AP Chemistry Question (Millikan's Oil-Drop Experiment) charge of electrons?

They gave you a hint to look at the smallest difference between drops. Here are the differences (all x 10^-19 coulombs): 1.64, 3.30, 1.63, 3.35, 1.60, 1.63, 3.18, 3.29. They look to be around 1.6 x 10^-19 coulomb difference (or double that). If I average them out, then I get 1.635 x 10^-19 coulombs as a candidate for smallest charge.

Based on that, I get close to the following for number of electrons on each droplet: 4, 5, 7, 8, 10, 11, 12, 14, 16

I do not think that the charge could be half that, because since the difference is a multiple, then we should expect an odd multiple (of that half-charge) on some of the differences, which we do not. I hope this helps.

Ignore the 10^-19 portion.

Divide through by 6.563.

8.204 / 6.563 = 1.25

11.50 / 6.563 = 1.75

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.

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26.18 / 6.563 = 3.989 = 4

You're looking for the smallest whole-number ratio between the charges.

That should get you started.

Here's a problem like yours:

/question/index?qid=20120...

spend some time and review this problem/solution in Zumdahl's chemistry 9th ed.

http://i57.tinypic.com/2ex5xdy.png

http://i59.tinypic.com/2uj2r85.png

http://i58.tinypic.com/1z5mb74.png

your problem is nearly identical.

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now look at your data set... and yes.. chemteams suggestion of ignoring the exponents is a good one for formatting purposes.. here they are in excel

http://i62.tinypic.com/35iwtmt.png

The basic idea is the drops must contain whole numbers of electrons. And therefore whole number multiples of the charge of 1 electron. In the left two columns (F and G) I assumed the charge on 1 electron was 6.536 (x10^-19) C and divided all the values by that number to see how many electrons were present in each drop. BUT.. I didn't get whole numbers of electrons! So that meant the 6.536 was an integer multiple of the true charge on the electron. So I next assumed that whole number multiple was "2" making the true charge of 1 electron = 6.536 / 2.. I then divided all my values by that number in columns I and J. Notice again, some of the values did not yield whole numbers. Since I'm assuming electrons cannot exist "partially" in any drop.. only as whole numbers.. I then assumed the 6.536 must have been 4x the true value of an electron. And that gave the true value = 6.536 / 4 = 1.634 (x10^-19) C. Dividing all the values by 1.634 gave whole numbers for all the drops

This tells me that the "true charge" on 1 electron is either 1.634x10^-19C.. or some whole number dividend of that value. 1.634x10^-19 / 3.. .or 1.634x10^-19 / 2.. or 1.634x10^-19 / 7... for example.

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so to answer the questions

.. (a) column M of that spreadsheet

.. (b) 1.634x10^-19 C...... 6.536x1^-19 / 4C

.. (c) absolutely.. I have no way of knowing if that 1.634x10^-19 is the "true" value

.. .. ..or an integer multiple of it. The way to find out if that is the case is the way

... ....Millikan did it.. and that is to collect more and more data.

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now.. silly as it sounds.. I have to copy and paste your question here to prevent YA's auto anti-spam software from flagging my answer as spam and hiding it.

AP Chemistry Question (Millikan's Oil-Drop Experiment) charge of electrons? Update : The question is #24 on pg 32 of this link: I've been stuck on this problem for a long time now, and any insight into the process of how to do it would be a lot of help. Thanks!