Molecular formula from isotopic peaks

TABLE 6.2. Exact Masses of Isotopes of Some Common Elements 

Element Average Mass Nuclide Isotopic Mass 

TABLE 6.3. Relative Abundances of Some Common Elements 

A striking feature of a mass spectrum is the existence of satellite peaks. Those peaks are isotopically shifted lines that appear at masses one or more units higher than the main peak M the mass of M is calculated using the atomic masses of the most abundant isotopic species (i.e., the primary isotope). The satellite peaks, designated as M +1, M + 2, and so on, reflect the differences in the natural abundances of the isotopes. From the elemental composition of a molecular ion or fragment ion, the abundances of its satellite peaks (i.e., the isotopic pattern) can be predicted (see Example 6.1). Consider a compound of the general formula CxHj,NjO the abundance of the [M - -1] ion relative to [M] = 100% is given by 

If S and Si atoms are also present, their contribution to %[M - -1] is calculated by multiplying the number of S atoms with 0.8 and of Si atoms by 5.1, and adding these terms to Eq. (6.3). 

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If a high-resolution mass spectrum is not available, it is still possible to obtain information about the molecular formula from the low-resolution spectrum. In the mass spectrum of benzene, shown in Figure 15.5, the molecular ion appears at mlz 78. In addition, there is a smaller peak at mlz 79, called the M + 1 peak, that is 6.8% of the intensity of the Mt peak. The M + 2 peak, at mlz 80, is 0.2% of the Mt peak. The M + 1 and M + 2 peaks are caused by the presence of isotopic atoms of heavier mass in some of the molecules. Their intensities relative to the Mt peak can be used to deduce information about the formula. Let s look at how these peaks arise in more detail. 

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. 

One limitation on the use of isotope peak intensities to determine the molecular formula is that the molecular ion must be relatively intense, otherwise the isotope peaks will be too weak to be measured with the necessary accuracy. Difficulty may also arise from spurious contributions to the isotope peak intensities from the protonated molecular ion, from weak background peaks or from impurities in the sample. In any event the method is only reliable for molecules having molecular weights up to about 250-300. 

On average, the deviation between the measured and calculated masses was 0.0003 amu. Further support for the assigned formulae comes from the fact that an entire homologous series was observed where the mass of each homolog was measured with high accuracy. In many cases, final confirmation of the assigned molecular formula results from observation of the 34s isotope peak two amu higher at the correct precise... 

The minimum requirement for the organic chemist is the ability to record the molecular weight of the compound under examination to the nearest whole number. Thus, the spectrum should show a peak at, say, mass 400, which is distinguishable from a peak at mass 399 or at mass 401. In order to select possible molecular formulas by measuring isotope peak intensities (see Section 1.5.2.1), adjacent peaks must be cleanly separated. Arbitrarily, the valley between two such peaks should not be more than 10% of the height of the larger peak. This degree of resolution is termed unit resolution and can be obtained up to a mass of approximately 3000 Da on readily available unit resolution instruments. 

Since the advent of high-resolution mass spectrometers, it is also possible to use very precise mass determinations of molecular ion peaks to determine molecular formulas. When the atomic weights of the elements are determined very precisely, it is found that they do not have exactly integral values. Every isotopic mass is characterized by a small mass defect, which is the amount by which the mass of the isotope differs from a perfectly integral mass number. The mass defect for every isotope of every element is unique. As a result, a precise mass determination can be used to determine the molecular formula of the sample substance, since every combination of atomic weights at a given nominal mass value will be unique when mass defects are considered. For example, each of the substances shown in Table 1.4 has a nominal mass of 44 amu. As can be seen from the table, their exact masses, obtained by adding exact atomic masses, are substantially different when measured to four decimal places. 

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. 

In Subsection 8.8.2 we presented three match values for MS isotope peak comparison. Table 8.22 shows the number of formula candidates with at least one carbon atom as well as the absolute and relative ranking positions of the true formulas. Calculations are based on g and MS accuracy 5 = 10 ppm. Numbers in the row headers cross-reference with the compounds in Table 8.20. Using MS data alone, only two compounds have ARP = 1, of which one has only one possible candidate (i.e. a trivial result) and the other has only two possible candidates. Using the ARP, it is not possible to choose the best method for calculating MS match values. While NDP works best for most of the samples, NSSE yields the best result for maltopentaose, and NSAE is best for cyclosporin C. If we look at the RRP results, creatine is excluded (RRP is not defined for only one candidate), but calculation of the arithmetic mean is possible. Thus, it appears that NDP performs best on average for these ten samples, but the other methods deliver quite similar mean results, and considering the small number of samples, this conclusion is by no means a statistically reliable result. What is clear from these results, the MS match value alone is by no means sufficient to unambiguously determine the correct molecular formula in most cases without further information or restrictions. 

The isotopic ratio of the elements is fixed and predictable, as seen in Table 14.1. If the number of carbon atoms in a molecrde is known, the intensity of the M+1 peak that resrdts from the presence of is predictable. The ratio of to is 1.11, so the intensity of the M+1 peak is 1.11 x C. Using isotopic ratios, similar predictions are possible for other elements. This observation is important because the ability to predict the number of carbon atoms in an unknown molecule allows determination of the formula. The isotopic ratio is a known quantity, and measurement of the M+1 peak relative to M (i.e., M/ M+1) will calculate the number of carbon atoms found in the molecrdar ion. The molecular ion also provides the molecular weight if 2 = +1, so the molecular formula is available for a given sample. For molecules that contain only C, H, N, and 0, the two formulas used to calculate a formula are the following ... 

If possible, determine the elemental composition for M" and other important peaks using isotopic abundances. In particular, look for isotope peaks from "M + 2" elements like Cl, Br, S, and Si (Table 8.36 and Fig. 8.54). If you are able to establish a molecular formula, calculate "rings + tt bonds" ... 

So far only the magnetic sector has been considered the purpose of the electrostatic sector is to achieve high resolution. Many molecules share the same molecular weight. A spectrum in which all of the masses are accurate to one atomic mass unit (nominal mass) can therefore be very ambiguous, but more specificity can be obtained if the resolution of the instrument can be increased, so that species with the same nominal mass may be resolved on the basis of their exact mass. In addition, accurate mass measurement can be used in conjunction with accurate values of isotopic masses to assign molecular formulae. For example, in a FAB spectrum acquired from a matrix containing caesium iodide (used as an internal standard for mass calibration) a nominal mass peak at m/e = 404 could correspond to either of the following structures ... 

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