Chapter 2 - Sections 2.1-2.9 - Exercises - Challenge Problems - Page 84: 124b

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The strength of each signal in the mass spectrum of the sample is proportional to the natural abundance of that particular isotope. $x_{296_{Wt}}= \frac{N_{296_{Wt}}}{N_{total}}= \frac{36}{50}= 0.72$ $x_{297_{Wt}}= \frac{N_{297_{Wt}}}{N_{total}}= \frac{2}{50}= 0.04$ $x_{298_{Wt}}= \frac{N_{298_{Wt}}}{N_{total}}= \frac{12}{50}= 0.24$