For FeSO4: 0.75g/151.9mol g-1 = 0.00494mol. (3.s.f)
For H2O: 0.66g/18mol g-1 = 0.0367mol. (3.s.f)
So the molar ratio is 0.00494:0.0367, or by dividing throughout by the smallest to get an empirical whole ratio, 1:7.42 (3.s.f)
This would imply that the sample of the sample of iron(II) sulphate was in heptahydrate form, i.e. FeSO4.7H20.
Method 2: By performing a redox titration with the hydrated iron(II) sulphate mixed in dilute sulphuric acid as the analyte and potassium permanganate as the titrant. Some of these chemicals are hazardous, H2SO4 is corrosive and KMnO4 is lethal if consumed in large enough doses. The full ionic equation for this titration is:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l). The potassium permanganate acts as a oxidising agent accepting electrons from the iron(II) which is oxidised to iron(III).
Results:
Mass of FeSO4.nH2O used = 3.05g.
Average titre= (22.3+22.2+22.3+22.3)/4
=22.275cm3.
Analysis:
With these results it is possible to calculate the amount of water from the mass of water obtained from the mass of hydrated iron(II) sulphate. To work out the amount of hydrated iron(II) sulphate the amount of potassium permanganate it reacts with needs to be known. The average amount of 0.01mol dm-3 potassium permanganate that is needed to react with the hydrated iron(II) sulphate is 0.022275dm3 (22.275cm3/1000). So the number of moles used is 2.2275×10-4mol (concentration × volume). To find the number of moles of hydrated iron(II) sulphate that reacted, ionic equations are used. During the titration manganate(VII) ions from the potassium permanganate, due to the presence of the acid (H+), oxidise iron(II) ions, in the hydrated iron(II) sulphate, into iron(III) ions:
1. Fe2+(aq) → Fe3+(aq) + e-, and
2. MnO4-(aq) + 8H+(aq) +5e- → Mn2+(aq) + 4H2O(l).
To put these half equations into a full ionic equation the electrons must cancel out on either side. To do this the oxidisation equation (1) is multiplied by 5 and combined with the reduction equation (2) to give the complete redox ionic equation:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l). From this equation the molar ratio between potassium permanganate and iron(II) sulphate is shown to be 1:5, so (2.2275×10-4)×(5) moles of hydrated iron(II) sulphate were present in 25cm3, and (2.2275×10-3) × (5) were present overall in the whole 250cm3 solution.
Moles of iron(II) sulphate = (2.2275×10-3) × (5) = 1.11375×10-2 moles. The mass of the iron(II) sulphate that reacted was moles × molecular mass; (1.11375×10-2) × (151.9)
= 1.69g.(3.s.f)
As the water in the hydrated iron(II) sulphate weighed at the start played no part in the reaction, subtracting the mass of the anhydrous iron(II) sulphate from the initial hydrated form will give the mass of water present in the sample; 3.05g – 1.69g = 1.36g of water.
As in method one the molar ratio between water and iron(II) sulphate in hydrated iron(II) sulphate can be calculated by moles = mass/molar mass.
Moles of H20 = 1.36g/18mol g-1 = 0.0756mol (3.s.f). Moles of FeSO4 = 0.0111375mol.
Molar ratio = 0.0111375:0.0756, and by dividing by the smallest value = 1:6.79.
Evaluation:
Both method one and two give the number of water atoms in the hydrated iron(II) sulphate as round about seven, indicating that the sample was in heptahydrate form, i.e. had a formula of FeSO4.7H20, the most common form of hydrated iron(II) sulphate also known as the mineral melanterite.
Although both methods gave the answer as about 7 the results were quite different. The results suggest that a titration method is more accurate as the answer was closer to seven than dehydration, although it wasn’t spot on. The result of 22.2cm3 on the second titre was most likely an anomaly and the mode of the titre should have been taken instead of a mean to stop it having an effect. Human error due to reaction time is the cause of this anomaly and a mode would give a more accurate answer. However the anomaly is only off by 0.1cm3 and this would have a huge impact. Therefore the inaccurate answer must come from another factor. I think that the scales measuring only to the nearest hundredth of a gram produced only rough figures for the small scale the titration that was carried out. Scales with a greater resolution, or more realistically the experiment being carried out on a larger scale would reduce this (procedural) error margin although this is only practical in laboratories and not schools.
The good thing about method one is that there are less of these sums that can amplify errors and even though the same weighing scales were used the lack of sums with many decimal places reduces the error margin. This method generally relies on chemistry instead of maths and the only variable that could have caused the off answer was again the scales. Carrying out this experiment on a larger scale would be far too dangerous in a school due to the use of fire.
The problem with method one being more in touch with chemistry is the risk of procedural errors. On heating the melanterite the substance changed in colour from blue to white indicating the water had been eliminated. However there were also areas of dark red which indicates a further reaction. The heat was too strong and iron(III) oxide was formed with sulphur di- and trioxide: 2FeSO4(s) → Fe2O3(s) + SO2(g) + SO3(g). These gases also escaped with the water and gave the impression that there was more water present as some sulphur and oxygen escaped as well. Heating the melanterite more gently and patiently for less time would make sure this didn’t happen.
If this didn’t happen I am confident that this method would be more accurate than a titration as it is more purely chemistry instead of maths, which holds a greater chance of carrying errors over, despite my results suggesting otherwise.
