[antoinetta]

Hi, I'm just having trouble with this problem:  A mixture of CuSO4-5H20 [Copper II Sulfate Pentahydrate] and MgSO4-7H2O [Magnesium Sulfate Heptahydrate] is heated until all the water is lost.  If 5.020g of the mixture gives 2.988g of the anhydrous salts, what is the percent by mass of CuSO4-5H20 in the mixture?

Here is what I have done so far:

Molar Mass of CuSO4-5H20 = 249.7g/mol
Molar Mass of MgSO4-7H2O = 246.5g/mol

Mass of Water: 5.020 g - 2.988 g = 2.032 g

Now what?

[enahs]

This is not to you, but good lord why do I keep answering the same question over and over in a slightly different form?!


A percent is per 100, when you have a percent divided into two separate things they must add up to 100.

One of the salts of you choice becomes X g, the other becomes 2.988-X g


Now, you know how much water was lost, and so how many moles of water was lost.

For every moles of CuSO₄ there is 5 moles of H₂O, and for every moles of MgSO₄ there is 7 moles of H₂O.

Therefore:

5 * Moles of CuSO₄ + 7 * Moles of MgSO₄ = Moles of H₂O

This is where your grams come in, and the molar mass of the anhydrous salts.

So it simplifies down to:

5 * (Xg / 159.6096 mol)   +   7 * (2.988-Xg / 120.3686 mol)  =  Moles of H₂O

Solve for X and you know have your grams of anhydrous salt for both (X in how I have it written out is grams of CuSO₄ and 2.988 – X is grams of MgSO₄).


Once you have the grams (and thus moles) of your salts you can simply add the appropriate amount of water (5 moles to 1 moles ) This will give you your original weight of CuSO₄ . 5H₂O, and then it is just division for percentage.

That sounds like a lot, but I am trying to type up the steps so you can do it yourself and learn, it is really quite simples.

[xstrae]

great work! I am a quite new person to this forum. I feel such detailed explanations from the enlightened lot is much more helpful, helps us learn faster and also saves a lot of time. anyway, continue the good work.