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?

What's the percent H2O in CuSO4.2H2O?

%H2O x 3.48 = ?

To find the mass of the residue, we need to first determine the molar mass of the compound CaSO4 · 2H2O, which is Calcium Sulfate Dihydrate.

The molar mass is calculated by adding up the atomic masses of all the elements in the compound. The atomic masses from the periodic table are as follows:

- Calcium (Ca): 40.08 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol
- Hydrogen (H): 1.01 g/mol

Next, we have to consider that there are two water molecules (H2O) attached to each calcium sulfate molecule (CaSO4). The molar mass of water is 18.02 g/mol.

So, let's calculate the molar mass:

Molar mass of CaSO4 · 2H2O = (1 Ca × 40.08 g/mol) + (1 S × 32.07 g/mol) + (4 O × 16.00 g/mol) + (2 H2O × 18.02 g/mol)

Molar mass of CaSO4 · 2H2O = 40.08 + 32.07 + 64.00 + 36.04

Molar mass of CaSO4 · 2H2O = 172.19 g/mol

Now that we know the molar mass of the compound, we can convert the given mass of 3.48 g to moles.

Number of moles = mass / molar mass

Number of moles = 3.48 g / 172.19 g/mol

Number of moles = 0.0202 mol

Since the water of hydration was expelled, we are left with the anhydrous form of calcium sulfate (CaSO4). This means that our remaining sample is calcium sulfate without the water molecules.

To find the mass of the residue, we can multiply the number of moles of anhydrous calcium sulfate by its molar mass.

Mass = number of moles * molar mass

Mass = 0.0202 mol * 172.19 g/mol

Mass = 3.479 g

Therefore, the mass of the residue after heating is 3.479 g.