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When a hydrated salt such as $\ce{CuSO4.5H2O}$ is subjected to heat, an anhydrous salt ($\ce{CuSO4}$) is produced. However the theoretical molar ratio of $\ce{H2O:CuSO4}$ and experimental molar ratio of $\ce{H2O:CuSO4}$ are not the same. Instead there is a percentage error, why does this occur? And how can we minimise this in this experiment? The hydrated salt was left on hot plate at 70 degrees celsius for 25 mins.

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closed as off-topic by andselisk, Todd Minehardt, Jon Custer, Tyberius, Mithoron Mar 6 at 22:41

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    $\begingroup$ Why, all experiments contain errors. If you want to know more, you have to tell us more. What is your experimental molar ratio, to begin with? $\endgroup$ – Ivan Neretin Mar 6 at 11:36
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    $\begingroup$ My experimental molar ratio of H2O: CuSO4 is 4:1 $\endgroup$ – Huda Alnusairi Mar 6 at 11:49
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    $\begingroup$ Then perhaps your experiment resulted in a monohydrate, rather than an anhydrous salt? $\endgroup$ – Ivan Neretin Mar 6 at 12:05
  • $\begingroup$ The hydrate contained 4 moles of water for every 1 mole of CuSO4. The expected was 5 moles of water for every 1 mole of CuSO4. It was not a monohydrate as there was more than 1 mole of water per mole of compound. $\endgroup$ – Huda Alnusairi Mar 6 at 12:12
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    $\begingroup$ Are you saying that maybe the hydrated salt was not heated enough, therefore not all water turned into steam therefore we were left with CuSO4.H2O + 4H2O (gas) instead of the intended CuSO4 + 5H2O (gas) $\endgroup$ – Huda Alnusairi Mar 6 at 12:26
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You can mention experimental details and how you achieved 1:4 ratio, which may explained if we know the maximum temperature of heating. CuSO4.5H2O loses water of hydration in steps as a function of temperature showing that those five water molecules are not equivalent. In this figure from German Wiki, follow the green line (ignore blue). The y-axis is % loss in weight and x-axis is the temperature. It clearly shows that water is lost in steps. The labels read Mass change (loss of crystal water). If your heating temperature is not enough you are left with CuSO4.H2O.

Loss of water

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  • $\begingroup$ Also I would mention that it might be difficult to keep the sample anhydrous both in the • sense and in a more trivial one. As the answer explain this matter is really linked to experimental details not to mention even humidity in the lab. $\endgroup$ – Alchimista Mar 6 at 15:10

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