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burtwood Offline
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Chemistry Help?!?!!? - April 26th 2009, 12:32 AM

Grade 11 Chemistry Help! Determining Percentage Yield Lab?

Ok, I'm having great difficulty with this lab. I'll write down the questions and the facts that I have, any help would be greatly appreciated.

Question: What mass of copper is formed when excess aluminum is reacted with a given mass of copper(II) chloride dihydrate.

Prediction: Calculate the theoretical yield for this experiment

A known mass of a copper salt is dissolved in water and is reacted with an excess of aluminum. The mass of copper is formed in the reaction is determined. The balanced equation is:

3 CuCl2.2H2o (aq) + 2 Al (s) ---> 3 Cu (s) + 2 AlCl3(aq) + 6 H2O(l)

We have 2.00 g of copper (II) chloride dihydrate.

Analysis: Identify the limiting and excess reagent in this reaction. What visible evidence is there to confirm your identification?

Determine the actual yield of copper

Determine the percentage yield of copper

If your percentage yield is less than 100% suggest scientific techniques or equipment the percentage yield that may account for the loss of product.

PLEASE HELP!!!!!!!!!!!!
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Re: Chemistry Help?!?!!? - April 26th 2009, 03:53 AM

First, convert CuCl2.2H2O to moles. 2.00g/170.48256 g/mol = 0.01173 moles.

Compare the molar ratio between the given (CuCl2.2H2O) and the unknown (Cu):

Coefficient of CuCl2.2H2O: 3
Coefficient of Cu: 3
# moles of CuCl2.2H2O: 0.01173 moles
# moles of Cu: x

Use the following ratio:

# moles given / coefficient of given = # moles of unknown / coefficient of unknown

so, 0.01173 moles / 3 * 3 = # moles of the unknown

# moles of the unknown = 0.01173 moles.

Now, change the number of moles of Cu to a mass by multiplying by the molar mass of Cu:

0.01173 moles * 63.54 g/mol = 0.7453 g copper

So the expected mass of copper yielded is 0.7453 g.

The percentage yield is obtained by dividing the amount of copper you actually get by the amount of copper you would expect to get. Did you do an experiement in class? Your question doesn't list an experimental yield. But lets say that you did the experiment and only got 0.5 g Cu.

% yield = actual amount of product / expected amount of product * 100
% yield = 0.5 g / 0.7453 g * 100
% yield = 37.265 %

If you have an excess of Al, that means you have as much as you need to react all of the CuCl2.2H2O, so the limiting reagent would be CuCl2.2H2O. The reaction can only carry on as long as you have some. However, if you had a limited amount of Al, you would solve it like this:

Convert both reactants to moles. We already know we have 0.01173 moles of CuCl2.2H2O, so we would need to calculate how many moles of Al we have. Then, you would divide the moles by the coefficients of their respective molecules:

for CuCl2.2H2O: 0.01173 moles /3
for Al: # moles / 2

The limiting reagent is the smallest number.

There could be several reasons for getting a percentage yield less than 100. The reactants may not be pure. Some of the products may be left behind in the reaction vessel, or it may be too difficult to completely isolate the yielded copper from the rest of the products. Then there's the obvious, human error - maybe the scientist measure the reactants wrong etc.

Wow...now I remember why Chemistry bores me so much.

Not around so much now that school's started

"Live a good life.
If there are gods and they are just,
then they will not care how devout you have been,
but will welcome you based on the virtues you have lived by.
If there are gods, but unjust, then you should not want to worship them.
If there are no gods, then you will be gone, but will have lived a noble life
that will live on in the memories of your loved ones."
Marcus Aurelius

Last edited by Grizabella; April 26th 2009 at 04:09 AM.
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burtwood Offline
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Re: Chemistry Help?!?!!? - April 27th 2009, 12:55 AM

It bores me too, but thank you very much.
Your answer was very helpful
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Katrina Offline
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Re: Chemistry Help?!?!!? - April 29th 2009, 12:32 PM

Originally Posted by burtwood View Post
It bores me too, but thank you very much.
Your answer was very helpful
Sweet, I'm glad you've figured that out. [: As such, I'm going to close this thread.

If you have additional questions in the future, feel free to ask!

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