Final answer:
The mass of water lost from the hydrated copper(II) sulfate after heating is 4.51 g. This corresponds to approximately 0.250 moles of water. After calculating the moles of anhydrous CuSO4, we find that x, the number of water molecules per formula unit of CuSO4, is 5, resulting in the formula CuSO4·5H2O for the hydrated compound.
Step-by-step explanation:
To find the mass of the water that was lost due to heating, we subtract the mass of the anhydrous copper(II) sulfate from the original mass of the hydrated copper(II) sulfate. Thus:
- Mass of hydrated copper(II) sulfate: 12.50 g
- Mass of anhydrous copper(II) sulfate: 7.99 g
- Mass of water lost = 12.50 g - 7.99 g = 4.51 g
To convert the mass of water to moles, we use the molar mass of water (H2O), which is 18.015 g/mol:
Moles of water = Mass of water / Molar mass of H2O = 4.51 g / 18.015 g/mol ≈ 0.250 moles
Next, we convert the mass of anhydrous CuSO4 to moles. We use the molar mass of CuSO4, which is approximately 159.61 g/mol:
Moles of CuSO4 = Mass of anhydrous CuSO4 / Molar mass of CuSO4 = 7.99 g / 159.61 g/mol ≈ 0.050 moles
Now we can find the value of x, the number of moles of water per mole of anhydrous CuSO4:
x = Moles of water / Moles of anhydrous CuSO4 = 0.250 moles / 0.050 moles = 5
Therefore, the hydrated form of copper(II) sulfate is CuSO4·5H2O, known as copper(II) sulfate pentahydrate.