MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4e Mass Spectrometry Atomic Identification

What this quiz covers

This quiz focuses on 4e Mass Spectrometry Atomic Identification, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Fewer ions were produced or transmitted, but the ions that were detected had the same mass-to-charge ratios as before (correct answer)
2. The molecular masses increased, but the instrument automatically corrected the $m/z$ axis to keep peaks fixed
3. The sample concentration increased, causing detector saturation that shifts peaks to lower $m/z$ values
4. The compound absorbed less infrared light after the settings change, reducing the apparent mass signal

Explanation: This question tests understanding of the relationship between ion production and mass spectral features. In mass spectrometry, the m/z values of ions are intrinsic properties determined by their mass and charge, independent of how many ions are produced, while peak intensity reflects the number of ions detected. When ionization efficiency decreases, fewer ions are produced and detected, reducing signal intensity, but the m/z values of the ions that are formed remain unchanged because their mass-to-charge ratios are not affected by the ionization conditions. The correct answer A accurately describes this phenomenon: fewer ions but unchanged m/z values. Option C incorrectly suggests that detector saturation from increased concentration would shift m/z values, but detector response affects only intensity, not the measured m/z positions. When troubleshooting mass spectrometry experiments, remember that changes in ionization efficiency affect peak heights but not peak positions. A key diagnostic is that m/z shifts indicate calibration issues or instrumental problems, while intensity changes alone typically indicate ionization or transmission efficiency variations.

Question 2

1. The sample contains two isotopes of the same element with masses 35 u and 37 u in a ~3:1 abundance ratio (correct answer)
2. The sample contains two different elements whose atomic numbers differ by 2, producing peaks 2 units apart
3. The 37 peak must be instrument noise because true isotope peaks must have equal intensity
4. The peaks indicate two charge states of a 70 u ion: $z=+2$ at 35 and $z=+1$ at 37

Explanation: This question tests the ability to identify isotope patterns in mass spectrometry for atomic identification. Mass spectrometry separates ions by their mass-to-charge ratio, and isotopes of the same element appear at different m/z values corresponding to their different masses when they have the same charge. The observed peaks at m/z=35 and m/z=37 with a 3:1 intensity ratio perfectly match the natural isotopic distribution of chlorine, which has two stable isotopes: Cl-35 (75.8% abundance) and Cl-37 (24.2% abundance), giving approximately a 3:1 ratio. The correct answer A accurately identifies this as an isotopic pattern of chlorine. Option D incorrectly suggests these could be different charge states of a 70 u ion, but this would require z=+2 at m/z=35 (70/2=35) and an impossible non-integer charge at m/z=37. When analyzing atomic mass spectra, look for peaks separated by small integer mass differences with intensity ratios matching known isotopic abundances. Remember that isotopes of the same element will have the same charge state under identical ionization conditions.

Question 3

1. Neon, because its isotopes produce peaks at $m/z=20$, 21, and 22 for $z=+1$ (correct answer)
2. Helium, because $m/z=20$ indicates a doubly charged ion of mass 40
3. Argon, because $m/z=20$ corresponds to a fragment of Ar formed in EI
4. Krypton, because heavier noble gases show multiple isotope peaks

Explanation: The skill being tested is identifying elements using isotope patterns in mass spectrometry. In mass spectrometry, atomic identification relies on the unique masses and relative abundances of isotopes for each element, where m/z corresponds to isotopic mass for singly charged ions. Here, the EI mass spectrum of a noble gas shows peaks at m/z=20, 21, and 22 with abundances 90%, 0.3%, and 9.7%, respectively. Choice A is correct because these match Neon's isotopes: ^20Ne (90.5%), ^21Ne (0.3%), ^22Ne (9.2%), consistent with the data for z=+1. Choice B fails because Helium's main isotope is ^4He, and m/z=20 would imply a mass 40 for z=+2, but Helium does not have an isotope at 40, reflecting a misconception about charge states. To verify similar patterns, compare observed abundances and m/z spacings to standard isotopic tables. Remember that noble gases often form singly charged ions in EI, so assume z=+1 unless evidence suggests otherwise.

