This is an experiment to find out how much CuO (copper oxide) is left and how much CO2 (carbon dioxide) is given off after heating a certain amount of CuCO3 (copper carbonate).
The table below shows the results: Mass Of Crucible (g)| Mass of CuCO3 added (g)| Mass of crucible and copper oxide (g)| Mass of CuO produced (g)| Mass of CO2 given off (g)| 11. 53| 0. 25| 11. 76| 0. 23| 0. 02| 15. 63| 0. 50| 15. 96| 0. 33| 0. 17| 11. 53| 0. 75| 12. 03| 0. 50| 0. 25| 15. 63| 1. 00| 16. 35| 0. 72| 0. 28| 11. 53| 1. 25| 12. 46| 0. 93| 0. 68|
I used two crucibles alternately so that I can use one while the other is cooling. The third column is measured after the copper carbonate is heated. I filled in the fourth column by subtracting the mass of the crucible itself from the mass of the crucible and the copper oxide after it is heated, leaving only the mass of copper oxide. The mass of carbon dioxide given off can also be found by subtracting the amount of copper oxide produced from the original mass of copper carbonate. After the experiment, I found that the mass of copper carbonate decreases when it is heated.
This happens because, when copper carbonate is heated, it produces carbon dioxide, and as carbon dioxide is a gas, it escapes from the crucible into the surrounding air, therefore causing a decrease in mass. CuCO3 CuO + CO2 If the mass of CuCO3is doubled, the amount of CuO would double as well because the ratio of the equation is 1:1. The table below shows the number of moles of carbon carbonate I started with and the number of moles of copper oxide I ended up with. The RAM of CuCO3 is 124, [64 + 12 + (16 x 3)], and the RAM of CuO is 80, (64+16).
The Essay on Percentage Composition By Mass Of Magnesium Oxide
Abstract: The purpose of this lab was to test the law of definite proportions for the synthesis reaction of combusting magnesium. In this lab, the polished magnesium ribbon was placed in covered crucible and was heated in order for it to react with Oxygen presented in air and in water provided. The result showed that Magnesium oxide formed through chemical reaction was made up of 60.19% magnesium ...
Mass of CuCO3 (g)| Number of Moles of CuCO3 (round to 3 dp. )| Mass of CuO (g)| Number of Moles of CuO (round to 3 dp. )| 0. 25| 0. 002| 0. 23| 0. 003| 0. 50| 0. 004| 0. 33| 0. 004| 0. 75| 0. 006| 0. 50| 0. 006| 1. 00| 0. 008| 0. 72| 0. 009| 1. 25| 0. 010| 0. 93| 0. 012| This table shows the number of moles and the volumes of CO2 (carbon dioxide) produced in each experiment. The RAM of CO2 is 44, [12 + (16 x 2)]. Mass of CO2 given off (g)| Moles of CO2 given off (round off to 4dp. )| Volume of CO2 given off(dm3)| 0. 02| 0. 0005| 0. 0120| 0. 17| 0. 0039| 0. 0936| 0. 25| 0. 057| 0. 1368| 0. 28| 0. 0064| 0. 1536| 0. 32| 0. 0073| 0. 1752| After the experiment and organising my results, I noticed some anomalies. The results for the number of moles of copper oxide was a bit strange, because the number of moles is larger than the number of moles of carbon carbonate we started off with, and this does not really make sense as I expected it to have a one to one ratio and that did not happen. This may have been caused by some errors throughout the experiment such as not heating the crucible of copper carbonate long enough for it to decompose completely.
Also, the results may not be as accurate as it could be because I was using the same two crucibles for every experiment, so the contents of the crucible must be got rid of before it can be used again, but the problem is, it is not easy to get rid of every grain of copper oxide left in the crucible without washing it, meaning that some of it must have been left inside, therefore affecting the results of the following experiments. Lastly, the next time I do this experiment, I should repeat it at least once in order for the results to be more accurate.
