Mass Spectrometry and Atomic Identification (4E) - Chem & Physics

A researcher analyzes a sample containing chlorine atoms using a mass spectrometer that produces singly charged atomic ions ( $z=+1$ ). Two dominant peaks appear at $m/z=35$ and $m/z=37$ with an intensity ratio of approximately 3:1. (Principle: isotope patterns reflect natural abundance.) Which conclusion is most consistent with the data?
Answer: The sample contains chlorine, with $^{35}$ Cl about three times as abundant as $^{37}$ Cl The skill being tested is recognizing halogen isotope patterns in mass spectrometry. In mass spectrometry, atomic identification uses natural isotope abundances, where peak intensities reflect relative isotopic prevalence for singly charged ions. The spectrum shows peaks at m/z=35 and 37 with a 3:1 intensity ratio for chlorine. Choice B is correct because chlorine's isotopes ^35Cl (75.8%) and ^37Cl (24.2%) match this ~3:1 ratio, consistent with the principle. Choice A fails because bromine's isotopes ^79Br and ^81Br have ~1:1 abundance, not 3:1, misconstruing the ratio for a different halogen. For similar tasks, calculate expected ratios from known abundances and compare to observed data. Always confirm charge state, as z>1 would alter m/z interpretations.
An instrument uses electrospray ionization (ESI) to analyze a small peptide. Two adjacent charge states are observed at $m/z=500$ and $m/z=400$ . (Principle: for the same molecule, increasing charge decreases $m/z$ .) Which interpretation is most consistent with these two peaks (ignore the small mass of added protons)?
Answer: They represent the same peptide with different charge states, with the $m/z=400$ peak having higher charge The skill being tested is interpreting charge states in electrospray ionization mass spectrometry. In mass spectrometry, for the same molecule, higher charge states result in lower m/z values since m/z = mass/charge. The ESI spectrum shows two peaks at m/z=500 and 400 for a peptide, interpreted as adjacent charge states. Choice B is correct because the m/z=400 peak has a higher charge (lower m/z) than m/z=500, consistent with the principle ignoring proton mass. Choice C fails by reversing the charge assignment, misconstruing that higher charge decreases m/z. To check similar problems, calculate possible masses assuming different z values and find consistent pairings. Remember that ESI often produces multiple charge states for peptides, spaced inversely with z.
An environmental chemistry group uses a mass spectrometer to compare two ionization methods for the same volatile compound: (i) electron impact (EI), which imparts high energy and causes extensive fragmentation, and (ii) a softer method that produces mostly intact molecular ions. The principle involved is that ionization energy affects fragmentation patterns but does not change the intrinsic mass of the intact molecule. Which observation is most consistent with switching from EI to the softer ionization method? (Constants: $1\ \text{u}=1.66\times10^{-27}\ \text{kg}$ .)
Answer: A higher relative intensity of the molecular ion peak (intact $M^+$ ) and fewer fragment peaks This question tests understanding of how ionization energy affects fragmentation patterns in mass spectrometry. Mass spectrometry ionization methods vary in the amount of energy they impart to molecules, with high-energy methods like electron impact (EI) causing extensive fragmentation, while softer methods preserve the molecular ion. When switching from EI to a softer ionization method, the molecular ion peak (M+) becomes more prominent because less energy is available to break chemical bonds, resulting in fewer fragment peaks. The correct answer A accurately describes this phenomenon: higher relative intensity of the intact molecular ion and fewer fragments. Option B incorrectly suggests that ionization energy changes the measured mass values, but m/z values depend only on the actual mass and charge of ions, not the ionization energy used to create them. When interpreting mass spectra, remember that fragmentation patterns depend on ionization method, but the m/z values of specific ions remain constant. A key strategy is to look for the molecular ion peak intensity as an indicator of ionization harshness.
A researcher analyzes a peptide by ESI-MS and observes a peak at $m/z = 750$ assigned to the $z=+2$ charge state. (Principle: $m \approx (m/z)\times z$ when the mass contribution of added protons is neglected for approximate reasoning.) Which is the most consistent approximate neutral mass of the peptide? Constants: none needed.
