Additional Materials Required

Teacher Tips

• “Mass Spectrometry” is supplied as one of three learning activities in Spectroscopy Guided-Inquiry Learning Activities for AP^® Chemistry, Catalog No. AP7710, from Flinn Scientific. The other two guided-inquiry activities in the AP7710 learning activity package are “Photoelectron Spectroscopy” and “Electron and Vibrational Spectroscopy.”
• We recommend a collaborative, peer-learning environment for students to complete this guided-inquiry learning activity. Effective group sizes may include 3−4 students. To ensure active participation by all members of the group, it is helpful to assign rotating roles to the students and call on different students in each group to be the “reporter” for class discussion and explanation of key parts of the activity.
• Mass spectrometry may not be covered in your general chemistry textbook. Online computer access is required for students to answer several questions in this guided-inquiry activity. Students will also need periodic tables for reference.
• The organization of this activity into three parts provides convenient stopping points to discuss key questions and explain relevant learning goals. See, for example, Questions 7 and 11 in Part A and Questions 3 and 6 in Part B.
• The Background section and Part A describe the history of mass spectroscopic evidence for isotopes. Aston’s confirmation of the existence and properties of neon isotopes had been preceded by an earlier line of studies on lead isotopes generated by radioactive decay of uranium. When the atomic mass of lead deposits in radioactive uranium minerals was analyzed, it was found to be statistically different from the atomic mass of lead in lead ore. The actual composition of the lead atoms seemed to be different, depending on their origin. In 1913, Frederick Soddy, professor of chemistry at the University of Glasgow, coined the term isotope to define atoms of the same element that have the same chemical properties but different atomic masses. The word isotope was derived from Greek words meaning “same place” to denote the fact that isotopes occupy the same place in the periodic table (they are the same element) even though they have different masses. Soddy received the Nobel Prize in Chemistry in 1921 for his investigations into the nature and origin of isotopes.
• The evidence and reasoning in this learning activity should be used to recall, review and reinforce learning objectives related to the development of atomic theory. Students should be familiar with major experiments and the timeline of discoveries in our knowledge of the atom. The AP Chemistry Learning Objectives also stress how various events depended and built upon earlier discoveries. Examples include Rutherford’s discovery of the properties of the nucleus, which would not have been possible without the earlier work on radioactivity and alpha particles. Similarly, Thomson’s work with gas discharge tubes and the discovery of the electron led the way for Aston to prove the existence of isotopes. Opportunities for further inquiry in this vein include how the atomic number of an element was “proved” and how unanswered questions concerning the charge and mass of the nucleus led to and were subsequently rectified by the discovery of neutrons.
• How do you calculate the probability of generating molecules containing one each ⁷⁹Br and ⁸¹Br? The probability of EITHER of two events C or D occurring is obtained by adding their individual probabilities, p (C) + p (D). p (C) = (0.50)(0.5) = 0.25. p (D) is also 0.25, so the probability of C or D is equal to 0.25 + 0.25 = 0.50.
• The remarkable accuracy of modern mass spectrometry, which is continually being improved, has contributed to regular revisions of reference values for the standard atomic weights and isotopic abundance of elements. Every two years the International Union of Pure and Applied Chemistry (IUPAC) reviews and updates these standard values as needed in light of ever more accurate data. In 2013 IUPAC announced revisions for the atomic weights of 19 elements.
• Common misconceptions and conceptual hurdles or stumbling blocks in this unit on mass spectrometry include:
  • How are the actual mass of a nuclide (single atom), mass number, and average atomic mass of an element related?
  • What is the difference between atomic mass units (amu) and molar mass expressed in grams per mole?
  • What atom is used as the reference mass in mass spectrometry and atomic mass measurements?
• Although chemical ionization (CI) may seem to be preferred over electron ionization (EI) mass spectra because the molecular ion is more prevalent in the former, CI has some disadvantages. CI spectra often contain peaks for protonated molecular and fragment ions, such as MH⁺. Also, reactions between sample molecules (M) and reagent gas ions (e.g., CH₄⁺, CH₅⁺, C₂H₅⁺) in CI produce charged adducts (complex ions) that complicate the interpretation of fragment ions. Most reference spectra of organic compounds were generated by electron ionization.
• Electrospray ionization has revolutionized the use of mass spectrometry for studying proteins and has led to a new field of research called proteomics. Mass spectrometry can be used to determine the molecular weight of proteins, identify the sequence of amino acids in the primary structure of a protein, and obtain a distribution profile of various proteins in cells and organelles.
• Mass spectra for thousands of organic compounds may be found on the NIST Chemistry WebBook at http://webbook.nist.gov/chemistry/. This searchable online database contains many kinds of spectral data, including IR, UV and mass spectra, as well as thermochemistry data. NIST is the National Institute of Standards and Technology.
• The coupling of chromatographic and spectroscopic techniques has had a great influence on the applications of mass spectrometry for forensic analysis. The most powerful instruments are gas chromatography−mass spectrometry (GC/MS) and liquid chromatography−mass spectrometry (HPLC/MS).

