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

Hi, I'm just having trouble with this problem: A mixture of CuSO45H20 [Copper II Sulfate Pentahydrate] and MgSO47H2O [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 CuSO45H20 in the mixture?
Here is what I have done so far:
Molar Mass of CuSO45H20 = 249.7g/mol
Molar Mass of MgSO47H2O = 246.5g/mol
Mass of Water: 5.020 g  2.988 g = 2.032 g
Now what?

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 2.988X g
Now, you know how much water was lost, and so how many moles of water was lost.
For every moles of CuSO_{4} there is 5 moles of H_{2}O, and for every moles of MgSO_{4} there is 7 moles of H_{2}O.
Therefore:
5 * Moles of CuSO_{4} + 7 * Moles of MgSO_{4} = Moles of H_{2}O
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 * (2.988Xg / 120.3686 mol) = Moles of H_{2}O
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 CuSO_{4} and 2.988 – X is grams of MgSO_{4}).
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 CuSO_{4} . 5H_{2}O, 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.

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