Question 4

1. Flight time increases because the ion experiences more repulsion in the flight tube
2. Flight time decreases because the ion gains more kinetic energy at the same voltage (correct answer)
3. Flight time is unchanged because only mass affects TOF
4. Flight time becomes undefined because doubly charged ions cannot be detected

Explanation: This question tests understanding of charge effects on ion velocity in time-of-flight mass spectrometry. When an ion's charge doubles while mass remains constant, it gains twice the kinetic energy (K = qV) when accelerated through the same potential difference. Since K = ½mv², doubling the kinetic energy increases velocity by √2, thereby decreasing flight time. The correct answer B recognizes that higher charge leads to more kinetic energy, higher velocity, and shorter flight time. Answer A incorrectly suggests increased repulsion affects flight time, while C wrongly claims only mass matters in TOF. For TOF-MS calculations, remember that flight time is proportional to √(m/z), so doubling charge (halving m/z) reduces flight time by a factor of √2.

Question 5

1. The instrument can distinguish $\text{CO}^+$ from $\text{N}_2^+$ at the same $m/z$ using only mass analysis
2. Both gases can contribute to the same $m/z=28$ peak, so MS alone may not differentiate them here (correct answer)
3. Only $\text{N}_2$ can appear at $m/z=28$ because $\text{CO}$ is neutral and cannot be ionized
4. The peak at $m/z=28$ must be a hydrogen fragment because small fragments dominate spectra

Explanation: 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.

Question 6

1. The compound's molecular mass decreased due to the higher ionization energy
2. Higher ionization energy increased fragmentation, reducing molecular ion abundance without shifting its $m/z$ (correct answer)
3. The analyzer recalibrated, shifting all peaks to lower $m/z$
4. The observed changes are due to fluorescence quenching at higher energy

Explanation: 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.

Question 7

1. The $m/z=24$ peak is $M^{2+}$ and the $m/z=12$ peak is $M^{+}$ for the same metal
2. The $m/z=24$ peak is $M^{+}$ and the $m/z=12$ peak is $M^{2+}$ for the same metal (correct answer)
3. The two peaks must be two different elements because charge state does not affect $m/z$
4. The peaks indicate two different electronic transitions measured by atomic absorption spectroscopy

Explanation: This question tests understanding of how charge states affect m/z values in mass spectrometry and the ability to deduce charge states from peak patterns. The fundamental principle is that m/z = M/z, so doubling the charge halves the m/z value for the same mass M. The observation of peaks at m/z = 24 and m/z = 12, where one is exactly half the other, strongly suggests the same metal ion in different charge states. Since 12 = 24/2, the m/z = 12 peak must have twice the charge of the m/z = 24 peak, meaning if m/z = 24 is M⁺ (z=+1), then m/z = 12 is M²⁺ (z=+2), both from a metal with mass M = 24 amu (likely magnesium). Choice C incorrectly claims charge state doesn't affect m/z, contradicting the fundamental m/z = M/z relationship that defines mass spectrometry. To identify charge state relationships, look for m/z values that are simple fractions of each other (1/2, 2/3, 3/4). When peaks show identical isotopic patterns at different m/z values with integer relationships, they likely represent the same species at different charge states.

Question 8

1. The $m/z=14$ peak is most consistent with a fragment that has approximately half the mass of the $m/z=28$ parent ion (correct answer)
2. The $m/z=14$ peak must represent the doubly charged form of the $m/z=28$ ion
3. The $m/z=28$ peak is an isotope peak of the $m/z=14$ ion caused by $^{14}$C labeling
4. The two peaks indicate two different retention times from gas chromatography rather than ion masses

