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# Relative abundance watch

1. 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.

Equation is

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.

Why?!
2. 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.

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Updated: January 26, 2010
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