Further Extensions

Alignment with the Curriculum Framework for AP^® Chemistry

Essential Knowledge
1D1: As is the case with all scientific models, any model of the atom is subject to refinement and change in response to new experimental results. In that sense, an atomic model is not regarded as an exact description of the atom, but rather a theoretical construct that fits a set of experimental data.
1D2: An early model of the atom stated that all atoms of an element are identical. Mass spectrometry data demonstrate evidence that contradicts this early model.

Learning Objectives
1.13 Given information about a particular model of the atom, the student is able to determine if the model is consistent with specified evidence.
1.14 The student is able to use data from mass spectrometry to identify the elements and the masses of individual atoms of a specific element.

Answers to Questions

Part A. Isotopic Abundance and the Mass Spectra of Elements

1. The modern definition of isotopes is based on the subatomic particle structure of atoms. Complete the following table to show the number of protons, neutrons, and electrons in neutral atoms of Cl–35 and Cl–37.
   {13775_Answers_Table_1}
2. Write a 2−3 sentence definition for isotopes that includes all of the following terms: protons, neutrons, atomic number and mass number.

   Isotopes of an element have the same number of protons but different numbers of neutrons. The atomic number is the number of protons in the nucleus of an element, which defines the element. The mass number is equal to the sum of the number of protons and the number of neutrons in an atom. Isotopes thus have the same atomic number but different mass numbers.

3. Which property of an atom defines the identity of an element? Explain why isotopes form the same compounds and undergo the same reactions.

   The identity of an element depends on the number of protons (the atomic number). Isotopes thus have identical chemical properties and are chemically indistinguishable from one another—they form the same compounds, undergo the same reactions, etc.

4. Radioactive isotopes (radioisotopes) are widely used in medicine. Because isotopes have identical chemical properties, the reaction and distribution of radioisotopes in the body is similar to that of their natural isotopes. Iodine–131, for example, is an artificial radioisotope that is used to diagnose thyroid disorders. When administered to a patient, I–131 is taken up by the thyroid gland, where it is incorporated into the thyroid hormone, just as iodine in the diet would be. Based on where the following elements are likely to be found in the body, match each radioisotope with its medical use.
   {13775_Answers_Table_2}
5. Regardless of its source on earth, the element chlorine always contains 75.8% chlorine-35 atoms and 24.2% chlorine-37 atoms. The atomic mass of an element is the weighted average of the masses of the isotopes. Fill in the values of x and y in the following equation to show how the atomic mass (z) of chlorine is calculated.

   (0.758) (x amu) + (y) (37 amu) = z amu
   x = 35 amu, y = 0.242 and z = 35.5

6. Figure 3 shows an expanded mass spectrum of silicon. How many isotopes of silicon exist in nature? List the mass number and natural abundance of each isotope.
   {13775_Answers_Figure_3_Mass spectrum of silicon}

   Silicon has three naturally occurring isotopes, Si-28, Si-29 and Si-30. Their natural abundances are 92.2%, 4.7% and 3.1%, respectively. Note: Remind students that the percent abundance of all isotopes for an element must total 100%.

7. Calculate the average atomic mass of silicon using the natural abundances obtained above and the mass number of each isotope. Compare the calculated value with the literature or reference value for the atomic mass of silicon.

   (0.922)(28 amu) + (0.047)(29 amu) + (0.031)(30 amu) = 28.1 amu
   The atomic mass of silicon is reported to be 28.085 amu.

