Does copper (II) sulfate tetrahydrate exist

we did a lab and the ratio of water molecules to copper (II) sulfate molecules was approximately 3.9: 1. However, why would this happen as the most common hydrate seems to be copper (II) sulfate pentahydrate not tetrahydrate. By the way, we used 2 grams of copper (II) sulfate * xH20 to start, and ended up with like 1.4 grams.

Could this be becaues the number of water molecules to copper (II) sulfate ions is not precise and more like an average, and we happened to get the odd one out?

Very interesting and a triumph for your technique! Four of the H2O molecules in CuSO4.5H2O are bound to the Cu^2+ ion, the fifth is H-bonded to SO4^- ions. Here is what happens when you heat CuSO4.5H2O: "On being warmed the pentahydrate looses water to give first the trihydrate, then the monohydrate; above about 200 °C the virtually white anhydrous sulfate is obtained and then this then forms CuO by loss of SO3 above about 700 °C."[1]
So what you have determined is
CuSO4.5H2O →Δ→ CuSO4.H2O + 4H2O
Presumably it is the H2O not bound to the Cu^2+ that is more tightly bound.
How to avoid this: use slightly higher T (you won't get near 700 °C with a Bunsen burner) and heat to a constant weight. I hope you are using a balance that measures to at least two decimal places (three would be nice).
[1]N. N. Greenwood, A. Earnshaw, Chemistry of the Elements 2nd ed. (1997) p 1190.

So on mild heating (<200°C; check T on apparatus?) only 4 H2O will be lost and the 3.9 maybe is because you assumed final product was CuSO4 not CuSO4.H2O (redo calc). Or maybe there was still some of the 4H2O's left. Hence reheat-cool-weigh, repeat until const weight. (You can email me.)

Report Abuse

The only hydrate is the pentahydrate. It's possible you did not heat enough to drive off all the water.