Gravimetric Analysis

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Recycling Aluminum Chemically

Question:How does theoretical yield of Aluminum compare with the percent yield?


Mass of aluminum foil1.033g
Mass of filter paper0.472g
Mass of Al crystals, filter paper, weight boat20.823g
Mass of weigh boat2.064g
Mass of Al crystals18.287g

Claim:Based on my calculations, the theoretical yield of Aluminum was 18.160g, and the actual yield was 18.287g. This gives a percent yield of 100.699%.

Reasoning:By looking at the overall reaction involved in the lab, we can see that two moles of Aluminum are needed to react in the reaction, producing 2 moles of KAl(SO4)2 ·12H2O. This lets us know that however many moles of Aluminum there are in the original sample will be equal to the number of moles of the produced hydrate. By dividing the grams of Al added by the molar mass of aluminum, we find that there are 0.03828 moles of aluminum. This means that there are also 0.03828 of KAl(SO4)2 ·12H2O. By totaling the molar masses of each individual element, we find that the molar mass of KAl(SO4)2 ·12H2O is 474.3902 grams. By multiplying this number by 0.03828, we find the theoretical yield of Aluminum is 18.160 grams. The mass of the crystals produced from the lab was 18.287 grams. By dividing this actual yield by the theoretical yield, we find the percent yield to be 100.699%. My percent error was –0.699%, meaning the measured value for the crystals was slightly higher than the accepted value.

Improvements: One source of error within the lab was that when the KOH and Aluminum solutions was being heated, some of the dissolved Aluminum may have been lost. This would alter the calculations of the lab by lowering the yield of Aluminum below the theoretical value. Another source of error within the lab was the possibility that not all of the Aluminum dissolved in the KOH solution. When the solution was put through the vacuum filtration system for the first time, the undissolved Aluminum would have...
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