Multidimensional Chromatography

Question 1

Why use multidimensional chromatography?

Answer 1

Naturally occurring mixtures are usually very complex, containing many components. It generally exceeds the capacity of any single method, causing the chromatogram to show a raised baseline at higher m/z due to some material not being analysed by the method.

Multidimensional chromatography offers an increase on conventional GC analysis.

Question 2

What is resolution (R_S) and efficiency?

Answer 2

Resolution is given by the equation below.

A higher efficiency means that there are more theoretical plates and that the system is better at resolving many components. A higher resolution is better, with 1.5 being the baseline resolution.

• R ∝ N^1/2 - increased resolution with very long columns
• R ∝ (L/H)^1/2 - but increased L means peak broadening

Question 3

What is peak capacity (n)?

Answer 3

The maximum number of components which a system can resolve, assuming even peak distrubution.

Question 4

Why is peak capacity an optimistic view?

Answer 4

It only gives the maximum number of components that can be resolved and assumes evenly distributed peaks. However, peaks are usually randomly distributed and so often overlap.

Question 5

What is a more realistic view than peak capacity?

Answer 5

The statistical method of overlap (SMO). This shows how peak capacity (n) is related to number of resolved simple peaks (s) and actual number of components in the mixture (m).

Question 6

What is orthogonal separation in 2D and why is it important?

Answer 6

Orthogonal separation is where the elution times in the two dimensions can be treated as statistically independent. The coupled separations of two columns must be based on different separation methods e.g. polarity, boiling point, chirality, etc.

Question 7

How is peak capacity described in 2D?

What is the peak capacity for full orthogonal separation and part-orthogonal separation?

Answer 7

Peak capacity, in 2D, is described by a number of squares which represents the number of compounds that can theoretically be separated.

Full orthogonal separation gives an overall peak capacity of n₁ x n₂

Part-orthogonal separation (where the two columns are related/correlated) gives an overall peak capacity of n₁ + n₂

Question 8

Describe separation via heartcut zones.

Answer 8

They’re useful in interested in certain components.

Small heartcut zones are taken from the 1D chromatogram and transferred to the second column to get selective zones more spread out in the 2D chromatogram.

This is a linear but non-comprehensive separation because only portions are transferred.

Question 9

Describe separation by simply taking contiguous zones and transferring them to 2D.

Answer 9

Interested in all components and just want to increase peak capacity.

If all of these are passed to 2D then we get severely overlapping zones.

Question 10

Describe how contiguous zones could give a comprehensive separation.

Answer 10

If the contiguous zones are transferred to separate traps, each being independently analysed to give many 2D chromatograms, then a comprehensive separation is obtained.

However this is very expensive and so not commonly used.

Question 11

Describe separation by modulation.

Answer 11

If the right modulation process is used on small contiguous zones from 1D, with an orthogonal column choice, then a comprehensive GC x GC chromatogram is obtained.

Question 12

Compare separation by contiguous zones and by modulation.

Answer 12

• Contiguous zones - the zones are severely overlapping because the 2D analysis takes longer than the 1D analysis for the heartcut zones.
• Modulation - this is essentially the same but here the 2D analysis takes the same/less time than the 1D analysis for the heartcut zones, also the transfer of solute from 1D to 2D involves zone compression of the chromatographic band

Question 13

What practical criteria need to be followed to obtain orthogonal 2D chromatograms?

Answer 13

1. The zone to be transferred from 1D to 2D must be compressed in space.
2. The zone must be delivered to 2D rapidly and as a sharp pulse (about 10 ms).

These can both be achieved by a modulator.

1. 2D must be capable of producing fast GC results - the analysis time of column 2 must be the same/less than the trapping time of the modulator, allowing for no overlap of adjacent transferred zones.

Question 14

How does cryogenic modulation work?

Answer 14

In position A, the trap cools and therefore traps solute in the top part of the 2D column.

In position B, the trap has moved down therefore allowing the solute at position A to warm up by the heat of the oven and pass it to 2D.

This is useful as it makes sure that no analyte is lost.

Question 15

What is a consequence of cryogenic modulation?

Answer 15

The peaks produced at detector 2 will be increased in peak height response and all of the first column solute will be transferred to the second column.

Due to the conservation of mass, the peak areas in 1D = peak areas in 2D. Compression upon transfer causes the width to decrease so the peaks must become taller.

Question 16

What is the equation for peak height and what does increased peak height result in?

Answer 16

It results in increased sensitivity and therefore increased signal to noise, which is a big advantage of the modulation.

Question 17

What are the typical GC x GC column choices?

