Chemistry - mole and emprical formulae calculations

Moles

Divide the mass of water12.23 grams) by its formula mass(18 atomic mass units)

12.23/18 =0.68mole

Moles

Divide the mass of copper sulfate (16.84grams) by its formula mass (160 atomic mass units)

16.84/160=0.105mole

Ratio

Divide both the mole of water and the mole of copper sulfate by the lowest number. In this case it is 0.105.
6.5/0.105 : 0.105/0.105
6.5 : 1

Whole number ratio

Since we can not have 0.5 molecules of water we multiply both sides of the ratio by 2 to remove all decimals.

6.5 X 2 : 1 X 2

13 : 2

Calculations

Record the results

(1)Weight of crucible	(2) Weight of crucible and hydrated copper sulfate	(3) Weight of crucible and dehydrated copper sulfate
35.03 grams	64.10 grams	51.87 grams

Calculate the mass of hydrated copper sulfate .
(2)-(1) = 64.10 -35.03 grams = 29.07 grams (this includes copper sulfate and water)
Calculate the mass of dehydrated copper sulfate
(3)-(1) = 51.87-35.03 grams = 16.84 grams ( pure copper sulfate)
Calculate the mass of water removed from the sample.
Mass of hydrated copper sulfate minus the mass of dehydrated copper sulfate.
29.07-16.84 grams=12.23 grams (water) Click on the coloured cells for more information. Double click to remove the information.

	Water(grams)	:	Copper sulfate(grams)
Mass	12.23	:	16.84
Moles	0.680	:	0.105
Ratio	6.5	:	1
Simplest ratio	13	:	2

So for every 2 units of copper sulfate there are 13 molecules of water.

Now conduct your own investigation with your own sample (approximately 20grams) of hydrated copper sulfate.

Continue with the empirical formula of magnesium oxide.

Home