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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: antoinetta on September 23, 2006, 08:48:51 PM

Title: Question about Stoichiometry
Post by: antoinetta on September 23, 2006, 08:48:51 PM
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?
Title: Re: Question about Stoichiometry
Post by: enahs on September 23, 2006, 11:31:22 PM
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.


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.
Title: Re: Question about Stoichiometry
Post by: xstrae on September 25, 2006, 11:44:22 AM
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. ;D