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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/as-level/chemistry/edexcel/as-mass-spectrometry/flashcards","label":"Mass spectrometry","subjectCode":"as_level_chemistry","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"3832c48a-20bc-468f-a86e-32b8746bf3d8","title":"Mass spectrometry Flashcards","topicId":"5309ece0-1cd8-53e7-b311-f0de72e0ac43","topicSlug":"as-mass-spectrometry","cards":[{"id":"b27b79e4-cd66-4c6a-beab-7b3cdeefdb91","cardType":"basic","points":1,"front":"What does a mass spectrum plot?","back":"Relative abundance against $m/z$.","mcqOptions":[{"text":"Relative abundance against $m/z$.","correct":true,"explanation":"A mass spectrum shows how much of each detected ion is present as relative abundance, plotted against its mass-to-charge ratio, $m/z$."},{"text":"$m/z$ against relative abundance.","correct":false,"explanation":"Reversing the axes is a common graph-reading error. Relative abundance is plotted on the vertical axis against $m/z$ on the horizontal axis."},{"text":"Absolute abundance against $m/z$.","correct":false,"explanation":"Mass spectra use relative, not absolute, abundance because the largest peak is standardised to 100%. The correct plot is relative abundance against $m/z$."},{"text":"Relative abundance against ionisation energy.","correct":false,"explanation":"Ionisation energy is not represented on a standard mass spectrum. The spectrum plots the relative abundance of detected ions against their $m/z$ values."}]},{"id":"b43c1655-18e3-43e0-9c7f-0ae35f4b368a","cardType":"cloze","points":null,"front":"In most A-Level organic mass spectra, ions have charge [ ], so $m/z$ equals the ion’s [ ].","back":"In most A-Level organic mass spectra, ions have charge +1, so $m/z$ equals the ion’s relative mass.","mcqOptions":[{"text":"In most A-Level organic mass spectra, ions have charge +1, so $m/z$ equals the ion’s relative mass.","correct":true,"explanation":"Electron impact usually produces singly charged positive ions, so $z=1$. Dividing the ion’s relative mass by one makes its $m/z$ numerically equal to its relative mass."},{"text":"In most A-Level organic mass spectra, ions have charge +2, so $m/z$ equals the ion’s relative mass.","correct":false,"explanation":"A charge of +2 would make $m/z$ equal to half the ion’s relative mass, not the relative mass itself. Most organic ions considered at A-Level have charge +1."},{"text":"In most A-Level organic mass spectra, ions have charge +1, so $m/z$ equals the ion’s relative charge.","correct":false,"explanation":"The denominator $z$ is charge, but $m/z$ does not equal relative charge. For the usual +1 ions, it is numerically equal to the ion’s relative mass."},{"text":"In most A-Level organic mass spectra, ions have charge −1, so $m/z$ equals the ion’s relative mass.","correct":false,"explanation":"Standard electron impact mass spectrometry detects positive ions rather than negative ions. Most organic ions have charge +1, making $m/z$ equal to their relative mass."}]},{"id":"6ec12668-048b-418b-adab-59bbb5d87b6d","cardType":"basic","points":2,"front":"Why must the sample be ionised before detection?","back":"The detector records only charged particles.","mcqOptions":[{"text":"The detector records only charged particles.","correct":true,"explanation":"Mass spectrometer detectors respond to the movement or arrival of ions, which carry charge. Neutral particles therefore must be ionised before they can produce a signal."},{"text":"The detector records only neutral particles.","correct":false,"explanation":"Neutral particles do not respond appropriately to electric or magnetic fields and do not generate the required detector signal. The detector records only charged particles."},{"text":"The detector records only radical particles.","correct":false,"explanation":"The detector is not restricted to particles containing unpaired electrons; many detected ions have paired electrons. Detection requires particles to carry an electrical charge."},{"text":"The detector records only gaseous particles.","correct":false,"explanation":"Vaporisation produces gaseous particles, but being gaseous is not sufficient for detection. They must also be ionised because the detector records only charged particles."}]},{"id":"801f91cb-1406-4150-b322-801105684323","cardType":"basic","points":3,"front":"Put the four essential mass spectrometer stages in order.","back":"

