a student heated 3.48 g of a pure sample of CaSO4 2 H2O until all the water of hydration was expelled. what was the mass of the residue?
See the other post.
To find the mass of the residue, we can calculate the mass of water lost and subtract it from the initial mass of the sample.
First, let's calculate the molar mass of CaSO4 · 2H2O:
- The molar mass of Ca is 40.08 g/mol.
- The molar mass of S is 32.07 g/mol.
- The molar mass of O is 16.00 g/mol.
- The molar mass of H2O is (2 x 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol.
The molar mass of CaSO4 · 2H2O is:
(1 x 40.08 g/mol) + (1 x 32.07 g/mol) + (4 x 16.00 g/mol) + (2 x 18.02 g/mol) = 172.17 g/mol.
Now, let's find the number of moles of CaSO4 · 2H2O in the given sample:
Mass of CaSO4 · 2H2O = 3.48 g.
Number of moles = Mass / Molar mass = 3.48 g / 172.17 g/mol = 0.020 mol.
Since the compound has 2 moles of water per mole of the compound, the moles of water lost would be:
0.020 mol × 2 = 0.040 mol.
Now, let's calculate the mass of water lost:
Mass of water lost = Moles of water lost × Molar mass of water = 0.040 mol × 18.02 g/mol = 0.72 g.
Finally, we can find the mass of the residue:
Mass of residue = Initial mass - Mass of water lost = 3.48 g - 0.72 g = 2.76 g.
Therefore, the mass of the residue is 2.76 g.