Percent Yield Of Aluminum Lab Report
Title: Using Four Reactions to Calculate the Percent Yield of Recycled Aluminum Author Information: *Mia Ingram, Tyler Snody (*Primary Author) CHEM 111, Section 511, Experiment 4: The Chemistry of Recycling Introduction Aluminum is a common and crucial metal used in consumer’s everyday lives. It is mainly used in making containers and packaging, car parts, aircrafts and other household items. Recycling aluminum, usually in the form of cans, requires very low energy in order to make a useful product. In this experiment, aluminum is converted into potassium aluminum sulfate, or what the paper will now refer to as “alum” in order to see how much of the can is recyclable. The purpose of the abstract will be to go over the process of producing alum, writing net ionic equations for each step in the production of alum, and calculating the percent yield of alum. Materials and Methods The first step was adding 1.1013 g of aluminum can pieces to a 250. mL beaker. The same beaker was then filled with 50. mL of 1.4 M KOH solution, and the solution was heated by a hot …show more content…
Because the coefficients are the same in the balanced equation, the moles of aluminum equalled the moles of alum. Calculating the theoretical yield of alum required the moles of alum and the molar mass of alum. The molar mass of alum is 474.3884 g/mol. So the theoretical yield of alum was calculated as: (0.04082 mol alum) * (474.3884 g/mol alum) = 19.36 g of alum. Lastly, the percent yield of alum was found by using the percent yield formula: [(Actual Yield) / (Theoretical Yield)] * 100 = Percent Yield The actual yield of alum was found by subtracting the mass of the watch glass itself, 51.8004 g, from the overall mass of the watch glass and alum, 66.2013 g. So the actual yield of alum was 66.2013 g - 51.8004 g = 14.4009 g of alum actually yielded. After plugging in all the calculated values into the percent yield
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615 Words | 3 PagesA thermometer was also placed in at an angle. After ten minutes the thermometer was adjusted until it was completely straight in the middle of the beaker. Then the hot plate was turned on. One group member monitored the hot plate, while a second group member retrieved a stirring stick to stir in the beaker, this was to help the copper sulfate to dissolve. Two different group members observed what was occurring so that notes could be taken.
O-Vanillin Lab Report
840 Words | 4 PagesAs seen in table 1, the theoretical yield was .712 g of C_17 H_19 NO_3. The % yield of this experiment was 7.51 % of C_17 H_19 NO_3. . This low yield can be explained from a poor recrystallization technique combined with potential contamination. Throughout the experiment, the mixture changed color from green, orange, to yellowish lime, and eventually clear.
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Notion Of Moles And Avogadro's Number
501 Words | 3 PagesObjective The foremost purpose of this laboratory experiment is to comprehend notion of moles and Avogadro’s number; and also to grasp the idea of converting among mass, moles, and number of atoms in a sample (Hammerschlag 1). Importance Pace University Professor Carroll Zahn allocated a novel assignment to his introductory computer class, in which the results of this assignment greatly astounded both the professor and his class. The subsequent description that was stimulated by the unanticipated results of this miraculous event is the inspiration for the following case study. Perplexing his class, Professor Zahn told them to calculate the cost of a single aluminum atom in a roll of aluminum foil that he had recently purchased.
Nucleophilic Substitution Reactions Lab
1039 Words | 5 Pages(150.22g/mol)(3.5 x 10^-3 mol of nucleophile) = 0.525 g Actual yield = 0.441 g, Percent Yield = (0.441g/0.525g) x 100% = 84% 10. Percent recovery from recrystallization = (0.172g/0.441g) x 100% = 38% 11.
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1125 Words | 5 PagesThe final product weight for percent yield was only the solid E product, which missed one half of the final product produce. If both products were weight, the percent yield would have been larger that it was. Instead of 22.33%, it could have been 44.66%. To prove that both products were obtained, but only one of the two products was analyze, a TLC plate of the DCM layer, that contains both products, and of the final product, was obtain.
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453 Words | 2 Pages51 g and 87.2875 + 6.78= 94.01 g Between 80.51 g and 94.01 g there are exactly 37/80 x 100= 46.25%. This is under value for one standard deviation which leads to speculation about if there was a large harvest during the year creating large apples. 2 standard deviations from the mean is 87.2875 - (2 x 6.78)= 73. 73g and 87.625 + (2x 6.78)= 100.85 g.
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