(Long Question) Please help if you have time.

Question

Percentage Yield Lab

Note to students: This is only a practice lab (No Submissions). Feel free to get extra help if needed. Before the final one is given out. Have Fun!

Introduction

The reaction between aluminum and copper (II) chloride dihydrate produces copper metal. In this investigation you will analyze this reaction with excess aluminum, assuming you collected the copper and determined the percentage yield of copper metal produced.

Prelab questions

1. Write the balanced chemical equation for this reaction. Use the metal reactivity series to explain why this reaction occurs.

1. If 4.00 g of copper (II) chloride dihydrate is used, calculate the exact mass of aluminum needed to react completely with the copper compound. To ensure that aluminum is in excess, you will use 5 times this mass of aluminum in your experiment.

1. Calculate the theoretical yield of copper for this reaction.

Hypothesis:

Write out a proper hypothesis for this Lab.

Materials:

List all the materials that you would need to perform this experiment. Include all equipment and chemicals necessary. Assume you have access to all the equipment in our school.

Procedure:

Write a numbered step by step procedure to carry out this lab. Be specific enough that someone could come in and repeat what you did exactly. Include the specific amounts of materials you need to use, how you will decide the reaction is complete, and how you will collect your copper metal at the end. (some research may be required here)

Results:

Make a properly formatted data table for this investigation to record any data that you need to collect for this lab. You do not need to fill it in.

Analysis:

Suppose you collected 0.75g – 1.75 g (Select any value to 2 decimal places) of dry copper metal when everything was done. Determine the % yield.

Discussion:

In a few paragraphs, explain any sources of error that you might encounter in the performance of this lab. Differentiate between human and experimental sources of error. Suggest possible improvements that you could make to your procedure to improve the results.

Conclusion:

Write out a proper conclusion to the lab based on the amount of product you collected.

Answer 1

The balanced chemical equations is 2 Al + 3 CuCl₂H₄O₂ = 2 AlCl₃ + 3 Cu + 6 H₂O.

This tells us we need 2 moles of Al for every three moles of CuCl₂H₄O₂. for a balanced reaction that will produce 2 moles of AlCl3, 3 moles of copper, and 6 moles of water.

We are limited to 4.00 g of copper (II) chloride and have excess aluminum (more than needed). To determine the amount of aluminum needed for a balanced reaction, we first calculate the number of moles of CuCl₂H₄O_2:

The molar mass of CuCl₂H₄O₂ is 170.48 g/mole, We have 4.00 g. To calculate moles:

= 4.00 g /170.48g/mole = 0.0234 moles

The calculation is rounded to the same number of significant digits (3) as the CuCl₂H₄O₂ .

We have excess aluminum, so to calculate the theoretical yield of copper, we assume that all of the copper chloride dihydrate is consumed in the reaction. This can be predicted from the metal reactivity table. Aluminum is more reactive than copper and will thus replace it in this compound to form AlCl₃ and knock the copper out. The equation says we’ll get 2 moles of Cu for every 3 moles of CuCl₂H₄O₂. We take that ratio and multiply it times the moles of CuCl₂H₄O₂  we reacted to calculate the moles of Cu produced, in theory.

0.0234 moles * 2/3 = 0.0156 moles of Cu

The molar mass of Cu is 63.546 g/mole. To calculate the number of grams that is contained in 0.0156 moles of Cu, do the conversion:

(0.0156 moles Cu) x (63.546 g/mole) = 0.994 grams Cu

Assuming one obtained 0.80 g of dried copper from the experiment, the % yield would be calculated:

(Actual/Theoretical) times 100%

Here:

(0.80/0.994) x 100% = 0.80 or 80%

You got 80% of what would have been expected, if everything went perfectly. A lab report usually has a section for discussing what errors may have occurred, as is requested here. There are no clear answers, unless your labmate dropped the sample of the lab table and used his hand to brush it back into the beaker, as mine did one year. That was easy. Other possibilities include scale not in calibration, weighing and pouring errors, contaminants (e.g., not entirely dried or excess reactant present) and calculation error.  The presence of contaminants can lead to recovery values of over 100%. [editor's note: 120% recovery sounds like a sure “A,” but the teacher knows the deal.]