Determination of the formula of hydrated Iron (II) Sulphate crystals (FeSO4xH2O)

4.d)40.34 - 39.670.676FeSO43 - 51.48 - 0.670.81To work out the formula of the FeSO4xH2O:Moles of FeSO4Mr = 56 + 32 + (16 x 4) = 152number of moles = massMrn = 0.81 = 0.00533 moles152Moles of H2OMr = 18n = mass n = 0.67 = 0.0372 molesMr 18Empirical formulaTo work out molar ratio of FeSO4 to H2O (value of x):1 x 0.0372 = 6.980.00533x = 6.98therefore the formula is = FeSO4(H2O)7Method 21. 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 titrationOriginal mass of FeSO4xH2O = 3.02 gRough titration result = 23.10cm3Titration numberAmount of KMnO4 (cm3) amount121.75221.70323.71421.70Average titration21.72The 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+ + 4H2OAbove 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+ + 4H2O5 : 1The equation also gives the molar ratio of Iron to MnO4- = 5 : 1This 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 gMolesMnO4-number of moles = v c = 21.7 x 0.01 = 2.17 x 10-4 moles1000 1000Therefore moles of Fe2+ = 2.17 x 10-4 x 5 = 0.00109 moles= 1.09 x 10-3 molesIn the whole 250 cm3 solution there is 1.09 x 10-2 moles of FeSO4xH2O.Empirical formulaMr of FeSO4xH2O:Mr = mass Mr = 3.02 = 277number of moles 1.09 x 10-2Mr of FeSO4 = 56 + 32 + (16 x 4) = 152Mr of H2O = (1 x 2) + 16 = 18Value of x277 - 152 = 125125 18 = 6.94Round 6.94 value up to 7Therefore the formula of FeSO4xH2O is:FeSO4(H2O)7EvaluationMethod 1The main measurement error in this method was the balance as it was only accurate to two decimal placesBalancepercentage error = 0.05 x 100 = 3.4%1.48This 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%1.48gPercentage error with an increased mass = 0.05 x 100 = 0.34%14.80gThe 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 2Anomalous resultTitration 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 cm3Burettepercentage error = 0.5 x 100 = 2.3%21.70cm3This 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%217.0cm3The 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.