Chemistry Honors Lab: Composition of a Copper Sulfate Hydrate

Lab: Composition of a Copper Sulfate Hydrate

Copper Sulfate Prior to Heating:

Copper Sulfate After Heating:

Amount of Hydrate Used:

46.09 g (Mass of Dish + Hydrate) - 45.21 g (Mass of Dish) = 0.88 g (Copper Sulfate Hydrate)

Amount of Water Lost:

45.67 g - 45.21 g = .46 g CuSO₄ (Copper Sulfate Anhydrate)

0.88 g (Copper Sulfate Hydrate) - 0.46 g (Copper Sulfate Anhydrate) = 0.42 g (Water)

Percentage of Water in Hydrate:

0.42 g (Water) / 0.88 g (Copper Sulfate Hydrate) * 100 = 48% (Water)

Percent of Error:
(Actual percentage of water in CuSO_4 Hydrate: 36%)

((0.48 - 0.36) / 0.36) * 100 = 33% of Error



Predicted Empirical Formula:

a) Moles of Water Evaporated:

0.42 g (Water) * (1 mole / 18.01 g (Water)) = 0.023 mol (Water Evaporated)

b) Moles of Copper Sulfate Anhydrate Remaining:

0.46 g (Copper Sulfate Anhydrate) * (1 mole / 159.6 g (Copper Sulfate Anhydrate) = 0.0029 mol (Copper Sulfate Anhydrate

c) Ratio of Copper Sulfate to Water:

0.0029 g / 0.023 g = 1 (Copper Sulfate) : 8 (Water)

d) Empirical Formula:

1 CuSO₄ * 8 H₂O


Due to my percent of error being significantly higher than anticipated, I believe that my prediction of 8 water molecules for every CuSO₄ molecule is way to high.  After doing the math (see below), I have determined that, in actuality, there are supposed to be 5 water molecules for every CuSO₄ molecule.

Math to prove above statement:

Step 1) Percent to Grams

36% = 36 grams of water
64% = 64 grams of Copper Sulfate

Step 2) Grams to Moles

36 g * (1 mole / 18.01 g (water)) = approx 2 moles

64 g * (1 mole / 159.6 g (CuSO₄)) = approx. 0.4 moles

Step 3) Divide by Smallest

2 moles of water / 0.4 = 5 moles of water
0.4 moles of copper sulfate / 0.4 moles = 1 mole of copper sulfate

5 water molecules for every 1 copper sulfate molecule

Correct empirical formula for the hydrate: 1 CuSO₄ * 5 H₂O