## Workbook Page 11 Number 5

Allison_Eminhizer_3A
Posts: 16
Joined: Wed Sep 21, 2016 2:56 pm
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### Workbook Page 11 Number 5

Given is Copper(II)Sulfate . 5water
I'd type out the formula but I'm not sure how to do subscripts

My first question is why there is a period between copper(II)sulfate and water, I don't know what to make of it or if its just a typo?

The rest of the problem is determine the mass of Copper(II)Sulfate . 5water that must be used to prepare 250mL of 0.20 M Copper(II)Sulfate(aq).
I'm really rusty on this topic and I'm not sure how to carry out the steps.

Preston_Dang_1B
Posts: 22
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Workbook Page 11 Number 5

The placement of the period is a little bit off in the equation. The dot is meant to be a little higher up and in between Copper(II) Sulfate and 5 Water and is meant to be a sort of "multiplication dot" that indicates that the compound is a type of hydrate, hence the additional 5 water molecules that are also part of the formula and mass of the compound. The formula that is implied is CuSO4 $\cdot$ 5H2O. From there, multiply 0.20 M by .250 L and you'll get the number of moles of Copper(II) Sulfate Pentahydrate that are required to make that solution. After that, multiply the number of moles you get by the molar mass of CuSO4 $\cdot$ 5H2O and you'll get the mass needed to make a solution of that molarity.

Isayan_Natalie_3L
Posts: 14
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Workbook Page 11 Number 5

That weird period in between water and copper(II) sulfate is not a typo, it means that it's a hydrate. The rest of the problem is pretty simple, you just use the M=moles of solute/liter $M= \frac{n}{L}$ formula, plug in the values for molarity and volume, which are already given, and solve for n. After getting n you just multiply number of moles by the molar mass of copper(II) sulfate and 5 H2O to get the answer.

$M= \frac{n}{L}$

$.20M= \frac{n}{.250L}$

n=.05 moles hydrated copper(II) sulfate

$n=\frac{mass}{molar mass}$

$.05 moles=\frac{m}{239.609g}$
$m=12.0 g {CuSO_{\: 4}}^{}\cdot 5{H_{2}}^{}$O

Hopefully that makes sense & equation editor is working.