Experimental Determination of Avogadro's Number

Anode mass lost: 0.3554 grams (g)
Current(average): 0.601 amperes (amp)
Time of electrolysis: 1802 seconds (s)

Remember:
one ampere = 1 coulomb/second or one amp.s = 1 coul
charge of one electron is 1.602 x 10^-19 coulomb

1. Find the total charge passed through the circuit.
   (0.601 amp)(1 coul/1amp-s)(1802 s) = 1083 coul

2. Calculate the number of electrons in the electrolysis.
   (1083 coul)(1 electron/1.6022 x 10¹⁹coul) = 6.759 x 10²¹ electrons

3. Determine the number of copper atoms lost from the anode.
   The electrolysis process consumes two electrons per copper ion formed. Thus, the number of copper (II) ions formed is half the number of electrons.
   Number of Cu²⁺ ions = ½ number of electrons measured
   Number of Cu²⁺ ions = (6.752 x 10²¹ electrons)(1 Cu²⁺ / 2 electrons)
   Number of Cu²⁺ ions = 3.380 x 10²¹ Cu²⁺ ions

4. Calculate the number of copper ions per gram of copper from the number of copper ions above and the mass of copper ions produced.
   The mass of the copper ions produced is equal to the mass loss of the anode. (The mass of the electrons is so small as to be negligible, so the mass of the copper (II) ions is the same as the mass of copper atoms.)
   mass loss of electrode = mass of Cu²⁺ ions = 0.3554 g
   3.380 x 10²¹ Cu²⁺ ions / 0.3544g = 9.510 x 10²¹ Cu²⁺ ions/g = 9.510 x 10²¹ Cu atoms/g

5. Calculate the number of copper atoms in a mole of copper, 63.546 grams.
   Cu atoms/mole of Cu = (9.510 x 10²¹ copper atoms/g copper)(63.546 g/mole copper)
   Cu atoms/mole of Cu = 6.040 x 10²³ copper atoms/mole of copper
   This is the student's measured value of Avogaro's number!

6. Calculate percent error.
   Absolute error: |6.02 x 10²³ - 6.04 x 10²³ | = 2 x 10²¹
   Percent error: (2 x 10 21 / 6.02 x 10 23)(100) = 0.3 %

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