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Topic: Calculating Avogadro's number  (Read 2387 times)

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Offline thetada

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Calculating Avogadro's number
« on: October 14, 2015, 11:13:10 AM »
Hi everyone,

I'm trying to develop a resource that would allow my students to determine Avogadro's number from first principles, but all is not going to plan. My understanding is that Avogadro's number is the constant that enables a substance's molar mass in grams to equal its relative mass. To that end, I thought I could get my students to divide the relative mass of different elements by the actual mass of a single atom of each respective element. For example, Carbon 12 would be 12/2.0090761x10-23g. The numbers for oxygen, nitrogen, hydrogen and carbon consistently come out at 5x1021 lower than Avogadro's number (5.98x1023). I dimly recall someone mentioning that the actual mass of sub-atomic particles varies depending on the element because of relativistic effects. Is that what's going wrong?

Thanks

Offline mjc123

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Re: Calculating Avogadro's number
« Reply #1 on: October 14, 2015, 11:30:55 AM »
Where did you get that value for carbon-12? Here:
https://en.wikipedia.org/wiki/Atomic_mass
it is given as 1.998467052 x 10−26 kilogram. Were you using an isotopically averaged value for C? NA is defined as the number of 12C atoms in exactly 12g of 12C. For other elements, (apart from 12C) their relative atomic masses are not integers, because of variations in binding energy per nucleon (E = mc2), so you must use both an accurate absolute atomic mass and an accurate relative atomic mass (e.g. 7.016, not 7, for 7Li).

Offline thetada

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Re: Calculating Avogadro's number
« Reply #2 on: October 14, 2015, 11:42:30 AM »
That's interesting. I used values for individual sub-atomic particles and multiplied according to the different element. I was using C-12 because I wanted to use relative isotopic masses. I see now that the relative isotopic mass for a particular element would not be an integer, even though it would probably be a different non-integer value to the relative atomic mass. I'll try it with relative atomic masses and accurate absolute atomic masses.

Thanks!

Offline thetada

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Re: Calculating Avogadro's number
« Reply #3 on: October 14, 2015, 11:50:52 AM »
Perfect, they're all dead on now, except carbon, which for some reason comes out as 6.01x1023

Much obliged!

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