You have read the exam question carefully and decided that the following are the key bits of information:
- 3.69 g of soluble M2CO3 is made into a 250 cm3 solution
- A 25 cm3 sample of the solution is titrated with 26.7 cm3 of 0.2 mol dm-3 HNO3.
- Calculate the relative atomic mass of M, and hence, identify the element M.
| Step 1 We know that moles = concentration x volume: n HNO3 = 0.2 x (26.7 / 1000) = 5.34 x 10-3 mol.
Step 2 The balanced symbol equation isn’t given so we need to work it out. The only thing that we know about M is that it has a 1+ charge. You might guess that it is probably a Group 1 metal, especially as we’re told it’s a soluble carbonate. We can therefore write down the balanced symbol equation: 2 HNO3 + M2CO3 –> 2 MNO3 + H2O + CO2
Step 3 The balanced symbol equation tells us that HNO3 and M2CO3 react in a 2:1 ratio so the M2CO3 moles in the sample is: n M2CO3 = 0.5 x 5.34 x 10-3 mol = 2.67 x 10-3 mol
Step 4 The total is 10x bigger than the sample (i.e., 250 / 25) so we can work out the initial M2CO3 moles: n M2CO3 = 10 x 2.67 x 10-3 = 2.67 x 10-2 mol
Step 5 We have the mass of M2CO3 (3.69 g) and can assume that it is pure. This means that the Mr of M2CO3 is: Mr M2CO3 = 3.69 / 2.67 x 10-2 = 138.2
Step 6 The Ar of M can then be found by subtracting the mass of CO3 (60) from the Mr to find the mass of just M2 (138.2 – 60 = 78.2). The Ar of M is therefore 78.1 / 2 = 39.1 The closest match on the periodic table is potassium (K). |
You might want to think about why this approach wouldn’t work for Group 2 carbonates?… The answer is that they are all insoluble. In this case, we would need to do a back titration.