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

Tetraamminecopper(II) sulfate hydrate Write-up EssayPurposeThe purpose of this experiment is to form tetraamminecopper(II) sulphate hydrate and determine the yield.MaterialsCuSo45 weeweeNH3 (concentrated) fermentation alcohol50 cm3 measuring cylinder250 cm3 beakerSpatulaEquipment for hoover filtrationProcedureWeigh out approximately 5.0g of CuSo45H2ODissolve it in 30 cm3 water in the beaker wreak 10 cm3 concentrated ammonium hydroxide (NH3) and stir the answerAdd 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 weighing boat and leave to dry. Procedure and observations in kindFirst 5.01g of CuSo45H2O was weighed out. After it was dissolved 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 col our bright blue. Then the solution was filtered through a Buchner flask and the last(a) 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 number of moles CuSo45H2O used.To find out the number of moles the formula 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 density 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 finish be calculate 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 reactants is in unembellished? Which is the limiting reagent?CuSo45H2ONH3Number of moles (n)0.020.134 assign 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 reagentLimiting 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 tidy sum be seen that the ratio between CuSO4 . 5H2O and Cu(NH3)4SO4 . H2O is 1 1. Therefore 0.02 moles of CuSO4 . 5H2O lead 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 character 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 exp eriment can be considered successful. The difference could have 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.