Mass Spectrometry Questions and Answers

Answer: A

Every analyser separates ions by the ratio of mass to charge, so a doubly charged ion of mass two hundred appears at a hundred. Molecular weight follows only once the charge is known, atom counts are inferred from the pattern, and ionisation energy is a separate quantity. That the axis is a ratio rather than a mass is what makes multiply charged ions both a complication and, in protein work, an opportunity. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

Every analyser works by accelerating and deflecting particles with fields, and a neutral molecule feels no force from either. Mass is irrelevant to whether a field acts at all, fragmentation is a consequence of some ionisation methods rather than the purpose, and the detector requirement follows from the same fact. That the technique's first step exists only to make the second possible is worth keeping in mind when a fragile molecule is destroyed by it.

Answer: C

A beam of electrons accelerated through about seventy volts knocks an electron out of each molecule, leaving a radical cation with a great deal of excess energy. The other options describe field ionisation, photoionisation and electrospray. Seventy volts is far more than the ten or so needed, and that surplus is precisely what produces the rich fragmentation the technique is valued for. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: D

Above about fifty volts the ionisation efficiency and the fragmentation pattern both become insensitive to the exact energy, so spectra recorded anywhere in that region are comparable and can be matched against a library. Ionisation itself needs only ten volts or so. The molecular ion still forms, though weakly for some compounds. A standard chosen because the result stops depending on it is a good standard, and it is why libraries of hundreds of thousands of spectra are usable.

Answer: A

With so much excess energy a labile molecule fragments completely and no molecular ion appears, so the molecular weight cannot be read from the spectrum. The fragments are informative rather than uninterpretable, the sample must be a vapour, and efficiency is good. Losing the one peak that gives the molecular weight is what drove the development of the softer ionisation methods.

Answer: B

Electrons ionise a reagent gas such as methane, and the resulting ions transfer a proton to the sample with far less excess energy than a direct electron impact delivers. The electron energy is unchanged, cooling is not involved, and the source is still under vacuum. Interposing an intermediate that moderates the energy transfer is the general idea, and it typically leaves a strong protonated molecular ion with little fragmentation. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: C

Proton transfer gives an even-electron ion one unit heavier than the molecule, which is why a chemical ionisation spectrum's principal peak sits at one more than the electron impact molecular ion. The radical cation is the electron impact product. Knowing which ion a source produces is essential before a molecular weight is read off a spectrum, since the two methods differ by exactly one unit. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: D

Spraying a solution from a charged capillary produces ions from proteins and other large polar molecules without heating them at all, which is why it made biological mass spectrometry possible. Small volatile molecules are better served by electron impact, and involatile solids by laser desorption. An ionisation method that works from solution couples naturally to liquid chromatography, which is much of why it spread. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: A

A protein acquires many protons and appears as a family of peaks differing by one charge, each at a much lower mass-to-charge ratio than the true mass. Deconvoluting that family gives the molecular weight with great precision. Little fragmentation occurs and the isotope pattern is a separate matter. That multiple charging brings a hundred-kilodalton protein within the range of an ordinary analyser is the technique's decisive advantage. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: B

The analyte is embedded in a large excess of a small organic compound that absorbs strongly at the laser wavelength and expands into the vapour, carrying the intact analyte with it. It does contribute protons, which is a secondary role, but the energy absorption is the point. Interposing a sacrificial absorber between an energy source and a fragile target is a general strategy, and here it delivers proteins into the gas phase undamaged. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.