Answer: 1500 u This question tests calculation of molecular mass from m/z and charge state information. In ESI-MS, the relationship between neutral mass and m/z for multiply charged species is approximately m = (m/z) × z when the small mass contribution of added protons is neglected. For a peak at m/z = 750 assigned to z = +2, the neutral mass is approximately 750 × 2 = 1500 u. The correct answer C calculates this product correctly. Answer A incorrectly divides instead of multiplying, while B uses only the m/z value without accounting for charge. When determining molecular masses from ESI-MS data, multiply the observed m/z by the charge state for a quick approximation, then subtract the mass of added protons (z × 1 u) for more precision.
A lab compares two ionization methods for the same small molecule ( $M$ ): electron-impact (EI) and matrix-assisted laser desorption/ionization (MALDI). The core principle is that the ionization method affects fragmentation and charge state distribution, which changes how many peaks appear and where. EI produces a spectrum with many smaller peaks, while MALDI produces one dominant peak near the molecular ion mass with minimal fragments. Which outcome is most consistent with switching from EI to MALDI for the same analyte? Constants: $e = 1.60\times 10^{-19}\ \text{C}$ ; $1\ \text{amu} = 1.66\times 10^{-27}\ \text{kg}$ .
Answer: A dominant intact-ion peak with fewer fragment peaks This question tests understanding of how different ionization methods affect fragmentation patterns and spectral appearance in mass spectrometry. The key principle is that 'soft' ionization methods like MALDI deposit less energy into molecules compared to 'hard' methods like electron-impact (EI), resulting in different fragmentation patterns. MALDI typically produces intact molecular ions with minimal fragmentation because the matrix absorbs most of the laser energy and gently transfers charge to the analyte, while EI bombards molecules with high-energy electrons causing extensive bond breaking. Therefore, switching from EI to MALDI for the same molecule would show a dominant intact-ion peak (usually [M+H]⁺ or [M+Na]⁺) with few or no fragment peaks, contrasting with EI's multiple fragment peaks. Choice A incorrectly reverses this relationship, suggesting MALDI causes more fragmentation than EI, which contradicts the fundamental differences between soft and hard ionization. To predict ionization outcomes, remember the energy hierarchy: EI > CI > ESI ≈ MALDI, where higher energy means more fragmentation. Choose MALDI or ESI for intact molecular weight determination and EI for structural information from fragmentation patterns.
A chemist compares two ionization methods for analyzing a thermally labile organic compound: electron ionization (EI) and electrospray ionization (ESI). Under EI, the spectrum shows many fragment peaks and a weak molecular ion peak. Under ESI, the spectrum shows a strong peak at $m/z=301$ and additional peaks at $m/z=151$ and $m/z=101$ that correspond to multiply charged ions of the same molecule. Which outcome is most consistent with the difference in ionization methods?
Answer: EI is more likely to produce extensive fragmentation, whereas ESI is more likely to preserve the intact molecular ion and generate multiple charge states This question tests understanding of different ionization methods in mass spectrometry and their effects on molecular ion preservation and charge state formation. Electron ionization (EI) uses high-energy electrons that often cause extensive fragmentation of molecules, while electrospray ionization (ESI) is a "soft" technique that preserves intact molecular ions and commonly produces multiply charged species. The ESI spectrum showing peaks at m/z = 301, 151, and 101 represents the same molecule with charges +1, +2, and +3 respectively (since 301/1 = 301, 301/2 = 150.5 ≈ 151, and 301/3 = 100.3 ≈ 101). The correct answer A accurately describes these characteristic differences between EI and ESI. Answer B reverses the typical behavior of these techniques, incorrectly attributing multiple charge states to EI rather than ESI. When choosing an ionization method, consider that harsh methods like EI cause fragmentation useful for structure elucidation, while soft methods like ESI preserve molecular ions and enable analysis of large, fragile biomolecules through multiple charging.
An analyst observes a peak at $m/z=14$ while running MS on a nitrogen-containing compound. The instrument is set to detect singly charged ions, and common fragments include atomic ions. (Principle: fragment ions can correspond to atomic masses when $z=+1$ .) Which assignment is most consistent with the $m/z=14$ peak? Constants: none needed.
