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of Synthesis Zinc Iodide Lab Parker Ashbaugh Chem 104A Section 47 Mike Spinner 2/7/06 I. Introduction The objectives for this lab are to react iodine with an excess of zinc to make zinc iodide, measure the amounts reacted, isolate and weigh the product, determine the apparent formula for zinc iodide, and lastly determine the percent yields for the entire reaction. The experiment will proceed as follows. First a .5 g sample of Iodine should be obtained and placed in a test tube, with which 2 mL of water and 4 drops of 2M acetic acid will be added. To this will be added a .5 g sample of zinc. After the reaction a gravity filter will be used to separate the un-reacted zinc. Then dry the un-reacted zinc on a hot plate. Next recover the zinc iodide by heating up the solution. Lastly dehydrate the zinc iodide by reheating and re-massing the crystals to confirm dehydration. Data: *The data to this lab can be found on the last page of the report. Calculations Moles of Iodine: (Mass of Iodine sample) x (molar mass of Iodine) = .5013g I x 1 mol = .00395mol Iodine 126.9g Weight of Zinc: mass of Zinc sample mass of un-reacted Zinc = .5046g - .267g= .2376g zinc reacted Moles of Zinc: (mass of zinc reacted) x (molar mass f zinc) = .2376g Zinc x 1 mol = .0036 mol Zinc 65.39g Molar Ratio: Moles Zinc = .0036mol Zn = .911 2mol I Moles Iodine .00395mol I 1 mol Zn xZn + yI ZnxIy = 1Zn + 2I ZnI2 Mass of Zinc Iodide: [mass of product(after reaction)] x (molar mass of zinc iodide) = .3832g ZnI2 x 1 mol ZnI2 = .00119mol ZnI2 319.16g Percent Yield: Moles of Iodine Reacted: .00395 mol I2 Theoretical Yield= (moles of iodine reacted) x (ratio of moles between ZnI 2 and I2 ) ( .00395) ( 1mol ZnI2 / 1mol I2)= .00395 Percent Yield= experimental yield = .00119 mol ZnI2 = 30.1% based on Iodine theoretical yield .00395 mol ZnI2 Results: Weights of Substances Substance Iodine Used Zinc Used Unreacted Zinc Anhydrous Zinc Iodide Moles of Substances Substance Iodine Reacted Zinc Reacted ZnI2 Weight(g.) .5013 .5046 .267 .3832 Moles .00395 .0036 .00119 Calculated Ratio= 2 mol I / 1 mol Zn Proposes equation= xZn + yI ZnxIy = 1Zn + 2I ZnI2 Percent and Theoretical Yield based on the Substances Theoretical Yield based on Iodine .00395 mol ZnI2 Theoretical Yield based on Zinc Percent Yield of Iodine Percent Yield of Zinc .0036 mol ZnI2 30.1% 33.1% Discussion After completing the experiment the resulting formula from the reaction between zinc and iodine to form zinc iodide found was to be 1Zn + 2I ZnI2. The formula created from the data was done by dividing the moles of zinc by the moles of iodine to generate a mole ratio. Based on the known formula, it should have been close to 1:2, which is what we were able to calculate. Of course our results weren't 100% accurate. So in order to have better results we determined that iodine was a limiting reactant meaning that the reaction was run with excess zinc in order to react the entire mass of iodine. Therefore, since the moles of iodine reacted is a fixed number, the only way to get a desired mole ratio of zinc to iodine is to have recorded less amounts of zinc reacted. The percent yield, calculated by dividing the experimental yield by the theoretical yield and multiplying by 100, was calculated twice; once based on iodine and once based on zinc. In order to have generated higher percent yields, the experimental moles of ZnI 2 would have to be greater and/or the moles of zinc reacted would need to be lower. On top of that, there is a reason for a lower value of moles of ZnI2 and once again it comes from the filtration process. This reason is that possibly a portion of zinc iodide in the solution absorbed into the filter paper, never to be evaporated and massed. Thus producing a lower mass and less moles of zinc iodide. Likewise a small amount of zinc clung to the filter paper and couldn't be transferred into the watchglass. Also a small amount of the zinc iodide could have been lost when heating the solution in the crucible. The reason for the difference between the percent yields based on iodine and zinc is due to the empirical formula. The mole ratio of zinc to iodine is 1:2 and therefore the mole ratio of zinc iodide to zinc is 1:1 and of zinc iodide to iodine is 1:2. This means that the moles of iodine reacted had to be divided by two to calculate the theoretical mole yield of zinc iodide. That will then affect the percent yield because the theoretical yield based on zinc will differ from iodine while the experimental yield for both remains constant. In conclusion, this lab would have produced near perfect data and conclusions without slightly flawed laboratory practices. If this experiment between zinc and iodide had been able to be performed without the filtration procedure, it is possible that the data would have turned out much nicer. This experiment demonstrated the law of constant composition and was good practice for percent yields and stoichiometry. ... View Full Document

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