Mass Spectrometry Questions and Answers

Answer: C

An absent molecular ion means the excess energy of electron impact has destroyed it, so chemical ionisation or electrospray should be tried to obtain the molecular weight. More energy would make matters worse, and neither range nor resolution addresses a peak that was never formed. Changing the ionisation rather than the analyser is the remedy when the problem lies in what happened before the ions were sorted. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: D

A ladder of peaks decreasing in spacing is the classic electrospray pattern of one molecule carrying different numbers of protons, and deconvoluting it gives a single accurate molecular weight. Three proteins would not space so regularly, fragments are uncommon in electrospray, and isotope peaks differ by one unit. Recognising the charge-state envelope is what turns an apparently complex spectrum into one number. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: A

Two isomers can give nearly identical fragmentation, so the chromatographic dimension must settle it: running an authentic standard and comparing retention times distinguishes them. Vacuum, electron energy and gain affect the spectrum's quality rather than the ambiguity. Using the second dimension of a hyphenated technique when the first cannot decide is exactly why the two were coupled together. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: B

Involatile material deposits on the source's surfaces over many injections, charging up and defocusing the ion beam, so sensitivity falls gradually. A calibration drift would shift masses rather than intensities, an improving vacuum would help, and overload distorts peak shapes. A slow monotonic decline across a sequence points at accumulation rather than at any single event, and source cleaning is the routine answer. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: C

The pairing is A-P, B-Q, C-R. Each analyser occupies a different corner of the trade between resolution, speed, size and mass range: a sector resolves finely but scans slowly and fills a room, a quadrupole is compact and quick at roughly unit resolution, and a time-of-flight instrument measures every ion rather than filtering, so nothing bounds its mass range but patience. The application decides which corner is wanted, and that no single design dominates is why all three remain in production decades after they first appeared.

Answer: D

The first three are correct and the fourth is not, so three match. A loss of thirty-one is a methoxy group, characteristic of methyl esters and ethers; hydroxyl is seventeen. A short table of common losses accounts for most of what a spectrum shows, and reading differences between peaks rather than their absolute masses is what makes the table usable.

Answer: A

An M plus two peak about a third the height of the molecular ion is chlorine's three to one isotope ratio, and the loss of thirty-five to give eighty-seven confirms it by showing the chlorine departing. Bromine would give a nearly equal pair, two chlorines a three-peak cluster, and sulphur about four per cent. Reading the isotope ratio and the neutral loss together makes the assignment certain. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: B

This is the molecular ion, whose mass gives the molecular weight directly. The base peak is simply the tallest, a fragment is any piece produced by breaking, and a precursor is the ion selected for fragmentation in a tandem experiment. Four terms for four different peaks, and the molecular ion is the one that must be identified before anything else in a spectrum can be interpreted.

Answer: C

Collision-induced dissociation fragments ions that already exist rather than creating them, so it belongs to the tandem experiment rather than to the source. The other three are genuine ionisation methods. Distinguishing what makes ions from what breaks them is essential when a method is being designed, since the two steps address quite different problems.

Answer: D

Knocking out one electron leaves a singly charged positive ion, and doubly charged species are a small minority. Negative ions form in a separate mode of operation, and a neutral particle would not be analysed at all. That the charge is almost always one is what lets the mass-to-charge axis be read as a mass in routine work, and why a doubly charged peak at half the expected value can be so confusing. Most errors here come from carrying a result across from a neighbouring case where it does not hold.