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How to measure concentration in g/dm³ and mol/dm³, and how the limiting reactant decides how much product forms.
Specification references this topic covers: AQA 5.3.2; Edexcel C1.49, C1.52, C1.53.
A cup of squash can be weak or strong. The difference is how much squash is dissolved in the same amount of water. Chemists measure that with concentration.
It matters because labs and factories need exact amounts. A medicine that is too strong can be harmful. A reaction with too little of one chemical makes less product than you hoped for.
By the end you will be able to work out concentration in g/dm³ and mol/dm³. You will also find which reactant runs out first and use it to calculate how much product forms.
Course note: concentration in g/dm³ is for both tiers. Limiting reactants are Higher tier in Combined Science. Concentration in mol/dm³ is Separate Chemistry content, so use it if your teacher has taught it.
Concentration tells you how much solute is dissolved in a volume of solution. A solute is the substance that dissolves. In the bars, solution C is the most concentrated. We often give concentration in g/dm³, which is grams of solute in each dm³.
1 dm³ is 1000 cm³. To change cm³ into dm³, divide by 1000. So 250 cm³ is 0.25 dm³.
Concentration (g/dm³) = mass (g) ÷ volume (dm³). If 10 g of salt is dissolved in 250 cm³, the concentration is 10 ÷ 0.25 = 40 g/dm³. Rearranged, mass = concentration × volume.
A mole (mol) is the unit chemists use for the amount of a substance. Chemists often give concentration in mol/dm³. Concentration (mol/dm³) = amount of solute (mol) ÷ volume (dm³). A 2.0 mol/dm³ solution has 2.0 moles in each dm³.
Rearrange it to find the moles: moles = concentration × volume. Rearrange again to find the volume: volume = moles ÷ concentration. Always change the volume into dm³ first.
Example: how many moles are in 25 cm³ of a 0.20 mol/dm³ solution? 25 cm³ = 0.025 dm³. Moles = 0.20 × 0.025 = 0.0050 mol.
To change g/dm³ into mol/dm³, divide by the Mr. Mr is the relative formula mass, which is the mass of one mole. To change mol/dm³ into g/dm³, multiply by the Mr. Think of Mr as an exchange rate between grams and moles.
A 40 g/dm³ solution of sodium hydroxide has Mr = 40. So it is 40 ÷ 40 = 1.0 mol/dm³.
Example: 4.0 g of NaOH is dissolved to make 500 cm³ of solution. Moles = 4.0 ÷ 40 = 0.10 mol. Volume = 0.50 dm³. Concentration = 0.10 ÷ 0.50 = 0.20 mol/dm³.
In many reactions one reactant is added in excess, which means more than is needed. This makes sure the other reactant reacts completely. The reactant used up first is the limiting reactant. It limits how much product can form.
Think of sandwiches. Each one uses 2 slices of bread and 1 slice of cheese. With 10 slices of bread and 20 of cheese, the bread runs out first. Bread is limiting and the spare cheese is in excess.
In industry the cheaper reactant is used in excess. Then none of the expensive one is wasted.
First work out the moles of each reactant. Then use the balanced equation to see how much of one is needed to react with all of the other. The one there is less of, compared with the ratio, is the limiting reactant.
Take Mg + 2HCl → MgCl₂ + H₂. Suppose you have 2.4 g of magnesium, which is 0.10 mol. You also have 0.15 mol of hydrochloric acid.
0.10 mol of Mg needs 0.20 mol of HCl, but only 0.15 mol is there. So HCl is the limiting reactant and Mg is in excess.
The amount of product is decided by the limiting reactant, not the excess. In the last example, 0.15 mol of HCl makes 0.075 mol of H₂. The ratio of HCl to H₂ is 2 : 1.
Another example: 2Mg + O₂ → 2MgO with 4.8 g of Mg (0.20 mol) and 4.8 g of O₂ (0.15 mol). 0.20 mol of Mg needs only 0.10 mol of O₂. So O₂ is in excess and Mg is limiting.