Answer: $\text{N}^+$ This question tests identification of fragment ions in mass spectrometry of nitrogen-containing compounds. When nitrogen-containing molecules undergo electron ionization, they can fragment to produce atomic nitrogen ions. A peak at m/z = 14 for a singly charged ion corresponds to a mass of 14 u, which matches atomic nitrogen (¹⁴N). The correct answer A identifies N⁺ as the fragment ion at m/z = 14. Answer B (N₂⁺) would appear at m/z = 28, while C (CH₄⁺) would appear at m/z = 16, and D (O⁺) at m/z = 16. When analyzing fragments from nitrogen compounds, expect to see N⁺ at m/z = 14 as a common fragment, particularly under hard ionization conditions.
A time-of-flight (TOF) mass spectrometer accelerates ions from rest through a potential difference of $V=2.0\,\text{kV}$ before they drift a fixed distance to the detector. Two ions are produced from the same compound: Ion 1 is $\text{M}^+$ and Ion 2 is $\text{M}^{2+}$ . Both have the same mass $m$ . Constants: $e=1.60\times10^{-19}\,\text{C}$ . What change in the measured mass-to-charge ratio ( $m/z$ ) would be expected for Ion 2 relative to Ion 1?
Answer: Ion 2 has half the $m/z$ of Ion 1 because its charge magnitude is doubled This question tests understanding of how charge state affects m/z measurements in time-of-flight mass spectrometry. In mass spectrometry, m/z is calculated as mass divided by charge magnitude, so for ions with the same mass but different charges, the m/z values will differ inversely with charge. For Ion 1 (M+) with charge +1e, the m/z equals m/1 = m, while for Ion 2 (M2+) with charge +2e, the m/z equals m/2, which is half that of Ion 1. The correct answer A recognizes this fundamental relationship: doubling the charge halves the m/z ratio. Answer B incorrectly suggests m/z would double, confusing the effect of charge on acceleration with its effect on the m/z calculation itself. When comparing ions of the same mass but different charge states, remember that m/z varies inversely with charge number: M2+ appears at half the m/z of M+, M3+ at one-third, and so on.
A sample contains a mixture of $\text{CO}$ and $\text{N}_2$ , both of which have molecular mass 28 u. The MS is operated to detect singly charged molecular ions only, and both species produce a peak at $m/z=28$ . (Principle: MS separates by $m/z$ ; isobars can be indistinguishable without additional information.) What is most consistent with the mass spectrometry result? Constants: none needed.
Answer: Both gases can contribute to the same $m/z=28$ peak, so MS alone may not differentiate them here This question tests understanding of mass spectrometry's limitations in distinguishing isobaric species. Isobars are different chemical species with the same mass, and when they carry the same charge, they appear at identical m/z values in standard mass spectrometry. CO and N₂ both have molecular mass 28 u (C=12, O=16, N=14), so their singly charged molecular ions both appear at m/z = 28. The correct answer B recognizes that MS alone cannot differentiate these isobaric species without additional techniques like high-resolution MS or tandem MS. Answer A incorrectly claims MS can distinguish them using only mass analysis, while C wrongly states CO cannot be ionized. When analyzing mixtures by MS, be aware that different molecules with identical masses produce overlapping peaks and require additional methods for differentiation.
A time-of-flight (TOF) mass spectrometer accelerates ions of equal charge through the same potential difference $V$ before they drift a fixed distance $L$ to the detector. (Constants: $q=1.60\times10^{-19}\ \text{C}$ .) If two singly charged ions have masses 12 amu and 16 amu, which ion will reach the detector first, assuming equal kinetic energy after acceleration?
Answer: The 12 amu ion, because lower mass gives higher velocity at fixed energy The skill being tested is understanding time-of-flight principles in mass spectrometry. In TOF mass spectrometry, ions accelerated through the same potential gain kinetic energy qV, so lighter ions achieve higher velocities and arrive sooner. The two ions have masses 12 amu and 16 amu, both singly charged, accelerated equally. Choice B is correct because the 12 amu ion has lower mass, thus higher velocity, reaching the detector first. Choice A fails by stating higher mass gives higher velocity, misconstruing the inverse mass-velocity relationship at fixed energy. For similar problems, calculate velocity from KE = (1/2) $mv^2$ to predict arrival order. Remember that TOF separates by mass due to velocity differences post-acceleration.
