Advanced Separations

Question 1

Describe the general apparatus used in chromatography.

Answer 1

• a mobile phase containing an analyte is loaded into a sample injector where it’s injected into a column with a stationary phase attached
• the sample and mobile phase make it to a detector which creates the chromatogram
• injector and detector are as close to the column as possible so all the separation is just due to the interactions happening

Question 2

Describe the fundamental principal that chromatography is based on.

Answer 2

Different molecules have different affinities for the stationary phase, causing different molecules to spend different amounts of time in it. This means the molecules with a lower affinity for the s.p. will elute faster.

Question 3

What is retention time?

Answer 3

t_R

The time an analyte spends in the s.p. and therefore the elution time from injection. A larger retention time means more time was spent in the s.p.

Question 4

What is resolution?

Answer 4

The degree of separation between peaks, defined in terms of the retention time and peak widths.

For good baseline resolution, R_s > 1.5 = full peak separation.

Question 5

What is the capacity factor?

Answer 5

k’

An experimental parameter describing the amount of time a solute spends in the stationary phase relative to the time in the mobile phase. The longer a component is retained by a column, the higher the capacity factor.

The ideal range is 1-5, but in reality it can be 0.5-20.

Question 6

What is the selectivity factor?

Answer 6

α

A ratio of the capacity factors.

For separation to occur, the analytes must have different capacity factors.

α usually > 1.

Question 7

How is the capacity factor (k’) related to the equilibrium constant (K)?

Answer 7

Because the ratio of phase volumes is constant for a given column and mobile phase, the capacity factor for any analyte is directly proportional to its equilibrium constant.

This means that if one analyte (B) is retained loner than another (A) and it’s at equilibrium, then C_S (B) > C_S (A).

Question 8

How can the capacity factor be found from a chromatogram?

Answer 8

Migration velocity through a column depends on the distribution of analyte between the m.p. and s.p.

Question 9

What is band broadening?

Answer 9

The change in band shape with time.

The band of analyte at time = zero is narrow, but the band at time = t is much broader.

Question 10

Why does band broadening occur?

Answer 10

Rate theory of chromatoraphy:

A random walk in 1 dimension to give a symmetrical spread of velocities around the mean value.

• transferring between the phases takes energy from the surroundings, but the gain of energy is random
• some molecules speed up while some don’t = symmetrical spread

Question 11

What is a Gaussian shaped peak?

Answer 11

Gaussian band shape occurs when the partition coefficient, K, is independent of the [solute] on the column. It’s where band broadening is statistically random and creates a normal distribution.

σ = standard deviation, length units

τ = standard deviation, time units = Width/4

σ² = τ² = variance

Question 12

Describe the relationship between peak area and standard deviation for a Gaussian shaped peak.

Answer 12

• 68% of the peak area is in the range t_R ± τ
• 95% of the peak area is in the range t_R ± 2τ

Question 13

How does diffusion affect band spreading?

Answer 13

• diffusion is the main cause of band spreading
• the diffusion coefficient, D, measures the rate a substance moves randomly from a region of [high] to [low]

σ² = 2Dt

Question 14

Describe how asymmetric band shapes can occur.

Answer 14

When K is dependent on the [solute] on the column, skewing can occur:

• fronting - overloading too much solute, correct by using smaller/more dilute samples
• tailing - small quantities of solute and retained more strongly than large quantities, correct by masking strong adsorption sites on the station phases

Question 15

What is plate height?

Answer 15

H

The constant of proportionality between the variance of the band (σ²) and the distance it has travelled (length, L).

σ² = HL

H = D/u

(u = flow rate)

Different solutes have different plate heights because they have different diffusion coeffcients (D). Smaller H = narrower bandwidth.

Question 16

What is efficiency?

Answer 16

N, the theoretical number of plates. It’s a measure of the dispersion of a peak. Narrow peaks take up less space and therefore allow for more peaks to be separated.

N = L/H

A bigger N value is better, as it indicates that more peaks can be separated.

Question 17

What can be changed to ensure that resolution will be greater than 1.5?

Answer 17

Resolution can be described by the equation below, containing both a selectivity term and an efficiency term. When peaks get closer (less separation) the selectivity term decreases. To ensure fully resolved peaks (R_s > 1.5) the selectivity and efficiency might need to be changed.

Question 18

What is the van Deemter equation and what does it contain?

Answer 18

H = A + B/u + Cu

A = multiple flow paths

B = logitudinal diffusion term

C = mass transfer term

Question 19

How do the terms of the van Deemter equation contribute in different situations?

Answer 19

• in packed columns, all three terms contribute
• in open tubular columns, A = 0
• in capillary electrophoresis, A and C = 0 aka very high resolution

Question 20

What is multiple flow paths (A)?

How can it be reduced?

