Madeline
Registered: 10/10/09
Posts: 38
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Reply with quote | #1 |
When Cl, Br, or S is in a compound formula, you figure out the ratio of M+2 to M when determining the relative intensities on the mass spectrum. For example, if the molecule has one Cl, you observe that there is a 3 : 1 ration of M : M+2 and thus infer that the M+2 is 33%. Why don't you do this for C or H as well?
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PurplePpleEator
Registered: 10/04/09
Posts: 41
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Reply with quote | #2 |
I think with today's lecture you might have a better understanding of the reasoning now; if so this will just be repetitive information and you can ignore it.
When you look at the ion abundances table on page 91 of the lecture supplement you can see that both C and H have such small natural abundances as M+2 contributors that we ignore them. As for M+1 we do use this percentage to determine the number of C atoms within the molecular formula, but once again H has a very small natural abundance. It is more accurate to use the Hydrogen/Halogen Rule (2C + N + 2) to determine the maximum number of hydrogens+halogens instead.
Hope this helps.
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jamar
Registered: 10/12/09
Posts: 17
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Reply with quote | #3 |
I understand the Hydrogen/Halogen Rule. What still throws me is, once this number is obtained, what's the immediate next step in determining how many of these are hydrogens and how many are halogens?
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PurplePpleEator
Registered: 10/04/09
Posts: 41
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Reply with quote | #4 |
The mass spectrum tells us the amount of halogens and hydrogens through the M, M+1, and M+2 peaks. I believe that Dr. Hardinger said that F and I will be specifically specified if they are present because there is no mass spec indicator for either of them (lecture supplement pg 99). As for S, Cl, and or Br, they are determined from the M+2 peak and C is determined from the M+1 peak. Lecture supplement page 99 also shows that:
MW (molecular weight) - mass due to C, S, Cl, Br, F, and I = the mass due to O, N, and H.
This should help you in determining how many are hydrogens and how many are halogens. Please let me know if you need clarification.
Michael
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Hoffy90
Registered: 10/09/09
Posts: 12
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Reply with quote | #5 |
Another thing that helped me, when looking at the Hydrogen/Halogen Rule, is that often times the size of the Halogens makes the total number allowed limited.
For example, Br has a mass of 79 amu, so unless your M is very large, you can probably safely assume that there is only 1 Br. The hydrogen/halogen rule has also helped me a lot with checking my answers, as it reminds me if I am miscounting the number of hydrogens. |
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brittsies
Registered: 10/24/09
Posts: 3
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Reply with quote | #6 |
I'm really confused about where the 3:1 ratio of Cl comes from. The relative intensity is 33% for M+2 but I don't understand how this is gotten. Don't you just use the relative abundance of Cl (isotope 37) of: 24.23% to calculate the relative intensity?
(The example in practice problem 5(f) says "the relative intensity of the M+2 peak is
2 x 33= 66. Why don't you multiply by 24.23??)
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PurplePpleEator
Registered: 10/04/09
Posts: 41
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Reply with quote | #7 |
The 33% ratio comes from the M and M+2 contributor. The M contributor for 35Cl has a natural abundance of 75.77 and the M+2 contributor for the 37Cl has a natural abundance of 24.23. So the M+2 contributor is about 1/3 the total of the M or 33%
Hope this helps.
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claireharrison
Registered: 10/19/09
Posts: 13
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Reply with quote | #8 |
If M+1 is supposed to be an isotopic atom with a mass number one unit higher than the molecular ion, then why in Q5c (p. 111) does the answer use the natural abundance of carbon 13, hydrogen 2, and oxygen 17? Wouldn't this add up to M+3? |
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aokeke
Registered: 10/25/09
Posts: 1
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Reply with quote | #9 |
M+1 refers to EACH isotopic atom in that state. M+1 takes into account each isotopic atom with a mass number one unit higher than the molecular ion. Two or 3 different M+1 values do not "stack" to make M+2 or M+3. I hope this helped.
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