adamconner
  • adamconner
question is attached. Please Help! Fan and Medal!!
Chemistry
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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adamconner
  • adamconner
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aaronq
  • aaronq
so, the 2 peaks observed are 44 amu and 42 amu, the % abundance of 42 is about twice that of 44. so, \(A_{42}= 100\%\; and\; A_{44}=50\%\;\rightarrow A_{42}=\dfrac{2}{3}\; and \;A_{44}=\dfrac{1}{3}\) we know that: \(A_{average\;atomic\;weight} = 40.5929 amu = A_{42}*A_{42}\%+A_{44}*A_{44}\%\) But, \(A_{44}*(\dfrac{2}{3})+A_{42}*(\dfrac{1}{3})=42.6667 amu\) So the calculated and observed average atomic values are not the same. Can you take it from here?
adamconner
  • adamconner
first of all how did u get A_42 is 2/3 and A_44 is 1/3??

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aaronq
  • aaronq
I used the height of the peaks which is their relative abundance. for example, if you have A = 100% and B=50%, in total of 150 %. In terms of absolute abundance A=100/150=2/3 B=50/150=1/3
aaronq
  • aaronq
btw, i just noticed i wrote "\(A_{44}∗(2/3)+A_{42}∗(13)\)=42.6667amu", wrong. I switched the abundance around, BUT when i plugged into the calc i did it right. it should read: " \(A_{42}∗(\dfrac{2}{3})+A_{44}∗(\dfrac{1}{3})\)=42.6667 amu "
adamconner
  • adamconner
ok, now after that what do i need to do??
aaronq
  • aaronq
It says in the question, "Sketch a best guess for what the spectrum should be".
adamconner
  • adamconner
|dw:1378077935940:dw| ??
adamconner
  • adamconner
how high would it be?
aaronq
  • aaronq
we'll i think that at least one of the isotopes should be less than 42, because you won't be able to have the average of atomic weight at 40.5929 amu. so i would first work with this equation: 40.5929 amu =\(A_{x}Abundance_x\% +A_{y}Abundace_y\% +A_{z}Abundace_z\%\) and find what isotopes you should have.
aaronq
  • aaronq
The height of the peaks you will estimate from the % abundance.
adamconner
  • adamconner
ok i lost u for a second what is the Ax Abundancex %+Ay Abundacey %+Az Abundacez %
adamconner
  • adamconner
Sorry, its just that we didnt take this stuff yet. but if we can do it we get extra points and i need them
aaronq
  • aaronq
well i took \(A_x\) to represent an isotope, like \(A_{44}\) would be isotope with mass of 44 amu. Abundance % would represent it's absolute abundance.
aaronq
  • aaronq
In the question, it's telling you that two isotopes IS NOT the answer, therefore there must be more than 2 (or less than 2). So, basically want to plug in values into the formula above, and find the masses of the isotopes and their abundance that would be equal to the average atomic mass known (40.5929 amu) first, then graph that information.
adamconner
  • adamconner
omg this is so hard! what are the values that i plug in?
aaronq
  • aaronq
the values is what you have to find, they're not given.
adamconner
  • adamconner
ok then how do i find them?
adamconner
  • adamconner
u said i plug in values in the formula to find the mass and the abundance
aaronq
  • aaronq
yeah, plug in the values you make up (that are reasonable) because that's what the question is saying. ^that's how you find them. remember that it has to be equal to 40.5929 amu
adamconner
  • adamconner
so basically any 3 values that equal to 40.9529? and in which part do i plug them? the \[A _{x} \] or the abundance % or is it all one part?
aaronq
  • aaronq
we'll \(A_x\) would be a mass, say 40 amu and percent abundance of that isotope would be whatever, like 50% (0.5) so 40.5929 amu = 40 amu*0.5+ ... etc
aaronq
  • aaronq
remember, make it reasonable
adamconner
  • adamconner
ok, is this close enough: 40amu*25% + 41amu*30% + 39*47% ?? so thats about 10 + 12.3 + 18.33= 40.63
adamconner
  • adamconner
wait do i need to use the 42 and 44 ??
aaronq
  • aaronq
It doesn't say you need to, explicitly. I wouldn't worry about using them. If it adds up, what you wrote seems reasonable to me.
aaronq
  • aaronq
The only problem now, is that you have to sketch the percent and 47% might be hard to do.
adamconner
  • adamconner
i mean its gonna be hard to get it exactly to 40.9529amu but i guess 40.63 is pretty close? and why would it be hard to draw??
aaronq
  • aaronq
what i would do is find 2 isotopes, and then solve for the other algebraically. for example, if you have: 40amu*25% + 41amu*30%+39*x=40.5929 solve for x
adamconner
  • adamconner
ok, it turns out to be 47.82%
aaronq
  • aaronq
that's okay, at least it's more accurate with the average atomic mass. Just try to aim below 50%
adamconner
  • adamconner
so now i just put these values on a graph and im done??
aaronq
  • aaronq
yeah, i don't know if they want you to normalize them though. If so, then you have to let the largest abundance be equal to 100% then scale the others relatively.
adamconner
  • adamconner
|dw:1378081670039:dw|
adamconner
  • adamconner
so roughly it would look something like this
aaronq
  • aaronq
yeah, exactly, if you wanna normalize it, it's not that bad, you're essentially multiplying by 2. |dw:1378081906450:dw| don't forget to label the axes.
adamconner
  • adamconner
yes of course, the x would be mass and the why would be relative abundance right??
adamconner
  • adamconner
and if i decided to make the points 42,43, adn 44 would it still work?
aaronq
  • aaronq
yep, remember it's atomic mass (not just mass). It would work but you'd have to change all the % abundance values
adamconner
  • adamconner
yes ok. Thanks so much again
aaronq
  • aaronq
no problem, dude. I would say that, that was a fairly difficult problem (assuming you're in Highschool).

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