, so neither is in excess — they are in the exact ratio. Moles of MgO = 0.2 mol. Mass of MgO = 0.2 × 40 = 8.0 g. Doubling the magnesium doubles the product because the same amount of oxygen is available and both are now fully used up.
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