The key to reacting mass calculations at A level is to have a foolproof method for setting them out so that as they get more complicated, you can always follow your working and your logic if you get lost (and so can the examiner – it’s impossible to give credit to an answer space full of random numbers).
I explain my method in the video – no different from GCSE – and then I finish with an exam question that relies more on general maths skill and logic to solve.
There are a few practice questions below (beware: question 4 is nasty!) but you should complete all the questions in your text books and past papers if you need to build confidence in this area.
Practice questions
1. What mass of glucose must be fermented to give 5.00 kg of ethanol?
C6H12O6(aq) ⇾ 2C2H5OH(aq) + 2CO2(g)
2. What mass of oxygen would be produced by completely decomposing 4.25g of sodium nitrate?
2NaNO3(s) ⇾ 2NaNO2(s) + O2(g)
3. When 0.27g of aluminium is added to excess copper sulphate solution, 0.96g of copper is precipitated. Deduce the equation for the reaction.
4. A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate weighing 10.000g was heated until it reached a constant mass of 8.708g. Calculate the composition of the mixture in grams of each component.
2NaHCO3(s) ⇾ Na2CO3(s) + H2O(g) + CO2(g)
Answers:
1. mol of ethanol = 5/46 = 0.109mol; ratio of glucose to ethanol is 1:2;
mol of glucose = 0.0543; mass of glucose = 0.0543 x 180 = 9.77kg
2. mol of NaNO3 = 4.25 / 85 = 0.05 mol; ratio of NaNO3 to NaNO2 = 1:1;
mass of NaNO2 = 0.05 x 69 = 3.45g
3. mol of aluminium = 0.27/27 = 0.01 mol;
mol of copper = 0.96/63.5 = 0.015 mol;
this tells us that the ratio of the reactant Al: product Cu is 0.01:0.015 (1:1.5) but since we don’t tend to have half numbers in balanced equations, let’s double up … we also know that the ratio of CuSO4 : Cu is 1:1
2Al + 3CuSO4 ⇾ 3Cu + Al2(SO4)3
4. the 8.708g of solid ‘product’ is a mixture of undecomposed Na2CO3 from the original mixture and Na2CO3 from the decomposition of NaHCO3.
10.000 – 8.708 = 1.292g is the gaseous H2O and CO2 from the decomposition of NaHCO3, which were produced in a 1:1 molar ratio.
Mr of H2O (18) + Mr of CO2 (44) = 62.
18/62 x 1.292g = 0.375g of H2O which is 0.375/18 = 0.208 mol.
1.292 – 0.375g = 0.917g of CO2 which is 0.917/44 = 0.208 mol.
Ratio of either H2O or CO2 : NaHCO3 = 1:2, so number of mol in original mixture = 0.0416 mol x 84 = 3.494g of NaHCO3
10.000 – 3.494 = 6.506g of Na2CO3 in original mixture.