Explanation: This question tests understanding of fragmentation patterns in electron-impact ionization mass spectrometry, where high-energy electrons cause molecules to break into smaller pieces. In EI-MS, fragmentation produces ions with masses that are fractions of the parent ion mass, typically maintaining z=+1 charge, so fragment m/z values reflect actual fragment masses. The observation of peaks at m/z = 28 and m/z = 14, where one is exactly half the other, suggests that m/z = 14 is a fragment ion with approximately half the mass of the m/z = 28 parent ion. This pattern is consistent with molecules like N₂ (mass 28) fragmenting to N⁺ (mass 14) or CO (mass 28) fragmenting to C⁺ or O⁺, though the latter requires more energy. Choice B incorrectly interprets the m/z = 14 peak as a doubly charged form of m/z = 28, but EI typically produces singly charged ions, and fragmentation is the more common explanation for such patterns. To distinguish fragmentation from multiple charging, consider the ionization method: EI favors fragmentation and single charges, while ESI favors intact ions with multiple charges. When m/z values show simple integer relationships in EI-MS, fragmentation is usually the cause, not charge state differences.

Question 9

1. Copper, due to isotopes near 63 and 65 amu with ~7:3 abundance (correct answer)
2. Zinc, because its dominant isotopes at 64 and 66 amu shift to 63 and 65 after ionization
3. Gallium, because isotopes at 69 and 71 amu appear at 63 and 65 when singly charged
4. Nickel, because its multiple isotopes always produce exactly two peaks at 63 and 65

Explanation: This question tests atomic identification through isotopic abundance patterns in mass spectrometry, requiring recognition of characteristic isotope ratios. In mass spectrometry with singly charged ions, m/z values directly represent isotopic masses, and the intensity ratio reflects natural isotopic abundances unique to each element. The peaks at m/z = 63 and 65 with a 7:3 intensity ratio match copper's isotopic distribution: Cu-63 (~69.2%) and Cu-65 (~30.8%), giving approximately a 7:3 ratio. This specific combination of masses separated by 2 amu with this abundance ratio is diagnostic for copper, as it's the only common element with significant isotopes at exactly these masses with this ratio. Choice B incorrectly suggests zinc isotopes shift mass during ionization, but ionization only removes electrons without changing nuclear mass - Zn isotopes are at 64, 66, 67, 68, and 70, not 63 and 65. When identifying elements by isotope patterns, memorize key signatures: Cl (35/37, ratio 3:1), Br (79/81, ratio 1:1), and Cu (63/65, ratio 7:3). Always verify that both the m/z values AND intensity ratios match known isotopic data for conclusive identification.

Question 10

1. The $+2e$ ion has lower kinetic energy and arrives later
2. The $+2e$ ion has higher kinetic energy and arrives earlier (correct answer)
3. Both ions have the same kinetic energy because they experience the same voltage
4. Arrival times depend only on ionization energy, not on charge

Explanation: 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.

Question 11

1. Neon, because natural isotopes near 20 u and 22 u can produce peaks at $m/z=20$ and $m/z=22$ for $z=+1$ (correct answer)
2. Helium, because a 4 u atom commonly forms a $+2$ ion giving an apparent $m/z$ near 2
3. Oxygen, because $\mathrm{O_2}$ fragments into two atoms that each appear at $m/z=16$ and $m/z=8$
4. Argon, because its most abundant isotope is 40 u and would appear at $m/z=20$ if doubly charged

Explanation: This question tests the ability to identify atoms using mass spectrometry by analyzing isotope patterns. Mass spectrometry separates ions based on their mass-to-charge ratio (m/z), and for singly charged ions (z=+1), the m/z value directly equals the atomic mass in atomic mass units (u). The observed peaks at m/z=20 and m/z=22 with a 9:1 intensity ratio match the natural isotopic distribution of neon, which has two stable isotopes: Ne-20 (90.5% abundance) and Ne-22 (9.3% abundance). The correct answer A accurately identifies neon based on these isotope masses and their relative abundances. Option D incorrectly suggests argon-40 would appear at m/z=20 if doubly charged, but the question states the ions are primarily singly charged, making this explanation inconsistent with the given conditions. When analyzing mass spectra for atomic identification, always compare both the m/z values and intensity ratios to known isotopic patterns. Remember that for singly charged ions, the m/z value directly corresponds to the isotope mass in atomic mass units.