Answer 17

• 1D stationary phase - non polar ‘boiling point’
• 2D stationary phase - polarity, different solvents have different polarities based on the analyte

Question 18

Why use HPLC-GC?

Answer 18

The LC and GC separations can be based on different physical mechanisms, giving high peak capacity.

Question 19

What analytes ar suitable for coupled LC-GC?

Answer 19

Analytes must be suitable for both separation steps i.e. medium range volatility and must be soluble in non-polar LC mobile phase.

Question 20

What is the main problem of on line LC-GC?

Answer 20

The transfer of solvent from LC to GC. The technique must selectively remove the solvent without removing the analytes, therefore leaving the solute in a sharp band at the entrance to the separation column. If all the solvent from LC is injected into GC it could ruin the column and the chromatography.

Question 21

What kind of HPLC does coupled LC-GC use?

Answer 21

It uses normal phase HPLC.

This involves a non-polar mobile phase and a polar stationary phase. This is because a non-polar (organic) mobile phase is much easier to remove than a polar mobile phase (e.g. water).

Since the solvent from LC must be evaporated at the beginning of GC prior to the second separation, volatile HPLC solvents are required.

Question 22

How does the transfer from LC to GC work?

Answer 22

Only portions are transferred from LC to GC.

1. When the end of the desired fraction appears in the LC detector, the valve is switched and the sample loop fraction is pushed into the GC by the carrier gas.
2. The solvent is immediately evaporated during transfer through an uncoated pre-column in a GC oven and into an early vapour exit, prior to the coated analytical column. The loop interface is limited to low compounds of low volatility.
3. After solvent evaporation, the early vapour exit is closed so the analytes can travel through to the separation column.

Question 23

What three primary characteristics of compounds can be used to create HPLC separations?

Answer 23

• Polarity
• Electrical charge
• Molecular size

Question 24

How does elution occur in normal phase HPLC?

Answer 24

This technique typically uses adsorption chromatography of bare silica particles/aminosilane coated particles for the stationary phase.

Solvent molecules compete with solute molecules for sites on the stationary phase. Elution can be described as a displacement of the solute from the stationary phase by the solvent.

Question 25

How can selectivity in NPLC for characteristics in addition to polarity be achieved?

Answer 25

By using extra groups bonded to the silica particles, e.g.

• separation of aromatics and/or unsaturated compounds - phenyl and biphenyl phases
• separation of halogenated/oxygenated/nitro/basic compounds - cyano and fluorinated phases

Question 26

Why can’t you just inject gases into GC?

Answer 26

Gases are very dilute so you can’t just inject the gaseous mixture. There won’t be enough mass to achieve any signal. You need pre-concentrate the sample via a thermal desorption tube.

Question 27

How do you inject pre-concentrated gases into GC?

Answer 27

• evacuated canisters collect gas samples
• the gas is pumped through a thermal desorption tube containing a polymer sample (cooled to -30^oC if the components are volatile)
• carbon based molecules will stick to the polymer sample
• heat the tube very rapidly to cause desorption to occur rapdily to give sharp peaks

Question 28

How do you inject a high concentration liquid analyte into GC?

Answer 28

(> 0.1%)

The norm in capillary GC is to use a splitter, which injects about 1% of the sample into the column. 100 mL/min is split off while 1 mL/min is injected (split ratio of 100:1).

Question 29

How do you inject a low concentration liquid analyte into GC?

Answer 29

(< 0.01%)

Perform a splitless injection. Inject about 2 μL of the sample, then use either solvent trapping or cold trapping to trap the analyte at the head of the column - ensuring short injection length. Then, rapidly raise the temperature.

Question 30

What is solvent trapping and cold trapping when injecting liquids into GC?

Answer 30

• solvent trapping - column is cooled to about 40^oC below the boiling point of the solvent.
• cold trapping - column is cooled to about 150^oC below the boiling point of the analyte.

Question 31

Under what conditions must mass spectrometers function under and why?

Answer 31

Under high vacuum.

This is necessary because you want to avoid collisions of your ions with other molecules, possibly changing ion trajectory or leading to unwanted side reactions.

Question 32

Give the equation for mean free path.

Answer 32

Question 33

What is the upper limit of tolerable pressure for a mass spectrometer?

Answer 33

A pressure of about 10^-4 Torr (0.01 Pa or 10^-7 atm) ensures a mean free path of 50 cm, so this is the upper limit.

Question 34

Describe the transfer from vacuum pumps to a mass spectrometer.

