Determination of the formula of hydrated Iron (II) Sulphate crystals (FeSO4xH2O)
Analysis
Method 1
1. Using a balance that weighs to two decimal places, weigh a crucible. Add between 1.30 - 1.50g of hydrated iron (II) sulphate crystals. Record the masses.
2. Place the crucible containing the hydrated iron (II) sulphate crystals on the pipe-clay triangle and gently heat for two minutes.
3. Allow to cool and weigh the crucible and the iron (II) sulphate.
4. Repeat steps 2 and 3 until the masses after heating are consistant.
5. Record all the masses.
Use the results to calculate the moles of FeSO4 and the moles of water and hence deduce the formula of the hydrated iron(II) sulphate crystals, FeSO4xH2O.
Masses
Mass number
Items being weighed
Mass (g)
1
Crucible and lid
38.86
2
Crucible, lid and FeSO4xH2O
40.34
3
FeSO4xH2O
1.48
4.a)
Crucible, lid and FeSO4xH2O after heating
after first heating
39.77
4.b)
second heating
39.67
4.c)
third heating
39.67
4.d)
fourth heating
39.67
After heating four times, the weight of the crucible, lid and FeSO4xH2O was consistant showing that all of the H2O has evaporated.
Therefore the following masses can be calculated:
Mass number
Substances
Mass number calculation
Mass calculation (g)
Mass (g)
5
H2O
2 - 4.d)
40.34 - 39.67
0.67
6
FeSO4
3 - 5
1.48 - 0.67
0.81
To work out the formula of the FeSO4xH2O:
Moles of FeSO4
Mr = 56 + 32 + (16 x 4) = 152
number of moles = mass
Mr
n = 0.81 = 0.00533 moles
152
Moles of H2O
Mr = 18
n = mass n = 0.67 = 0.0372 moles
Mr 18
Empirical formula
To work out molar ratio of FeSO4 to H2O (value of x):
1 x 0.0372 = 6.98
0.00533
x = 6.98
therefore the formula is = FeSO4(H2O)7
Method 2
1. Weight between 2.85 and 3.10 g of hydrated iron(II) sulphate crystals, FeSO4xH2O. Record the mass and then dissolve the crystals in 50.0cm3 of 1 mol dm-3 H2SO4 (aq) and make up to 250cm3 in a volumetric flask with distilled water. Invert the volumetric flask several times to ensure that the solution is evenly mixed.
2. Using a pipette filler, pipette 25.0cm3 of the solution of iron(II) sulphate into a conical flask and add approximately 20.0cm3 of the 1 mol dm-3 H2SO4 (aq) solution provided.
3. Titrate the acidified Fe2+ (aq) solution with 0.0100 mol dm-3 potassium permanganate, KMnO4 (aq), and continue the titration to the normal end point.
4. Repeat the titration until the results are consistant.
Use the results to calculate the moles of Fe2+ and hence deduce the formula of the hydrated iron(II) sulphate crystals, FeSO4xH2O.
Results of titration
Original mass of FeSO4xH2O = 3.02 g
Rough titration result = 23.10cm3
Titration number
Amount of KMnO4 (cm3) amount
1
21.75
2
21.70
3
23.71
4
21.70
Average titration
21.72
The average was taken excluding the anomalous result.
To calculate the moles of Fe2+ and deduce the formula of the FeSO4xH2O:
number of moles = v c where v = volume (cm3)
1000 and c = concentration (mol dm-3)
Equations
(1) 5Fe2+ 5Fe3+ + 5e-
(2) MnO4- + 8H+ + 5e- Mn2+ = 4H2O
(3) 5Fe2+ + MnO4- = 8H+ Mn2+ + Fe3+ + 4H2O
Above the Iron is multiplied by 5 in equation (1) to equal the electrons needed in equation (2). The electrons then cancel each other out when the equations are put together in (3) to give the overall reaction during the titration.
5Fe2+ + MnO4- = 8H+ Mn2+ + Fe3+ + 4H2O
5 : 1
The equation also gives the molar ratio of Iron to MnO4- = 5 : 1
This ratio can be used to work out the number of moles of Fe2+ in the reaction also giving the number of moles of FeSO4xH2O.
21.70cm3 of 0.01M KMnO4 were used to neutralise 25cm3 of FeSO4xH2O. The original mass of the FeSO4xH2O was 3.02 g
Moles
MnO4-
number of moles = v c = 21.7 x 0.01 = 2.17 x 10-4 moles
1000 1000
Therefore moles of Fe2+ = 2.17 x 10-4 x 5 = 0.00109 moles
= 1.09 x 10-3 moles
In the whole 250 cm3 solution there is 1.09 x 10-2 moles of FeSO4xH2O.
Empirical formula
Mr of FeSO4xH2O:
Mr = mass Mr = 3.02 = 277
number of moles 1.09 x 10-2
Mr of FeSO4 = 56 + 32 + (16 x 4) = 152
Mr of H2O = (1 x 2) + 16 = 18
Value of x
277 - 152 = 125
125 18 = 6.94
Round 6.94 value up to 7
Therefore the formula of FeSO4xH2O is:
FeSO4(H2O)7
Evaluation
Method 1
The main measurement error in this method was the balance as it was only accurate to two decimal places
Balance
percentage error = 0.05 x 100 = 3.4%
1.48
This is a relatively small percentage but could be reduced further by increasing the amount of the substance used or by using a balance accurate to three decimal places.
For example:
Percentage error with more accurate balance = 0.005 x 100 = 0.34%
1.48g
Percentage error with an increased mass = 0.05 x 100 = 0.34%
14.80g
The main procedural error that may have occurred in method one is that the crucible and the FeSO4(H2O)7 may not have cooled completely before they were weighed. If this happened for every time that the crucible was weighed after heating, the end measurement may not be correct as the substances would weight a different amount when hot as they would when cool. To counter this error the crucible and the FeSO4(H2O)7 must be completely cooled before they are weighed.
I think that the overall accuracy of this method was good as no obvious anomalous results occurred.
However, some aspects of the method could be changed to improve the accuracy of the overall experiment.
Method 2
Anomalous result
Titration 3 was 0.01 cm3 higher than the result for titration 2. The result for titration 4 was then concordant with the other results. I think that this result was anomalous as it went against the trend in the measurements. The anomaly was probably caused by a measurement error.
The main measurement error for this method was the burette. This was accurate to 0.05 cm3
Burette
percentage error = 0.5 x 100 = 2.3%
21.70cm3
This is a smaller percentage than for the balance in method 1 but only because the amount of the substance used was larger. This percentage can be reduced by increasing this amount.
For example:
percentage error = 0.5 x 100 = 0.23%
217.0cm3
The main procedural error for this method is the inaccuracy that could occur when using the pipette filler. It is very easy to make an inaccurate measurement when using the pipette filler as it is hard to keep the meniscus of the liquid at the 25cm3 point on the pipette so therefore quite easy to inadvertently use too much or too little of the liquid.
I think that overall method one was more accurate even though the percentage error of the balance was more than that of the burette as there are easily made procedural errors in method two and as there were no anomalous results in method 1.
