One method for every multi-step calculation: the mole map, reacting masses, limiting reactants, titrations, yield and atom economy, with marks, units and rounding.
Objectives
Use one method, the mole map, to plan any calculation and show every step for the marks.
Combine reacting masses, limiting reactants and percentage yield in a single question.
Work through a titration from titres to a concentration in mol/dm³ and g/dm³.
Give answers with the right units and number of significant figures, and check they are sensible.
Exam calculations: moles, solutions, titration and yield explained
Start here
Quantitative chemistry questions look like many different things, but almost all of them are the same few moves in a different order. Marks are lost in the same places every year: a missed cm³ to dm³ change, the wrong ratio, or a unit left off.
This lesson gives you one method that always works, then practises it on mixed exam-style questions. It pulls together the moles, concentration, titration and yield lessons.
Course note: moles, reacting masses and limiting reactants are Higher tier in Combined Science, and concentration in g/dm³ is for both tiers. Concentration in mol/dm³, titration calculations, percentage yield and atom economy are Separate Chemistry content. Use the parts your course has taught.
The mole map
Think of moles as the middle of a map. Mass is on one side and solutions are on the other. To go from mass to moles, divide by Mr. To go back, multiply by Mr. To go from concentration to moles, multiply by the volume in dm³. To go back, divide by the volume.
Every calculation starts by turning what you are given into moles. Then, if there is a second substance, you cross using the ratio from the balanced equation. Last, turn the moles of the thing you want into the unit asked for.
The unit trap: concentration is per dm³, so change cm³ to dm³ by dividing by 1000 before you multiply or divide. For example, 25.0 cm³ is 0.0250 dm³.
A map linking mass, moles and concentration. Dividing mass by Mr gives moles and multiplying moles by Mr gives mass. Multiplying concentration by volume gives moles and dividing moles by volume gives concentration. Volume is in dm cubed.
Reacting masses in four lines
Use the same four lines every time. One, write the balanced equation. Two, turn the known mass into moles. Three, use the ratio of the balancing numbers to find the moles of the substance you want. Four, turn those moles into a mass.
The ratio line is where marks vanish. Write it out in words: for 2Al₂O₃ → 4Al + 3O₂, "2 mol of aluminium oxide make 4 mol of aluminium", so the moles of aluminium are twice the moles of oxide.
Always finish with a sense check. A product made from a single reactant cannot weigh more than the reactant, and the masses of all products must add up to the mass of the reactants.
Limiting reactants
When the question gives amounts of two reactants, one will run out first. Find the moles of each. Then use the ratio to decide which one is limiting: ask how much of the second reactant the first would need.
Example: Mg + 2HCl → MgCl₂ + H₂ with 0.30 mol of Mg and 0.50 mol of HCl. 0.30 mol of Mg needs 0.60 mol of HCl, but there is only 0.50 mol. So HCl is limiting and Mg is in excess.
Everything after that uses the limiting reactant, never the one in excess. The HCl, 0.50 mol, makes 0.50 ÷ 2 = 0.25 mol of H₂.
Percentage yield and atom economy
A long question often carries on from the limiting reactant. Use the limiting reactant to find the theoretical yield, which is the most product the equation allows. Then percentage yield = actual yield ÷ theoretical yield × 100.
Example: 5.0 g of calcium carbonate reacts with 100 cm³ of 0.50 mol/dm³ hydrochloric acid in CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. CaCO₃ is 5.0 ÷ 100 = 0.050 mol. HCl is 0.50 × 0.100 = 0.050 mol. 0.050 mol of CaCO₃ needs 0.10 mol of HCl, so HCl is limiting. CO₂ = 0.050 ÷ 2 = 0.025 mol, which is 0.025 × 44 = 1.1 g. If 0.88 g is collected, the percentage yield is 0.88 ÷ 1.1 × 100 = 80%.
