Reacting Masses Calculations

Reacting masses calculations are the backbone of quantitative chemistry at IGCSE level. They connect the mass of one substance in a reaction to the mass of another, using the balanced chemical equation as the bridge. Master this method and you have the framework for almost every stoichiometry question on the paper.

The method

1. Write a balanced chemical equation for the reaction.
2. Identify the substance whose mass you know (the “given”) and the substance whose mass you need to find (the “target”).
3. Calculate the Mr of both the given and target substances.
4. Find the moles of the given substance: moles = mass / Mr.
5. Use the mole ratio from the balanced equation to find the moles of the target substance.
6. Calculate the mass of the target substance: mass = moles x Mr.

The formula

moles = mass / Mr

mass = moles x Mr

Where:

• mass is in grams (g)
• Mr is the relative formula mass (no units)
• moles is in mol

The mole ratio comes from the coefficients in the balanced equation.

Worked examples

Example 1: Magnesium and oxygen

Question: What mass of magnesium oxide is produced when 12 g of magnesium burns in excess oxygen? (Ar: Mg = 24, O = 16)

Working:

• Balanced equation: 2Mg + O2 -> 2MgO
• Mr of Mg = 24, Mr of MgO = 24 + 16 = 40
• Moles of Mg = 12 / 24 = 0.5 mol
• Mole ratio: 2Mg : 2MgO = 1 : 1
• Moles of MgO = 0.5 mol
• Mass of MgO = 0.5 x 40 = 20 g

Answer: 20 g of magnesium oxide

Examiner note: The 1:1 ratio here is straightforward. Always write the mole ratio step even when it is 1:1, as the examiner needs to see it for full marks.

Example 2: Calcium carbonate decomposition

Question: What mass of calcium oxide is produced when 50 g of calcium carbonate is heated? (Ar: Ca = 40, C = 12, O = 16)

Working:

• Balanced equation: CaCO3 -> CaO + CO2
• Mr of CaCO3 = 40 + 12 + (3 x 16) = 100
• Mr of CaO = 40 + 16 = 56
• Moles of CaCO3 = 50 / 100 = 0.5 mol
• Mole ratio: 1 CaCO3 : 1 CaO
• Moles of CaO = 0.5 mol
• Mass of CaO = 0.5 x 56 = 28 g

Answer: 28 g of calcium oxide

Examiner note: This is a thermal decomposition. The equation has a 1:1 ratio, so the mole conversion is direct.

Example 3: Iron and hydrochloric acid

Question: What mass of iron(II) chloride is formed when 5.6 g of iron reacts with excess hydrochloric acid? (Ar: Fe = 56, Cl = 35.5, H = 1)

Working:

• Balanced equation: Fe + 2HCl -> FeCl2 + H2
• Mr of Fe = 56, Mr of FeCl2 = 56 + (2 x 35.5) = 127
• Moles of Fe = 5.6 / 56 = 0.1 mol
• Mole ratio: 1 Fe : 1 FeCl2
• Moles of FeCl2 = 0.1 mol
• Mass of FeCl2 = 0.1 x 127 = 12.7 g

Answer: 12.7 g of iron(II) chloride

Examiner note: “Excess hydrochloric acid” tells you that iron is the limiting reagent. All the iron reacts.

Example 4: Sodium and water (ratio is not 1:1)

Question: What mass of hydrogen gas is produced when 4.6 g of sodium reacts with excess water? (Ar: Na = 23, H = 1)

Working:

• Balanced equation: 2Na + 2H2O -> 2NaOH + H2
• Mr of Na = 23, Mr of H2 = 2
• Moles of Na = 4.6 / 23 = 0.2 mol
• Mole ratio: 2Na : 1H2, so Na : H2 = 2 : 1
• Moles of H2 = 0.2 / 2 = 0.1 mol
• Mass of H2 = 0.1 x 2 = 0.2 g

Answer: 0.2 g of hydrogen

Examiner note: The mole ratio is 2:1, not 1:1. This is the step where most errors occur. Always write the ratio explicitly.

Example 5: Aluminium and iron oxide (thermite reaction)

Question: What mass of aluminium is needed to react completely with 32 g of iron(III) oxide? (Ar: Al = 27, Fe = 56, O = 16)

Working:

• Balanced equation: 2Al + Fe2O3 -> Al2O3 + 2Fe
• Mr of Fe2O3 = (2 x 56) + (3 x 16) = 160
• Mr of Al = 27
• Moles of Fe2O3 = 32 / 160 = 0.2 mol
• Mole ratio: 2Al : 1Fe2O3, so Al : Fe2O3 = 2 : 1
• Moles of Al = 0.2 x 2 = 0.4 mol
• Mass of Al = 0.4 x 27 = 10.8 g

Answer: 10.8 g of aluminium

Examiner note: Here you are working backwards from product to reactant. The method is identical regardless of direction.

Example 6: Zinc carbonate and hydrochloric acid

Question: What mass of zinc chloride is produced when 25 g of zinc carbonate reacts with excess hydrochloric acid? (Ar: Zn = 65, C = 12, O = 16, Cl = 35.5, H = 1)

Working:

• Balanced equation: ZnCO3 + 2HCl -> ZnCl2 + H2O + CO2
• Mr of ZnCO3 = 65 + 12 + (3 x 16) = 125
• Mr of ZnCl2 = 65 + (2 x 35.5) = 136
• Moles of ZnCO3 = 25 / 125 = 0.2 mol
• Mole ratio: 1 ZnCO3 : 1 ZnCl2
• Moles of ZnCl2 = 0.2 mol
• Mass of ZnCl2 = 0.2 x 136 = 27.2 g

Answer: 27.2 g of zinc chloride

Examiner note: This reaction produces three products. Identify the target substance clearly in your working so you calculate the Mr of the right compound.

Common mistakes

1. Using an unbalanced equation. The mole ratio is only correct when the equation is balanced. Always check coefficients before using them.
2. Forgetting to convert the mole ratio. When the ratio is not 1:1, students often skip the ratio step and use the same number of moles for both substances.
3. Calculating Mr of the wrong substance. In a multi-product reaction, make sure you are calculating Mr for the substance the question asks about.
4. Not converting units. If a mass is given in kilograms or tonnes, convert to grams before dividing by Mr.
5. Ignoring “excess.” When the question says a reagent is “in excess,” it means the other reagent is the limiting reagent and determines the amount of product.

When this appears

Reacting masses questions appear regularly on Paper 4 (structured questions) and occasionally on Paper 2 (multiple choice). They build directly on moles from mass and relative formula mass. The same method extends to more advanced calculations such as percentage yield and titration calculations. See the stoichiometry topic for the underlying theory.