m/z molecular peak and empirical formula

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#1
In the mass spectrum of an organic compound, the molecular ion occurs at m/e = 86.
Which of the following could be the empirical formula of the compound?

A C6H14
B C5H10N
C C5H12O
D C5H7F
why???
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1 month ago
#2
(Original post by universalcj)
In the mass spectrum of an organic compound, the molecular ion occurs at m/e = 86.
Which of the following could be the empirical formula of the compound?

A C6H14
B C5H10N
C C5H12O
D C5H7F
why???
Its impossible to be B , so you can eliminate that as the Mr of the whole molecule is <86 which is the m/z peak on the spectrum.

It can't be A as the only way you could get Mr 86 is by having C6H12 but this requires breaking 2 C-H bonds, however fragments can only form by breaking C-C bonds. So it can't be C (Also you can only break one bond in a molecule to form a fragment, i.e I can't break 2 C-C bonds , only one.)

It can't be D due to the same logic as for why its not A.

So it has to be C.

Pigster is what I'm saying right ?
0
1 month ago
#3
(Original post by universalcj)
In the mass spectrum of an organic compound, the molecular ion occurs at m/e = 86.
Which of the following could be the empirical formula of the compound?

A C6H14
B C5H10N
C C5H12O
D C5H7F
why???
So work out the relative molecular mass of each compound using a periodic table :
C6H14= 86
C5H10N= 84
C5H12O= 88
C5H7F= 86

as the mass spectrum shows 86 is the most common Mr it has to be A or D. Then we use the fact we are looking for the empirical formula (the simplest ratio) to rule out A, as C6H14 can be simplified and is therefore not an empirical formula.
so the answer is D, C5H7F!
it cannot be simplified and has an Mr of 86
0
1 month ago
#4
(Original post by seals2001)
Pigster is what I'm saying right ?
As ra.chel has spelt out so clearly, it turns out you aren't right.
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1 month ago
#5
(Original post by Pigster)
As ra.chel has spelt out so clearly, it turns out you aren't right.
My bad. I was confusing it with fragment ions. But was what I said about fragment ions actually correct ? At the AS level , you assume that no C-H bonds break and that only one bond breaks in the molecule to form a fragment , right ?

Thanks.
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1 month ago
#6
(Original post by seals2001)
My bad. I was confusing it with fragment ions. But was what I said about fragment ions actually correct ? At the AS level , you assume that no C-H bonds break and that only one bond breaks in the molecule to form a fragment , right ?

Thanks.
At A level (as far as I know all exam boards will be the same) only one bond breaks.

Although that could be a C-H bond. After all, what would methane do?
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#7
(Original post by seals2001)
Its impossible to be B , so you can eliminate that as the Mr of the whole molecule is <86 which is the m/z peak on the spectrum.

It can't be A as the only way you could get Mr 86 is by having C6H12 but this requires breaking 2 C-H bonds, however fragments can only form by breaking C-C bonds. So it can't be C (Also you can only break one bond in a molecule to form a fragment, i.e I can't break 2 C-C bonds , only one.)

It can't be D due to the same logic as for why its not A.

So it has to be C.

Pigster is what I'm saying right ?
even though you were wrong it was really helpful cuz at least you tried to help out other people who faced difficulty
have a good dY
1
#8
(Original post by ra.chel)
So work out the relative molecular mass of each compound using a periodic table :
C6H14= 86
C5H10N= 84
C5H12O= 88
C5H7F= 86

as the mass spectrum shows 86 is the most common Mr it has to be A or D. Then we use the fact we are looking for the empirical formula (the simplest ratio) to rule out A, as C6H14 can be simplified and is therefore not an empirical formula.
so the answer is D, C5H7F!
it cannot be simplified and has an Mr of 86
what a clear answer and sincerely express my gratitude have a good day
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