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Author Topic: How much would 1 H atom, 2He atoms, 4 Li atoms, 8 Be atoms... weigh?  (Read 1810 times)

Offline chiralSPO

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Disclaimer: this is a question I have already worked out (it's not very hard to do in excel, or to work out a rough answer in your head). I just thought it was an interesting thought experiment...

If you had a vessel that contained 1 H atom, 2 He atoms, 4 Li atoms, 8 Be atoms, 16 B atoms etc. where the number of atoms in the vessel is determined as 2^(atomic # minus one). What is the total mass of the contents of the vessel for all elements up to Uranium (atomic # 92)?

Hint #1 For those doing back-of the envelope calculations: Avagadro's # is almost exactly 279 (less than 0.5% error)

Hint #2 Also for envelope (or head) calculations: recall that 41726e33375eff39b99128e4c4a277d7.gif = 2i + 1 1

Hint #3 There will be less than 1 g of Technetium, so no need to worry about whether it's part of the sample or not.


 

Offline Colin2B

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I like it, a science take on the 'penny on a chess board' question.
Mmmm, in this paperless society where do I find an envelope?
 

Offline chiralSPO

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Thanks Colin.


Follow-up: what would it be worth?
 

Offline Colin2B

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Thanks Colin.

Follow-up: what would it be worth?
Now that would require some research.
But it is going to be another very very big number.
Best not include antimatter.
 

Offline chiralSPO

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Here is the spreadsheet I used to solve the mass problem (unfortunately had to upload as a pdf, so no built in equations to fool around with...). If you're trying to solve it on your own, don't look. If you "ain't got time for that" but still interested in the answer, check it out.
 

Offline chiralSPO

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Thanks Colin.

Follow-up: what would it be worth?
Now that would require some research.
But it is going to be another very very big number.
Best not include antimatter.

Yeah, I found this: https://en.wikipedia.org/wiki/Prices_of_elements_and_their_compounds but it is not very up-to-date.

Definitely should not include antimatter, and probably not any short-lived isotopes...

Overall price is almost certainly essentially determined by the heaviest 18 (5.8 g of Rhenium and up), and is totally dominated by the last 9 (and I wouldn't want to be any where near this vessel, by the way, if it has 29 kg of francium, let alone 238 kg of thorium!)
 

Offline Bored chemist

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Reading this thread another thought struck me.
Very broadly- the more expensive elements are the heavier ones because they are rarer.
So, perhaps it would be interesting to do the same calculation "the other way round"
So 1 atom of Uranium, two of protactinium and so on, then find out which element was the "most expensive" in this list.
obviously, you get lots of hydrogen, but that's very common and thus very cheap. you wouldn't have much gold or platinum, but they are rather expensive (per atom or per gram).
If you do this are there any elements that stand out? (presumably Tc At Fr etc but, apart from them)

having said that the function 2^n goes up so fast that the light elements would dominate to an extreme extent.

A related question is which is the most valuable element in the earth's crust?
If you multiply the price per gram by the % in the Earth, which element gives the highest product?
 

Offline chiralSPO

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Reading this thread another thought struck me.
Very broadly- the more expensive elements are the heavier ones because they are rarer.
So, perhaps it would be interesting to do the same calculation "the other way round"
So 1 atom of Uranium, two of protactinium and so on, then find out which element was the "most expensive" in this list.
obviously, you get lots of hydrogen, but that's very common and thus very cheap. you wouldn't have much gold or platinum, but they are rather expensive (per atom or per gram).
If you do this are there any elements that stand out? (presumably Tc At Fr etc but, apart from them)

having said that the function 2^n goes up so fast that the light elements would dominate to an extreme extent.

A related question is which is the most valuable element in the earth's crust?
If you multiply the price per gram by the % in the Earth, which element gives the highest product?

Yes, I rather like turning it around like that. Makes the solution much more reasonable. You end up with:
4.14 kg H
8.23 kg He
7.13 kg Li (this might dominate price depending on how well Tesla does in the next year)
4.63 kg Be
2.78 kg B
1.54 kg C (if it's diamond this might dominate the price, especially if it's a diamond ;))
899 g N
513 g O
305 g F
162 g Ne
Na, Mg, Al, Si, S, P, Cl, Ar, K, and Ca are all pretty common
Sc is the first "rare" element (today $18 per gram, compare to $16.7 per gram of Ir), but since wer're down to 176 mg Sc, that's just over $3...
« Last Edit: 12/10/2015 03:51:53 by chiralSPO »
 

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