A researcher uses a mass spectrometer to identify an unknown compound that can form either a molecular ion $M^+$ or a sodium adduct $[M+Na]^+$ . Under the same conditions, two peaks appear at $m/z=100$ and $m/z=123$ , and both are singly charged. The principle involved is that adduct formation increases the mass of the detected ion while keeping charge the same. Which assignment is most consistent with the data? (Constants: $m(\mathrm{Na})\approx 23\ \text{u}$ .)
Answer: $m/z=100$ is $M^+$ and $m/z=123$ is $[M+Na]^+$ This question tests understanding of adduct formation in mass spectrometry and its effect on m/z values. In mass spectrometry, molecules can form adducts with metal ions like sodium, increasing the detected mass by the mass of the adducting species while maintaining the same charge. The 23-unit difference between the peaks at m/z=100 and m/z=123 exactly matches the mass of sodium (23 u), indicating one peak is the molecular ion and the other is the sodium adduct. Since [M+Na]+ has higher mass than M+, it must appear at the higher m/z value, making m/z=100 the molecular ion M+ and m/z=123 the sodium adduct [M+Na]+. The correct answer B accurately identifies this assignment. Option A reverses the assignment, incorrectly placing the heavier sodium adduct at the lower m/z value, which violates the fundamental principle that higher mass produces higher m/z for the same charge. When analyzing adduct peaks, remember that metal adducts always appear at higher m/z than the molecular ion by the mass of the metal. A reliable strategy is to look for mass differences matching common adducting species: +23 for Na+, +39 for K+, or +1 for H+.
A biochemistry lab uses mass spectrometry to confirm the identity of a purified protein by measuring its intact mass. Using ESI, the protein appears as a series of multiply charged peaks. One peak is at $m/z=2000$ and is assigned charge state $z=+10$ . The principle involved is that the neutral mass is approximately $M\approx (m/z)\cdot z$ when the added proton masses are small relative to $M$ . What is the most consistent estimate of the protein's mass? (Constants: $m(\mathrm{H^+})\approx 1\ \text{u}$ .)
Answer: $\approx 20{,}000\ \text{u}$ , because $2000\times 10=20{,}000$ This question tests the calculation of protein mass from multiply charged ions in electrospray ionization mass spectrometry. For a multiply charged ion, the relationship between observed m/z, charge state z, and neutral molecular mass M is: m/z = (M + z×1)/z ≈ M/z when M >> z. Given m/z=2000 and z=+10, we can solve for M: M ≈ (m/z) × z = 2000 × 10 = 20,000 u. The correct answer C accurately performs this calculation. Option A incorrectly divides m/z by charge instead of multiplying, yielding 200 u, which would be far too small for any protein and represents a fundamental misunderstanding of the m/z relationship. When analyzing multiply charged protein ions, remember that the neutral mass is approximately the product of m/z and charge state. A useful check is that proteins typically have masses in the range of thousands to hundreds of thousands of daltons, so results outside this range suggest calculation errors.
A protein digest is analyzed by ESI-MS. A peptide produces peaks at $m/z=600$ and $m/z=300$ that are assigned to the same peptide at different charge states. (Principle: $m/z$ scales inversely with charge for the same mass.) Which charge assignments are most consistent with this observation (ignore proton mass)?
Answer: $m/z=600$ is $z=+1$ and $m/z=300$ is $z=+2$ The skill being tested is assigning charge states to peptide peaks in ESI mass spectrometry. In mass spectrometry, for multiply charged ions, m/z decreases as charge z increases for a fixed molecular mass. The peaks at m/z=600 and 300 are from the same peptide at different charges. Choice A is correct because m/z=600 as z=+1 gives mass 600, and m/z=300 as z=+2 gives mass 600, consistent with the principle. Choice B fails by reversing assignments, misconstruing that higher m/z corresponds to lower z. For verification, solve for mass = m/z * z and find consistent values across peaks. Note that ESI charge states are typically higher for lower m/z in a series.
A forensic lab uses mass spectrometry to identify an unknown gas. The instrument produces primarily singly charged ions. The spectrum shows a dominant peak at $m/z=44$ and smaller peaks at $m/z=45$ and $m/z=46$ . The principle used is that isotopic substitution changes mass (and thus $m/z$ for $z=+1$ ) without changing chemical identity. Which molecular identification is most consistent with this pattern? Constants: $1\ \text{amu} = 1.66\times 10^{-27}\ \text{kg}$ ; $e = 1.60\times 10^{-19}\ \text{C}$ .
