Decomposition of Copper Carbonate on Heating

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Decomposition of Copper Carbonate on Heating


Copper can form two oxides, Cu2O and CuO. I am to devise and perform an experiment to determine which of these two oxides is formed when Copper Carbonate is thermally decomposed. Because Copper can form two possible oxides, there are two possible equations for the reaction:

2 CuCO3 (s)          Cu2O (s)   +   2CO2  (g)   +   ½ O2 (g)


                      CuCO3 (s)        →    CuO (s)   +   CO2  (g)


I know that one mole of gas will occupy 24 dm3 at room temperature and pressure (r.t.p.). In the first equation, 1.25 moles (1 mole of CO2 and ¼ of a mole of O2) of gas are produced for every mole of CuCO3 that reacts, however, in the second equation only 1 mole of gas (CO2) is produced. Thus by measuring the volume of gas given off for a specific mass of Copper Carbonate, I can determine which reaction is taking place. If 1 mole of Copper Carbonate were reacted, then the reactions would produce 30 dm3 and 24 dm3 respectively. However, using these quantities would be impractical. One mole of Copper Carbonate would weigh 123 grams, and would most likely not fit inside the conical flasks we have available. The gas syringes available are also too small, with a maximum capacity of 100 cm3 or 0.1 dm3.

I will aim to collect a maximum of 80cm3 of gas, to allow some error before the maximum capacity of the gas syringe is reached. Due to the fact that the first reaction will produce a greater volume of gas, and the figure being taken into consideration is the maximum volume produced, the first reaction is what shall be used to calculate sensible quantities for the reactant.

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80 cm3  = 0.08 dm3

Volume / Volume Of One Mole = Number of Moles1

0.08/24 = 0.0033

In the first reaction, there is a mole ratio of 1:1.25 for Copper Carbonate: Gas.

Moles Of Gas / Moles of Gas Produced per Mole of CuCO3 = Moles of CuCO3

  • 0.0033 / 1.25 = 0.0026

No. Of Moles * Relative Atomic Mass = Mass

  • 0.0026  * 123 = 0.328

Thus, to obtain 80 cm3 of I would need to use roughly 0.328 grams of Copper Carbonate, assuming the first equation to be correct. Following is ...

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