Atom economy is separate. It uses only the equation: Mr of the useful product ÷ total Mr of the reactants × 100, with each Mr multiplied by its balancing number. For C₂H₄ + H₂O → C₂H₅OH it is 46 ÷ (28 + 18) × 100 = 100%, because there is only one product.
The titration chain
Use the concordant titres only, then work in five links. One, the mean titre. Two, the moles of the solution whose concentration you know: concentration × volume in dm³. Three, the moles of the other solution, from the equation ratio. Four, its concentration: moles ÷ volume in dm³. Five, convert to g/dm³ by multiplying by Mr, if asked.
Example: 25.0 cm³ of sodium hydroxide is titrated with 0.100 mol/dm³ hydrochloric acid. The titres are 24.10 (rough), 23.40, 23.50 and 23.45 cm³. The mean of the concordant titres is 23.45 cm³. Moles of HCl = 0.100 × 0.02345 = 0.002345. The ratio is 1 : 1, so NaOH is 0.002345 mol. Concentration = 0.002345 ÷ 0.0250 = 0.0938 mol/dm³. In g/dm³ it is 0.0938 × 40 = 3.75 g/dm³.
Watch the ratio. Sulfuric acid is 1 : 2 with sodium hydroxide, so the moles of acid are half the moles of alkali.
Marks, units and rounding
Show every line of working. In a 4 mark calculation, usually 1 mark is for the conversion to moles, 1 for the ratio, 1 for the next conversion and 1 for the answer with its unit. A correct answer with no working may still get full marks, but a wrong answer with good working keeps most of them.
Keep full calculator values between steps and round only at the end. Give the final answer to the same number of significant figures as the least precise number in the question, or to the number the question asks for. 3.752 g/dm³ to 3 significant figures is 3.75 g/dm³.
Write the unit every time: mol, g, g/dm³, mol/dm³, cm³. A percentage yield cannot be over 100%. If an answer is not sensible, go back and look for the cm³ and dm³ slip or the ratio.
Exam technique and recap
Command words. Calculate: work out a number and show working. Show that: finish with the value you were given and show every step. Explain which is limiting: use the ratio and compare the moles needed with the moles present. Suggest: use the numbers to give a reasoned reason.
Recap. Convert to moles first. Use the balanced-equation ratio to cross. Convert cm³ to dm³. Limiting reactant decides the product. Percentage yield compares actual with theoretical. Atom economy uses Mr values from the equation only. Give units and sensible rounding.
Practise the whole method on mixed questions below, and always check the answer is sensible before you move on.
Key points
Moles = mass ÷ Mr, and moles = concentration × volume in dm³. Divide cm³ by 1000.
Use the balancing numbers to cross from one substance to another.
The limiting reactant runs out first and decides the amount of product.
Percentage yield = actual ÷ theoretical × 100. Atom economy = Mr of useful product ÷ total Mr of reactants × 100.
Worked examples
Example 1
Mg + 2HCl → MgCl₂ + H₂. Calculate the mass of magnesium chloride made when 4.8 g of magnesium reacts with excess hydrochloric acid. Ar of Mg = 24 and Cl = 35.5.
Moles of Mg = 4.8 ÷ 24 = 0.20 mol.
The ratio Mg : MgCl₂ is 1 : 1, so moles of MgCl₂ = 0.20 mol. Mr of MgCl₂ = 24 + 2 × 35.5 = 95.
Mass of MgCl₂ = 0.20 × 95 = 19 g.
Answer: 19 g
Common slips
Slip: Moles of Mg = 24 ÷ 4.8 = 5.0. Moles is mass divided by Mr, so the mass goes on the top: 4.8 ÷ 24.
Slip: Mr of MgCl₂ = 24 + 35.5 = 59.5. There are two chlorine atoms in MgCl₂, so use 2 × 35.5.
Slip: Mass = 0.20 × 2 = 0.40 g. Multiply by the Mr of the product you want, MgCl₂, which is 95.