Vaporisation
Ionisation
Separation by $m/z$
Detection

","mcqOptions":[{"text":"

Vaporisation
Ionisation
Separation by $m/z$
Detection

","correct":true,"explanation":"The sample is first vaporised, then ionised to form charged particles. These ions are separated according to $m/z$ before reaching the detector."},{"text":"

Ionisation
Vaporisation
Separation by $m/z$
Detection

","correct":false,"explanation":"Ionisation cannot normally precede vaporisation in the standard sequence because the sample must first enter the gas phase. The correct order starts with vaporisation, then ionisation."},{"text":"

Vaporisation
Ionisation
Detection
Separation by $m/z$

","correct":false,"explanation":"Detection must occur after separation so individual $m/z$ values can be measured distinctly. The ions are vaporised, ionised and separated before being detected."},{"text":"

Vaporisation
Separation by $m/z$
Ionisation
Detection

","correct":false,"explanation":"Particles must be ionised before separation because electric or magnetic fields separate charged ions, not neutral molecules. Ionisation therefore precedes separation by $m/z$."}]},{"id":"f41cb0a4-17fb-4feb-ba03-6cfa4a1baf28","cardType":"cloze","points":null,"front":"In electron impact ionisation, high-energy electrons remove [ ], forming a [ ].","back":"In electron impact ionisation, high-energy electrons remove one electron, forming a positive radical cation.","mcqOptions":[{"text":"In electron impact ionisation, high-energy electrons remove one electron, forming a positive radical cation.","correct":true,"explanation":"A high-energy electron knocks one electron out of a molecule. The resulting molecular ion is positive and has one unpaired electron, so it is a radical cation."},{"text":"In electron impact ionisation, high-energy electrons remove one proton, forming a positive radical cation.","correct":false,"explanation":"Removing a proton is not the electron impact process and would not produce the stated ion. Electron impact removes one electron, forming a positive radical cation."},{"text":"In electron impact ionisation, high-energy electrons remove one electron, forming a negative radical anion.","correct":false,"explanation":"Removing a negatively charged electron leaves the molecule positively charged, not negatively charged. The product is a positive radical cation because it also has an unpaired electron."},{"text":"In electron impact ionisation, high-energy electrons remove two electrons, forming a positive dication.","correct":false,"explanation":"The standard molecular ion forms when one electron is removed, not two. Removing two electrons would produce a doubly charged ion rather than the usual positive radical cation."}]},{"id":"e7821173-5e8b-4753-ba52-c0a94cbe23c7","cardType":"basic","points":1,"front":"What does the dot in $\\text{M}^{+\\cdot}$ show?","back":"An unpaired electron.","mcqOptions":[{"text":"An unpaired electron.","correct":true,"explanation":"The dot represents a single unpaired electron. Together, the superscript plus and dot show that the molecular ion is both positively charged and a radical."},{"text":"A positive charge.","correct":false,"explanation":"The plus sign indicates the positive charge, while the dot has a different meaning. It represents an unpaired electron, showing that the ion is a radical."},{"text":"A paired electron.","correct":false,"explanation":"An electron pair would be represented by two electrons, not a single dot. The dot in the molecular-ion notation specifically indicates one unpaired electron."},{"text":"A removed proton.","correct":false,"explanation":"The dot is electronic notation rather than a count of removed protons. It shows that the molecular ion contains an unpaired electron and is therefore a radical."