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The Magnificent LAB 4C :)

Hey, this is Jessica, and TODAY we did something that will blow your mind. We were all assigned a hydrate of an unknown composition, and we used a bunsen burner to evaporate the water molecules. WOO it was so exciting.
Before I tell you about the lab, ther first thing you need to know is what a hydrate is. A *hydrate* is a compound that has a definite number of water molecules incorporated into its crystal structure. The crystals appear dry, but when these compouds are heated strongly, water is given off, leaving the *anyhydrous* form of the compound.
The first mass we found was the empty crucible. We weighed the crucible on a centigram scale. The mass my partner and I found was 25.23g. Then we found the mass of the crucible and the hydrate, which was 30.11g.
To find the mass of the hydrate alone, this is the formula:
Mass of crucible and hydrate - Mass of empty crucible = Mass of hydrate
30.11g - 25.23g = 4.88g
Next, we heated the crucible and anhydrous salt using a bunsen burner and weighed it again on the centigram scale. The mass we found was 27.79g. After that was found, we heated the same objects again for a second heating and came up with a mass of 27.78g
To find the mass of the anhydrous salt alone, this is the formula:
Mass of the first heating - Mass of the empty crucible = Mass of anhydrous salt
Lastly, to find the mass of the water given off, this is the formula:
Mass of the hydrate - Mass of the anhydrous salt = Mass of the water given off
**ANALYSIS OF RESULTS **(whopee)
1. Calculate the percentage of water in a hydrate
%H20 = 2.33g
---------- x 100% = 47.7%
4.88g
2. Calculate the number of moles of the anhydrous salt left behind.
2.55g x 1mole
--------------- = 0.02 mol
120.4g/mol
3. Calculate the number moles of water removed by heat from your sample of hydrate.
2.33g x 1 mol
----------- = 0.13 mol
18g/mol
4. Calculate the moles of water per mole of the anhydrous salt.
0.02
------ -> 1
0.02
0.13
------ -> 7
0.02
5. What is the empirical dormula of the hydrate?
AB . ** **7H2O
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