You have read the exam question carefully and decided that the following are the key bits of information:
- 3.70 g of insoluble MCO3 is added to 250 cm3 of 0.5 mol HNO3.
- A 20 cm3 sample of the solution is titrated with 19.9 cm3 of 0.3 mol dm-3 NaOH.
- Calculate the relative atomic mass of M, and hence, identify the element M.
| Step 1 We know that moles = concentration x volume: n HNO3 = 0.5 x (250 / 1000) = 0.125 mol. n NaOH = 0.3 x (19.9 / 1000) = 5.97 x 10-3 mol.
Step 2 The balanced symbol equations aren’t given so we need work them out. The only thing that we know about M is that it has a 2+ charge. It is therefore likely to be a Group 2 metal or possibly a transition metal. In any case, we can write down the balanced symbol equations: Reaction 1: 2 HNO3 + MCO3 –> M(NO3)2 + H2O + CO2 Reaction 2: HNO3 + NaOH –> NaNO3 + H2O
Step 3 The balanced symbol equation for Reaction 2 tells us that HNO3 and NaOH react in a 1:1 ratio so the HNO3 moles at the start of Reaction 2 was: n HNO3 = 1 x 5.97 x 10-3= 5.97 x 10-3 mol
Step 4 The total is 12.5x bigger than the sample (i.e., 250 / 20) so we can work out the HNO3 moles at the end of Reaction 1: n HNO3 = 12.5 x 5.97 x 10-3= 7.46 x 10-2 mol
Step 5 We know that there was 0.125 mol HNO3 at the start and 7.46 x 10-2 mol HNO3 at the end of Reaction 1. By subtraction, we can work out the HNO3 moles used up in Reaction 1: 0.125 – 7.46 x 10-2 = 5.04 x 10-2 mol
Step 6 From Reaction 1, we know that MCO3 and HNO3 react in a 1:2 ratio, so: n MCO3 = 0.5 x 5.04 x 10-2 = 2.52 x 10-2 mol
Step 7 We have the mass of MCO3 (3.70 g) and can assume that it is pure. This means that the Mr of MCO3 is: Mr MCO3 = 3.70 / 2.52 x 10-2 = 147
Step 8 The Ar of M can then be found by subtracting the mass of CO3 (60) from the Mr to find the Ar of M: 147 – 60 = 87 The closest match on the periodic table is Strontium (Sr). Remember to always check that element identified forms ions with the correct charge. You know the ion forms +2 ions, so the answer is unlikely to be a Group 1 metal. |