Homework Help: chemistry 11

Posted by Daniel on Sunday, November 22, 2009 at 3:37pm.

Hi, I am stuck on a couple questions in chemistry 11. The questions are

1. Calculate the number of moles of the anhydrous salt left behind

2. Calculate the number of moles of water removed by heat from your sample of hydrate.

3. Calculate the moles of water per mole of anhydrous salt.

My problem with the first question is I am not sure of how to find the amount of anhydrous salt left behind. The second one, I am not sure of how to calculate the number of moles of water that were removed by heat, and I am not sure of how to calculate the moles of water per mole of anhydrous salt :\

I have a data table.
Mass of empty crucible and lid = 30.93g
Mass of crucible, lid, and hydrate = 33.78g
Mass of hydrate = 2.85g
Mass of crucible, lid, and anhydrous salt = 32.39g
Mass of anhydrous salt = 1.46g
mass of water given off = 1.39g
Mass of one mole of anhydrous salt = 111.1g

The anhydrate that we used is CoCl2

Any help will be greatly appreciated.

chemistry 11 - DrBob222, Sunday, November 22, 2009 at 4:19pm
I really don't understand your problem. You have all of the data you need to finish the problem.
1. 1.46g which is in the data part.
2. 1.39g which is in the data part.
3.Convert grams to moles.
1.46/molar mass H2O = ??
1.39/molar mass CoCl2 = ??
Now divide moles CoCl2 by itself (to get 1.00) and divide moles H2O by the same small number (about 0.011 or so), round the water to the nearest whole number and you will have x in CoCl2.xH2O.
chemistry 11 - J, Tuesday, November 30, 2010 at 11:07pm
mrs. lawsons chem ll? tfox?

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