Methods of analysis and detection, a2 chem.
Relative abundance of the M peak is 8.8
Calculate the relative abundance of the M+1 peak.
n = height of (M+1) peak x 100 / height of m peak x 1.1
There's a mass spectrum etc.
The peak stuff makes sense. I just don't get how you figure out what n is? According to the mark scheme it's = 2.
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Relative abundance watch
- Thread Starter
- 26-01-2010 16:46
- 26-01-2010 18:09
The peak of the (m+1) peak is going to be [n x 1.1/100] as large as the m peak. Where n is the number of carbons.
This is beacuse 13C is 1.1% of carbon.
So if you have C40H70 then the (m+1) peak is going to be 44% as high as the m peak.
Or C2H4 then (m+1) peak is going to be 2.2% as high as the m peak.