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AS Mass Spectrometry
The mass spectrum of CH3Br contains two molecular ion peaks of equal abundance with m/z values of 94 and 96. Deduce the number of molecular ion peaks in the spectrum of the compound CH2Br2. Give their relative abundances and the m/z value of the molecular ion with the greatest mass.
The answer is the first one on here:
http://www.aqaanswers.co.uk/Chemistry/chapter17/Chapter_17_Answers_to_examination_style_questions.pdf
What on earth is it going on about?!
Thank you!
The answer is the first one on here:
http://www.aqaanswers.co.uk/Chemistry/chapter17/Chapter_17_Answers_to_examination_style_questions.pdf
What on earth is it going on about?!
Thank you!
Scroll to see replies
No bloody idea. But, thanks for the link: thats the answers for our textbook right!?
Laura373
The mass spectrum of CH3Br contains two molecular ion peaks of equal abundance with m/z values of 94 and 96. Deduce the number of molecular ion peaks in the spectrum of the compound CH2Br2. Give their relative abundances and the m/z value of the molecular ion with the greatest mass.
The answer is the first one on here:
http://www.aqaanswers.co.uk/Chemistry/chapter17/Chapter_17_Answers_to_examination_style_questions.pdf
What on earth is it going on about?!
Thank you!
The answer is the first one on here:
http://www.aqaanswers.co.uk/Chemistry/chapter17/Chapter_17_Answers_to_examination_style_questions.pdf
What on earth is it going on about?!
Thank you!
OMG, you are a genius. I've been looking bloody everywhere for those answers!
Bromine has two common isotopes in about 50% split with masses 79 and 81, so for a molecule with two bromines you can get:
81 + 81
81 + 79
79 + 81
79 + 79
81 + 81
81 + 79
79 + 81
79 + 79
Hedgehunter
No bloody idea. But, thanks for the link: thats the answers for our textbook right!?
lmao xD
indeed. www.aqaanswers.co.uk provides the answers for the Biology, Chemistry and Physics textbooks.
EierVonSatan
Bromine has two common isotopes in about 50% split with masses 79 and 81, so for a molecule with two bromines you can get:
81 + 81
81 + 79
79 + 81
79 + 79
81 + 81
81 + 79
79 + 81
79 + 79
Aha thank you. I still don't understand the m/z bit though?
and I can't take credit for that link; some guy posted it on here earlier this week.
Laura373
lmao xD
indeed. www.aqaanswers.co.uk provides the answers for the Biology, Chemistry and Physics textbooks.
indeed. www.aqaanswers.co.uk provides the answers for the Biology, Chemistry and Physics textbooks.
HOW STUPID am I: I've been marking things with the first link to answers that has the answers all weirdly set out.
Laura373
Aha thank you. I still don't understand the m/z bit though?
and I can't take credit for that link; some guy posted it on here earlier this week.
and I can't take credit for that link; some guy posted it on here earlier this week.
The m/z value is just the mass:charge to ratio (the charge is normally +1), the m/z peak with the highest mass will include the 81 + 81 arrangement.
Laura373
Aha thank you. I still don't understand the m/z bit though?
and I can't take credit for that link; some guy posted it on here earlier this week.
and I can't take credit for that link; some guy posted it on here earlier this week.
Just add the mass numbers for each.
Malsi101
could someone explain quickly to me why the relative abundances are 1: 2 :1 and what for
for the masses
Br + Br + CH2 = mass of molecule
81 + 81 + 14 = 176
79 + 81 + 14 = 174
81 + 79 + 14 = 174
79 + 79 + 14 = 172
so 1: 2 :1 ratios since the middle two are the same (peak twice as high compared to the lowest and highest peaks)
EierVonSatan
Bromine has two common isotopes in about 50% split with masses 79 and 81, so for a molecule with two bromines you can get:
81 + 81
81 + 79
79 + 81
79 + 79
81 + 81
81 + 79
79 + 81
79 + 79
I actually knew the answer to the OP!
Trust EVS to get there first
Champagne Supernova
I actually knew the answer to the OP!
Trust EVS to get there first
Trust EVS to get there first
EierVonSatan
as EierVonSatan said, there are two isotops of Br - 79 and 81. For CH2Br2 there are two bromine atoms, each of which are either going to be 79 or 81.
so the possible combos are:
15 (Mr CH3) + 79 + 79 = 172
15 + 79 + 81 = 174
15 + 81 + 79 = 174
15 + 81 + 81 = 176
so theres going to be twice as many 174s than the others. Hence 1:2:1
so the possible combos are:
15 (Mr CH3) + 79 + 79 = 172
15 + 79 + 81 = 174
15 + 81 + 79 = 174
15 + 81 + 81 = 176
so theres going to be twice as many 174s than the others. Hence 1:2:1
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