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.
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
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
I believe it comes out to CaSO4 x 2H20
20.9%
2:1 ratio
I think haha
To determine the mass percent of water in the hydrate, we need to calculate the percentage of water lost from the total mass of the hydrate.
1. Calculate the mass of water lost:
Mass of water lost = 0.523 g
2. Calculate the mass of the anhydrous salt remaining:
Mass of anhydrous salt = Total mass of the hydrate - Mass of water lost
= 2.50 g - 0.523 g
= 1.977 g
3. Calculate the mass percent of water in the hydrate:
Mass percent of water = (Mass of water lost / Total mass of the hydrate) * 100
= (0.523 g / 2.50 g) * 100
= 20.92%
The mass percent of water in the hydrate is 20.92%.
Next, let's determine the formula of the hydrate.
4. Calculate the number of moles of the anhydrous salt:
Moles of anhydrous salt = Mass of anhydrous salt / Molar mass of the anhydrous salt
The molar mass of calcium sulfate (CaSO4) is approximately:
Ca: 40.08 g/mol
S: 32.07 g/mol
O: 16.00 g/mol (from 4 oxygen atoms)
Molar mass of CaSO4 = (40.08 g/mol) + (32.07 g/mol) + (4 * 16.00 g/mol)
= 136.14 g/mol
Moles of anhydrous salt = 1.977 g / 136.14 g/mol
≈ 0.0145 mol (rounded to 4 decimal places)
5. Calculate the number of moles of water lost:
Moles of water lost = Mass of water lost / Molar mass of water
The molar mass of water (H2O) is approximately:
H: 1.01 g/mol (from 2 hydrogen atoms)
O: 16.00 g/mol
Molar mass of water = (2 * 1.01 g/mol) + (16.00 g/mol)
= 18.02 g/mol
Moles of water lost = 0.523 g / 18.02 g/mol
≈ 0.0290 mol (rounded to 4 decimal places)
6. Determine the ratio of the anhydrous salt to the water:
Divide the number of moles of the anhydrous salt by the number of moles of water lost.
This will give us the ratio of the anhydrous salt to the water.
Ratio of anhydrous salt to water ≈ Moles of anhydrous salt / Moles of water lost
≈ 0.0145 mol / 0.0290 mol
≈ 0.5
From this ratio, we can deduce that the formula of the hydrate is CaSO4 . 2H2O.
Therefore, the mass percent of water in the hydrate is 20.92%, and the formula of the hydrate is CaSO4 . 2H2O.