Stoichiometry and the MoleCatherine Lozier02/05/2013Chem 1310 Section C02TA Ryan BucherLab Partners:Chelsey Arnold Sunya MorinTran GHonor Pledge: I did not copy this work from any others student(s), current students in lab or old lab reports ________________________________________________SignatureDATA AND OBSERVATIONSPart ADataReagents Formula mL or g Used Moles Limiting?PotassiumHydroxide, 1.4 MKOH 35 mL 0.049 NoSulfuric Acid, 9 M H2SO415 mL 0.135 NoAluminum Al 0.511 g 0.019 YesObservationsCrystals took longer than expected to form. The first crystals were observed after ~10 minutes. Decided against induced crystallization, but allowed an extra 5 minutes in the ice bath to allow for crystal formation. A significant amount of alum crystal was lost during the drying process, as they would not rinse out of the beaker. Approximately 25% of the crystals could not be rinsed from the beaker using ethanol, and did not make it into the vacuum filtration device. An extra 5 minutes was allowed for drying since the crystals were still damp at the end of 10 minutes.CalculationsVolume KOH for 10 g Alum10 g K[Al(SO4)2] / 258.2 gmol-1 = 0.0387 mol1 KOH : 1 K[Al(SO4)2]  0.0387 mol KOH0.0387 mol * 56.108 gmol-1 = 2.17 g KOH1.4 M KOH  78.5512 gL-12.17g/78.5512 gL-1 = 0.0276 L KOH27.6 mL KOHVolume H2SO4 for 10 g Alum10 g K[Al(SO4)2] / 258.2 gmol-1 = 0.0387 mol2 H2SO4 : 1 K[Al(SO4)2]  0.0774 mol H2SO40.0774 mol * 98.076 gmol-1 = 7.59 g H2SO49 M H2SO4  882.684 gL-1 H2SO47.59 g H2SO4/882.684 gL-1 = 0.0086 L H2SO48.6 mL H2SO4Theoretical Yield0.511g Al  0.511g / 26.89gmol-1 = 0.0189 mol Al1 Al : 1 K[Al(SO4)2]  0.0189 mol K[Al(SO4)20.0189 mol * 258.2 gmol-1 = 4.88 g alumY100% = 4.88 gActual YieldYactual = (0.860g / 4.88 g) * 100%Yactual = 17.6%Part BHypothesisIf the product of the experimental reaction has excess iron, the iron is type Fe2+. If the product is relatively clear, it is type Fe3+. The experimental reaction uses exactly enough copper sulfate to react with 1.0 g of Fe3+. If the iron is type Fe2+, the iron will be in excess.Procedure1. Obtain 1.0 g of sample Fe, 50 mL H2O, and ~4.387 g CuSO4.2. Mix the copper sulfate and water with a stirring rod in a 300 mL beaker until all of the copper sulfate is dissolved.3. Add the iron sample and stir for 5 minutes or until the solution becomes clear. Make sure that all of the ironcomes in contact with the copper sulfate mixture.4. Record observations on the physical appearance of the product.Results and ObservationsReagents Amount UsedCopper Sulfate (CuSO4) 4.31 gIron (Fe) 1.002 gWater (H2O) 50 mLAfter stirring for 5 minutes, the aqueous solution had gone from blue to clear. There was still a significant amount of unreacted iron remaining in the beaker. Most of the iron wool used has broken up into small particles of iron. Because there was an excess of iron left in the products of this reaction, the unknown iron used in the steel wool is likely Fe2+.DISCUSSIONPart AThe purpose of Part A of this experiment is to use stoichiometry to calculate the amount of reagents needed to yield a certain amount of a product, as well as become familiar with crystal synthesizing techniques. The goal of the experiment was to synthesize 10 g of alum. In order to do this, it was necessary to first calculate the amounts of potassium hydroxide and sulfuric acid needed to react with the given amount of aluminum. It was calculated that 27.6 mL KOH and 8.6 mL H2SO4 would be needed to synthesize 10 g of alum. To ensure that the aluminum was the only limiting reagent, 35 mL and 15 mL of mL KOH and H2SO4 were used, respectively. Although the original intentwas to synthesize 10 g of alum, only 0.511 g of aluminum were used, which would theoretically yield 4.88 g. The actual yield was 0.860 g, which is only 17.6%. This indicates that there were likely several sources of error in the experiment. One of the biggest error contributions was human error. After chilling the crystals, they were to be rinsed into the vacuum apparatus. Although the glass was rinsed with ethanol to help remove the crystals, a significant number of the crystals would not leave the beaker and did not make it into the vacuum filtration system. As much as 25% percent of the alum yield was discarded. The abnormally low yield could also be due to the fact that the crystals took longer to develop. The directions stated that the crystals should be given 15 minutes to synthesize, however crystal formation was not observed for at least 10 minutes. It is possible that the solution was not chilled properly, which was not conducive to crystal formation. Overall, there were many flaws in the experiment, which produced a very small yield.Part BThe purpose of Part B of the experiment was to promote creative thinking and apply stoichiometry to determine the specific ion of iron in a material. The goal of this part of the experiment was to design an experimental process to determine if the iron used in a sample of steel wool was of type +2 or +3. The experiment could use 50 mL of water and CuSO4, and no other materials. The experiment made use of the fact that CuSO4 is blue in solution, and turns clear when it reacts with iron. Depending on the type of iron, there were two possible reactions that could take place between iron and CuSO4.Fe2+: Fe + CuSO4 · 5H2O  FeSO4 + Cu + 5H2OFe3+: 2Fe +3(CuSO4 · 5H2O)  Fe2(SO4)3 + 3Cu + 15H2OTherefore the amount of CuSO4 for the Fe3+ reaction was used. If the product was clear, then all of the iron would have reacted, so it would have been type Fe3+. However, if there was an excess of iron at the conclusion of theexperiment, then it would be Fe2+. For 1.002g of iron, 4.31 g of CuSO4 was used, according to the Fe3+ reaction chemical equation. After stirring the reagents for several minutes, the solution had gone clear; however there was an excess of iron particles. Thus it was not Fe3+, and instead the iron in the steel wool was Fe2+. Some sources of error would be human error in the measurements, as well as uneven mixing of the solution. However, these sources of error are not significant enough to have such a drastic effect on the product. There was a significant amount of iron remaining in the solution, enough that the bottom of the beaker was covered.DISCUSSION QUESTIONS1. If the paint and lacquer had not been removed from the aluminum, the yield would have decreased significantly. The piece of aluminum can initially weighed 0.560 g; after removing the