Question

Step 1. Procedure Obtain a weighted amount of the Copper (II) Sulfate hydrate (say 30 g), place in an 100 ml beaker

Step 2. Heat compound with bunsen burner until all of the water is driven-off (you can verify this be weighing the beaker while heating it, when the weight stops declining you’ve removed all the water.

Step 3. Weigh final compound. write your observations

Observations:\

Initial Weight: 30 grams

Final Weight: 19.1773 grams

Weight of Water driven off (Initial weight - final weight): 30g–19.1773g=10.8227grams of water

moles of water loss: Weight of water / GMW H₂O (18.016 )

moles of CuSO4: final weight/ GMW CuSO4 (159.61)

X = moles of water loss/ moles of CuSO4

Given that hydrate is: CuSO₄ (XH₂O)

Formula of hydrate:?????

Answer 1

first calculate % composition of CuSO₄ and H₂O

30 gm of sample = 100 % then 19.1773 gm of CuSO₄  = 19.1773 100 / 30 = 63.92 %

% of CuSO₄ = 63.92%

% of water = 100 - 63.92 = 36.08 %

calculate retio by dividing % composition by molecular weight

for CuSO₄  = 63.92 / 159.61 = 0.4

for H₂O = 36.08 / 18. 016 = 2.0

calculate simplest retion by dividing above retion by smallest retio

for CuSO₄  = 0.4 / 0.4 = 1

for H₂O = 2.0 / 0.4 = 5

thus retio of CuSO₄ and H₂O is 1 : 5

Formula of hydrate is = CuSO₄.5H₂O