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Spectroscopy q

The mass spectrum of CH3Br contains two molecular 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 answers are 79Br and 81Br, and CH2Br2 has 3 peaks with relative abundances 1:2:1, the m/z values are 172, 174 and 176.

I don't understand how am I supposed to know how many peaks there are? How do I work out the abundances too?
Original post by Magenta96
The mass spectrum of CH3Br contains two molecular 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 answers are 79Br and 81Br, and CH2Br2 has 3 peaks with relative abundances 1:2:1, the m/z values are 172, 174 and 176.

I don't understand how am I supposed to know how many peaks there are? How do I work out the abundances too?


You are expected to recognise that bromine has two isotopes with approximate abundances of 50% ...

Statistically with two atoms of bromine you have double the possibility of 79Br81Br than 79Br79Br or 81Br81Br, so the peak height is 1:2:1
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Magenta96
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Original post by charco
You are expected to recognise that bromine has two isotopes with approximate abundances of 50% ...

Statistically with two atoms of bromine you have double the possibility of 79Br81Br than 79Br79Br or 81Br81Br, so the peak height is 1:2:1


Thank you :smile:

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