So in the first equation, two moles of copper carbonate is needed to produce one mole of cuprous oxide, two mole of carbon dioxide and half a mole of oxygen gas. In the second equation, one mole copper carbonate is needed to produce one mole of cupric oxide and one mole of carbon dioxide.
I will also be using the theory of relative atomic mass (RAM) to find out, which reaction takes place in the lab. The definition of relative atomic mass is the average mass of the naturally occurring isotopes of an atom, taking in to account relative abundance, compared to carbon-12 (which is exactly 12).
The molar mass (the mass of one mole) is the RAM in grams. By calculating the molar mass, and starting with a known mass of copper carbonate, I will be able to know how much gas will be produced in each case and therefore, which reaction takes place in the lab.
Variables
My independent will be the volume of gas produced during the reaction, as this is what I will be measuring, however I have no dependent variable as there is nothing that I am changing.
Calculations
Equation 1: 2CuCO3 (s) → CuO (s) + 2CO2 (g) + 1/2O2 (g)
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Start with a known volume of gas – 24cm3 of CO2
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Moles (for any gas at room temperature and pressure) =
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moles of CO2 = = 0.001 moles
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The ratio of moles of CO2:O2 is 2:½ or 1:¼, therefore if there is 0.001 moles of CO2 there is 0.00025 moles of O2
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Volume in cm3= moles × 24000 = 0.00025 × 24000 = 6cm3
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Volume of gas produced = 6 + 24 = 30cm3
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As 2 moles of CO2 : 2 moles of CuCO3, 0.001 moles of CO2 : 0.001 moles of CuCO3
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Mass of CuCO3 required = moles × Relative Atomic Mass (RAM)
= 0.001 × 123.5
= 0.1235g of copper carbonate
Equation 2: CuCO3 (s) → CuO (s) + CO2 (g)
- Start with the same mass of copper carbonate: 0.1235g
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Moles of CuCO3 = = = 0.001 moles (as in the first equation)
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1 mole of CuCO3 : 1 mole of CO2
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0.001 moles of CuCO3 : 0.001 moles of CO2
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0.001 moles of CO2 gas at r.t.p = volume = moles × 24000 = 0.001 × 24000 =24cm3
Therefore if equation 1 is correct, I should end up with 30cm3 of gas and if equation 2 is correct, I should collect 24cm3 of gas. This is what I will use to ascertain, which equation is correct. However, a limitation of this experiment is that if the experiment is not carried out at room temperature and pressure, then the calculations will be incorrect, therefore I will not collect either of the expected volumes of gas, which would mean I wouldn’t be able to establish, which one is the correct equation.
Apparatus
Procedure
The copper carbonate will be measured out using a scale that measures to 2 decimal places. It will then be placed inside the boiling tube and the apparatus setup as above. The Bunsen burner will then be lit and switched off once all the copper carbonate has thermally decomposed. I will know the reaction has finished when the gas syringe stops moving out. The gas trapped inside the gas syringe will then be allowed to cool to room temperature (so that the amount of gas can be accurately measured) and the volume measured. This will then be repeated until two results have been collected that are within 0.2 cm3 of each other. I will be careful when handling the copper carbonate as it is harmful if swallowed or inhaled, and I will also need to take precautions, because I am using a Bunsen burner. As a result, I will need to tie my hair back, wear safety glasses and wear a lab coat at all times. Also, I will use a heat proof mat and gauze (so that the equipment doesn’t become damaged), and be careful when the Bunsen burner is on, which includes leaving the flame on orange when I am not using it. Furthermore, I will clean my workspace thoroughly afterwards.
Bibliography