Answer: $\text{CO}_2$ , because $^{13}\text{C}$ and $^{18}\text{O}$ isotopes can yield $m/z=45$ and 46 near a main $m/z=44$ peak This question tests molecular identification through isotopic patterns in mass spectrometry, specifically recognizing how isotopic substitution creates characteristic peak patterns. In mass spectrometry of molecules, isotopic variants appear as satellite peaks at higher m/z values, with intensities reflecting natural isotopic abundances. The dominant peak at m/z = 44 with smaller peaks at 45 and 46 is characteristic of CO₂, where the main peak represents ¹²C¹⁶O₂ (mass 44), while m/z = 45 includes ¹³C¹⁶O₂ or ¹²C¹⁶O¹⁷O, and m/z = 46 includes ¹²C¹⁶O¹⁸O or ¹³C¹⁶O¹⁷O variants. This pattern is diagnostic for CO₂ because it's one of the few common gases with mass 44 that shows significant M+1 and M+2 peaks due to carbon and oxygen isotopes. Choice A incorrectly suggests N₂ produces m/z = 44, but N₂ has mass 28 (2 × 14), and its isotopes would appear near m/z = 29, not 44. When identifying molecules by isotope patterns, consider which elements have significant minor isotopes: carbon (~1% ¹³C), oxygen (~0.2% ¹⁸O), and remember that multiple substitutions become less probable. The M+1 and M+2 pattern with decreasing intensity is characteristic of molecules containing C and O.
In a TOF mass spectrometer, ions are accelerated from rest through a potential difference $V=2000\ \text{V}$ . Two ions have the same mass but charges $+e$ and $+2e$ . (Constants: $e=1.60\times10^{-19}\ \text{C}$ .) (Principle: $qV$ sets kinetic energy after acceleration.) Which statement is most consistent with the expected TOF results?
Answer: The $+2e$ ion has higher kinetic energy and arrives earlier The skill being tested is analyzing charge effects on time-of-flight in mass spectrometry. In TOF mass spectrometry, kinetic energy is qV, so higher charge leads to higher energy and velocity, thus shorter flight time. The ions have same mass but charges +e and +2e, accelerated through V=2000 V. Choice B is correct because +2e ion gains twice the energy, arriving earlier. Choice A fails by stating lower energy for +2e, misconstruing qV relationship. To verify, calculate KE = qV and v = sqrt(2KE/m) for arrival times. Remember charge multiplies energy in acceleration, affecting TOF.
In a time-of-flight (TOF) mass spectrometer, ions are accelerated through the same potential difference $V$ and then drift a fixed distance. The principle is that ions with larger $m/z$ travel more slowly when given the same kinetic energy: $KE = z e V = \tfrac{1}{2}mv^2$ . An unknown sample produces two singly charged peaks at $m/z=35$ and $m/z=37$ with an intensity ratio of about 3:1. Based on the spectra, which atom is most likely present? Constants: $e = 1.60\times 10^{-19}\ \text{C}$ ; $1\ \text{amu} = 1.66\times 10^{-27}\ \text{kg}$ .
Answer: Chlorine, due to isotopes at 35 and 37 amu with ~3:1 abundance This question tests atomic identification through isotopic patterns in time-of-flight mass spectrometry, where ions are separated based on their velocity after acceleration through a fixed potential. In TOF-MS with singly charged ions (z=+1), the m/z value directly corresponds to the isotopic mass, making isotope pattern recognition key to element identification. The observed peaks at m/z = 35 and 37 with a 3:1 intensity ratio precisely match chlorine's natural isotopic distribution: Cl-35 (~75.8%) and Cl-37 (~24.2%), giving approximately a 3:1 ratio. This isotopic signature is unique to chlorine among common elements, as it's one of the few with two stable isotopes separated by 2 amu with this specific abundance ratio. Choice B incorrectly suggests sulfur isotopes shift mass during ionization, which violates mass conservation principles - ionization removes electrons but doesn't change nuclear mass. When identifying elements by mass spectrometry, focus on both the mass difference between isotopes (here, 2 amu) and their relative intensities. Remember that chlorine and bromine are the two halogens with distinctive two-peak patterns, but bromine's isotopes (79 and 81) appear at much higher m/z values with nearly 1:1 intensity.
