The example above is simple, but the same methods can be applied to determine isotope peaks in more complicated molecules as well. The molecule C 4 Br 1 O 2 H 5 has several isotope effects: 13 C, 2 H, 81 Br, 17 O, and 18 O all must be taken into account. First we will look at the (M+1) + peak in comparison with the M + peak. Only isotopes that will increase the value of M by 1 must be taken into consideration here – since 81 Br and 18 O would both increase M by 2, they can be ignored (the most abundant isotopes for Br and O are 79 Br and 16 O). Like the previous example, there are 1.08 13 C atoms for every 100 12 C atoms. However, there are 4 carbon atoms in our molecule, and any one of them being a 13 C atom would result in a molecule with mass (M+1). So it is necessary to multiply the probability of an atom being a 13 C atom by the number of C atoms in the molecule. Therefore, we have: 4C * 1.08 = 4.32 = molecules with a 13 C atom per 100 molecules We can repeat this analysis for 2 H and 17 O: 5H * 0.015 = 0.075 = molecules with a 2 H atom per 100 molecules 2O * 0.04 = 0.08 = molecules with a 17 O atom per 100 molecules Any of the three isotopes, 13 C, 2 H, or 17 O occurring in our molecule would result in an (M+1) + peak. To get the ratio of (M+1) + /M + , we need to add all three probabilities: 4.32 + 0.075 + 0.08 = 4.475 = (M+ 1) + molecules per 100 M + molecules We can say then that the (M+1) + peak is 4.475% as high as the M + peak. A similar analysis can be easily repeated for (M+2) + : 1Br * 98 = 98 = molecules with an 81 Br molecule per 100 molecules 2O * 0.2 = 0.4 = molecules with an 18 O molecule per 100 molecules 98 + 0.4 = 98.4 = (M+2) + molecules per 100 M + molecules The (M + 2) + peak is therefore 98.4% as tall as the M + peak. This method is useful because using isotopic differences, it is possible to differentiate two molecules of identical mass numbers.