Tuesday, July 30, 2019
Find out the compounds that would get formed when heating copper carbonate
The colour of CuO and Cu2O are black and red respectively. Heating copper carbonate strongly will produce copper (ll) oxide and carbon dioxide that will be given off so basically the equation that results from this is: CuCO3 (s) ? CuO (s) + CO2 (g). By heating for about 3g of the green powder of copper carbonate, I should obtain a new compound with the black colour proving the presence of copper (ll) oxide. The volume of the carbon dioxide that will result from heating copper carbonate depends on the mass of copper carbonate. Actually, it is proportional to it: the bigger the mass of copper carbonate the bigger the volume of gas given off and the bigger the mass of the product formed. The time of heating is very important as well because the copper carbonate isn't completely burnt, it will affect the quantity of gas and the mass of the compound formed. In the preliminary experiment, I just identified which compound that was formed knowing the colours. Using the same apparatus as in the proper experiment, I heated 1.00gram of a green powder of copper carbonate and obtained 0.30gram of copper (ll) oxide. That experiment was limited in the fact that I couldn't measure directly the volume of gas that was given of in the reaction and, considering the accuracy of the chemical balance used, that mass used was small providing an error of ? 1% in the mass of copper carbonate. So, to improve this I used a much bigger mass in the proper experiment for the accuracy of the balance couldn't be improved. * Crucible and lid. * Pipe clay triangle. * Tripod. * Heatproof mat. * Bunsen burner. * Tongs * Chemical balance. * Green powder of copper carbonate. * Bell jar. (Eye protection required: WEAR SAFETY GOGGLES ?TAKE CARE TO AVOID BURNS. WEIGH (to the nearest 0.01g) EVERYTHING TWICE AT LEAST TO AVOID ERRORS. 1. Set the tripod, Bunsen burner (switched off), heatproof mat and pipe clay triangle as above. 2. Weigh the crucible and lid and record the measurement. 3. Letting the crucible on the balance, add the powder of copper carbonate for a little more than 3.00g. 4. Put the lid back and record the measurement. 5. Place the set onto the pipe clay triangle. 6. Switch the Bunsen burner on and heat the crucible strongly. 7. Using the tongs, lift the lid slightly from time to time to check whether the colour of the copper carbonate has completely changed or not. 8. When the colour has changed totally (after about 10 minutes), switch the Bunsen burner off and remove the crucible and lid using tongs form the pipe clay triangle. 9. Allow it to cool into a Bell jar. 10. Re-weigh the crucible and lid and copper (ll) oxide formed in and record the measurement. 11. Range the apparatus back. Mass of crucible + lid = 17.86g Mass of crucible + lid + copper carbonate = 21.58g Mass of copper carbonate = 3.72g Mass of crucible + lid + copper oxide formed = 20.45g CuCO3 (s) ? CuO (s) + CO2 (g). n CuCO3 (s) = n CO2 (g). M CuCO3 (s) = V CO2 (g). Mr Vm Mr = 63.5 + 12 + 3 x 16 = 123.5 gmol-1 M CuCO3 (s) = 3.72 g Vm = 24 dm-3 3.72 g = V CO2 (g). 123.5 gmol-1 24 dm-3 So V CO2 (g) = 0.723 dm-3 M CuO (s) = Mass of (crucible + lid + copper oxide formed) ââ¬â Mass of (crucible + lid) so M CuO (s) = 2.59g. If my method and my results are right then the volume of CO2 given up was 0.723 dm-3 and the mass of CuO obtain was 2.59g. This method could only enable us to calculate the volume. The total uncertainties in that volume is the same of one of the mass of copper oxide formed for they depend quantitatively to the mass of copper carbonate used. The chemical balance was accurate to 0.01g. That error is [?(0.01/3.72) x 100] ? 0.27% then the order of proportionality of the results are: V CO2 (g) = (0.723 ? 0.0027) dm-3 and M CuO (s) = (2.59 ? 0.0027) g. If I had to repeat this experiment, I would use a gas inch well greased (to enable the pressure of gas to push it) by which I can just measure the volume of gas directly using a similar mass.
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