8. High-resolution mass spectrometers provide masses of atoms and molecules that are precise to 4−6 decimal places. The actual mass of a Si-28 nuclide is 27.97693 amu. Define the nuclear binding energy of an atom and explain how it accounts for the difference in mass between the actual mass and the mass number. Hint: E = mc²

   Nuclear binding energy is expressed as the positive energy that would be required to break the strong nuclear force holding protons and neutrons together in the nucleus of an atom. According to Einstein’s famous equation (E = mc²) linking mass and energy, the binding energy released in forming a nucleus is associated with a mass defect. The mass of the nucleus is thus less than the sum of the masses of isolated or individual protons and neutrons. To a first approximation, the latter is equal to the mass number.

9. “The atomic mass of silicon represents the mass of the most common naturally occurring isotope.” Explain why this statement is false.

   ,em>For any element that occurs as a mixture of isotopes, there is NO single atom having the average atomic mass of the element. The atomic mass is the weighted average of the masses of the isotope.

10. Copper occurs in nature as a mixture of two isotopes, 69.2% Cu-63 and 30.8% Cu-65. Sketch the expected mass spectrum of copper ions.
    {13775_Answers_Figure_4_Mass spectrum of copper}
11. The mass spectrum of elemental bromine, a red liquid composed of diatomic molecules (Br²), is shown. What is the isotopic composition of individual molecules giving rise to the three ion peaks at m/z 158 (M⁺), 160 (M+2) and 162 (M+4)?
    {13775_Answers_Figure_7}

    The isotopic composition of the molecular ion responsible for the M⁺ peak at 158 is ⁷⁹Br−⁷⁹Br. The M+2 peak at 160 is due to ⁷⁹Br−⁸¹Br while the M+4 peak at 162 is due to ⁸¹Br−⁸¹Br.

12. Use the natural abundance of bromine isotopes to explain the 1:2:1 height ratio for the m/z 158, 160 and 162 peaks in the mass spectrum of bromine. Hint: Consider the probability of forming each molecule.

    The simplest way to illustrate the height ratio is to imagine or count the number of ways you can construct Br−Br molecules from individual isotopes available in their natural abundance. Assuming Br−79 and Br−81 are present in equal numbers, there is one chance in four (p = 0.25) of getting ⁷⁹Br−⁷⁹Br (m/z 158) and a second chance in four of ⁸¹Br−⁸¹Br (m/z 162).There are two ways of generating m/z 160, ⁷⁹Br−⁸¹Br and ⁸¹Br−⁷⁹Br, and the remaining probability is thus two chances in four, or p = 0.5.
    
    Mathematically, the probability of two events A and B both occurring is equal to the probability of A times the probability of B. Let A be the chance of picking a particular isotope for the “first” atom in a molecule of Br₂, and B that of picking an isotope to be the second atom in the molecule. p (A) = p (B) = 0.50. The likelihood of forming ⁷⁹Br−⁷⁹Br (m/z 158) is (0.5)(0.5) = 0.25, which is also the probability of forming ⁸¹Br−⁸¹Br (m/Mz 162). The combined probability of all possible events or combinations must equal one, so p (160) = 0.5.

13. (Optional) Estimate and explain the expected height ratio of the m/z 70, 72 and 74 molecular ion peaks in the mass spectrum of chlorine gas (Cl₂). Recall the natural abundance of chlorine isotopes, 75% Cl–35 and 25% Cl–37.

    The peaks at m/z 70 and 74 are due to ³⁵Cl−³⁵Cl and ³⁷Cl−37Cl, respectively. If we round p = 0.75 for Cl−35 and p = 0.25 for Cl−37, the estimated probabilities are: p = (0.75)(0.75) = 0.5625 for ³⁵Cl−³⁵Cl (m/z 70); and p = (0.25)(0.25) = 0.0625 for ³⁷Cl−³⁷Cl (m/z 74). The remaining probability [1 − (0.5625 + 0.0625)] = 0.375 is assigned to molecules containing one each Cl−35 and Cl−37 (m/z 72). Dividing each probability by the smallest value (0.0625) gives an expected height ratio of 9:6:1 for the m/z 70, 72 and 74 molecular ion peaks.

Part B. Ionization Methods and Fragmentation Patterns in Mass Spectrometry

1. Compare and contrast electron ionization (EI) and chemical ionization (CI) in terms of the relative energies of the ions produced in mass spectrometry, the intensity of the molecular ion, and the number of fragment ions that may appear in the mass spectrum of a compound.

   Electron ionization gives rise to more energetic ions than chemical ionization. The higher energy results in more fragment ions forming in EI than in CI. As a consequence, the intensity or relative abundance of the molecular ion peak is usually greater in CI than in EI.

