For a school project, I am in a group tasked with calculating Avogadro's Number in multiple manners. Other than the Millikan Oil Drop Experiment, we know of no ways to compute it. A Google search yielded no methods that could be of use (either complex beyond the high school level, or through using a mole of materials, which defeats the point of deriving the number). What methods can we (intelligent high schoolers on a limited budget) use to compute Avogadro's Number?

• You can use Faraday's electrolysis method. Find the amount of electricity in couloumbs required for the electrodeposition of one equivalent of any substance, and then divide the amount by the charge on one electron. This gives you the avogadro's number – stochastic13 Jun 6 '14 at 4:13
• Thanks! Can you please explain in more detail or link me to a website that explains in more detail? – Dave Coffman Jun 7 '14 at 1:09

Placing a drop of oleic acid on the surface of water contained in a pan is a relatively straightforward experiment that has been widely used to determine Avogadro's number. Here are two links that provide good background on a) how to run the experiment, b) how to analyze the data, and 3) the science behind the experiment. The two links are just variations on the same experiment, but if you read them both it will really enhance your understanding of the experiment.

Here is a crude method of performing the experiment:-

1. Take two copper electrodes, weigh them, and dip them in $\ce{CuSO4}$ solution

2. Use a constant current source or a fixed-EMF battery.

3. Connect an ammeter in series with the battery and the electrolytic setup
4. Note down the time of closing the switch. Pass a current for a fixed time and also note down the current magnitude.
5. After say some time $t$ of passing the current $I$, open the switch and remove the cathode and weigh it.
6. Say the increase in mass of the cathode is $m$. The equivalent weight of copper is $63.54/2$, hence the equivalents of copper deposited is $2m/63.5$.
7. The amount of electricity passed is $It$. This must be equal to one $2m/63.5$ equivalents of electrons. Since the charge on one electron is $\approx 1.6\times 10^{-19}=e$, the number of electrons that passed is $It/e%$. Therefore the number of electrons in one equivalent or one mole, (which is the avogadro's number) will be $(It/e)/(2m/63.5)$.

You will have to make appropriate arrangements for accurate electrolysis to ensure correct weight measurement.

• One issue with this approach is that it requires the charge of the electron which is determined by the Millikan Oil Drop Experiment. – Spencer Jul 20 '14 at 4:34