Where's my experimental error coming from? I’m teaching chemistry to a home-ed 15 year old.  Occasionally we do some practical work, since I believe very strongly that physical science learning should have some real lab work to bring it alive, and also to teach things like observational skills and attention to detail.

My background is a PhD in organic chemistry, so I’m familiar with hazard assessment, and all the lab work I do with my student is something I’ve done myself already.  Any time she touches the apparatus it’s under close supervision by myself.

So the practical involved taking hydrated copper sulphate, heating it to drive off the water, weighing before and after, and thus calculating the number of water molecules of crystallisation, based on the respective molecular weights of the anhydrous salt and water.

I have a porcelain evaporating dish to contain the hydrate during heating, electronic scales that will weigh to 300 g in increments of 10 mg, and commercially bought copper sulphate that is allegedly 99.5 % pure.

I started with the hydrate, heated it in the evap dish over a flame (using a gas cooker ring) until it had all turned greyish white, then allowed it to cool and weighed again.

The results were:

Starting wt hydrate: 13.03 g

Wt after heating: 8.22 g

Hence water lost: 4.81 g

So that’s 4.81/18 = 0.267 mol water, and 
8.22/(63.5 + 32 + 64) = 0.0515 mol anhydrous copper sulphate

Hence 0.267/0.0515 = 5.18 mol water per mol copper sulphate.

So my question: why not 5.0?

The electronic scales (quite cheap) were an obvious first candidate for a source of error.  I’ve compared its readings at weights of about 2, about 20, and about 50 g, with another similar model, and they agree to within 10 mg (one digit in the final decimal place) at all weights.

I could perhaps have not heated the hydrate enough, and not driven off all the water – but then I’d have less than 5, not more, waters per mol.

Or the hydrate, as supplied, could be a little wet.  I haven’t investigated that yet, but I plan to weigh a sample before and after leaving in a warm airing cupboard overnight.

If you’ve read this far, I’m already grateful, but I’d be even more so, if anyone could suggest where the difference between 5 and 5.18 is coming from.

I did it a second time, and got 5.16.

The difference between 5.00 and 5.16 is a weight error of about 200 mg, and I was really careful, so this seems unlikely.

The weight after cooling of the evap dish is constant.

Any ideas?

Edit (2):


There was no sign of $\ce{CuO}$ after the heating.  No trace of black, and indeed we dissolved the anhydrous salt in water (for growing some seed crystals later), and there was no insoluble residue.
I’ve now heated the dish over a hot blue natural gas flame for ten minutes.  Its weight before cooling was 0.02 g (on my scales) less than its empty weight cold.


Edit (3)

I weighed out some of the hydrate into another dry dish, flattened it out as much as possible within the confines of the dish to expose as much surface area as possible, and left it in the airing cupboard for 24 h at 26 degrees Celsius.  Then I re-weighed.

Pre drying wt: 42.68 g

Post drying wt: 41.98 g

Assumed water content: 42.68 − 41.98 = 0.7 g (!!!)

So assuming that 26 degrees is not warm enough to cause partial loss of water of crystallisation, that suggests that at best my hydrate is only 98.36 % pure, not 99.5 as it says on the bottle (typo corrected above).

So repeating my calculation above:

Wt before: 13.03 × 0.9836 = 12.82 g hydrate
Wt after: 8.22 g
Water lost: 4.60 g

Moles water: 4.60/18 = 0.256

Moles $\ce{CuSO4}$: 8.22/159/5 = 0.052

Ratio: 4.96

I’m much happier with that!

Many thanks.
I would go with a full fledged error propagation analysis on this one, because without systematic approach you might be guessing forever. So if I look at the calculation process you have the following equation to get from known quantities to the answer:

$$X=\left(\frac{W_0}{W_e}-1\right)\left(\frac{M_{CuSO_4}}{M_{H_2O}}\right) \tag{1}$$

where $W_0$ is the start weight, $W_e$ the end weight and $M_i$ is the respective molecular mass. $X$ is the desired answer.

To find the expected error in the answer, $\Delta X$, assuming no error in the molecular masses, we calculate:
$$\Delta X=\sqrt{\left(\frac{\partial X}{\partial W_0}\Delta W_0\right)^2+\left(\frac{\partial X}{\partial W_e}\Delta W_e\right)^2} \tag{2}$$
which works out to:
$$\Delta X=\frac{\sqrt{(\Delta W_0)^2 W_e^2+(\Delta W_e)^2 W_0^2}}{W_e^2}\left(\frac{M_{CuSO_4}}{M_{H_2O}}\right)  \tag{3}$$

So now the final thing left to do is estimate how big $\Delta W_0$ and $\Delta W_e$ are. For $\Delta W_0$ I think it is safe to say that this is only the inaccuracy of the scale, so perhaps $0.01$ g. For $\Delta W_e$ I would go for the worst case scenario: that all of the 0.6% impurity in the original copper sulfate has evaporated, meaning that the weight will be off by 0.6% times $W_0$ i.e. $0.08$ g.

Now plugging in the numbers in Eq.$(3)$ I find $\Delta X=0.14$ mol water per mol copper sulphate. Which is very close to the actual error you find. 

This error is almost completely determined by the error in $W_e$. If, like @brinnb suggests, an additional reduction in $W_e$ could be expected due to adhered water say that the error in $W_e$ is $0.1$ g you already get to $\Delta X=0.17$ mol water per mol copper sulphate.

Conclusion

The error you find falls within the worst-case expected error due to impurity and potentially adhered water.