Answer 34

• at 10^-4 Torr (10^-7 atm), 1 cm³ of gas at atmospheric pressure becomes 10⁷ cm³
• therefore a 1 cm³ min^-1 flow requires a pumping flow of 10⁷ cm³ min^-1 or 166 L s^-1
• the actual capacities of pumps in MS range from 50 to 1000 L s^-1 so that the maximum gas flow is about 5 cm³ min^-1 under 1 atm pressure

Question 35

Describe the transfer from GC (or GCxGC) to a mass spectrometer.

Answer 35

• in GC, capillary columns have a typical flow rate of 1-2 cm³ min^-1 and therefore are within acceptable values so it can be coupled directly to MS
• however this doesn’t allow for elimination of solvent

Question 36

Describe the transfer from HPLC to a mass spectrometer.

Answer 36

Coupling of LC to MS is more difficult because:

• gas phase ions must be produced (LC uses liquid) for MS
• flow rates are much higher and the elution solvent must be eliminated

Many different methods can be used to couple LC to MS and ionise the sample, the most common being electrospray ionisation (ESI) and atmospheric pressure chemical ionisation (APCI).

Question 37

What are the advantages of TOF analysers over other mass spectrometers?

Are there any disadvantages?

Answer 37

• rate of data (spectral) acquisition is extremely high
• all ions formed are analysed
• ion transmission is high - good sensitivity in TOF
• the main disadvantage is that mass resolution can be poor without proper correction but it’s usually corrected

Question 38

How does linear TOF-MS work?

Answer 38

Positive ions are expelled from the source in bundles that are either produced by an intermittent process such as laser desorption or by pulsing the potentials of extraction plates. These ions are accelerated, by the acceleration voltage V_S, into the field-free region.

Ideally, all ions enter this region with the same kinetic energy (E_k). The detector at the end of this region is negatively charged, attracting the positive ions to it.

Question 39

Write the equation for kinetic energy.

Answer 39

The higher the extraction potential (V_S) the faster the ions will be accelerated and the higher their E_k. As it leaves the source, an ion with mass m and charge q = ze has a kinetic energy of

Question 40

Show how t² is proportional to m/z.

Answer 40

Question 41

Give the equation for mass resolution for TOF.

Answer 41

Mass resolution for TOF is dependent on the time between two ions of m₁/z and m₂/z arriving, Δt) and the total flight time (t).

Question 42

What is mass resolution affected by in TOF?

Answer 42

It’s affected by factors that create a distribution in flight times among ions with the same m/z ratio, which ideally would be expected to arrive at exactly the same time.

The factors:

• the variation of the initial kinetic energy of the ions - variability in velocity of the ions
• the size of the volume where ions are formed - space distribution

Question 43

Describe how the variation of the initial kinetic energy of the ions can affect mass resolution in TOF.

Answer 43

Even if the ions have the same m/z ratio, if one ion starts off with more kinetic energy then it will reach the detector quicker, as this greater velocity will be maintained in the field-free region.

Consequently, the ions have different flight times and arrive at the detector at different times.

Question 44

Describe how the size of the volume where ions are formed can affect mass resolution in TOF.

Answer 44

Ions with the same m/z ratio and velocity but some are formed closer to the exit of the extraction region, allowing them to reach the detector first.

Question 45

The size of the volume where ions are formed in TOF can affect mass resolution.

What factors can we take advantage?

Answer 45

We can take advantage of two opposing factors:

• ions in the source that are more distant from the detector have further to travel
• ions formed more distant from the detector experience a larger extraction potential and therefore reach a higher velocity in the drift tube

Question 46

How can spacial distributions in TOF be corrected?

Answer 46

They can be corrected by introducing a second extraction region (V₂) and changing the position of the space focus plane - the point where the spacial distribution is minimised for ions of any given m/z.

The space focus plane can be moved to the detector by changing the rtio of the potentials V₁ and V₂.

Question 47

What is a reflectron?

Answer 47

It is an electrostatic reflector. It creates a retarding field (D) that acts as an ion mirror by deflecting the ions and sending them back through the flight tube.

Question 48

How does a reflectron correct energy dispersion?

What does this mean?

Answer 48

The reflectron corrects the energy dispersion of the ions leaving the source, as the ions with larger kinetic energy will travel further into the retarding field. With properly chosen voltages, path lengths and fields, ions with different kinetic energy but the same m/z ratios reach the detector at the same time.

This means that reflectron TOFs offer much better mass resolution than quadrupole MS.

Question 49

Define E_K‘/E_K = a²

Answer 49

Question 50

Define t/t’, t_r/t_r’ and t_tot/t_tot’.

Answer 50

Question 51

How do the variations of the flight times compensate each other?

Answer 51

If a > 1, then the ion with an excess of kinetic energy will have a shorter flight time out of the reflectron (t/a) but a longer flight time in the reflectron (at_r), and vice verse. This means that the flight times compensate each other.

