How do you find the number of water molecules in a hydrate?

Question

The question is:

The aluminum sulfate hydrate [Al2(SO4)3*x H2O] contains 8.50 percent Al by mass. Calculate x, that is, the number of water molecules associated with each Al2(SO4)3 unit.

Answer 1

This is how I would solve it....

Assume you have one mole of the ANYHYDROUS substance (it actually doesn’t matter what amount you assume, but 1 mole makes the math slightly easier). If you have 1 mole of Aluminum Sulfate anhydrate, you should be able to calculate how many grams of Aluminum you have in the sample and then by dividing this value by the molar mass of Al2(SO4)3, you can get the percentage Aluminum of the anhydrate.

When the sample absorbed water, its molar mass will increase and the percentage of Aluminum in the sample by mass will decrease. You are told the new % Aluminum by mass in the problem. You previously calculated the old % Aluminum. By what factor did this change?

For example,
If the molar mass increases by a factor of 2, the percent Aluminum will drop to 1/2 its original value.

Take the new % Aluminum and divide it by the old % Aluminum. This is the factor that the molar mass increased by. Since you already know the molar mass of the anhydrate, you should then easily be able to calculate the molar mass of the hydrated sample. The difference in their masses is the amount of water (in grams) that the compound absorbed.

Take the number of grams of water absorbed and divide by the molar mass of water to find the number of moles of water absorbed. Since we assumed we had 1 moles of Aluminum Sulfate to start out with, the number of moles of water absorbed would be "per mole" of Aluminum Sulfate, and would thus be the value of x we are searching for.


The molar mass of Al2(SO4)3 is: 342.17 g/mol.
The percentage of Aluminum in this 1 mole sample is: 53.96 g / 342.17 g = 15.77% Aluminum by mass.

We are told that the new % Aluminum is 8.50%, so the factor the molar mass increased by is:
15.77 % / 8.50% = 1.86

The new molar mass of the Hydrated Aluminum Sulfate must then b 1.86 times the molar mass of the anhydrous Aluminum Sulfate.
342.17 g/mol * 1.86 = 636.43 g/mol

The change in mass of one mole of the Anhydrous Aluminum Sulfate going to 1 mole of hydrated Aluminum Sulfate is thus:
636.43 g – 342.17 g = 294.26 grams,
This means that it absorbed 294.26 grams of water, or 16.3 moles of water (= 294.26 g / 18.02 g/mol water).

That means that the value of x = 16,
We are talking about Aluminum Sulfate * 16 H2O.
Which is a known Hydrate of Aluminum Sulfate if one was so inclined to look it up (although the 18 H2O is more common appearently).

Answer 2

complex task. search onto google and yahoo. just that might help!