Moles, Formulae and Reacting Mass Calculations
Reacting mass calculations use the mole as a 'chemical amount' that links the mass of a substance to the number of particles it contains, via molar mass (Mr, in g/mol) and the ratios in a balanced symbol equation. At IGCSE, balanced equations are used to calculate the mass of a reactant needed or a product formed, identify a limiting reactant when reactants are supplied in the wrong ratio, and calculate percentage yield when a reaction does not produce the full theoretical amount of product.
Before you start
Make sure you're comfortable with these topics first:
Method
- Learn the core mole equation: number of moles = mass (g) / molar mass (g/mol), and be able to rearrange it to find mass (moles x molar mass) or molar mass (mass / moles).
- Calculate the molar mass of any formula by adding the relative atomic masses (Ar) of every atom in it, remembering to multiply by any subscript - for example the molar mass of CaCO3 is 40 + 12 + (16 x 3) = 100 g/mol.
- For a reacting mass question, always start from a correctly balanced symbol equation, since the balancing numbers give the mole ratio between reactants and products.
- Convert the given mass to moles, use the mole ratio from the balanced equation to find the moles of the substance you want, then convert that back into a mass using its own molar mass.
- To find a limiting reactant, convert the mass of each reactant given into moles, then divide each by its balancing number in the equation; the reactant giving the smaller value is used up first and limits the amount of product formed, so use only its moles for the rest of the calculation.
- Calculate percentage yield using: percentage yield = (actual yield / theoretical yield) x 100, where the theoretical yield is the mass calculated from the equation assuming the reaction goes to completion with no losses, and the actual yield is the mass actually obtained (always less than 100%, since practical processes lose product through incomplete reactions, side reactions, and transfer or purification losses).
Worked example
Calcium carbonate decomposes on heating: CaCO3(s) -> CaO(s) + CO2(g). Calculate the maximum mass of calcium oxide that can be produced by heating 25 g of calcium carbonate. (Ar: Ca = 40, C = 12, O = 16)
- Calculate the molar mass of CaCO3: 40 + 12 + (16 x 3) = 100 g/mol.
- Calculate moles of CaCO3: 25 / 100 = 0.25 mol.
- Use the equation's mole ratio (1:1 between CaCO3 and CaO) to find moles of CaO: 0.25 mol.
- Calculate the molar mass of CaO: 40 + 16 = 56 g/mol.
- Calculate the mass of CaO: 0.25 x 56 = 14 g.
- Final answer: 14 g of calcium oxide.
Practice questions
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Q1Calculate the molar mass of magnesium oxide, MgO. (Ar: Mg = 24, O = 16)Show answer
Answer: 40 g/mol (24 + 16)
Q2Calculate the number of moles in 11 g of carbon dioxide, CO2. (Ar: C = 12, O = 16, so Mr = 44)Show answer
Answer: 0.25 mol (11/44)
Q3Calculate the mass of 0.5 mol of sodium chloride, NaCl. (Ar: Na = 23, Cl = 35.5, so Mr = 58.5)Show answer
Answer: 29.25 g (0.5 x 58.5)
Q4In the equation 2Mg + O2 -> 2MgO, state the mole ratio of magnesium to magnesium oxide.Show answer
Answer: 1:1 (2 mol of Mg produces 2 mol of MgO).
Q5A student calculates a theoretical yield of 8.0 g of copper but obtains only 6.4 g in the experiment. Calculate the percentage yield.Show answer
Answer: 80% ((6.4/8.0) x 100)
Q6In the reaction N2 + 3H2 -> 2NH3, 2 mol of nitrogen and 3 mol of hydrogen are mixed. Identify the limiting reactant.Show answer
Answer: Hydrogen. Dividing moles by the balancing number gives nitrogen 2/1 = 2 and hydrogen 3/3 = 1, so hydrogen gives the smaller value and is used up first.
Q7Give two reasons why the actual yield of a reaction is usually less than the theoretical yield.Show answer
Answer: Any two of: the reaction may not go to completion; some product may be lost when transferred between containers or during filtration/purification; unwanted side reactions may occur, producing different products.
Exam-style questions
Written in the style of a IGCSE Science exam paper, with a full mark scheme.
Magnesium reacts with hydrochloric acid: Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g). Calculate the mass of magnesium chloride produced when 6.0 g of magnesium reacts completely. (Ar: Mg = 24, Cl = 35.5, so Mr of MgCl2 = 95)
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A student reacts 5.0 g of calcium carbonate with excess hydrochloric acid and collects 1.65 g of carbon dioxide gas. The theoretical yield of carbon dioxide from 5.0 g of calcium carbonate is 2.2 g. Calculate the percentage yield.
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Iron reacts with sulfur to form iron(II) sulfide: Fe(s) + S(s) -> FeS(s). A student mixes 11.2 g of iron with 9.6 g of sulfur and heats the mixture until it reacts completely. (Ar: Fe = 56, S = 32). (a) Determine which reactant is in excess. (b) Calculate the maximum mass of iron(II) sulfide that could be produced.
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Free printable worksheet
Want more practice on paper? Download the moles, formulae and reacting mass calculations worksheet pack - 5 pages of exam-style questions with a full mark scheme. One email opens every download in this browser for 14 days - no account, no card. Print it for personal and classroom use.
This topic is chapter 17 of IGCSE Science Workbook, the whole course as one free printable PDF.
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