0.20 mol of Mg makes 0.20 mol of MgO. Its mass is 0.20 × 40 = 8.0 g. In an exam, name the limiting reactant, then use its moles.
Concentration (g/dm³) = mass ÷ volume in dm³. Divide cm³ by 1000 to get dm³.
Concentration (mol/dm³) = moles ÷ volume in dm³. Divide g/dm³ by Mr to get mol/dm³.
The limiting reactant is used up first and decides how much product forms.
Compare the moles using the equation ratio to find which reactant is limiting.
Convert 750 cm³ into dm³.
Answer: 0.75 dm³
8.0 g of sodium hydroxide is dissolved to make 250 cm³ of solution. Calculate the concentration in g/dm³.
Answer: 32 g/dm³
6.0 g of magnesium is added to 100 cm³ of hydrochloric acid of concentration 2.0 mol/dm³. Mg + 2HCl → MgCl₂ + H₂. Show which reactant is limiting. Ar of Mg = 24.
Answer: HCl is the limiting reactant. 0.25 mol of Mg needs 0.50 mol of HCl and only 0.20 mol is present.
Concentration and limiting reactants practice questions for GCSE Combined Science Chemistry. Try each one, then open the answer. Easy questions come first.
Convert 500 cm³ into dm³.
Answer: 0.5 dm³
Convert 0.25 dm³ into cm³.
Answer: 250 cm³
A solution has 12 g of solute in 1.0 dm³. What is the concentration in g/dm³?
Answer: 12 g/dm³
Which is the limiting reactant: the one used up first or the one left over?
Answer: the one used up first
Convert 2000 cm³ into dm³.
Answer: 2.0 dm³
Convert 5.0 dm³ into cm³.
Answer: 5000 cm³
10 g of salt is dissolved to make 250 cm³ of solution. Calculate the concentration in g/dm³.
Answer: 40 g/dm³
A solution has a concentration of 20 g/dm³. What mass of solute is in 50 cm³ of this solution?
Answer: 1.0 g
Longer questions in the style of a written paper, with the marking points. These are Yumio practice questions.
A student dissolves 15.0 g of sodium chloride in water to make 500 cm³ of solution. (a) Calculate the concentration of the solution in g/dm³. [2 marks] (b) The student takes 50 cm³ of this solution. Calculate the mass of sodium chloride in it. [2 marks]
Answer: (a) 30 g/dm³ (b) 1.5 g
4.8 g of magnesium is added to 100 cm³ of hydrochloric acid of concentration 1.0 mol/dm³. Mg + 2HCl → MgCl₂ + H₂. Ar of Mg = 24. (a) Calculate the moles of magnesium and of hydrochloric acid. [2 marks] (b) Explain which reactant is limiting. [2 marks] (c) Calculate the moles of hydrogen formed. [1 mark]
Answer: (a) 0.20 mol Mg and 0.10 mol HCl (b) 0.20 mol of Mg needs 0.40 mol of HCl, but only 0.10 mol is present, so HCl is limiting. (c) 0.050 mol
Using the volume in cm³ in a concentration calculation
Divide by 1000 to get dm³ first.
Choosing the limiting reactant by comparing masses
Compare moles and use the ratio in the balanced equation.
Saying the excess reactant decides how much product forms
The limiting reactant decides the amount of product.
Forgetting to convert g/dm³ to mol/dm³ by dividing by Mr
Divide a concentration in g/dm³ by Mr to get mol/dm³.
Using the amount of the reactant in excess to work out the product
Find the limiting reactant first. The product amount comes from the limiting reactant only.
Concentration and limiting reactants is on both Foundation and Higher papers for GCSE Combined Science Chemistry, aimed at around grade 6.
Using the volume in cm³ in a concentration calculation. Divide by 1000 to get dm³ first.
Open Show answer under each question on this page. The questions are Yumio practice questions, not past exam questions.
See how the papers are set: AQA Combined Science papers and Edexcel Combined Science papers.
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