[davon806]

Aim- To determine the percentage content of water (H2O) in hydrated copper(II) sulphate crystals (CuSO4 . 5H2O).
 Equipment- Approximately 1 gram of copper sulphate crystals, evaporating dish, tongs, Bunsen burner, gauze, tripod, spatula, weighing balance.
 Method- a) Set up the Bunsen burner, gauze and tripod.
 c) Weigh the evaporating dish and approximately 1 gram of copper sulphate crystals.
 d) Gently heat the copper sulphate crystals in the evaporating dish, until they appear to be white (If heated too much, the crystals will break down further and turn yellow.) e) Take the evaporating dish away from the heat with the tongs, and weigh the evaporating dish containing the now white crystals.
 f) To determine the percentage water, you subtract the weight of the crucible from the final weight, and then you subtract the weight of the substance left from the initial weight of the copper sulphate crystals.
 
Please see the bold words.I wonder why yellow solid is formed.I have no idea coz I don't think any yellow substance will be formed among Cu,S,O and H.At first I thought it is because sulphur is formed but CuO is black in colour and Cu(OH)2 is pale blue and so they would not give a yellow appearance.Can anyone help me?Thx.

[Arkcon]

If you overheat the solid, it just seems to char.  Like you said, it doesn't contain anything the should "char" or turn yellow.  Maybe a trace of the copper sulfate decomposes to CuO, add a tiny trace of black in a mass of white looks yellow, then brown, before it turns black?