Question 52

What conditions must be met to provide the quantitatively correct flight time compensation?

Answer 52

A correct compensation yielding the same flight time for all of the same m/z but different kinetic energies requires choosing the proper values of E (electron field in the reflectron, related to x).

Question 53

What is proton transfer reaction mass spectrometry (PTR-MS)?

Answer 53

It’s a direct technique (no GC front end) where the molecular ion is maintained due to it being a softer ionisation technique.

Can be used by protonating water (used as the chemical ionisation agent) which then goes on to protonate the analyte.

H₂O -> H₃O⁺

Question 54

PTR-MS was developed fromt SIFT MS, but it differs from it in two ways. What are these?

Answer 54

1. There is no mass selection of chemical ionisation ions, i.e. the ion source is directly connected to a drift tube allowing a greater intensity of chemical ionisation ions to enter the flow tube (sometimes at the expense of ‘contamination’ ions)
2. There is no introduced carrier gas therefore the analytes are not diluted - if the sample is already a gas then it’s not needed (oxygen and nitrogen in air aren’t protonated by typical ionisation agents like H₃O⁺

Question 55

How can you predict which compounds will react with H₃O⁺ as the chemical ionisation agent?

Answer 55

By using proton affinities.

• compounds with lower proton affinities will not be ionised, whereas compounds with higher proton affinities will
• most VOCs have higher proton affinities
• generally, larger molecules have larger proton affinities
• proton affinity is a measure of exothermicity of the gas phase proton transfer

Question 56

What are the thermodynamics of proton transfer to R by H₃O⁺?

Answer 56

Typically, differences in proton affinities are small compared to dissociation energies of the protonated ions, so soft chemical ionisation occurs with little or no fragmentation.

When R is protonated, you’re not giving much energy to it and therefore it’s not likely to undergo fragmentation.

Question 57

What are the kinetics of proton transfer to R by H₃O⁺?

Answer 57

Under pseudo-first order conditions (H₃O⁺ in excess):

[RH⁺] = [R] exp (-k[H₃O⁺]t)

Question 58

How else could protonation occur in PTR-MS using H₂O?

Answer 58

• Proton transfers by water clusters may also occur:

H₃O⁺-H₂O + R –> (H₂O)₂ + RH⁺

This is only energetically possible if:

PA(R) > PA(H₂O) + bond energy(H₂O-H₃O⁺)

• Or it could occur by ligand switching:

H₃O⁺-H₂O + R –> H₂O + RH⁺H₂O

Followed by dissociation to RH⁺

Question 59

Describe charge transfer in the PTR reactor.

Answer 59

• need to induce sufficient collisions - H₃O⁺ and analyte mixed in a react tube (drift cell)
• needs to be a voltage gradient across reactor
• aim is to have dilute analyte so analyte molecules only collide with H₃O⁺ - keeps ionisation scheme simple
• pressure is about 1 mbar, approx. 2000 collisions down a 10 cm length reactor

Question 60

Describe the PTR drift tube (reactor) and the E/N ratio.

Answer 60

• E is the voltage gradient down the drift tube in V/cm
• N is the gas density in molecules/cm³ (1mbar typically)
• E/N in Townsend; 1 Townsend = 10^-17 V cm²
• 120-140 Td is ideal for non-fragmentation of organic molecules
• If E/N is too high - fragmentation
• If E/N is too low - clustering of water molecules occurs

Question 61

Why does fragmentation occur at high E/N?

Answer 61

• energy balance for an ion is a sum of drift velocity (v_d) and T-dependent term (T_drift)
• v_d usually dominates, except at low v_d where T_drift term will have a non-negligible effect
• μ₀ and N₀ are both constants so v_d ∝ E/N
• higher E/N = higher v_d = higher energy collisions making fragmentation more likely

Question 62

When do water clusters form?

Answer 62

More water clusters form because of:

• humid conditions
• low E/N ratio

Question 63

What are the desired conditions for PTR-MS?

Answer 63

• low E/N - less analyte fragmentation and more reaction time
• pure H₃O⁺

Question 64

Why is pure H₃O⁺ desirable?

Answer 64

• cleaner spectrum
• H₃O⁺ and its water clusters may react at different rates with the analyte, which would make the senstivity a function of water vapour

Question 65

Describe the energy balance of the ions as a function of the E/N ratio.

Answer 65

The amount of energy input to the ion during a collision can be related to cluster ratio.

• at high E/N, there’s very little dependence of I₃₇/I₄₀ on temperature (T_drift) - high drift velocity so velocity term in the equation dominates
• at low E/N, there’s a higher dependence on I₃₇/I₄₀ (T_drift)