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# Thermal Decomposition of copper carbonate

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Introduction

Thermal 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 equation is correct. This can be done by volumetric analysis i.e. calculating the volume of gas produced. This is then compared to the calculated volume of gas produced in each equation and the equation with the nearest volume of gas is correct. This is a thermal decomposition reaction and when an element like copper can form two oxides, which one forms is based on the stability of the compound formed. The more stable the compound, the more likely it is to form. The stability of a compound with respect to its elements can be predicted by the ?Hf (molar heat of formation). This is the energy change when 1 mole of a compound is formed from its elements. If it is exothermic (negative), then the compound is stable with respect to its elements. If it is endothermic (positive), then the compound is unstable with respect to its elements. In general, the lower the value of ?Hf, the greater the energetic stability of the compound with respect to its elements. ...read more.

Middle

For a compound the mass of one mole is the same as the relative formula mass or molecular mass in grams. The molecular or formula mass of a compound is found by adding the relative atomic masses of its constituent elements, as found in the periodic table. Substance Mr/Ar - and hence, mass of 1 mole (g) Copper Carbonate (CuCO3) 123.5 Copper Oxide (Cu2O) 143 Copper Oxide (CuO) 79.5 Carbon Dioxide (CO2) 44 Oxygen (O2) 32 So using the equation of a reaction, it is possible to predict the masses of products that will be made by a given mass of reactants. In this experiment, from the two equations given, it is possible to calculate how much gas would be given off by each. Avogadro's law states that 1 mole of any gas occupies 24dm3 at room temperature and pressure (rtp), so it is possible to calculate the volume of gas given off. The experiment can then be carried out, and the volume of gas produced compared with the predictions for each equation. Whichever equation best predicts the volume given off is therefore shown to be the correct one. Firstly I must deduce the amount of gases produced if 10g of copper carbonate is produced. No. of moles = Mass s Relative Molecular Mass Relative Molecular mass of Copper Carbonate = Relative atomic mass of: Cu (63.5) + C (12) ...read more.

Conclusion

This also guarantees that the results obtained are accurate 4. Light Bunsen burner and place under tube. 5. Wait until the copper carbonate has completely decomposed. This will be indicated by a color change, and the fact that the copper carbonate would have stopped bubbling. Copper carbonate is a green solid. Copper oxide is a jet black solid. Green Black 6. When you are sure the copper carbonate has completely decomposed, measure and note down the volume of gas obtained 7. Repeat stages 1 - 6 three to four times, and then take the average. This reduces the risk of error in your results. Also ensure that all variables capable of influencing the results are kept constant. For example, the same amount of copper carbonate is used in each experiment and using the same apparatus again helps reduce the risk of error Results Table 1 Measurement of boat Measurement of boat and Copper Carbonate Measurement of boat after Final measurement Table 2 Measurement of copper carbonate used g Volume of Gases produced cm� In this experiment if the volume produced is nearer to 242.9 cm� then It will mean that equation 1 is correct. If the volume of gases produced is nearer to 194.4 cm� then equation 2 is correct. I predict that the most probable equation that is correct is equation 2. Previous knowledge may also help to support this prediction as when anything combusts oxygen gas is never given off as a product because the oxygen is needed for the reaction. ...read more.

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