Question 12

1. The $m/z=14$ peak most likely corresponds to a doubly charged molecular ion $\text{M}^{2+}$
2. The molecule most likely fragments into two equal-mass pieces, each producing a singly charged ion at $m/z=14$ (correct answer)
3. The $m/z=28$ peak must represent a singly charged fragment ion rather than the molecular ion
4. The $m/z=14$ peak indicates the molecule contains an atom with atomic number 14

Explanation: This question tests understanding of molecular fragmentation patterns in mass spectrometry for atomic identification. Mass spectrometry can break molecular ions into fragments, and the m/z values of these fragments provide clues about molecular structure - here, a diatomic molecule with molecular ion M+ at m/z = 28 produces a fragment X+ at m/z = 14. Since the molecular mass is 28 u and the fragment mass is 14 u (exactly half), the most logical explanation is that the diatomic molecule splits into two equal parts during fragmentation. The correct answer B recognizes that a homonuclear diatomic molecule (like N2 with mass 28 u) would fragment into two identical atoms (each N with mass 14 u), producing the observed pattern. Answer A incorrectly interprets the m/z = 14 peak as a doubly charged molecular ion, which would require the molecule to have mass 28 u and produce M2+ at m/z = 14, but this contradicts the assignment of m/z = 28 to M+. To analyze fragmentation patterns, compare the m/z values of the molecular ion and fragments - simple integer ratios often indicate loss of specific atomic or molecular units.

Question 13

1. The samples must be identical compounds because the same $m/z$ peak uniquely identifies molecular structure
2. The samples could be different species with the same nominal $m/z$ (isobars), so additional information is needed (correct answer)
3. The samples must be different because identical $m/z$ cannot occur for two different molecules
4. The samples are best distinguished by measuring their absorbance at 280 nm rather than $m/z$

Explanation: This question tests understanding of the limitations of mass spectrometry for compound identification, specifically the concept of isobaric species. Mass spectrometry measures m/z ratios, and different molecular species can have the same nominal mass (isobars) - for example, H₂O (mass 18.015 u) and NH₄ (mass 18.039 u) both appear at m/z = 18 in low-resolution spectra. The observation of identical m/z = 18 peaks does not prove the samples are identical because multiple compounds can have the same nominal mass. The correct answer B recognizes that additional information (such as high-resolution mass, fragmentation patterns, or complementary analytical techniques) is needed to distinguish isobaric species. Answer A incorrectly assumes m/z uniquely identifies molecular structure, failing to consider that isomers and isobars share the same mass. When using mass spectrometry for identification, remember that m/z provides molecular mass information but cannot distinguish between different compounds with the same mass - structural information requires additional data from fragmentation, high resolution, or other analytical methods.

Question 14

1. The sample contains only $^{35}\text{Cl}$ because the $m/z=37$ peak is instrument noise
2. The sample contains chlorine with two common isotopes, and $^{35}\text{Cl}$ is about three times as abundant as $^{37}\text{Cl}$ (correct answer)
3. The sample contains bromine because bromine isotopes also differ by 2 u
4. The 3:1 ratio indicates the ions are triply charged at $m/z=35$ and singly charged at $m/z=37$

Explanation: This question tests the ability to identify isotopes using mass spectrometry peak patterns and intensity ratios. In mass spectrometry, different isotopes of the same element appear as separate peaks because they have different masses - chlorine has two stable isotopes, 35Cl and 37Cl, which differ by 2 mass units. The 3:1 intensity ratio of the m/z = 35 and m/z = 37 peaks directly reflects the natural abundance ratio of these chlorine isotopes (approximately 75% 35Cl and 25% 37Cl). The correct answer B recognizes both the isotope identification and abundance information from the spectrum. Answer D incorrectly interprets the intensity ratio as indicating different charge states, failing to recognize that isotope abundance is the primary factor determining peak intensity ratios for the same element. When analyzing mass spectra for isotope patterns, look for peaks separated by integer mass units with intensity ratios matching known natural abundances - this is a powerful tool for element identification.

