Asked by thao nguyen
A 2.50 g sample of a hydrate of calcium sulphate losses 0.523 g of water when heated. Determine the mass percent of water in the hydrate and the formula of the hydrate.
Answers
Answered by
bobpursley
CaSO4.XH2O
find the mole ratio of water to CaSO4.
moleswater=.523/18
molesCaSO4=(2.5-.523)/molmassCaSO4
divide moles water by the moles CaSO4, that is X
find the mole ratio of water to CaSO4.
moleswater=.523/18
molesCaSO4=(2.5-.523)/molmassCaSO4
divide moles water by the moles CaSO4, that is X
Answered by
DrBob222
percent water = (mass H2O/mass sample)*100= (0.523/2.50)*100 = 20.92 which I would round to 20.9%.
To find the formula, take a 100 gram sample which provides
20.9 g H2O
100-20.9 = 79.1 g CaSO4
moles water = 20.9/18 = 1.16
moles CaSO4 = 79.1/136 = 0.582
Now find the ratio of water to 1 atom of the CaSO4. The easy way to do that is to divide 0.582 by 0.582 = 1.0000 CaSO4
Then 1.16/0.582 = ?? H2O
To find the formula, take a 100 gram sample which provides
20.9 g H2O
100-20.9 = 79.1 g CaSO4
moles water = 20.9/18 = 1.16
moles CaSO4 = 79.1/136 = 0.582
Now find the ratio of water to 1 atom of the CaSO4. The easy way to do that is to divide 0.582 by 0.582 = 1.0000 CaSO4
Then 1.16/0.582 = ?? H2O
Answered by
Chris
I believe it comes out to CaSO4 x 2H20
20.9%
2:1 ratio
I think haha
20.9%
2:1 ratio
I think haha