Was the mass lost equal to the mass gained? From the mass of copper lost, calcul
ID: 524773 • Letter: W
Question
Was the mass lost equal to the mass gained? From the mass of copper lost, calculate the moles of copper lost. From the total Coulombs used in the experiment and the moles of copper lost, calculate the charge in Coulombs required to produce one mole of copper(II) ions. The formation of copper(II) ions in (3) requires two moles of electrons which would have a value equal to two Faraday (2F). What is the value of one F or charge in Coulombs per mole? The accepted value Faraday (to 3 significant figures) is 96500 C/mol. Calculate the percent error in your determination of Faraday's constant. Avogadro's number can also be calculated from this experiment. (a) Calculate the number of electrons that were lost from the total Coulombs and the fact that the charge on an electron is 1.602 times 10^-19 Coulombs. (b) Two electrons are lost for each atom of Cu, so the number of Cu atoms is _____. (c) This number of atoms corresponds to the measured mass of Cu lost. If 63.54 grams of copper (one mole) is lost, how many atoms of Cu would be lost? (d) The accepted value for Avogadro's number is 6.02 times 10^23. Calculate the percent error in your determination of Avogadro's number.Explanation / Answer
2) Molar mass of copper = 63.543 g mol-1.
Mass of copper lost = 0.043 g.
Therefore, moles of copper lost = (0.043 g)/(6.543 g mol-1) = 6.767*10-4 mole 6.77*10-4 mole (ans).
3) 6.77*10-4 mole copper uses 7.765*10 Coulombs of electricity.
Therefore, 1 mole of copper will use (7.765*10 Coulombs)*(1 mole/6.77*10-4 mole) = 114697.2 Coulombs of electricity (ans).
4) The reduction reaction is
Cu2+ (aq) + 2 e- ------> Cu (s)
2 moles of electrons are used per mole of copper (II) reduced and that corresponds to 2 F of electricity.
Therefore, 2 F = 114697.2 Coulombs/mol (combining 3 and 4)
Thus, 1 F = (114697.2 Coulombs/mol)*(1 F/2 F) = 57348.6 Coulombs/mol (ans).
5) The accepted value of 1 F = 96500 C/mol; the value determined experimentally = 57348.6 C/mol.
Percent error in the determination = [(96500 – 56348.6) C/mol]/(96500 C/mol)*100 = 41.6076 41.61 (ans).
6a) Charge of 1 electron = 1.602*10-19 Coulombs.
Total Coulombs of electricity used = 7.765*10 Coulombs.
Therefore, number of electrons lost = (7.765*10 Coulombs)/(1.602*10-19 Coulombs/e) = 4.847*1020 4.85*1020 (ans).
6b) 1 atom of copper loses 2 electrons.
Therefore, number of copper atoms = (4.85*1020 electrons)*(1 Cu atom/2 electrons) = 2.425*1020 (ans).
6c) 2.425*1020 atoms of copper = 0.043 g copper
Therefore, 1 mole of copper = 65.543 g copper = (2.425*1020 atoms)*(63.543 g/0.043 g) = 3.583*1023 atoms of copper (ans).
6d) The accepted value of Avogadro’s number = 6.02*1023; the value determined = 3.583*1023.
Percent error = (6.02*1023 – 3.583*1023)/(6.02*1023)*100 = 40.48 (ans).
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