Answer 20

This is where different molecules may take different length paths through the stationary phase due to encountering different sized s.p. particles.

This can be reduced by useing smaller s.p. particles to make the pathways more similar in length.

Question 21

What is longitudinal diffusion (B)?

How can it be reduced?

Answer 21

This is where the longer the column, the more diffusion that can take place which causes the peak to broaded.

This can be reduced by faster linear flow, so less time is spent in the column therefore less opportunity for diffusion.

Question 22

What is mass transfer/equilibration time (C)?

How can it be reduced?

Answer 22

This is where a finite amount of time is needed for the analyte to equilibrate between the mobile and stationary phases.

This can be reduced by a decrease in s.p. thickness or column radius, or an increase in temperature.

Question 23

How does plate height (H) vary with flow rate (u)?

Answer 23

Extent of band broadening (due to H) depends on the length of time in m.p. is in contact with the s.p. (u). An ideal flow rate, giving the lowest H, needs to be found to get the least amount of band broadening.

• A doesn’t change with flow rate
• B/u gives high H for low u as there’s more time for diffusion
• Cu gives high H at high u as there’s not enough time for equilibration

Question 24

What is the order of elution in GC?

Answer 24

Within a compound class, analytes elute in the order of their boiling points. When comparing different compound classes, analytes with greatest similarity to the s.p. are more strongly retained (elute later).

Question 25

What is the equation linking capacity factor with temperature?

What does it show?

Answer 25

It shows that for a homologous series of analyters, k’ increases as T_bp increases.

It also shows that for all analytes, k’ decreases as T increases.

Question 26

How can temperature be used to control retention time?

Answer 26

If the component you’re interested in takes a long time to elute (or there’s a wide range of boiling points) then gradually increasing the temperature of the experiment (T) will cause k’ to decrease. This makes the analytes elute faster and also gives better peak spacing and peak shape.

Question 27

What is the order of elution in reverse phase HPLC?

Answer 27

The analytes elute in order of their hydrophobicity i.e. the most hydrophobic analyte elutes last.

(In normal HPLC the opposite is true).

Question 28

How can capacity factor be controlled in RP-HPLC?

Answer 28

The s.p. in RP-HPLC isn’t really changed but the m.p. is (usually polar/water based).

An organic modifier can be added to the m.p.:

log k’ = log k’_w - SΦ

Log k’ decreases linearly with the volume fraction of OM (Φ) which will make the analytes elute faster. Therefore, we can change the amount of OM (S) and the volume fraction of OM (Φ).

Question 29

What is eluent strength in RP-HPLC?

Answer 29

ε^o

It is a measure of the solvent adsorption energy. As eluent strength increases solutes elute more rapidly, meaning that in RP less polar solvents have higher eluent strength.

Question 30

What happens when you change the organic modifier (S) in RP-HPLC?

Answer 30

Differents organic modifiers have different S values. As the S value increase, log k’ decreases.

A nomograph shows the volume percentage of solvents with the same eluent strength.

Question 31

Varying the m.p. composition in RP-HPLC can help elution. How?

Answer 31

Similar to changing T in GC, it allows slow eluting analytes to elute faster while maintaining good peak separation (decreasing k’).

• isocratic elution - m.p. composition stays the same throughout the experiment
• gradient elution - m.p. composition is slowly changed throughout the experiment

Question 32

How does porous vs. non-porous silica affect RP-HPLC?

Answer 32

Non-porous silicas have much lower surface areas and carbon loadings. This means the phase ratio, V_S/V_M is lower and therefore so is k’.

Question 33

Describe how plate heights differ in LC compared to GC.

Answer 33

Fast equilibration is needed in the direction transverse to flow to counteract zone broadening.

In GC, transverse diffusion is fast but so is diffusion in all other directions. In LC, transverse diffusion is slower meaning that plate heights in LC are usually smaller.

Question 34

How does plate height change with particle diameter and flow rate in LC?

Answer 34

When particle diameter is smaller plate height decreases. When flow rate increases plate height also increases. However, at smaller particle diameters the gradient for increasing H is lower.

Question 35

Why does plate height change with increasing particle diameter?

Answer 35

Plate height increases with increasing particle diameter for two reasons:

• A term increases with increasing d_p - less uniform flow through the column
• C term increases with increases d_p - takes longer to diffuse through inner pores

Question 36

Compare LC plate height and efficiency with GC.

Answer 36

In LC, plate height is lowered by a factor of 10-100 (slower diffusion) but efficiency is higher in capillary GC.

Question 37

How can you optimise efficiency in LC?

Answer 37

• Use packed columns with particle diameters as small as possible, consistent with pressure limitations
• minimise any extra broadening effects:

total = injector + column + detector

• the key factor is reducing column to detector distance