A researcher increases the ionization energy in an EI source while analyzing the same compound. The molecular ion peak intensity decreases, and several lower- $m/z$ fragment peaks increase in intensity, while the molecular ion $m/z$ value stays the same. (Principle: fragmentation changes the distribution of ions but not the $m/z$ position of a given ion.) Which conclusion about the ionization process is most consistent with the data? Constants: none needed.
Answer: Higher ionization energy increased fragmentation, reducing molecular ion abundance without shifting its $m/z$ This question tests understanding of how ionization energy affects fragmentation without changing molecular ion m/z values. In electron ionization MS, increasing ionization energy deposits more internal energy into molecules, increasing fragmentation probability while the m/z value of any given ion (including the molecular ion) remains unchanged. The observation of decreased molecular ion intensity with increased fragment peaks at lower m/z values indicates enhanced fragmentation. The correct answer B recognizes that higher ionization energy increases fragmentation, redistributing signal from the molecular ion to fragments without shifting the molecular ion's m/z position. Answer A incorrectly suggests the mass changes with ionization energy, while C implies a calibration shift affecting all peaks. Remember that m/z values are determined by mass and charge, not by the ionization energy used to create the ions.
A scientist calibrates a mass spectrometer using a standard that forms singly charged ions. The instrument reports a peak at $m/z=40$ for the standard. Later, an unknown sample shows a peak at $m/z=20$ under identical ionization conditions. (Principle: $m/z$ decreases if charge increases for the same mass.) Which scenario could explain the $m/z=20$ peak if the unknown has the same mass as the standard?
Answer: The unknown forms predominantly doubly charged ions ( $z=+2$ ) instead of singly charged ions The skill being tested is explaining m/z shifts due to charge state changes in mass spectrometry. In mass spectrometry, for the same mass, higher charge z decreases m/z since m/z = mass/z. The unknown at m/z=20 compared to standard at m/z=40 suggests same mass but different z. Choice A is correct because z=+2 for unknown halves m/z to 20. Choice B fails by suggesting double mass, which would double m/z, not halve it. For similar cases, calculate possible z values to match masses. Assume consistent conditions unless specified.
An analyst suspects a compound contains bromine. A mass spectrum of the molecular ion region shows two peaks separated by 2 amu with approximately equal intensities. (Principle: halogen isotope patterns in MS.) Which conclusion is most consistent with the spectrum?
Answer: The compound likely contains one bromine atom, because $^{79}$ Br and $^{81}$ Br occur in ~1:1 abundance The skill being tested is recognizing halogen isotope signatures in mass spectrometry. In mass spectrometry, halogens like Br show characteristic 1:1 doublets at M and M+2 for one atom due to ^79Br/^81Br. The molecular ion shows two peaks 2 amu apart with equal intensities. Choice A is correct because this matches one Br atom's ~1:1 abundance ratio. Choice B fails by attributing to Cl, which has 3:1 ratio, not 1:1, confusing halogen abundances. To confirm, recall binomial patterns for halogens and match ratios. Note spacing of 2 amu typically indicates Br or Cl isotopes.
A lab analyzes an unknown element by MS and observes peaks at $m/z = 63$ and $m/z = 65$ with an intensity ratio of about 69:31. (Principle: isotope abundances determine peak ratios for singly charged ions.) Based on the spectra, which atom is most likely present? Constants: none needed.
Answer: Copper, because $^{63}\text{Cu}$ and $^{65}\text{Cu}$ occur with ~69% and ~31% abundance This question tests element identification through isotopic abundance patterns in mass spectrometry. The observation of peaks at m/z = 63 and 65 with a 69:31 intensity ratio is characteristic of copper's natural isotopic distribution. Copper has two stable isotopes: ⁶³Cu (69.2%) and ⁶⁵Cu (30.8%), matching the observed ~69:31 ratio. The correct answer B identifies copper based on this distinctive isotopic signature. Answer A incorrectly describes zinc's isotope pattern, which has five stable isotopes with ⁶⁴Zn being most abundant, not a simple two-isotope pattern at 63 and 65. To identify elements by isotope patterns, compare both the m/z values and intensity ratios to known natural abundances, with copper showing its characteristic ~69:31 ratio at masses 63:65.

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