}]},{"id":"274850bb-bfe9-4af9-83aa-43bc4a614c3c","cardType":"basic","points":1,"front":"For a singly charged molecular ion, how do you get $M_r$ from the spectrum?","back":"Use the molecular ion’s $m/z$ value.","mcqOptions":[{"text":"Use the molecular ion’s $m/z$ value.","correct":true,"explanation":"For a molecular ion with charge +1, dividing its relative mass by the charge leaves the same numerical value. Its molecular-ion $m/z$ therefore gives $M_r$."},{"text":"Use the base peak’s $m/z$ value.","correct":false,"explanation":"The base peak is the most abundant ion and is often a fragment, so it need not represent the whole molecule. $M_r$ comes from the molecular ion’s $m/z$."},{"text":"Use the molecular ion’s relative abundance.","correct":false,"explanation":"Relative abundance describes signal intensity rather than molecular mass. For a singly charged molecular ion, the required $M_r$ is obtained from its $m/z$ value."},{"text":"Use the $M+1$ peak’s $m/z$ value.","correct":false,"explanation":"The isotope peak one unit above the molecular ion commonly represents a molecule containing a heavier isotope. The molecular ion’s own $m/z$ gives $M_r$."}]},{"id":"96c41b53-bdfd-404c-87da-7ae24e837faf","cardType":"cloze","points":null,"front":"The base peak is the [ ] and is assigned [ ].","back":"The base peak is the tallest peak and is assigned 100% relative abundance.","mcqOptions":[{"text":"The base peak is the tallest peak and is assigned 100% relative abundance.","correct":true,"explanation":"The base peak is the spectrum’s most intense, and therefore tallest, peak. Its intensity is defined as 100%, with every other peak measured relative to it."},{"text":"The base peak is the molecular ion peak and is assigned 100% relative abundance.","correct":false,"explanation":"The molecular ion may be weak and is not necessarily the base peak. The base peak is whichever peak is tallest, and it is assigned 100% relative abundance."},{"text":"The base peak is the tallest peak and is assigned 50% relative abundance.","correct":false,"explanation":"The tallest peak is used as the reference maximum, so its relative abundance is defined as 100%, not 50%. Other peak heights are scaled to it."},{"text":"The base peak is the highest-$m/z$ peak and is assigned 100% relative abundance.","correct":false,"explanation":"The peak with the greatest $m/z$ is not necessarily the most abundant ion. The base peak is the tallest peak and is assigned 100% relative abundance."}]},{"id":"4993c954-62ed-4400-bbab-7f6f38a1b044","cardType":"basic","points":1,"front":"A molecular ion peak is at $m/z=74$ with a small isotope peak at $m/z=75$. What is $M_r$?","back":"74.","mcqOptions":[{"text":"74.","correct":true,"explanation":"The molecular ion peak corresponds to the molecule containing its most abundant isotopes, so its $m/z$ value gives the relative molecular mass as 74."},{"text":"75.","correct":false,"explanation":"This mistakes the M+1 isotope peak for the molecular ion peak. The main molecular ion is at $m/z=74$, so the compound’s $M_r$ is 74."},{"text":"73.","correct":false,"explanation":"This subtracts one from the molecular ion peak, perhaps confusing isotope spacing with a correction. The molecular ion itself is at $m/z=74$, giving $M_r=74$."},{"text":"76.","correct":false,"explanation":"This adds the isotope-peak spacing twice rather than reading the molecular ion directly. The peak at $m/z=74$ is the molecular ion, so $M_r=74$."}]},{"id":"378a59d1-e440-45b2-b7f9-a0f4e5708c0f","cardType":"basic","points":2,"front":"Why might a small peak just to the right of the molecular ion not give $M_r$?","back":"It is often an isotope peak, such as M+1.","mcqOptions":[{"text":"It is often an isotope peak, such as M+1.","correct":true,"explanation":"Molecules containing a heavier isotope, commonly carbon-13, produce a small M+1 peak one mass unit above the molecular ion, so that peak does not give $M_r$."},{"text":"It is often a molecular ion peak, such as M.","correct":false,"explanation":"A molecular ion peak represents the whole ionised molecule and is normally used to determine $M_r$. A small adjacent peak is instead often an isotope peak such as M+1."