2. The mass spectrum of butyl methacrylate (molecular formula C₈H₁₅O₂) was obtained by both EI and CI. Which spectrum in Figure 5 was most likely obtained by EI? Explain your reasoning.
   {13775_Answers_Figure_5_Mass spectra of butyl methacrylate}

   The molecular mass of butyl methacrylate is (8 x 12) + (15 x 1) + (2 x 16) = 143 amu. Mass spectrum (B) does not have a molecular ion (M⁺) peak at m/z 143 and is thus more likely the mass spectrum generated by EI. CI mass spectra always exhibit the M⁺ peak because the ions are generated with less energy and are thus less likely to fragment.

3. “The mass spectrum of a compound is unique and characteristic of its structure, much like a fingerprint.” Using the spectrum of 1-bromobutane in Figure 2 (see the Background section) as an example, describe why this statement is true.

   The mass spectrum of bromobutane shows a set of peaks of varying intensities due to the molecular ion, the presence of bromine isotopes, and specific fragment ions associated with the way the atoms are joined together via chemical bonds. The pattern of peaks is highly dependent on and reflects both the molecular formula and the structure of the compound. The pattern would be different if either the formula or the structure were different. Thus even compounds that are closely related, such as isomers, would be expected to have unique and characteristic spectra, just like fingerprints are unique to an individual.

4. Comparing the mass spectra of butyl methacrylate in Figure 5, describe one very important limitation to the general statement quoted in italics in Question 3.

   In order to compare spectra and assign them to the same or different compounds, it is necessary for the spectra to be obtained using the same ionization technique.

5. Mass spectrometry is primarily a method of qualitative analysis. Why isn’t it useful for quantitative analysis?

   The intensity of a molecular ion peak in the mass spectrum of a compound depends on how the ions are generated and also on the relative ease of that ion fragmenting. Since intensity is not directly proportional to concentration, mass spectrometry is not well suited to quantitative analysis.

6. Mass spectrometry is a “workhorse” instrument in forensic analysis, toxicology and drug analysis. Its use is often portrayed on television shows, where the process of forensic analysis is simplified, making it appear that a computer can instantaneously identify the names and structures of all compounds in an evidence sample. What is the minimum information that must be stored in a computer library for this process to work?

   The computer must have stored digital spectral information for each compound that may be identified in the mass spectrum of a sample mixture. The spectra must be generated under the same conditions, although they don’t have to be from the same instrument.

7. Proteins and other large molecules are nonvolatile and degrade when heat or energy is applied, making them unsuitable for analysis by EI or CI mass spectrometry. In 2002, the Nobel Prize in Chemistry was awarded to three individuals for their discovery of electrospray ionization (ESI) as an alternative method of ionizing proteins and obtaining their mass spectra. Look up electrospray ionization online and briefly summarize how ions are generated using this technique.

   ESI is carried out by first dissolving a nonvolatile solute, such as a protein, in a solvent and then spraying the solution through a capillary tip into an electric field. The capillary flow produces an aerosol of small liquid droplets that become charged in the electric field. The solvent is evaporated from the droplets, leaving behind charged ions of the solute molecules, which are sent through the analyzer in a mass spectrometer.

8. Figure 6 shows the structural formula and mass spectrum of 4-heptanone. Verify that the highest mass peak in the mass spectrum corresponds to the molecular ion for this compound. Note: Use the mass numbers for the most common isotopes of C, H and O to calculate the molecular mass.
   {13775_Answers_Figure_6}

   The molecular formula of 4-heptanone is C₇H₁₄O, and its molecular mass is (7 x 12) + (14 x 1) + (1 x 16) = 114 amu. The m/z ratio for the highest mass peak is 114; this is the molecular ion M⁺.

9. Identify the two tallest (most intense) peaks in the mass spectrum of 4-heptanone. (a) How are the masses of these fragment ions related mathematically to the mass of the molecular ion? (b) What is the significance of this mathematical relationship in terms of bond breakage?

   The tallest peaks in the mass spectrum of 4-heptanone are found at m/z 71 and 43. The molecular ion peak M⁺ = 114. The m/z 71 fragment ion is equal to M⁺ − 43, and the m/z 43 fragment ion peak is equal to M⁺ − 71. Since 71 + 43 = 114, two fragments may arise via breakage of the same bond.