Example 2
5.0 g of calcium carbonate is added to 100 cm³ of hydrochloric acid of concentration 0.50 mol/dm³. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Mr of CaCO₃ = 100 and CO₂ = 44. Show which reactant is limiting. The student collects 0.88 g of carbon dioxide. Calculate the percentage yield.
Moles of CaCO₃ = 5.0 ÷ 100 = 0.050 mol. Moles of HCl = 0.50 × 0.100 = 0.050 mol.
The ratio CaCO₃ : HCl is 1 : 2, so 0.050 mol of CaCO₃ needs 0.10 mol of HCl. Only 0.050 mol is present, so HCl is limiting.
Answer: HCl is limiting, because 0.050 mol of CaCO₃ needs 0.10 mol of HCl and only 0.050 mol is present. The theoretical yield of CO₂ is 1.1 g, so the percentage yield is 80%
Common slips
Slip: The moles are equal, so neither is limiting. The equation needs 2 mol of HCl for each mol of CaCO₃, so the acid runs out first.
Slip: Moles of CO₂ = 0.050 mol, from CaCO₃. Use the limiting reactant, HCl, and its 2 : 1 ratio with CO₂.
Slip: Percentage yield = 1.1 ÷ 0.88 × 100 = 125%. Divide the actual yield by the theoretical yield, not the other way round.
Example 3
25.0 cm³ of sodium hydroxide solution is neutralised by hydrochloric acid of concentration 0.100 mol/dm³. The titres are 24.10 (rough), 23.40, 23.50 and 23.45 cm³. Calculate the concentration of the sodium hydroxide in g/dm³. Mr of NaOH = 40. HCl + NaOH → NaCl + H₂O
Mean of the concordant titres = (23.40 + 23.50 + 23.45) ÷ 3 = 23.45 cm³. Moles of HCl = 0.100 × 0.02345 = 0.002345 mol.
The ratio is 1 : 1, so moles of NaOH = 0.002345 mol. Concentration = 0.002345 ÷ 0.0250 = 0.0938 mol/dm³.
In g/dm³ the concentration is 0.0938 × 40 = 3.75 g/dm³.
Answer: 3.75 g/dm³
Common slips
Slip: Mean titre = (24.10 + 23.40 + 23.50 + 23.45) ÷ 4. The rough titre is not used in the mean. Use only the concordant results.
Slip: Moles of HCl = 0.100 × 23.45 = 2.345 mol. Change the titre to dm³ first: 23.45 ÷ 1000 = 0.02345.
Slip: Concentration in g/dm³ = 0.0938 ÷ 40 = 0.00235. To change mol/dm³ into g/dm³, multiply by the Mr.
Questions and answers
Exam calculations: moles, solutions, titration and yield practice questions for GCSE Combined Science Chemistry. Try each one, then open the answer. Easy questions come first.
Question 1 Easy
Calculate the relative formula mass (Mr) of calcium hydroxide, Ca(OH)₂. Ar of Ca = 40, O = 16 and H = 1.
Show answer
Answer: 74
Ca(OH)₂ has 1 Ca, 2 O and 2 H.
40 + 2 × 16 + 2 × 1 = 74.
Question 2 Easy
Calculate the number of moles in 11 g of carbon dioxide. Mr of CO₂ = 44.
Show answer
Answer: 0.25 mol
Moles = mass ÷ Mr.
11 ÷ 44 = 0.25.
Question 3 Easy
Convert 35 cm³ into dm³.
Show answer
Answer: 0.035 dm³
Divide by 1000.
35 ÷ 1000 = 0.035.
Question 4 Easy
Calculate the mass of 0.50 mol of sodium chloride. Mr of NaCl = 58.5.
Show answer
Answer: 29 g
Mass = moles × Mr.
0.50 × 58.5 = 29.25, which is 29 g to 2 significant figures.
Question 5 Easy
A student collects 6.3 g of a product. The theoretical yield is 7.0 g. Calculate the percentage yield.
Show answer
Answer: 90%
6.3 ÷ 7.0 = 0.90.
0.90 × 100 = 90%.
