G #17 b

[Yuchien Ma 2L]

Determine the mass of CuSO₄ • 5H₂O that must be used to prepare 250 mL of 0.20 M CuSO₄(aq).

The solutions manual says 12 g which is what I got if I set up the equation as followed and solve for x:
x grams * 1 mol/249.611 * 1 L/0.2 mol = 0.250 L

But isn't CuSO₄ • 5H₂O a hydrate? Therefore the mass of the water is included right? Wouldn't that also dilute the solution even further, diluting it to less than 0.20 M if we calculated it like that like that?

[Alexandra_Ivanova_1A]

CuSO4 • 5H2O is a hydrate, yes. We include the water through both accounting for its contribution to the total molar mass of CuSO4 • 5H2O, as well as through the final volume. This is a dilution problem, so it is helpful to think of these sorts of problems (dilutions with hydrates) through how they would actually be prepared. In a dilution, the compound is dissolved in a certain/small amount of water, and then more water is added to reach the final volume. Because of this addition of water after dissolving, the 5H2O is accounted for in the problem through the preparation of the solution, and you are not diluting the solution further than is asked for.

[Alex Dib 4H]

Hey Yuchien,
I solved this problem using dimensional analysis. First, I found the amount of moles of CuSO4 in 250 mL of a 0.20 solution, which is 0.05 moles of CuSO4. Then, we have to convert that into moles CuSO4.5H20 with a conversion factor of 1 mole CuSO4 for every 1 mol of CuSO4.5H20, which also gives 0.05 moles of CuSO4.5H20. Finally, convert those moles of CuSO4.5H20 into mass with the conversion factor 249.691g per 1 mole of CuSO4.5H20, giving 12 with correct significant figures.

(250 mL CuSO4 solution)(0.20 mol CuSO4/1000 mL CuSO4 solution)(1 mol CuSO4.5H20/1 mol CuSO4)(249.691g CuSO4.5H20/1 mol CuSO4.5H20)
=12g CuSO4.5H20

Not sure if this was tackling your question, but I hope it helps!

[Maggie Eberhardt - 2H]

Hey Yuchien,
I solved this problem using dimensional analysis. First, I found the amount of moles of CuSO4 in 250 mL of a 0.20 solution, which is 0.05 moles of CuSO4. Then, we have to convert that into moles CuSO4.5H20 with a conversion factor of 1 mole CuSO4 for every 1 mol of CuSO4.5H20, which also gives 0.05 moles of CuSO4.5H20. Finally, convert those moles of CuSO4.5H20 into mass with the conversion factor 249.691g per 1 mole of CuSO4.5H20, giving 12 with correct significant figures.

(250 mL CuSO4 solution)(0.20 mol CuSO4/1000 mL CuSO4 solution)(1 mol CuSO4.5H20/1 mol CuSO4)(249.691g CuSO4.5H20/1 mol CuSO4.5H20)
=12g CuSO4.5H20

Not sure if this was tackling your question, but I hope it helps!

Where did you get the number 249.691 when converting the moles into mass?

[claudia_1h]

The 249.61 comes from using the molar masses for each element in the compound CuSO4 5H2O, found on the periodic table, and multiplying each number by the number of atoms present in the compound.