10. (a) Draw the structure of the molecular ion for 4-heptanone and show by means of a dashed line which bond may be broken to produce the fragment ions identified in Question 10. (b) The molecular ion is a radical cation. Explain how bond breakage can lead to either fragment ion.
    1. {13775_Answers_Figure_8}
    2. Recall that a molecular ion is generated by loss of one electron from a neutral parent molecule and is thus a radical cation, meaning it has a positive charge and one unpaired electron. For convenience we can show the unpaired electron on the O atom. Breakage of the C(O)−CH₂ bond will generate two fragments, m/z 43 and 71, and either fragment may have the positive charge, leaving the other fragment with an unpaired electron and electrically neutral.
    Part C. Molecular Structures and Mass Spectra of Organic Compounds
    
    The names and structural formulas of six organic compounds are given (1−6). These are followed by a set of spectra A−F. Calculate the molecular mass for the most common isotopic composition of each compound, identify its mass spectrum, and explain the possible origin of at least one major fragment ion in the spectrum. All mass spectra were obtained by EI—note that spectra are not presented in order!
    {13775_Answers_Table_3}
    Assign each mass spectrum A–F to the correct compound, 1–6. The spectra are not in order!
    {13775_Answers_Figure_9}

References

AP^® Chemistry Curriculum Framework: 2013−2014; The College Board, 2011.

Introduction

Mass spectrometry was developed in the search for and discovery of isotopes almost 100 years ago, and has evolved into an incredibly powerful tool for analyzing the mass and structure of compounds. It is widely used in laboratories all across the world for forensic analysis of trace amounts of substances, and also in research to determine the structures of large and complex natural products, such as proteins and nucleic acids. What are the basic principles of mass spectrometry and how are they applied to determine the structure of a compound?

Concepts

• Mass spectrometry
• Ionization
• Atomic theory
• Isotopes
• Natural abundance
• Atomic mass
• Molecular ions
• Fragment ions
• Bonding and molecular structure

Background

The historical development of mass spectrometry can be traced back to the work of the British scientist J. J. Thomson, who obtained precise mass/charge measurements of electrons emanating from the negative electrode in a gas discharge tube. Thomson was awarded the Nobel Prize in Physics in 1906 for the characterization of negatively charged electrons as a universal constituent of matter. Further investigation led Thomson to also study the nature of positively charged streams of atoms generated when electrons are stripped away from a gas under high voltage. When these positive ions were “bent” or deflected in the presence of electric and magnetic fields and then allowed to strike a photographic film, they left curved spots on the film at an angle that depended on their mass-to-charge ratios. In 1912, Thomson found that using neon as the gas source produced two spots. These results implied that neon consisted of or contained two types of atoms having different masses.

Conclusive proof for the existence of isotopes came from the work of Francis Aston, Thomson’s protégé at Cambridge University. Aston built a modified, more accurate version of the gas-discharge apparatus, which he called a mass spectrograph. In 1919 he obtained precise measurements for major and minor isotopes of neon, corresponding to 20 and 22 atomic mass units (amu), respectively. Aston received the Nobel Prize in Chemistry in 1922 for his discovery of isotopes. Advances in instrumentation throughout the 1920s and 30s allowed scientists to measure the masses and natural abundances of isotopes for the known elements in the periodic table. In 1939−40, Alfred Nier of the University of Minnesota, a pioneering figure in the design and construction of even more powerful mass spectrometers, used one of his instruments to separate the U-235 and U-238 isotopes of uranium. Nier supplied microgram samples of the isotopes to Enrico Fermi and his colleagues working on the Manhattan project. The samples were used to prove that U-235 was the isotope responsible for fission.

From these early developments, mass spectrometry has evolved into a powerful method of chemical analysis based on the production, separation, and measurement of charged atoms and molecules according to their masses. While modern mass spectrometers incorporate many advanced features, all mass spectrometers share the following basic design elements (see Figure 1). Atoms and molecules in a sample are stripped of electrons and converted to positively charged ions in an ion source, and are then sent through a mass analyzer, where they are separated according to their mass-to-charge ratio (m/z) by the application of electric and magnetic fields. The masses and relative amounts of the separated ions are measured in a detector, and computer processing is used to analyze the results and display them in chart form. The entire process is carried out under high vacuum conditions.

