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

Tetraamminecopper(II) sulphate hydrate Write-up Essay Purpose The purpose of this experiment is to form tetraamminecopper(II) sulphate hydrate and determine the yield. Materials CuSo4à ¯Ã‚ ¿Ã‚ ½5H2O NH3 (concentrated) Ethanol 50 cm3 measuring cylinder 250 cm3 beaker Spatula Equipment for vacuum filtration Procedure Weigh out approximately 5.0g of CuSo4à ¯Ã‚ ¿Ã‚ ½5H2O Dissolve it in 30 cm3 water in the beaker Add 10 cm3 concentrated ammonia (NH3) and stir the solution Add 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 class First 5.01g of CuSo4à ¯Ã‚ ¿Ã‚ ½5H2O 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 colour bright blue. Then the solution 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 processing 1. Calculate the number of moles CuSo4à ¯Ã‚ ¿Ã‚ ½5H2O 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) = 250 m = 5.01 n = 5.01 / 250 = 0.02004 à ¯Ã‚ ¿Ã‚ ½ 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 = 9.1g Since only 25% of ammonia is NH3 , mass of NH3 : 9.1 x 0.25 = 2.275g From here the amount of moles can be calculated by the formula n = m / Mr. Mr = 14 + (1 x 3) = 17 m = 2.275g n = 2.275 / 17 = 0.134 moles (3 s.f.) 3. Which of the reactants is in excess? Which is the limiting reagent? CuSo4à ¯Ã‚ ¿Ã‚ ½5H2O NH3 Number of moles (n) 0.02 0.134 Divide by smallest ratio 0.02 / 0.02 = 1 0.134 / 0.02 = 6.7 Divide by stoichiometric co-efficient from equation (Equation below this table) 1 / 1 = 1 6.7 / 4 = 1.675 Reactant in excess or limiting reagent Limiting reagent Reactant in excess (1)CuSO4 . 5H2O + 4NH3 Cu(NH3)4SO4 . H2O + 4H2O 4. Calculate the theoretical yield of Cu(NH3)4SO4 . H2O From 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 calculated: n = 0.02 Mr = 246 m = 0.02 x 246 = 4.92 g Calculate 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 à ¯Ã‚ ¿Ã‚ ½ 97.9% (3 s.f) Because the difference is so small (2.1%) the experiment 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.