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In an experiment we conducted, we heated 1.6 grams of copper dust, and got 1.7 grams of oxidised copper, (it probably reacted with 0.1 grams of $\ce{O2}$) and our goal was to find which of the two forms, $\ce{CuO}$ or $\ce{Cu2O}$ we got.

What I did:

\begin{array}{lll} \hline & \ce{Cu} & \ce{O2} \\ \hline m\ \text{(in g)}& 1.61 & 0.1 \\ M\ \text{(in g/mol)}& 63.5 & 32 \\ n\ \text{(in mol)}& 0.025 & 0.003125 \\ \text{ratio}& 8 & 1 \\ \hline \end{array}

And the ratio between copper atoms to oxygen atoms is $4:1$.

Thus it seems the form is $\ce{Cu4O}$.

But where is my mistake?

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Your calculation looks good. Maybe the issue is not with the calculation, but with the experiment. Did you wait long enough so that all the copper is oxidized?

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  • $\begingroup$ And then only half of the material reacted? $8Cu + O_2 \rightarrow 2Cu_2O + 4Cu$? $\endgroup$ – NightRa Oct 21 '13 at 4:35
  • $\begingroup$ Or only 1/4 of the copper reacted, and formed CuO (which is the expected product for spontaneous oxidation of copper upon heating in air) $\endgroup$ – F'x Oct 21 '13 at 8:02
  • $\begingroup$ Whether Cu2O or CuO is formed depends on temperature. Above about 1100°C you get Cu2O. Another problem is the gas around/within your cruzible, depending on how you heat(gas/electic etc) $\endgroup$ – Georg Oct 21 '13 at 14:46
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When I was at university I used to oxidize metal oxides. In my case was difficult to fully oxidize a metal with only one step. This is what I should do.

  1. First oxidation
  2. weigh it
  3. crush with a mortar what you have oxidized
  4. weigh it
  5. second oxidation
  6. weigh it

If the weight 4. and 6. are the same you can do the calculation if not you have to re-oxidize again.

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  • $\begingroup$ Good suggestion on how to check for complete oxidation. That's a better answer than mine! $\endgroup$ – F'x Oct 21 '13 at 13:38

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