Tetraamminecopper(II) sulphate hydrate Write-up Essay Example for Free

Tetraamminecopper(II) sulphate furnish Write-up EssayPurposeThe purpose of this experiment is to form tetraamminecopper(II) sulphate hydrate and determine the yield.MaterialsCuSo45H2ONH3 (concentrated)Ethanol50 cm3 measuring cylinder250 cm3 beakerSpatulaEquipment for vacuum filt rationProcedureWeigh out approximately 5.0g of CuSo45H2ODissolve it in 30 cm3 water in the beakerAdd 10 cm3 concentrated ammonia water (NH3) and stir the baseAdd 40 cm3 ethanol and stir carefully for a couple of minutes.Filter the solution through equipment for vacuum filtration.Transfer the product to a clean weigh boat and leave to dry. Procedure and observations in classFirst 5.01g of CuSo45H2O was weighed out. After it was fade away in 30 cm3 water, in the beaker, the solution got the colour blue. Next was 10 cm3 concentrated ammonia (NH3), which was added into the solution and the colour dark blue was observed. Then 40 cm3 ethanol was added and the solution got the colour bright blue. Then the so lution was filtered through a Buchner flask and the final product was weighed in a plastic weighing boat. The total mass was 5.98g, from which the weight of the boat, 1.16g, has to be subtracted. So the mass of the final product was 5.98 1.16 = 4.82g.Data processing1. Calculate the return of moles CuSo45H2O used.To find out the number of moles the commandment n = m / Mr has to be used.Mr = 64 + 32 + (16 x 4) + (5 x 16) = 250m = 5.01n = 5.01 / 250 = 0.02004 0.0200 moles (3 s.f.)2. Concentrated ammonia contains 25% NH3 by mass. The tautness of concentrated ammonia is 0.91g/cm3 . Calculate the number of moles of NH3 .Density of con. ammonia = 0.91g/cm3 and in the procedure there was used 10 cm3, so therefore mass of ammonia used 0.91 x 10 = 9.1gSince only 25% of ammonia is NH3 , mass of NH3 9.1 x 0.25 = 2.275gFrom here the amount of moles layabout be calculated by the formula n = m / Mr.Mr = 14 + (1 x 3) = 17m = 2.275gn = 2.275 / 17 = 0.134 moles (3 s.f.)3. Which of the reactant s is in excess? Which is the limiting reagent?CuSo45H2ONH3Number of moles (n)0.020.134Divide by smallest ratio0.02 / 0.02 = 10.134 / 0.02 = 6.7Divide by stoichiometric co-efficient from equation(Equation below this table)1 / 1 = 16.7 / 4 = 1.675Reactant in excess or limiting reagent restrain reagentReactant in excess(1)CuSO4 . 5H2O + 4NH3 Cu(NH3)4SO4 . H2O + 4H2O4. Calculate the theoretical yield of Cu(NH3)4SO4 . H2OFrom the equation above it can be seen that the ratio between CuSO4 . 5H2O and Cu(NH3)4SO4 . H2O is 1 1. Therefore 0.02 moles of CuSO4 . 5H2O will give 0.02 moles of Cu(NH3)4SO4 . H2O. By using the formula m = Mr x n the theoretical yield can be calculatedn = 0.02Mr = 246m = 0.02 x 246 = 4.92 gCalculate the yield in percentage of the theoretical and comment on any difference.The yield in percentage can be calculated by the formula actual mass / expected mass. 4.82 / 4.92 97.9% (3 s.f)Because the difference is so small (2.1%) the experiment can be considered successful . The difference could be possessed of been caused by different things like a small measurement mistake, a little bit was spilt or not transferred when the solution was held in the Buchner flask.