The values in Table 14.1 can be used to calculate precise masses for each molecular ion. C 6 H 9 N: 6(12.0000) + 9(1.00783) + 14.0031 = 95.0736 C 5 H 5 NO: 5(12.0000) + 5(1.00783) + 14.0031 + 15.9949= 95.0372 A high resolution mass spectrum will distinguish these compounds based on this small difference in the expected m/z ratio of the molecular ions. Problem 14.10 What rule would you expect for the m/z values of fragment ions resulting from the cleavage of one bond in a compound with an odd number of nitrogen atoms? As stated in Problem 14.7, molecules having an odd number of nitrogen atoms have an odd-numbered m/z ratio for the molecular ion. Cleavage of one bond of a molecular ion would generate a cation and a radical, only the cation of which would be observed in the mass spectrum. If the cation fragment retained an odd number of nitrogen atoms, then the fragment would have an even m/z ratio. If the cation fragment contained an even number of nitrogen atoms (or zero), then the fragment would have an odd m/z ratio. Problem 14.11 Determine the probability of the following in a natural sample of ethane. (a) One carbon in an ethane molecule is 13 C? The relative abundance of 13 C listed in Table 14.1 is 1.11 atoms of 13 C for every 100 atoms of 12 C. This corresponds to a probability that any one atom is a 13 C atom of 1.11/(100 + 1.11) = 0.0110. The corresponding probability that a given carbon atom is 12 C is 100/(100 + 1.11) = 0.0989. Therefore the probability of having one carbon 12 C and the other 13 C is (0.989)(0.011) = 0.0109. There are two ways to have this situation since either of the carbon atoms could be the 13 C so the final answer is 2(0.0109) = 0.0218. Converting to percent by multiplying by 100 means that there is a 2.18% chance that one of the carbon atoms in ethane is 13 C. (b) Both carbons in an ethane molecule are 13 C? The probability that both of the two carbon atoms in ethane is 13 C is (0.0110) 2 = 1.21 x 10 -4 . Converting to percent by multiplying by 100 means that there is a 0.012% chance that both of the carbon atoms in ethane is 13 C . (c) Two hydrogens in an ethane molecule are replaced by deuterium atoms? The relative abundance of 2 H listed in Table 14.1 is 0.016 atoms of 2 H for every 100 atoms of 1 H. This corresponds to a probability of 0.016/(100 + 0.016) = 1.6 x 10 -4 that any given H atom is a 2 H atom. The corresponding probability that any given H atom is 1 H is 0.99984. The probability that a given two H atoms in ethane are 2 H is given by (0.99984) 4 (0.00016) 2 = 2.55 x 10 -8 . There are 15 possible combinations of having two 2 H atoms in ethane, so the overall probability is (15)(2.55 x 10 -8 ) = 3.8375 x 10 -7 . Converting to percent by multiplying by 100 means that there is a 3.84 x 10 -5 % chance that two of the hydrogen atoms in ethane are 2 H. Problem 14.12 The molecular ions of both C 5 H 10 S and C 6 H 14 O appear at m/z 102 in low-resolution mass spectrometry. Show how determination of the correct molecular formula can be made from the appearance and relative intensity of the M + 2 peak of each compound. As shown in Table 14.1, 16 O occurs in greater than 99.7% abundance, so no (M + 2) peak will be observable for C 6 H 14 O. On the other hand, sulfur has one isotope, 32 S, that is 95% abundant and another isotope 2 amu higher, namely 34 S, that has an abundance of 4.2%.
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