if you had obtained a percentage of copper which was much larger than the accepted value, what should you hav expected the appearance of the contents of the crucible to be after heating?
please answer me soon because i have a test
please answer me soon because i have a test
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CuO or Cu2O + C --> Cu + CO2
if CuO, 2CuO + C --> 2Cu + CO2,
1.7g CuO / 79.5g/mole = 0.0214moles CuO
0.15gC / 12g/mole = 0.0125moles C
for 0.0214moles CuO, you would require 0.0107moles C, CuO = limiting reactant
for 0.0214moles CuO, you would yield 0.0214moles Cu
0.0214moles Cu x 63.5g/mole = 1.36g Cu
1.36g / 1.7g x 100% = 80% Cu in CuO
% error = abs(accepted value - experimental value) / accepted value x 100%
abs(79.55 - 84) / 79.55 x 100% = 5.6% error
5.6% error is not very large for a high school chem lab. actually, it is pretty good. what happened is probably not all of the oxygen got driven off by the heat and not all of the C was used to make CO2. remember, the CuO in your reaction was the limiting reactant so there would be excess mass due to the unused C still present in the crucible.
if you spilled the contents of the crucible, you would have less mass of the CuO and C in the crucible. therefore, the mass of the resultant Cu would be less. the % of Cu from CuO would be much smaller than expected.
you take 1.7g CuO, mass it, spill it, don't remass it, heat it, get 1g Cu. your % Cu in CuO would be 1g / 1.7g x 100% = 58.82%
if CuO, 2CuO + C --> 2Cu + CO2,
1.7g CuO / 79.5g/mole = 0.0214moles CuO
0.15gC / 12g/mole = 0.0125moles C
for 0.0214moles CuO, you would require 0.0107moles C, CuO = limiting reactant
for 0.0214moles CuO, you would yield 0.0214moles Cu
0.0214moles Cu x 63.5g/mole = 1.36g Cu
1.36g / 1.7g x 100% = 80% Cu in CuO
% error = abs(accepted value - experimental value) / accepted value x 100%
abs(79.55 - 84) / 79.55 x 100% = 5.6% error
5.6% error is not very large for a high school chem lab. actually, it is pretty good. what happened is probably not all of the oxygen got driven off by the heat and not all of the C was used to make CO2. remember, the CuO in your reaction was the limiting reactant so there would be excess mass due to the unused C still present in the crucible.
if you spilled the contents of the crucible, you would have less mass of the CuO and C in the crucible. therefore, the mass of the resultant Cu would be less. the % of Cu from CuO would be much smaller than expected.
you take 1.7g CuO, mass it, spill it, don't remass it, heat it, get 1g Cu. your % Cu in CuO would be 1g / 1.7g x 100% = 58.82%