Formula of Copper(II) Sulfate Hydrate Lab

Question

Data:

Mass of empty dish: 50.58 g

mass of crystals and dish: 55.63

mass of crystals after first heating: 54.57

mass of crystals after second heating: 53.82

mass of crystals after third heating: 53.74

then i get questions about determining the mass of crystals in the evaporating dish- for all of them.

then i have to determine the mass of the water that evaporated from the crystals during heating. then, i have to use the mass of evaporated water from the crystals after he third heating, determine the percentage by mass of water in the copper hydrate crystals.

and find the accepted percentage by mass of water in the hydrate crystals fir CuSO4.5H2O.

then calculate percent error. help.

Answer 1

To find the mass of crystals in the dish, you MUST subtract the mass of the dish from the mass of the dish and crystals.

Original mass of crystals = 55.63 g - 50.58 g = 5.05 g of crystals originally

After first heating you have 54.57 g - 50.58 g = 3.99 g of crystals

After 2nd heating you have 53.82 g - 50.58 g = 3.24 g of crystals

After 3rd heating you have 53.74 g - 50.58 g = 3.16 g of crystals

Notice that mass of crystals decreases with each successive heating. This is due to the loss of water during the heating of the crystals. Assuming that all of the water is lost by the 3rd heating, we can find the mass of water that was originally present by comparing the final mass of crystals to the original mass of crystals and take the difference. That will be the mass of water that was lost.

original mass of crystals = 5.05 g

final mass of crystals = 3.16 g

mass of water = 5.05 g - 3.16 g = 1.89 g water

% by mass of water = mass water / mass of crystals (x100%) = 1.89 g / 5.05 g (x100) = 37.4% water

For CuSO₄•5H₂O we have a molar mass of 249.7 g/mol

The mass of water = 5 mol H₂O x 18 g/mol = 90 g

% by mass of water = 90 g / 249.7 g (x100%) = 36.0% = accepted % mass of water in the hydrate