Question

1) Determine how the following error would affect calculated the percent yield of alum. (increase, decrease, no effect). Briefly explain.

1a. The Aluminum foil did not completely react.

1b. Alum sample was left in the beaker and not transferred to the filter paper.

1c. Alum Sample was not dried completely in the oven.

2) A piece of an aluminum can was cut into small pieces. Then 0.422g of the pieces from the can was used to prepare potassium alum according to the procedure used in lab.   How many grams of Alum should you expect if you had a 83.8% yield? (alum MW = 474.41g/mol)

** please answer all the questions**

Answer 1

1a) The Al foil is converted to alum in this lab. There is a 1:1 molar ratio between Al and alum. If the Al foil do not react completely, the number of mole(s) of Al that reacts will be low and hence, the percent yield of alum will be low.

1b) The filtration step is required to remove unwanted impurities from the alum sample. It is important that the alum sample is quantitatively transferred to the filtered paper, washed and dried. The final weight of the dried alum sample is recorded to calculate the percent yield. If some alum was left in the beaker, the mass of alum recovered will be low and hence, the percent yield will be low.

1c) In order to calculate the percent yield of alum, it is necessary that the mass of the dried alum is recorded in the final weighing step of the reaction. If the alum sample is not properly dried in oven, moisture will remain trapped in the alum sample. The mass of the moisture adds up to the mass of alum and hence, the recorded weight of alum will be high and consequently, the percent yield will be high.