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Fundamentals G

Posted: Wed Oct 09, 2019 9:23 pm
by Hannah_1G
In the second part of G.17 it asks for the mass of CuSO4 5H20 that must be used to prepare 250 mL of .2 M CuSO4(aq). I am very confused by what to do with the water molecule in this problem, I don't know how to start solving this.

Re: Fundamentals G

Posted: Wed Oct 09, 2019 9:37 pm
by Trinity Vu 1D
You would need to find how many moles of CuSO4 are needed to make the .2M solution. Then using that number of moles you would multiply by the molar mass of CuSO4 5H20 instead of the molar mass of just CuSO4. From part (a) since you know that you need .05 moles of CuSO4, you'd multiply .05 by 249.68g (molar mass of CuSO4 5H2O) to get 12g CuSO4.

Re: Fundamentals G

Posted: Wed Oct 09, 2019 9:37 pm
by Emily_4B
Do the same steps you did for part (a) except now use the molar mass of CuSO4•5H20, which is 249.68. So you’d have to multiply (.2 M)(0.250 L)(249.68 g/mol)=12 g of CuSO4•5H20

Re: Fundamentals G

Posted: Wed Oct 09, 2019 9:43 pm
by kevinchang_4I
The 5H2O really is a curveball.To begin with, we need to know how many moles of CuSO4 in 250 mL of a .2 M solution. This will be .05 moles of CuSO4. This is going to be 7.98 grams CuSO4. Now we must calculate CuSO4's percent composition in CuSO4 * 5H2O. Which is molar mass CuSO4 over molar mass CuSO4 * 5H2O. We get 63.92 %. This tells us that for every 100 g of CuSO4 * 5H2O, you get 63.92g of CuSO4. If we make a conversion factor with the 7.98 grams of CuSO4 times (100 g of CuSO4 * 5H2O/63.92g of CuSO4), then we get 12.48 grams of CuSO4 needed to prepare 250 mL of .2 M CuSO4(aq).

Re: Fundamentals G

Posted: Wed Oct 09, 2019 9:55 pm
by Jialun Chen 4F
I personally use the mole to mole conversion: 1 mole of CuSO4.5H2O contains 5 moles of H2O.