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help with finding average relative atomic masses with isotopes


steakyfask

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im not sure if this is right but.....

 

 

i have been given an element which is Neon.

 

i am told that Ne consists of three isotopes: Ne-20 Ne-21 and Ne-22 (which i asume to be the number pf protons) in the proportions of 90.9% 0.3% and 8.8%

 

i need to find the average relative atomic mass of Ne...

 

do i just find the mass of each proportion of isotope and then add them all up and divide by 3 to get an average?

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im not sure if this is right but.....

 

 

i have been given an element which is Neon.

 

i am told that Ne consists of three isotopes: Ne-20 Ne-21 and Ne-22 (which i asume to be the number pf protons) in the proportions of 90.9% 0.3% and 8.8%

That is not a correct assumption. The number of protons in an element always stays the same. If you change the number of protons, you change what element you're working with (as in a nuclear reaction). Protons dictate element, electrons dictate ionization, and neutrons dictate isotope (they are, for the purpose of this thread, just added weight with no charge). The numbers (20-22) are protons plus neutrons, where the number of neutrons is what's changing.

 

To find the average mass:

.909(20) + .003(21) + .088(22) = 18.18 + .063 + 1.936

 

So your answer is 20.179 amu or g/mol (assuming you don't have to worry about significant figures).

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Toadie, please don't answer people's homework questions for them. Give them clues on how they can do it for themselves.

 

That is not a correct assumption. The number of protons in an element always stays the same. If you change the number of protons, you change what element you're working with (as in a nuclear reaction). Protons dictate element, electrons dictate ionization, and neutrons dictate isotope (they are, for the purpose of this thread, just added weight with no charge). The numbers (20-22) are protons plus neutrons, where the number of neutrons is what's changing.

 

To find the average mass:

.909(20) + .003(21) + .088(22) = 18.18 + .063 + 1.936

 

So your answer is 20.179 amu or g/mol (assuming you don't have to worry about significant figures).

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