When separated from a water solution,
CuSO4 forms a hydrate with 5 water
molecules per formula unit. If 37 g of anhydrous CuSO4 are dissolved in water, what
mass of the hydrate could be recovered from
the solution?
mm = molar mass
37 g CuSO4 x (mm CuSO4.5H2O/mm CuSO4) = ? g CuSO4.5H2O that will be formed.
To calculate the mass of the hydrate that could be recovered from the solution, we need to find the mass of water in the hydrate. The molar mass of anhydrous CuSO4 is 159.61 g/mol, and each formula unit of anhydrous CuSO4 is associated with 5 water molecules.
1. Determine the number of moles of anhydrous CuSO4:
- Convert the given mass of anhydrous CuSO4 to moles by dividing it by the molar mass:
Moles of anhydrous CuSO4 = mass / molar mass = 37 g / 159.61 g/mol
2. Calculate the number of moles of water:
Since 1 formula unit of anhydrous CuSO4 is associated with 5 water molecules, the number of moles of water will be 5 times the number of moles of anhydrous CuSO4.
Moles of water = Moles of anhydrous CuSO4 × 5
3. Calculate the mass of water:
- Convert the moles of water to grams by multiplying it by the molar mass of water (18.015 g/mol):
Mass of water = Moles of water × molar mass of water = Moles of water × 18.015 g/mol
4. Calculate the mass of the hydrate:
The mass of the hydrate is the sum of the mass of anhydrous CuSO4 and the mass of water.
Mass of hydrate = Mass of anhydrous CuSO4 + Mass of water
Now let's plug in the values and calculate the result:
- Mass of anhydrous CuSO4 = 37 g
- Molar mass of anhydrous CuSO4 = 159.61 g/mol
- Molar mass of water = 18.015 g/mol
1. Moles of anhydrous CuSO4 = 37 g / 159.61 g/mol = 0.2319 mol
2. Moles of water = 0.2319 mol × 5 = 1.1595 mol
3. Mass of water = 1.1595 mol × 18.015 g/mol = 20.864 g
4. Mass of hydrate = 37 g + 20.864 g = 57.864 g
Therefore, the mass of the hydrate that could be recovered from the solution is approximately 57.864 grams.