},{"text":"It is often a fragment ion peak, such as M-1.","correct":false,"explanation":"Fragment peaks come from bond breaking and usually occur at lower $m/z$ values than the molecular ion. A peak immediately to its right is often an M+1 isotope peak."},{"text":"It is often the base peak, such as the tallest peak.","correct":false,"explanation":"The base peak is simply the spectrum’s most intense peak and need not lie beside the molecular ion. A small peak just to the right is often M+1."}]},{"id":"e911e6c4-b4a2-419d-80ed-26a85c83accf","cardType":"basic","points":2,"front":"How can ionisation create lots of peaks in a mass spectrum?","back":"It can break covalent bonds, forming fragment ions.","mcqOptions":[{"text":"It can break covalent bonds, forming fragment ions.","correct":true,"explanation":"Ionisation can give the molecular ion enough energy to break covalent bonds. Different fragmentation pathways form ions with different $m/z$ values, producing numerous peaks."},{"text":"It can break intermolecular forces, forming separate molecules.","correct":false,"explanation":"Intermolecular forces act between molecules and breaking them does not produce the charged molecular fragments responsible for multiple peaks. Covalent bond breaking forms the fragment ions."},{"text":"It can form covalent bonds, creating larger molecular ions.","correct":false,"explanation":"Forming additional covalent bonds would combine atoms rather than generate the smaller ions seen across a mass spectrum. Ionisation produces many peaks by breaking covalent bonds into fragments."},{"text":"It can remove ionic charges, forming neutral fragments.","correct":false,"explanation":"Neutral particles are not deflected and detected by a mass spectrometer, so neutralisation cannot account for the many recorded peaks. Covalent bond breaking creates detectable fragment ions."}]},{"id":"b260ec95-e664-4e93-a568-48d2b82fe3b8","cardType":"cloze","points":null,"front":"When a molecular ion breaks apart, the detected particle is the [ ]; the unrecorded piece is a [ ].","back":"When a molecular ion breaks apart, the detected particle is the fragment ion; the unrecorded piece is a neutral fragment.","mcqOptions":[{"text":"When a molecular ion breaks apart, the detected particle is the fragment ion; the unrecorded piece is a neutral fragment.","correct":true,"explanation":"Only charged particles are accelerated and detected in a mass spectrometer. Bond cleavage therefore records the charged fragment ion, while the accompanying neutral fragment remains undetected."},{"text":"When a molecular ion breaks apart, the detected particle is the neutral fragment; the unrecorded piece is a fragment ion.","correct":false,"explanation":"This reverses the products’ roles. The mass spectrometer detects the charged fragment ion because it responds to electric and magnetic fields, while the neutral fragment is unrecorded."},{"text":"When a molecular ion breaks apart, the detected particle is the molecular ion; the unrecorded piece is a neutral fragment.","correct":false,"explanation":"The molecular ion is the original ion before fragmentation, not the smaller detected product. After cleavage, the charged fragment ion is detected and the neutral fragment is unrecorded."},{"text":"When a molecular ion breaks apart, the detected particle is the isotope ion; the unrecorded piece is a molecular fragment.","correct":false,"explanation":"An isotope ion differs in isotopic composition rather than arising necessarily as the charged product of bond cleavage. The detected product is a fragment ion, and the neutral fragment is unrecorded."}]},{"id":"ae687269-11a2-49e0-9f6a-c30535ff5a7a","cardType":"basic","points":1,"front":"What common fragment ion gives $m/z=15$?","back":"CH₃⁺.","mcqOptions":[{"text":"CH₃⁺.","correct":true,"explanation":"The methyl cation contains one carbon and three hydrogens, giving a nominal mass of $12+(3×1)=15$. Its single positive charge therefore gives $m/z=15$."