Table 1. Tabulated data from Figure 2. Rel. Abundance m/Z Proposed Identity ~1 45 100 44 ~11 28 ~2 22 ~10 16 ~8 12 The exercise of analyzing a mass spectrum is a process of applying logic and chemical intuition. Let's begin by identifying the peak with the greatest relative intensity, which has a m/Z value of 44. Since the parent molecule has a molecular weight of 44 g/mol, logic would dictate that this peak is the result of the positive ion of that species, CO2". While in this case, the parent ion fragment (CO2') is the one with the highest relative intensity, that is not always the case. Depending on the degree of fragmentation, the relative intensity of the parent ion on a mass spectrum can be high, low, or even zero if it was fragmented completely. Looking closely, there is a peak with a very low relative abundance (~1) and a m/Z value of 45. You may recall that atoms can possess different isotopes, which have well-defined abundances for naturally occurring samples. Naturally occurring carbon on earth consists of ~99% 1²C and ~1% 13C. A CO2 molecule in which the carbon atom is the isotope with an atomic mass of 13 g/mol would have a molecular weight of 45 g/mol, rather than 44. Since only ~1% of naturally occurring carbon contains this isotope, it should then make sense that the relative abundance of 13CO2* in the sample is approximately 1% of the abundance of 1?CO2*. The next fragment to investigate has a relative abundance of ~11 and a m/Z value of 28. One might cleverly notice that the difference in m/Z between this peak and the parent ion is equal to 16, the atomic mass of oxygen. It should then follow that in the process of fragmentation, some of the parent ion (C02") lost an O-atom, resulting in the presence of the fragment CO (which has a m/Z value of 28) in the mixture. Identifying the remaining three peaks will be left as a exercise.
Table 1. Tabulated data from Figure 2. Rel. Abundance m/Z Proposed Identity ~1 45 100 44 ~11 28 ~2 22 ~10 16 ~8 12 The exercise of analyzing a mass spectrum is a process of applying logic and chemical intuition. Let's begin by identifying the peak with the greatest relative intensity, which has a m/Z value of 44. Since the parent molecule has a molecular weight of 44 g/mol, logic would dictate that this peak is the result of the positive ion of that species, CO2". While in this case, the parent ion fragment (CO2') is the one with the highest relative intensity, that is not always the case. Depending on the degree of fragmentation, the relative intensity of the parent ion on a mass spectrum can be high, low, or even zero if it was fragmented completely. Looking closely, there is a peak with a very low relative abundance (~1) and a m/Z value of 45. You may recall that atoms can possess different isotopes, which have well-defined abundances for naturally occurring samples. Naturally occurring carbon on earth consists of ~99% 1²C and ~1% 13C. A CO2 molecule in which the carbon atom is the isotope with an atomic mass of 13 g/mol would have a molecular weight of 45 g/mol, rather than 44. Since only ~1% of naturally occurring carbon contains this isotope, it should then make sense that the relative abundance of 13CO2* in the sample is approximately 1% of the abundance of 1?CO2*. The next fragment to investigate has a relative abundance of ~11 and a m/Z value of 28. One might cleverly notice that the difference in m/Z between this peak and the parent ion is equal to 16, the atomic mass of oxygen. It should then follow that in the process of fragmentation, some of the parent ion (C02") lost an O-atom, resulting in the presence of the fragment CO (which has a m/Z value of 28) in the mixture. Identifying the remaining three peaks will be left as a exercise.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Kindly identify the three remaining fragments and explain the process. The first three fragments are already given with explanations but i simply do not understand. Kindly answer the highlighted part. Thank you
Transcribed Image Text:Table 1. Tabulated data from Figure 2.
Rel. Abundance
m/Z
Proposed Identity
~1
45
100
44
~11
28
~2
22
~10
16
-8
12
The exercise of analyzing a mass spectrum is a process of applying logic and chemical intuition.
Let's begin by identifying the peak with the greatest relative intensity, which has a
m/Z value of 44. Since the parent molecule has a molecular weight of 44 g/mol, logic would
dictate that this peak is the result of the positive ion of that species, CO2'. While in this case, the
parent ion fragment (CO2') is the one with the highest relative intensity, that is not always the
case. Depending on the degree of fragmentation, the relative intensity of the parent ion on a mass
spectrum can be high, low, or even zero if it was fragmented completely.
Looking closely, there is a peak with a very low relative abundance (~1) and a m/Z value of 45.
You may recall that atoms can possess different isotopes, which have well-defined abundances for
naturally occurring samples. Naturally occurring carbon on earth consists of ~99% 1²C and ~1%
13C. A CO2 molecule in which the carbon atom is the isotope with an atomic mass of 13 g/mol
would have a molecular weight of 45 g/mol, rather than 44. Since only ~1% of naturally occurring
carbon contains this isotope, it should then make sense that the relative abundance of 1³CO2* in the
sample is approximately 1% of the abundance of l?CO2*.
The next fragment to investigate has a relative abundance of ~11 and a m/Z value of 28. One
might cleverly notice that the difference in m/Z between this peak and the parent ion is equal to
16, the atomic mass of oxygen. It should then follow that in the process of fragmentation, some of
the parent ion (C02") lost an O-atom, resulting in the presence of the fragment CO* (which has a
m/Z value of 28) in the mixture. Identifying the remaining three peaks will be left as a
exercise.
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