Peak intensity, isotopic abundances

Figure Bl.25.9(a) shows the positive SIMS spectrum of a silica-supported zirconium oxide catalyst precursor, freshly prepared by a condensation reaction between zirconium ethoxide and the hydroxyl groups of the support [17]. Note the simultaneous occurrence of single ions (Ff, Si, Zr and molecular ions (SiO, SiOFf, ZrO, ZrOFf, ZrtK. Also, the isotope pattern of zirconium is clearly visible. Isotopes are important in the identification of peaks, because all peak intensity ratios must agree with the natural abundance. In addition to the peaks expected from zirconia on silica mounted on an indium foil, the spectrum in figure Bl. 25.9(a)...

A SSIMS spectrum, like any other mass spectrum, consists of a series of peaks of dif ferent intensity (i. e. ion current) occurring at certain mass numbers. The masses can be allocated on the basis of atomic or molecular mass-to-charge ratio. Many of the more prominent secondary ions from metal and semiconductor surfaces are singly charged atomic ions, which makes allocation of mass numbers slightly easier. Masses can be identified as arising either from the substrate material itself from deliberately introduced molecular or other species on the surface, or from contaminations and impurities on the surface. Complications in allocation often arise from isotopic effects. Although some elements have only one principal isotope, for many others the natural isotopic abundance can make identification difficult. 

Figure 8. Schematic outline of a second-generation MC-ICPMS instrument (Nu Instalments Nu Plasma), equipped with a multiple-Faraday collector block for the simultaneous measurement of up to 12 ion beams, and three electron multipliers (one operating at high-abundance sensitivity) for simultaneous low-intensity isotope measurement. This instmment uses zoom optics to obtain the required mass dispersion and peak coincidences in place of motorized detector carriers. [Used with permission of Nu Instruments Ltd.]...

Example The isotopic pattern of CI2 is calculated from Eq. 3.9 with the abundances a = 100 and = 31.96 as (100 + 31.96) = 10000 -1- 6392 + 1019. After normalization we obtain 100 63.9 10.2 as the relative intensities of the three peaks. Any other normalization for the isotopic abundances would give the same result, e.g., a = 0.7578, b = 0.2422. The calculated isotopic pattern of CI2 can be understood from the following practical consideration The two isotopes Cl and Cl can be combined in three different ways i) Cl2 giving rise to the monoisotopic composition, ii) Cl Cl yielding the first isotopic peak which is here X-i-2, and finally iii) Cl2 giving the second isotopic peak X+4. The combinations with a higher number of chlorine atoms can be explained accordingly. 

The isotopic contribution of various atoms is additive. For low molecular weight compounds, the isotopic contribution originates mainly from the carbon atom as long as no other element with a second isotope of significant abundance is present. For a molecule of Mr 192 the intensity of the m/z 194 ion represents 12% of the [M+H]+ peak (m/z 193 Fig. 1.3A). Chlorine (Cl) has two intense isotopes Cl and Cl (76% and 24% abundance, respectively). Replacing one H by a Cl atom results in a change of the isotopic distribution of the molecule... 

Due to the distinctive mass spectral patterns caused by the presence of chlorine and bromine in a molecule, interpretation of a mass spectrum can be much easier if the results of the relative isotopic concentrations are known. The following table provides peak intensities (relative to the molecular ion (M+) at an intensity normalized to 100%) for various combinations of chlorine and bromine atoms, assuming the absence of all other elements except carbon and hydrogen.1 The mass abundance calculations were based on the most recent atomic mass data.1... 

