For a split/splitless injector operating in the split mode, what will be the split ratio if the carrier gas inlet flow rate is 100 mL/min and the column flow rate is 2 mL/min?
1 : 50 The flow rate is 2 mL/min in the column for a flow rate of 100 mL/min at the inlet of the injector => 2 : 100 => 1 : 50
98 : 2
1 : 102
I am using a split mode injector and wish to obtain a split ratio of 1:40. The flow rate of the carrier gas in the column is 1.5 ml/min. What will the gas inlet flow rate be into the injector?
60 mL/min The injector is in split mode so for 1.5 mL/min in the column, if the split ratio is 1: 40, then the gas inlet flow rate into the injector is = 40 x 1.5 mL/min. (see the "Injector" tab of this sheet)
In a capillary column the stationary phase coats the inner wall of the column.
True (see the "Separation column" tab of this sheet)
The separation of compounds by gas chromatography is carried out:
Only based on the difference of their affinity for the stationary phase
Only on the basis of their difference in boiling point
Based on the difference in both their boiling point and their affinity for the stationary phase (see the "Separation column" tab of this sheet)
Two compounds A and B with very different boiling points (T°bp) (T°bp A > T°bp B) and a similar affinity for the stationary phase are injected in gas chromatography. What is the order of elution of compounds A and B?
Compound A is eluted before B
Compound B is eluted before A Since the affinity for the stationary phase is identical for both compounds, then the separation will be carried out on the basis of the T°bp of the compounds. The compound which has the lowest boiling point will be eluted first, here compound B.
Both compounds are eluted together
Two compounds A and B with a different affinity for the stationary phase and very close boiling points are injected in gas chromatography. What is the order of elution of compounds A and B?
Compound A has more affinity for the stationary phase, it is eluted before B
Compound A has less affinity for the stationary phase, it is eluted before B Yes, indeed if the compounds have very close T°bp then the separation will be carried out on the basis of the affinity of the compounds for the stationary phase. The compound with the least affinity for the stationary phase will be eluted first.
Whatever the affinity of the two compounds for the stationary phase, the two compounds are co-eluted. This happens when the affinity of the two compounds for the stationary phase is not sufficiently different.
The stationary phase in the capillary columns can be of different thicknesses. At a constant column diameter, what will be the influence of a greater stationary phase thickness on the retention times of the compounds?
no influence on the retention times of the compounds, it just allows for the injection of a larger amount of sample into the column. It is true that the amount of stationary phase in a column determines the amount of compound that can be injected (column capacity). However, if the diameter remains constant, the thickness of the stationary phase will have an influence on the retention times.
a longer retention time of the compounds. For a constant capillary column internal diameter, the greater the stationary phase thickness, the smaller the phase ratio β (Volmobile phase/Volstationary phase) will be. As the retention factor is inversely proportional to β then the retention times are higher when the phase ratio β decreases
a lower retention time of the compounds
Two compounds of identical boiling points are co-eluted during gas chromatography. To hope to separate them, the operator must:
Change the oven temperature The temperature has an influence on the equilibrium constant K (). If the influence is not identical for the two compounds then the separation of the two compounds can occur.
Change the column to use another stationary phase Indeed, in this case, selectivity is increased by favoring the affinity of one compound over the other for the stationary phase.
Change the column and choose a column with the same stationary phase but with a greater phase thickness.
The flame ionization detector requires two gases to operate. The two gases that can be used for this type of detector are:
Oxygen and dihydrogen This detector works thanks to a flame that allows for the combustion of the compounds to be analyzed. To produce the flame, you need an oxidizer (dioxygen) and a fuel (dihydrogen).
Air and dihydrogen In this case the oxidizer will be the oxygen contained in the air.
Helium and air
Helium and dihydrogen