Gas Chromatography Questions and Answers

Answer: C

Depositing the liquid inside the column itself removes the vaporisation step where discrimination occurs, so the composition reaching the stationary phase is the composition injected. Slower vaporisation and a zero split reduce the problem without removing it, and injector temperature is a related but separate control. Eliminating a step rather than optimising it is the more reliable fix when that step is the source of the error. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: D

At a temperature suiting the volatile components the involatile ones would take hours, and at one suiting the latter the former would be unresolved - ramping the oven serves both. Detector sensitivity, gas consumption and bleed are unaffected or worsened. Changing a condition during a run so that each part of the sample meets favourable conditions in turn is what makes a wide-boiling mixture tractable at all. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: A

A mixture spanning a wide range of volatility cannot be run isothermally without either crowding the early peaks or waiting indefinitely for the late ones. The other options describe real observations that are consequences rather than the problem itself. Naming the difficulty explains why programming exists, and the same problem in liquid chromatography is solved by gradient elution for exactly the same reason. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: B

A gentler ramp gives each compound longer at a temperature where it partitions appreciably, so the peaks separate further at the price of a longer run. Speed falls rather than rises, bleed depends on the final temperature, and capacity is a column property. The ramp rate is the principal lever in method development once the phase is chosen, and it trades the two things the analyst usually cares about most. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: C

The sample enters over many seconds, so a low initial temperature condenses it into a narrow band at the head, which then elutes as a sharp peak when the ramp begins. Liner temperature, gas flow and detector protection are separate matters. Refocusing a band that entered slowly is what makes splitless injection compatible with narrow peaks, and without it the technique would be unusable. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: D

Organic compounds burn in a hydrogen flame and produce ions in proportion to the carbon they contain, and the resulting current is measured. The other options describe thermal conductivity, electron capture and absorbance detectors. Because the response tracks carbon content, it is nearly universal for organic compounds and nearly blind to water, carbon dioxide and the permanent gases - which is convenient or limiting depending on what is sought.

Answer: A

Substances with no carbon-hydrogen bonds to burn - water, carbon dioxide, nitrogen, the noble gases - produce essentially no ions and give no signal. Heavy hydrocarbons, halogenated compounds and aromatics all respond, some a little less per carbon. That a detector's blindness can be an advantage, letting a wet sample be injected without a solvent peak swamping the trace, is worth recognising alongside the limitation. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

Measuring how the eluting gas changes a heated filament's cooling makes the detector universal, responding to anything whose conductivity differs from the carrier's. It is far less sensitive than most alternatives, though its non-destructive nature is a genuine secondary benefit. A detector that misses nothing is the right choice when the question is what is present rather than how much of a known analyte there is. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: C

The signal is the difference in thermal conductivity between the carrier and the eluting compound, so a carrier of very high conductivity maximises that difference for every analyte. There is no flame, low molecular weight is the reason for the high conductivity rather than the point, and filament reaction is a separate concern. Choosing a reference far from everything being measured is what makes a differential detector sensitive. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: D

A radioactive source produces a steady current of slow electrons, and electronegative molecules capture them, reducing the current in proportion to their concentration. Carbon counting is flame ionisation, nitrogen and phosphorus selectivity belongs to another detector, and universality to thermal conductivity. Extraordinary sensitivity to one chemical class made this detector the foundation of pesticide and environmental analysis. Most errors here come from carrying a result across from a neighbouring case where it does not hold.