Question 6 Medium
Mg + 2HCl → MgCl₂ + H₂. Calculate the mass of hydrogen made when 2.4 g of magnesium reacts with excess acid. Ar of Mg = 24 and Mr of H₂ = 2.
Show answer
Answer: 0.20 g
Moles of Mg = 2.4 ÷ 24 = 0.10.
The ratio Mg : H₂ is 1 : 1, so moles of H₂ = 0.10.
Mass = 0.10 × 2 = 0.20 g.
Question 7 Medium
Calculate the concentration in mol/dm³ of a solution made by dissolving 4.0 g of sodium hydroxide in water to make 250 cm³. Mr of NaOH = 40.
Show answer
Answer: 0.40 mol/dm³
Moles = 4.0 ÷ 40 = 0.10.
Volume = 0.250 dm³.
Concentration = 0.10 ÷ 0.250 = 0.40 mol/dm³.
Question 8 Medium
Calculate the number of moles of hydrochloric acid in 25.0 cm³ of a 0.200 mol/dm³ solution.
Show answer
Answer: 0.00500 mol
25.0 cm³ = 0.0250 dm³.
Moles = 0.200 × 0.0250 = 0.00500.
Exam-style questions
Longer questions in the style of a written paper, with the marking points. These are Yumio practice questions.
Exam-style question 14 marks
25.0 cm³ of sodium hydroxide solution is neutralised by 21.0 cm³ of hydrochloric acid of concentration 0.100 mol/dm³. HCl + NaOH → NaCl + H₂O
(a) Calculate the moles of hydrochloric acid used. [2 marks]
(b) State the moles of sodium hydroxide in the 25.0 cm³. [1 mark]
(c) Calculate the concentration of the sodium hydroxide in mol/dm³. [1 mark]
2.4 g of magnesium is added to 100 cm³ of hydrochloric acid of concentration 0.50 mol/dm³. Mg + 2HCl → MgCl₂ + H₂. Ar of Mg = 24 and Mr of H₂ = 2.
(a) Calculate the moles of magnesium and the moles of hydrochloric acid. [2 marks]
(b) Show which reactant is limiting. [2 marks]
(c) Calculate the mass of hydrogen formed. [2 marks]
Show answer and marking points
Answer: (a) 0.10 mol Mg and 0.050 mol HCl (b) 0.10 mol of Mg needs 0.20 mol of HCl but only 0.050 mol is present, so HCl is limiting (c) 0.025 mol of H₂, which is 0.050 g
(a) 0.10 mol Mg (1)
(a) 0.050 mol HCl (1)
(b) 0.10 mol of Mg needs 0.20 mol of HCl (1)
(b) only 0.050 mol present so HCl is limiting (1)
(c) 0.025 mol of H₂ (1)
(c) 0.050 g (1)
Common mistakes
Using a volume in cm³ in a concentration calculation
Divide by 1000 to change cm³ into dm³ before you multiply or divide.
Comparing moles of the two reactants without using the equation ratio
Write what the first reactant needs of the second, using the balancing numbers, and compare that with what is there.
Using the reactant in excess to calculate the product
Find the limiting reactant first. Only it decides how much product forms.
Treating every titration as 1 : 1
Read the equation. With sulfuric acid and sodium hydroxide the acid moles are half the alkali moles.
Including the rough titre in the mean
Use only concordant results, within 0.10 cm³ of each other.
Questions students ask
Is Exam calculations: moles, solutions, titration and yield on the Foundation or Higher tier?
Exam calculations: moles, solutions, titration and yield is Higher tier content for GCSE Combined Science Chemistry, aimed at around grade 8.
What is a common mistake on Exam calculations: moles, solutions, titration and yield?
Using a volume in cm³ in a concentration calculation. Divide by 1000 to change cm³ into dm³ before you multiply or divide.
Where are the answers to the Exam calculations: moles, solutions, titration and yield questions?
Open Show answer under each question on this page. The questions are Yumio practice questions, not past exam questions.