Assessed Practical Essays

Assessed Practical Essays Assessed Practical Essay Assessed Practical Essay Method 1Mass (g)Total Mass Loss (g)Original Mass1.440.00Measurement 11.050.39Measurement 20.900.54Measurement 30.880.56Measurement 40.860.58Measurement 50.860.58CalculationsIn theory the remaining mass after the heating will be only FeSO4, so from this the mass that was evaporated off would be entirely water. From this we can calculate:The number of moles of H2O is the n= m/MrH=1O=16H2O = 18 =Mr0.58/18 = 0.032 moles of H2OThe remaining mass should be entirely FeSO4 so:Fe=56S=32O=1656 + 32 + (416) = 152 = MrThe number of moles of FeSO4 = 0.86/152 = 5.657894737 x 10^-3 MolesTo find the ratio of H2O we need to use:0.032/5.657894737 x 10^-3 = 5.655813953This is approximately 6 so the Formula of the Hydrated Iron (II) Sulphate Crystals is FeSO4.6 H2O.Method 2Titration #Start Vol. (cmà ¯Ã‚ ¿Ã‚ ½)End Vol. (cmà ¯Ã‚ ¿Ã‚ ½)Difference (cmà ¯Ã‚ ¿Ã‚ ½)19.000030.950021.9500211.000033.150022.1500315.000037.250022.050044.000026.100022.1000Average (cmà ¯Ã‚ ¿Ã‚ ½)22.0625Equ ation5 Fe2+ + MnO4- + 8 H+ = 5Fe3+ + Mn2+ + 4H2OBy finding the number of moles of Fe2+ ions being reduced by the MnO4- ions we can calculate the Mr of the FeSO4.xH2O that was used in the experiment.Calculations22.0625 / 1000 = 0.0220625 dm à ¯Ã‚ ¿Ã‚ ½Using n=VxM0.0220625 x 0.01 = 2.2062510-à ¯Ã‚ ¿Ã‚ ½ moles of Fe2+From the equation you can see that there are 5 moles of Fe2+ taking part in the reaction so:5 x 2.09710-à ¯Ã‚ ¿Ã‚ ½ = 1.103125 molThis method has only calculated this for 25cm3 of solution but we need 250 cm3 so we multiply by 10.1.103125 x 10 = 0.01103125 molUsing this and the original mass we can calculate the Mr of the compound and deduce the number of H2O.Using Mr = n/m3.08 / 0.01103125 = 279.2067989Then take away the Mr of FeSO4 gives the amount of water in the compound279.2067989 152 = 127.2067989Then divide by the Mr of water to get how many are in the original compound127.2067989 / 18 = 7.067044381So the formula of the compound is FeSO4.7H2OEvaluationMeasuremen t ErrorsOne of the largest measurement errors is in the weighing of the compound. The scales used only weigh to 2 decimal places. For a titration this is not accurate enough as the titration can precise. The preferable accuracy would be to at least 3 decimal places as the lack of accuracy could greatly affect the results of the calculations.Also in Method 1 we do not know whether or not there was any absorption of moisture during the cooling period. This can be a large problem as the method relies on the fact that the reaction no longer takes place when there is no more water. To escape this problem this reaction could be carried out in a vacuum or have all the moisture evacuated from around the apparatus.The greatest problem being the accuracy scales due to the fact that it can affect both methods but the second problem is mainly to do with the first method.I would suggest the use of Method 2 due to there being less measurement inaccuracies.Procedural ErrorsWithout knowing whether or not a compound is being thermally decomposed is a big downside to Method 1. During the heating there could be gases being given off other than steam, as it is assumed only the water is being removed from the compound.If there is thermal decomposition there would be the mass of the water removed as well as some of the compound. This could prove a large error in the calculations to find out how much water there is in the formula.To avoid this either a different method could be used or a boiling tube, upturned burette and hose could be used to accumulate the gases given off. This may still be inaccurate as the steam given off will also be trapped within the burette. To avoid this it could travel through a condenser first in order to extract the steam from other products. With this the total mass of the compound could be found and calculations could be accurate enough to calculate the amount of H2O in the formula.There is also a problem in not knowing when the reaction is finished ex cept by measuring the compound every so often to find when the mass no longer changes. A more accurate set of scales would help with this problem as we could check for when the mass of the compound stops changing by such large amounts, so that we could assume it was the compound that was now thermally decomposing.These errors are mainly to do with Method 1 and so I would again recommend Method 2 as there are far fewer inaccuracies and procedural errors. The only problem factor in Method 2 is human and unpredictable.