Some copper (ll) sulphate pentahydrate (CuSO4 5H2O), was heated at 120 oC with the following results. Wt of crucible = 10.00 g; Wt of crucible + CuSO4 5H2O =14.98g; Wt of crucible + residue = 13.54 g. How many molecules of water of crystalization were lost?
[H = 1, Cu = 63.5, O = 16, S = 32]
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Explanation
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Discussions (8)
The answer is correct, but the explanation is wrong.
After calculating the weight of water lost, convert it to mole. The same thing for the weight of residue, then you express their moles in a ratio.
Weight of water lost = 1.44g
1 mol of water = 18g
x mol of water = 1.44g
x = 1.44/18 = 0.08 mol
Weight of residue = 3.54g
1 mol of CuSO4 = 162g
x mol = 3.54g
x = 3.54/162 = 0.022 mol
In molar ratio, 0.08/0.022 ≈ 4 mol
Therefore, 4 molecules of water is lost
The 1.44 is in grams not moles. Why then will you approximate the 1.44g to 1g and subtract it from 5 moles of the pentahydrate?
This doesnt make sense
molar mass of dehydrated mass of dehydrated
-------------------------------------- = -------------------------------
molar mass of hydrated molar mass of hydrated
159.5 + 18x 3.54
---------------- = ------
249.5 4.98
x=3.9, approximately 4
