When creating my stock solution of Copper (II) Sulphate I accurately weighed out approximately 24. 96g of Copper (II) Sulphate crystals on a 3 decimal place balance. I used the weighing by difference method to accurately obtain an exact weight for the crystals. This required me to weigh the mass of the weighing boat and the crystals, add the crystals to a beaker and then re-weigh the empty boat at the end.
The masses I gathered from the weighing are as follows: Mass of boat and crystals first = 26. 88g Mass of empty boat after = 1.926g Actual amount of crystals = Mass of boat and crystals first - Mass of empty boat after = 26. 88g - 1. 926g = 24. 954g Now that I have the mass of the crystals used I will be able to calculate the number of moles of CuSO4 that I have by using the equation: Number of moles = Mass Molar Mass n = m Mr n = 24. 954 249. 6 n = 0 eulogy. 099975961 mol Now I know how many moles of CuSO4 I have I can work out the concentration of the solution (CuSO4. 5H2O) I have made up with distilled water in the 1dm3 flask by using the equation:
Concentration = Number of Moles Volume c = n v c of CuSO4. 5H2O = 0. 099975961 mol 1 = 0. 099975961 mol dm-3 = 0. 100 mol dm-3 (3sf ) Due to the apparatus I used in the experiment there is a possibility that the concentration I have calculated is not exactly right; this is known as the precision error.
To calculate the precision error I need to calculate the percentage uncertainty of each piece of apparatus that I used and then add them all together to give a total percentage uncertainty: % uncertainty of the 3dp balance = (5 x 10-4) x 2 = 0.001 x 100 24. 954g = 4 x 10-3 % % uncertainty of the 1dm3 flask = 0. 8 x 100 1000 = 0. 08 % Therefore the total uncertainty = 0. 08 % + 4 x 10-3 % = 0. 084 % So, to work out the percentage error I need to work out the percentage of the concentration that could be incorrect: Percentage Error = Total uncertainty x Concentration 100 = 0. 084 x 0. 099975961 mol dm-3 100 = i?? 8. 397980724 x 10-5 Now that I have the percentage error I can show the possible errors that there could be in the calculation of the concentration of CuSO4.5H2O:
Highest possible c of CuSO4. 5H2O = 0. 10005994mol dm-3 Lowest possible c of CuSO4. 5H2O = 0. 099891981 mol dm-3 However I will represent the final concentration I have calculated of the CuSO4. 5H2O as: x 0. 099975961 i?? 8. 397980724 x 10-5 mol dm-3 0. 100 i?? 8. 40 mol dm -3 (3sf) The figures I used to calculate the errors from I obtained from the British Standard maximum permitted errors (tolerances) for volumetric glassware table from my chemistry laboratory. All of the glassware used is of grade B standard.
When creating my stock solution of Copper (II) Sulphate I accurately weighed out approximately 24. 96g of Copper (II) Sulphate crystals on a 3 decimal place balance. I used the weighing by difference method to accurately obtain an exact weight for the crystals. This required me to weigh the mass of the weighing boat and the crystals, add the crystals to a beaker and then re-weigh the empty boat at the end.
The masses I gathered from the weighing are as follows: Mass of boat and crystals first = 26. 88g Mass of empty boat after = 1.926g Actual amount of crystals = Mass of boat and crystals first - Mass of empty boat after = 26. 88g - 1. 926g = 24. 954g Now that I have the mass of the crystals used I will be able to calculate the number of moles of CuSO4 that I have by using the equation: Number of moles = Mass Molar Mass n = m Mr n = 24. 954 249. 6 n = 0 eulogy. 099975961 mol Now I know how many moles of CuSO4 I have I can work out the concentration of the solution (CuSO4. 5H2O) I have made up with distilled water in the 1dm3 flask by using the equation:
Concentration = Number of Moles Volume c = n v c of CuSO4. 5H2O = 0. 099975961 mol 1 = 0. 099975961 mol dm-3 = 0. 100 mol dm-3 (3sf ) Due to the apparatus I used in the experiment there is a possibility that the concentration I have calculated is not exactly right; this is known as the precision error.
To calculate the precision error I need to calculate the percentage uncertainty of each piece of apparatus that I used and then add them all together to give a total percentage uncertainty: % uncertainty of the 3dp balance = (5 x 10-4) x 2 = 0.001 x 100 24. 954g = 4 x 10-3 % % uncertainty of the 1dm3 flask = 0. 8 x 100 1000 = 0. 08 % Therefore the total uncertainty = 0. 08 % + 4 x 10-3 % = 0. 084 % So, to work out the percentage error I need to work out the percentage of the concentration that could be incorrect: Percentage Error = Total uncertainty x Concentration 100 = 0. 084 x 0. 099975961 mol dm-3 100 = i?? 8. 397980724 x 10-5 Now that I have the percentage error I can show the possible errors that there could be in the calculation of the concentration of CuSO4.5H2O:
Highest possible c of CuSO4. 5H2O = 0. 10005994mol dm-3 Lowest possible c of CuSO4. 5H2O = 0. 099891981 mol dm-3 However I will represent the final concentration I have calculated of the CuSO4. 5H2O as: x 0. 099975961 i?? 8. 397980724 x 10-5 mol dm-3 0. 100 i?? 8. 40 mol dm -3 (3sf) The figures I used to calculate the errors from I obtained from the British Standard maximum permitted errors (tolerances) for volumetric glassware table from my chemistry laboratory. All of the glassware used is of grade B standard.
