Experimental Determination of Avogadro's Number

Electrochemical Method to Measure Avogadro's Number

Portrait of Amedeo Carlo Avogadro (Turin, 1776-1856), Count of Quaregna and Cerreto, Italian chemist and physicist, Engraving
Amedeo Carlo Avogadro. DEA / CHOMON / Getty Images

Avogadro's number isn't a mathematically derived unit. The number of particles in a mole of a material is determined experimentally. This method uses electrochemistry to make the determination. You may wish to review the working of electrochemical cells before attempting this experiment.


The objective is to make an experimental measurement of Avogadro's number.


A mole can be defined as the gram formula mass of a substance or the atomic mass of an element in grams.

In this experiment, electron flow (amperage or current) and time are measured in order to obtain the number of electrons passing through the electrochemical cell. The number of atoms in a weighed sample is related to electron flow to calculate Avogadro's number.

In this electrolytic cell, both electrodes are copper and the electrolyte is 0.5 M H2SO4. During electrolysis, the copper electrode (anode) connected to the positive pin of the power supply loses mass as the copper atoms are converted to copper ions. The loss of mass may be visible as pitting of the surface of the metal electrode. Also, the copper ions pass into the water solution and tint it blue. At the other electrode (cathode), hydrogen gas is liberated at the surface through the reduction of hydrogen ions in the aqueous sulfuric acid solution. The reaction is:
2 H+(aq) + 2 electrons -> H2(g)
This experiment is based on the mass loss of the copper anode, but it is also possible to collect the hydrogen gas that is evolved and use it to calculate Avogadro's number.


  • Direct current source (battery or power supply)
  • Insulated wires and possibly alligator clips to connect the cells
  • 2 Electrodes (e.g., strips of copper, nickel, zinc, or iron)
  • 250-ml beaker of 0.5 M H2SO4 (sulfuric acid)
  • Water
  • Alcohol (e.g., methanol or isopropyl alcohol)
  • Small beaker of 6 M HNO3 (nitric acid)
  • Ammeter or multimeter
  • Stopwatch
  • Analytical balance capable of measuring to nearest 0.0001 gram


Obtain two copper electrodes. Clean the electrode to be used as the anode by immersing it in 6 M HNO3 in a fume hood for 2-3 seconds. Remove the electrode promptly or the acid will destroy it. Do not touch the electrode with your fingers. Rinse the electrode with clean tap water. Next, dip the electrode into a beaker of alcohol. Place the electrode onto a paper towel. When the electrode is dry, weigh it on an analytical balance to the nearest 0.0001 gram.

The apparatus looks superficially like this diagram of an electrolytic cell except that you are using two beakers connected by an ammeter rather than having the electrodes together in a solution. Take beaker with 0.5 M H2SO4 (corrosive!) and place an electrode in each beaker. Before making any connections be sure the power supply is off and unplugged (or connect the battery last). The power supply is connected to the ammeter in series with the electrodes. The positive pole of the power supply is connected to the anode. The negative pin of the ammeter is connected to the anode (or place the pin in the solution if you are concerned about the change in mass from an alligator clip scratching the copper).

The cathode is connected to the positive pin of the ammeter. Finally, the cathode of the electrolytic cell is connected to the negative post of the battery or power supply. Remember, the mass of the anode will begin to change as soon as you turn the power on, so have your stopwatch ready!

You need accurate current and time measurements. The amperage should be recorded at one minute (60 sec) intervals. Be aware that the amperage may vary over the course of the experiment due to changes in the electrolyte solution, temperature, and position of the electrodes. The amperage used in the calculation should be an average of all readings. Allow the current to flow for a minimum of 1020 seconds (17.00 minutes). Measure the time to the nearest second or fraction of a second. After 1020 seconds (or longer) turn off the power supply record the last amperage value and the time.

Now you retrieve the anode from the cell, dry it as before by immersing it in alcohol and allowing it to dry on a paper towel, and weigh it. If you wipe the anode you will remove copper from the surface and invalidate your work!

If you can, repeat the experiment using the same electrodes.

Sample Calculation

The following measurements were made:

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

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
  1. 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!
  2. 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 %