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Section 2B. ESI-MS Data

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  • One of the advantages of ESI is that it is a “soft” ionization technique in which little fragmentation of large, thermally fragile biomolecules occurs. Consequently, molecular weight information is readily obtained with this technique. Additionally, the ions formed are often multiply charged, which enables the analysis of molecular masses exceeding 100,000 Da because the multiple charges bring the m/z values into the mass range of conventional mass analyzers, such as ion traps. With proteins and peptides, the mass spectrum consists of a series of peaks, call the “peak envelope” which represents a distribution of multiply charged analyte ions.

    Ubiquitin is a small protein with a monoisotopic molecular weight of 8560 Da. Electrospray ionization of this small protein typically results in major charge states of +8, +9, +10, +11, +12, and +13.

    Reading Questions

    1. Using this information, complete the table below, assuming that the charges on each ion come from protonation rather than sodium or potassium adducts. Round masses and m/z values to the ones place.

    z mass of [M+zH]+z m/z


    z mass of [M+zH]+z m/z
    8 8568 1071
    9 8569 952
    10 8570 857
    11 8571 779
    12 8572 714
    13 8573 659

    2. Using the data you entered in the table above, sketch an expected ESI-MS spectrum for ubiquitin. Label each peak with its charge state. What do you notice about the spacing of the peaks along the x-axis?


    The spacing of peaks widens along the length of the x-axis as the charge state decreases, as seen in the predicted spectrum.

    Based on the m/z values and peak spacing observed in the charge envelope, we can determine the charge state, z, for each peak and the molecular weight of the analyte using Equations \(\ref{1}\) and \(\ref{2}\),

    \[z = \dfrac{M_2-A}{M_1-M_2} \label{1}\]

    \[MW= \dfrac{(M_1-A)(M_2-A)}{M_1-M_2} \label{2}\]

    where MW is the molecular weight of the analyte, M1 is the m/z value for the first ion, z is the charge state of the first ion, M2 is the m/z value for a second ion of lower m/z, and A is the mass of the adduct ion, which is usually a proton (H+) but can be sodium (Na+) or potassium (K+) ions from glassware or buffers used in the experiment.

    Discussion Questions

    1. Why do aqueous samples for electrospray typically include an acid and a low surface tension solvent such as methanol?

    A. The acidity of the solution helps to ionize analytes by protonation, and low surface tension solvent promotes droplet fission.

    2. Figure 2 shows an experimentally obtained mass spectrum for ubiquitin. Compare this spectrum to the spectrum you predicted in Reading Question 2. Are there any differences? If so, what might cause these differences?

    Figure 2. ESI-MS spectrum of bovine ubiquitin from Protea Biosciences (

    A. The experimental spectrum agrees with the predicted spectrum with respect to the m/z values of the major peaks, which differ only slightly from the predicted values. The spectra differ in two ways: first, the intensity of the peaks varies because some charge states are more favorable than others; second, the experimental spectrum has additional smaller peaks at higher m/z values arising from low abundances of ions with charges of less than +8.

    3. Using Equation (\(\ref{2}\)) and any pair of peaks from Figure 2, calculate the molecular weight of ubiquitin and its percent error compared to the theoretical monoisotopic mass of 8560 amu.

    A. Any pair of peaks can be used. An example solution is shown below using the peaks at 714.92 and 779.83.

    \[MW= \dfrac{(M_1-A)(M_2-A)}{M_1-M_2} =\dfrac{(779.83-1.01)(714.92-1.01)}{779.83-714.92}=8566\: amu\]

    \[\%\: error= \dfrac{8566-8560}{8560}×100\%=0.07\%\]

    Figure 3. ESI-MS spectrum of cytochrome C. Data obtained at Trinity College.

    4. Figure 3 shows the ESI-MS spectrum for cytochrome C electrosprayed from a mixture of water, methanol, and acetic acid with pH of 2.5. What is the charge state of the peak at m/z = 773.36?

    A. Using equation (\(\ref{1}\)), \(z= \dfrac{727.94-1.008}{773.36-727.94}=\dfrac{726.932}{45.42}=+16\)

    If students use the peak at m/z=824.85 the charge comes out negative. It is helpful to remind them that the sample is usually prepared in acid and that they should expect positive charges from protonation.

    5. Determine the MW of the analyte, cytochrome C, using the data in Figure 3.

    A. Any pair of peaks can be used. An example solution is shown below using the peaks at 773.36 and 727.94 and equation (\(\ref{2}\)).

    \[MW= \dfrac{(M_1-A)(M_2-A)}{M_1-M_2} =\dfrac{(773.36-1.01)(727.94-1.01)}{773.36-727.94}= 12360\: amu\]

    6. How would you expect the mass spectrum to change if the cytochrome C sample was electrosprayed from a solution of higher pH? Explain your answer.

    A. Using a higher pH sample results in less protonation of the cytochrome C analyte and therefore lower charge states. This shifts the charge envelope to higher m/z values. For example, at pH 4.2 the z=+9 charge state is the base peak, and at pH 6.4 the +8 charge state is the base peak. See L. Konermann and D. J. Douglas, Biochem., 1997, 36, 12296-12302 for example data.

    7. Note that in the ESI-MS spectra for ubiquitin and cytochrome C, each major peak is accompanied by a series of less intense peaks of slightly different m/z. What is the source of these peaks?

    A. These peaks represent the isotopic distribution of atoms in the protein. Because the m/z value of a mass spectrum represents the mass-to-charge ratio of an individual molecule, they reflect the exact isotopic composition of each molecule. The most intense peak arises from the most likely composition of isotopes. Less intense peaks arise from molecules that happen to contain a different isotopic distribution, with the relative intensities of these peaks representing the isotopic abundance of each element in the ion and the number of atoms.

    Note to instructors: For small molecules, the most intense peak arises from ions composed solely of the most abundant isotopes of each atom, such as carbon-12, oxygen-16, hydrogen-1, and nitrogen-13. For organic molecules composed of these elements, the most abundant isotope is also the lightest isotope, and the most intense peak gives the monoisotopic mass of the ion. For larger molecules, the most intense peak may include heavier isotopes, and for very large molecules the most intense peak may actually present the average mass as calculated from atomic weights for each atom. For example, 1.10% of all carbon atoms are carbon-13, so molecules that contain more than 100 carbon atoms are likely to include at least one carbon-13 atom, making their most intense peak heavier than the monoisotopic mass. More information on isotopic distributions in large molecules, such as proteins, can be found in the 1983 Yergey et al. reference.

    8. Compare your description of the ion trap mass analyzer video with your group mates’ descriptions. As a group, write a consensus explanation for how an ion trap works.

    A. The video shows that the trap starts out with some gas molecules already in the trap. (This buffer gas is usually helium. Collisions with the buffer gas “cool” the ions as they enter the trap. This reduction in kinetic energy makes them easier to trap.) Ions from the ESI source enter the trap and follow a complex path between the end cap electrodes and the ring electrode. Eventually, the ions are damped to the middle of the trap. Then ions are scanned out of the trap by m/z value and detected to generate a mass spectrum.

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