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Decomposition of copper carbonate - proving one of two equations.

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Introduction

AS Chemistry Coursework - Rosalind Brock Winter 2002/3 Decomposition of copper carbonate Aim Copper has two oxides, Cu2O, and CuO. Copper carbonate, CuCO3 decomposes on heating to form one of these oxides and an equation can be written for each possible reaction Equation 1: 2CuCO3 (s) ? Cu2O (s) + 2CO2 (g) + 1/2 O2 (g) Equation 2: CuCO3 (s) ? CuO (s) + CO2 (g) The aim of this experiment is to prove which of these two equations is correct. Background Theory It is possible to determine which equation is correct by measuring the volume of gas given off by the decomposition. This is volumetric analysis. The equation is written in moles. 1 mole of any substance contains the same number of particles as 12g of carbon-12. 1 mole of any element contains 6.01 x 1023 atoms. 1 mole of a molecular compound contains 6.01 x 1023 molecules. This means that in a reaction in which 2 molecules of one substance react with 1 molecule of another - for instance the formation of water: 2H2 + O2 ? 2H2O, 2 moles of hydrogen molecules will react with 1 mole of oxygen molecules to give 2 moles of water molecules. For an element, the mass of 1 mole is the same as the atomic mass in grams. ...read more.

Middle

To improve the reliability of the experiment, it will be necessary to repeat the reaction several times and take an average, excluding anomalous results (those not within 10% of the other results). One problem with the pilot method was ensuring that all the gas from the delivery tube was caught by the narrow opening of the gas burette. To overcome this, I will place an inverted funnel under the mouth of the gas burette to funnel the gas into it. Some excess gas was given out because of the expansion of the air in the test tube when heated. To measure the volume of gas caused by this is difficult - it would be possible using the gas laws, but this requires measuring the temperature inside the boiling tube, which is difficult using standard laboratory equipment. Instead a "control" experiment could be set up, in which no copper carbonate is used, but an empty boiling tube is heated for the same length of time, and the volume of the gas collected due to expansion of the air measured. This volume could then be subtracted from the volume obtained by the decomposition of the copper carbonate to give a more accurate result for the volume of gas given off. Hypothesis I predict that the CuO compound will be formed, because this is in line with the pilot results, and would be supported by the background theory. ...read more.

Conclusion

2. Set up the apparatus as shown in the diagram, and place the boiling tube into the clamp stand, ensuring that the bung is firmly in place and the clamp is holding the boiling tube as close to the bung end as possible to avoid setting the cork on the clamp alight with the bunsen burner 3. Take the initial reading from the gas burette, and ensure that the delivery tube is pointing straight up the funnel into the burette. 4. Heat the copper carbonate vigorously until all the green powder has turned black, and no more gas is being produced. 5. Stop heating, and immediately remove the delivery tube from the trough by lifting the clamp stand holding the boiling tube, and moving everything clear of the water to prevent suck back. 6. Take the final reading from the gas burette, and calculate the total volume of gas given off. Theoretical volumes of gas given off by 0.16g of copper carbonate using each equation Equation 1: 2CuCO3 (s) ? Cu2O (s) + 2CO2 (g) + 1/2 O2 (g) Mr of CuCO3 is 123.5 1 mole of any gas occupies 24dm3 (rtp) 2 moles CuCO3 ? 2.5 moles of gas, ie 60dm3 247g CuCO3 ? 60 000cm3 0.16g CuCO3 ? 38.8663... = 38.87cm3 (2dp) Equation 2: CuCO3 (s) � CuO (s) + CO2 (g) 1 mole CuCO3 � 1 mole gas, ie 24dm3 123.5g CuCO3 � 24000cm3 0.16g CuCO3 � 31.0931... = 31.09cm3 (2dp) ...read more.

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