},{"text":"CH₂⁺.","correct":false,"explanation":"This ion has a nominal mass of $12+(2×1)=14$, so a singly charged ion appears at $m/z=14$. The fragment at 15 is the methyl ion."},{"text":"CH₄⁺.","correct":false,"explanation":"This ion has a nominal mass of $12+(4×1)=16$, not 15. The common singly charged fragment at $m/z=15$ is the methyl cation."},{"text":"C⁺.","correct":false,"explanation":"A carbon ion has nominal mass 12, so its single positive charge gives $m/z=12$. Adding three hydrogens gives the methyl fragment at $m/z=15$."}]},{"id":"ef59502c-b3fe-4aad-b4e5-ed69cef1e330","cardType":"basic","points":1,"front":"What common fragment ion gives $m/z=29$?","back":"C₂H₅⁺.","mcqOptions":[{"text":"C₂H₅⁺.","correct":true,"explanation":"The ethyl cation has a nominal mass of $(2×12)+(5×1)=29$. With a single positive charge, its mass-to-charge ratio is therefore 29."},{"text":"C₂H₃⁺.","correct":false,"explanation":"This fragment has a nominal mass of $(2×12)+(3×1)=27$, not 29. The common hydrocarbon fragment at $m/z=29$ is the ethyl cation."},{"text":"CH₃⁺.","correct":false,"explanation":"The methyl cation has a nominal mass of $12+(3×1)=15$, so it produces a peak at 15. The ethyl cation produces the peak at 29."},{"text":"C₃H₅⁺.","correct":false,"explanation":"This fragment has a nominal mass of $(3×12)+(5×1)=41$, so it appears at $m/z=41$. The ethyl ion has the required mass of 29."}]},{"id":"6159c681-1176-4f4a-9f70-060ea6dc8186","cardType":"basic","points":1,"front":"Name one common fragment ion that can give $m/z=43$.","back":"One of: C₃H₇⁺ or CH₃CO⁺.","mcqOptions":[{"text":"One of: C₃H₇⁺ or CH₃CO⁺.","correct":true,"explanation":"The propyl ion has mass $(3×12)+7=43$, while the acylium ion has mass $12+3+12+16=43$. Either singly charged ion can produce this peak."},{"text":"One of: C₃H₅⁺ or CH₃O⁺.","correct":false,"explanation":"These fragments have nominal masses 41 and 31 respectively, so neither gives $m/z=43$. Propyl or the acylium ion CH₃CO⁺ commonly gives 43."},{"text":"One of: C₂H₅⁺ or HCO⁺.","correct":false,"explanation":"The ethyl and formyl ions have nominal masses 29, not 43. A peak at 43 commonly comes from C₃H₇⁺ or CH₃CO⁺."},{"text":"One of: C₄H₉⁺ or C₂H₅O⁺.","correct":false,"explanation":"These ions have nominal masses 57 and 45, so they are heavier than the required fragment. Propyl and CH₃CO⁺ each have nominal mass 43."}]},{"id":"da3935d7-c777-4901-9240-d0375f3b006e","cardType":"basic","points":1,"front":"Name one common fragment ion that can give $m/z=57$.","back":"One of: C₄H₉⁺ or C₂H₅CO⁺.","mcqOptions":[{"text":"One of: C₄H₉⁺ or C₂H₅CO⁺.","correct":true,"explanation":"The butyl ion has mass $(4×12)+9=57$, while C₂H₅CO⁺ has mass $24+5+12+16=57$. Either singly charged fragment can give this peak."},{"text":"One of: C₃H₇⁺ or CH₃CO⁺.","correct":false,"explanation":"The propyl and CH₃CO⁺ ions both have nominal mass 43, not 57. The corresponding common ions at 57 are butyl and C₂H₅CO⁺."},{"text":"One of: C₄H₇⁺ or C₂H₅O⁺.","correct":false,"explanation":"These fragments have nominal masses 55 and 45 respectively, so neither produces the peak at 57. C₄H₉⁺ or C₂H₅CO⁺ has the required mass."},{"text":"One of: C₅H₁₁⁺ or C₃H₇CO⁺.","correct":false,"explanation":"The pentyl ion has nominal mass 71, while the butanoyl ion has nominal mass 71 as well. The analogous fragments containing one fewer carbon give $m/z=57$."}]},{"id":"48f21d1d-8dde-4bf6-b88b-4642de708e4f","cardType":"basic","points":2,"front":"Why can one low-resolution $m/z$ peak be ambiguous?","back":"Different ions can have the same nominal mass.","mcqOptions":[{"text":"Different ions can have the same nominal mass.","correct":true,"explanation":"Low-resolution instruments usually resolve only whole-number nominal masses. Ions with different elemental compositions can therefore coincide at one nominal $m/z$ value."},{"text":"Every ion has an exact integer mass.","correct":false,"explanation":"Low resolution does not make every ion’s exact mass an integer; exact isotope masses are generally non-integer. The ambiguity arises because different ions can have the same nominal mass."