Figure 4.2 shows the SIMS spectrum of a promoted iron-antimony oxide catalyst used in selective oxidation reactions. Note the simultaneous occurrence of single ions (Si+, Fe+, Cu+, etc.) and molecular ions (SiO+, SiOH+, FeO+, SbO+, SbOSi+). Also clearly visible are the isotope patterns of copper (two isotopes at 63 and 65 amu), molybdenum (seven isotopes between 92 and 100 amu), and antimony (121 and 123 amu). Isotopic ratios play an important role in the identification of peaks, because all peak intensities must agree with natural abundances. Figure 4.2 also illustrates the differences in SIMS yields of the different elements although iron and antimony are present in comparable quantities in the catalyst, the iron intensity in the spectrum is about 25 times as high as that of antimony ... 

Because of the occurrence of 13c, and 1 0, and because mass spectrometry sorts ions according to m/e values, the peak intensities at m/e 252-256 do not represent the true proportion of iron isotopes. Table II is an example of how the abundances of the diligand ions deviate from the abxmdeuices of iron isotopes. Theoretically, the abundances of the diligand species can be calculated from the abundances of iron isotopes (IJ ). Calculated values may be used as a reference for the accuracy of the experimental values. 

A tetrahedral MCI4 molecule in which M is isotopically pure and Cl is in natural abundance consists of five isotopic species because of mixing of the Cl (75.4%) and Cl (24.6%) isotopes. Table 2.6c lists their symmetries, percentages of natural abundance, and symmetry species of infrared-active modes corresponding to the V3 vibration of the Tj molecule. It has been established [927] that these nine bands overlap partially to give a five-peak chlorine isotope pattern whose relative intensity is indicated by the vertical lines shown in Fig. 2.19b. If M is isotopically mixed, the spectrum is too complicated to assign by the conventional method. For example, tin is a mixture of 10 isotopes, none of which is predominant. Thus 50 bands are expected to appear in the V3 region of SnCL. It is almost impossible to resolve all these peaks, even... 

For example, the presence of bromine can be determined easily, because bromine causes a pattern of molecular ion peaks and isotope peaks that is easily identified. If we identify the mass of the molecular ion peak as M and the mass of the isotope peak that is two mass units heavier than the molecular ion as M -t- 2, then the ratio of the intensities of the M and M+2 peaks will be approximately one to one when bromine is present (see Chapter 8, Section 8.5, for more details). When chlorine is present, the ratio of the intensities of the M and M + 2 peaks will be approximately three to one. These ratios reflect the natural abundances of the common isotopes of these elements. Thus, isotope ratio studies in mass spectrometry can be used to determine the molecular formula of a substance. 

The example of ethane can illustrate the determination of a molecular formula from a comparison of the intensities of mass spectral peaks of the molecular ion and the ions bearing heavier isotopes. Ethane, C2H6, has a molecular weight of 30 when it contains the most common isotopes of carbon and hydrogen. Its molecular ion peak should appear at a position in the spectrum corresponding to a mass of 30. Occasionally, however, a sample of ethane yields a molecule in which one of the carbon atoms is a heavy isotope of carbon, This molecule would appear in the mass spectrum at a mass of 31. The relative abundance of in nature is 1.08% of the atoms. In the tremendous number of molecules in a sample of ethane gas, either of the carbon atoms of ethane will turn out to be a atom 1.08% of the time. Since there are two carbon atoms in ethane, a molecule of mass 31 will turn up (2 x 1.08) or 2.16% of the time. Thus, we would expect to observe a peak of mass 31 with an intensity of 2.16% of the molecular ion peak intensity. This mass 31 peak is called the M+ peak, since its mass is one unit higher than that of the molecular ion. 

The mass spectrum of an unknown compound had the following relative intensities for the M, (w/e = 86), M -i- 1, and M H- 2 peaks respectively 18.5, 1.15, and 0.074 (percentage of base peak). From the following partial list of isotopic abundance ratios, determine the molecular formula of the unknown. 

IV.la) Natural Isotopic abundances An element can be definitively identified if its correct isotopic abundance is found, either directly or indirectly after peak stripping. For instance, the peak height intensities in the 92-100 mass range in Fig. 17 closely match the abundance ratios of the... 

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