At first restricted to the study of volatile substances, modern mass spectrometry may be used to analyze solids, liquids or gases. Advanced ionization techniques have been developed to permit the formation of gaseous ions from any size molecule and any sample state, even very large molecular weight proteins with masses of thousands of dalton units (Da, where 1 Da = 1 amu). The two most common types of ionization for volatile samples (generally, gases or liquids) are electron ionization (EI) and chemical ionization (CI). In the EI process, a sample is inserted directly into an ionization chamber, where it is exposed to a beam of electrons that have been accelerated to approximately 70 eV of energy. The electron beam strips electrons away from sample molecules (M), producing positively charged radical ions (M⁺). See Equation 1. In the process, the molecular ions M⁺ also absorb some of the excess electron energy, making them susceptible to fragmentation. The molecular or parent ions break apart into different size fragment ions, as discussed below. Because all of the ions will be separated by the mass analyzer and produce signals at various m/z values, the pattern of peaks (signal intensity versus m/z) in the mass spectrum is unique and characteristic of the structure of a compound, much like a fingerprint. Chemical ionization (CI) is a milder ionization technique. In CI, a reagent gas such as methane (CH₄) is first ionized by electron impact, and the sample M to be analyzed is then exposed to the reagent ions. Collisions and reactions between M and the reagent ions produce molecular ions M⁺ as well as fragment ions. The extent of fragmentation is generally less in CI than in EI, making it easier to associate the heaviest major peak in the mass spectrum with the molecular ion. Fragment ions are produced by bond breakage in the structure of a compound, with certain bonds being more likely to break. For typical organic compounds with many carbon atoms and just a few heteroatoms (e.g., X=O, N, Cl, Br, S), breakage of a C−X bond is common, as is breakage at highly branched carbon atoms.

In almost all cases, a sample molecule will lose only one electron via either CI or EI, producing an ion with a charge z = +1. Because loss of more than one electron from a single molecule is rare, m/z reduces to m, the mass number of an ion. The mass analyzer and detector in a mass spectrometer detect the masses of individual ions. In calculating the mass of a molecular ion or fragment ion it is necessary to add up the actual mass or mass numbers of individual atoms or nuclides in the formula of a molecule, NOT the average atomic mass of an element. Consider methyl chloride (CH₃Cl)—its average molar mass is 50.45 amu. Chlorine exists, however, as a mixture of two isotopes, Cl-35 and Cl-37, with the natural abundance of each equal to 75.8% and 24.2%, respectively. The mass spectrum of CH₃Cl shows two peaks for individual molecular ions. A major peak occurs at m / z = 50 and is designated M⁺. This is due to CH₃−³⁵Cl. A second peak at m/z = 52 arises from CH<sub3−³⁷Cl and is designated M+2. The height ratio of the M⁺/M+2 peaks is 3:1, reflecting the natural abundance for the two chlorine isotopes. (The major isotopes of C and H are C-12 and H-1, and their mass numbers are used to determine the mass of the CH₃ fragment, 15 amu.)> Figure 2 shows the mass spectrum and fragmentation pattern of 1-bromobutane, an organic compound with four carbon atoms in a hydrocarbon chain and a bromine atom at the end of the chain. The mass spectrum can be interpreted as follows to identify the structure of 1-bromobutane.

• Two closely spaced peaks at m/z = 136 and 138 appear because bromine has two isotopes, Br-79 and Br-81, with natural abundances of 50.7% and 49.3%, respectively. The height ratio of these peaks, labeled M⁺ and M+2, respectively, is 1:1.
• The tallest or most intense peak in the mass spectrum occurs at m/z = 57. This fragment ion is labeled M−79 and is due to the loss of a Br atom from the M⁺ ion (136 − 79 = 57). The C−Br bond is easily broken, giving rise to the fragment ion CH₃CH₂CH₂CH₂⁺.
• Breakage of a C−C bond in the middle of the hydrocarbon chain leads to the loss of a CH₃CH₂ radical and produces major fragment ion peaks at 107 and 109 labeled M−29 (136 – 29 = 107).

Experiment Overview

The purpose of this guided-inquiry learning activity is to investigate the principles and applications of mass spectrometry. The activity is divided into three parts.

1. Isotopic abundance and the mass spectra of elements.
2. Ionization methods and fragmentation patterns in mass spectrometry.
3. Molecular structures and mass spectra of organic compounds.

Student Worksheet PDF

13775_Student1.pdf