Percent Yield - Unsure of Which Molar Mass to Use

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Homework Statement


A percent yield lab says to add 5.00 g of CuCl2•2H2O(s) to 50 mL of dissolved water. Then, add 1.0 g of steel wool (100% iron) to the beaker.

I am trying to find out how many moles of CuCl2 react here. I am confused about the dihydrate.

Homework Equations


The reaction given in the lab is

Fe(s) + CuCl2(aq) → FeCl2(aq) + Cu(s)

The Attempt at a Solution


To find out how many moles of CuCl2•2H2O(s) I used, do I calcuate the molar mass including the dihydrate? Or do I use just the molar mass of CuCl2?

I think I am supposed to include the dihydrate in the molar mass calculation, but I'm confused because the dihydrate doesn't appear in the reaction. It doesn't react with anything. So wouldn't I need to figure out how many moles of CuCl2 there were, not how many moles of CuCl2•2H2O(s)?

Also, I know the CuCl2•2H2O(s) is the excess reactant, but to show that, I need to find out how many moles of it I have first.
 

Answers and Replies

  • #2
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do I calcuate the molar mass including the dihydrate?
You're preparing the solution from the dihydrate. Does the balance know you're only interested in CuCl2?
 
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When you measure out 5.00 g onto the balance, the dihydrate is included. The number of moles of reactant sitting on the balance would be calculated using the molar mass of CuCl2 • 2H2O.

That much I know. But when you add the CuCl2•2H2O(s) to a beaker of water, what happens to the dihydrate? Does it stay attached to the CuCl2? I guess it doesn't matter... 1 mol of CuCl2•2H2O(s) would still correspond to 1 mol of CuCl2 right?
 
  • #4
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1 mol of CuCl2•2H2O(s) would still correspond to 1 mol of CuCl2 right?
Yes.

Hydration water can be important when you are interested in the amount of water present. Say, you prepare a solution by using 1 g of dihydrate and 10 g of water and you need to know the exact, final concentration. But in typical situations we add water to the mark - so the final amount doesn't depend on the presence of hydration water at all.
 
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Okay, that makes sense. In this case, the amount of water doesn't matter anyway. Thanks!
 

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