Question 15

1. Ion 2 has half the $m/z$ of Ion 1 because its charge magnitude is doubled (correct answer)
2. Ion 2 has double the $m/z$ of Ion 1 because it experiences greater acceleration
3. Ion 2 has the same $m/z$ as Ion 1 because both originate from the same molecule
4. Ion 2 has an undefined $m/z$ because multiply charged ions cannot be analyzed by TOF

Explanation: 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.

Question 16

1. 250, because doubling the charge approximately halves $m/z$ for the same molecule (correct answer)
2. 500, because the molecular mass does not change when charge state changes
3. 1000, because adding a second proton doubles the measured $m/z$
4. 501, because adding one additional proton increases $m/z$ by 1 regardless of charge

Explanation: This question tests understanding of how charge state affects m/z values in mass spectrometry. In mass spectrometry, the measured value is the mass-to-charge ratio (m/z), where mass includes the molecule plus any added protons, and charge is the number of elementary charges. For the singly protonated species [M+H]+, we have m/z = (M+1)/1 = 500, which means the neutral molecule has mass M = 499 u. For the doubly protonated species [M+2H]2+, we have m/z = (M+2)/2 = (499+2)/2 = 501/2 ≈ 250.5, which rounds to 250. The correct answer A recognizes that doubling the charge approximately halves the m/z value for the same molecule. Option C incorrectly suggests that adding a second proton doubles the m/z, failing to account for the charge in the denominator of the m/z ratio. When analyzing multiply charged ions, remember that m/z = (molecular mass + number of protons × 1 u) / charge state. A useful check is that higher charge states always produce lower m/z values for the same molecule.

Question 17

1. A higher relative intensity of the molecular ion peak (intact $M^+$) and fewer fragment peaks (correct answer)
2. A systematic shift of all peaks to higher $m/z$ because added ionization energy increases measured mass
3. Disappearance of all peaks because softer ionization prevents ion formation entirely
4. Appearance of absorption bands at characteristic wavelengths because softer ionization favors spectroscopy

Explanation: 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.

Question 18

1. A 28 u atom forming both $+1$ (at 28) and $+2$ (at 14) ions (correct answer)
2. A 14 u atom forming both $+1$ (at 14) and $+2$ (at 28) ions
3. Two different atoms of 14 u and 28 u, because charge state cannot affect $m/z$
4. A 42 u atom forming $+1$ and $+3$ ions, because $42/3=14$ and $42/1=28$

Explanation: This question tests understanding of how multiple charge states affect m/z values for the same atom in mass spectrometry. The principle that m/z = mass/charge means that doubling the charge halves the m/z value for a fixed mass. Given peaks at m/z=28 and m/z=14 from the same element, we need to find a mass that produces these values for z=+1 and z=+2. If the atom has mass 28 u, then with z=+1 we get m/z=28/1=28, and with z=+2 we get m/z=28/2=14, perfectly matching the observed peaks. The correct answer A identifies this as a 28 u atom (likely silicon) forming both +1 and +2 ions. Option B incorrectly suggests a 14 u atom would give m/z=28 with z=+2, but this would require m/z=14×2=28, which confuses multiplication with the correct division operation in m/z calculations. When analyzing peaks from multiple charge states, work backwards from m/z values: multiply each m/z by possible integer charges to find consistent mass values. A useful check is that the mass calculated from different charge states must be identical for peaks from the same species.

Question 19

1. $m/z=100$ is $[M+Na]^+$ and $m/z=123$ is $M^+$
2. $m/z=100$ is $M^+$ and $m/z=123$ is $[M+Na]^+$ (correct answer)
3. $m/z=100$ is $[M+2Na]^{2+}$ and $m/z=123$ is $[M+Na]^+$
4. The 23-unit difference must be due to a longer retention time in chromatography, not mass spectrometry

Explanation: 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+.

Question 20

1. $m/z=600$ is $z=+1$ and $m/z=300$ is $z=+2$ (correct answer)
2. $m/z=600$ is $z=+2$ and $m/z=300$ is $z=+1$
3. Both peaks must have $z=+1$ but represent two isotopes separated by 300 amu
4. Both peaks must have $z=+2$ because ESI cannot form singly charged ions

Explanation: 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.