},{"text":"Isotopic peaks are always completely separated.","correct":false,"explanation":"Low resolution may merge nearby peaks rather than separating all isotopic peaks. Different ions can consequently appear at the same nominal mass and produce an ambiguous peak."},{"text":"All detected ions have the same electrical charge.","correct":false,"explanation":"Different charge states change $m/z$, but low-resolution ambiguity is not caused by all ions having equal charge. It occurs when different ions share the same nominal mass."}]},{"id":"b0e5784e-403a-4991-8653-374448d61435","cardType":"cloze","points":null,"front":"An M+1 peak is caused by one [ ] atom; its size [ ] as carbon number increases.","back":"An M+1 peak is caused by one carbon-13 atom; its size increases as carbon number increases.","mcqOptions":[{"text":"An M+1 peak is caused by one carbon-13 atom; its size increases as carbon number increases.","correct":true,"explanation":"Carbon-13 is about 1.1% naturally abundant, so molecules containing more carbon atoms have more opportunities to include one carbon-13 atom and produce an M+1 ion."},{"text":"An M+1 peak is caused by one carbon-12 atom; its size decreases as carbon number increases.","correct":false,"explanation":"Carbon-12 forms the main molecular ion rather than the M+1 peak. One carbon-13 atom adds one mass unit, and its probability increases with the number of carbon atoms."},{"text":"An M+1 peak is caused by one chlorine-37 atom; its size increases as carbon number increases.","correct":false,"explanation":"Chlorine-37 is two mass units above chlorine-35, so it contributes to an M+2 peak rather than M+1. Carbon-13 causes M+1, whose size increases with carbon number."},{"text":"An M+1 peak is caused by one carbon-13 atom; its size decreases as carbon number increases.","correct":false,"explanation":"Carbon-13 correctly accounts for the extra mass unit, but adding more carbon atoms increases rather than decreases the chance of incorporating one carbon-13 atom."}]},{"id":"5c399862-e2a9-414e-b590-2ffb8d76ccf8","cardType":"cloze","points":null,"front":"M and M+2 peaks two units apart in an approximate 3:1 ratio suggest [ ].","back":"M and M+2 peaks two units apart in an approximate 3:1 ratio suggest chlorine.","mcqOptions":[{"text":"chlorine","correct":true,"explanation":"Chlorine-35 and chlorine-37 occur naturally in approximately a 3:1 abundance ratio. A molecule containing one chlorine atom therefore gives M and M+2 peaks in that ratio."},{"text":"bromine","correct":false,"explanation":"Bromine-79 and bromine-81 have nearly equal natural abundances, producing approximately equal M and M+2 peaks. A 3:1 pattern instead indicates chlorine."},{"text":"sulfur","correct":false,"explanation":"Sulfur-34 can produce an M+2 peak, but its low natural abundance gives a much smaller peak than this 3:1 pattern. The characteristic element is chlorine."},{"text":"carbon","correct":false,"explanation":"Carbon-13 mainly produces an M+1 peak, not a strong M+2 peak in a 3:1 pair. Chlorine-35 and chlorine-37 create the stated pattern."}]},{"id":"6d1d77bd-96a8-4967-ab48-634fe92eb43c","cardType":"cloze","points":null,"front":"M and M+2 peaks two units apart with roughly equal heights suggest [ ].","back":"M and M+2 peaks two units apart with roughly equal heights suggest bromine.","mcqOptions":[{"text":"bromine","correct":true,"explanation":"Bromine-79 and bromine-81 differ by two mass units and have nearly equal natural abundances. One bromine atom therefore produces M and M+2 peaks of similar height."},{"text":"chlorine","correct":false,"explanation":"Chlorine also produces peaks two mass units apart, but chlorine-35 is about three times as abundant as chlorine-37. Roughly equal peaks instead indicate bromine."},{"text":"sulfur","correct":false,"explanation":"Sulfur-34 has a much lower abundance than sulfur-32, so sulfur produces only a relatively small M+2 peak. Equal M and M+2 heights indicate bromine."},{"text":"iodine","correct":false,"explanation":"Naturally occurring iodine is almost entirely iodine-127, so one iodine atom does not produce a characteristic equal-height M and M+2 pair. That pattern indicates bromine."}]},{"id":"fc716509-70e4-4b73-9bb8-62d1f9e0ce93","cardType":"basic","points":2,"front":"Why can high-resolution MS distinguish formulae with the same nominal $M_r$?","back":"Atoms have different exact isotope masses, not exact integer masses.","mcqOptions":[{"text":"Atoms have different exact isotope masses, not exact integer masses.","correct":true,"explanation":"Isotopes do not have perfectly integer masses because nuclear binding energies affect their exact masses. Different formulae therefore have distinguishable accurate masses despite sharing a nominal $M_r$."},{"text":"Atoms have different proton numbers, not different exact isotope masses.","correct":false,"explanation":"Proton number identifies an element but is not what the mass spectrometer directly resolves between equal-nominal-mass formulae. Their different sums of exact isotope masses provide the distinction."},{"text":"Atoms have exact integer isotope masses, not non-integer masses.","correct":false,"explanation":"Treating isotope masses as exact integers would preserve the nominal-mass ambiguity rather than resolve it. High-resolution MS distinguishes the small differences between non-integer exact isotope masses."},{"text":"Ions have different electrical charges, not different exact masses.","correct":false,"explanation":"Molecular ions used for formula comparison are ordinarily singly charged, so charge does not explain their separation. Different formulae are resolved through their different exact isotope masses."}]},{"id":"9850894a-df38-4e89-8d99-d27aa7490f09","cardType":"basic","points":1,"front":"A high-resolution molecular ion peak at $m/z=60.0575$ best matches C₃H₈O, C₂H₄O₂ or C₂H₈N₂?","back":"C₃H₈O.","mcqOptions":[{"text":"C₃H₈O.","correct":true,"explanation":"Using exact masses gives $3(12.0000)+8(1.007825)+15.994915=60.0575$. This agrees with the measured molecular-ion value for C₃H₈O."},{"text":"C₂H₄O₂.","correct":false,"explanation":"Using exact isotope masses, C₂H₄O₂ has a mass of about $60.0211$, not $60.0575$. The measured value matches C₃H₈O."},{"text":"C₂H₈N₂.","correct":false,"explanation":"Using exact isotope masses, C₂H₈N₂ has a mass of about $60.0687$, which is too high. C₃H₈O gives approximately $60.0575$."},{"text":"C₂H₄S.","correct":false,"explanation":"C₂H₄S has the same nominal mass of 60 but an exact mass of about $59.9823$. The high-resolution value instead matches C₃H₈O."}]},{"id":"eff53212-eb47-4683-8383-473758e3eb1b","cardType":"basic","points":2,"front":"Why can high-resolution MS suggest a formula but not prove the exact structure?","back":"Structural isomers can share the same molecular formula.","mcqOptions":[{"text":"Structural isomers can share the same molecular formula.","correct":true,"explanation":"A molecular formula specifies the numbers of each type of atom but not how those atoms are connected. Structural isomers therefore have identical molecular-ion exact masses."},{"text":"Structural isomers always have different molecular formulae.","correct":false,"explanation":"Structural isomers have the same molecular formula but different atom connectivity, not different formulae. Consequently, an accurate molecular-ion mass alone cannot identify the exact structure."},{"text":"Empirical formulae always give the exact arrangement of atoms.","correct":false,"explanation":"An empirical formula gives only the simplest whole-number ratio and does not specify atom connectivity. Even a molecular formula cannot distinguish structural isomers with different arrangements."},{"text":"Molecular ions always reveal the arrangement of atoms.","correct":false,"explanation":"A molecular-ion peak reveals molecular mass and can support a formula, but it does not show how the atoms are bonded. Structural isomers can therefore share that peak."}]}]}],"examCountdown":{"examDate":"2027-05-11","paperLabel":"Paper 1"},"heading":"Mass spectrometry","title":"Mass spectrometry Flashcards","topbar":{"backHref":"/as-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"5309ece0-1cd8-53e7-b311-f0de72e0ac43","walled":"$L39"}],"$L3a","$L3b"]