6.3 Flashcards
6.3.1 a) Explain what chromatography is, and the requirements for carrying out chromatography.
Chromatography is used to separate and identify individual components from a mixture of substances.
It requires two phases:
- The stationary phase, which is fixed in place and does not move. This is usually a solid, or a liquid supported on a solid.
- The mobile phase, which moves in a definite direction. This is normally a liquid or a gas.
Two examples of chromatography include thin-layer chromatography and gas chromatography.
6.3.1 a) Outline what TLC is and how to carry it out.
Thin layer chromatography (TLC) is a technique used to indicate how many components are in a mixture. TLC involves a glass or plastic plate being coated in a thin layer of solid - the solid is an adsorbent substance (e.g. silica gel or aluminium oxide). This provides the stationary phase. The mobile phase is an organic solvent - it moves in one direction, up and over the TLC plate.
To carry out TLC:
1) Using a pencil, draw a line near the bottom of the plate. Then, using a capillary tube, spot a small amount of the mixture onto this base line.
2) Place the prepared TLC plate in a beaker containing a small amount of the solvent - ensure the solvent does not touch the spot of mixture.
3) Cover the beaker with a watch glass - this reduces the rate at which the solvent is absorbed. and allows for better separation.
4) When the solvent has nearly reached the top of the plate, remove it from the beaker and mark the distance the solvent has moved (i.e. the solvent front) with a pencil. Circle any other visible spots.
6.3.1 a) Outline how to calculate the retention factor in TLC, and explain why different chemicals have different Rf values.
The retention factor, Rf, for each component shows how far a chemical moves up a chromatogram - this is when compared to the solvent front.
Rf = distance moved by the component / distance moved by the solvent
Each component can be identified by comparing its Rf value with known values - these values have to be recorded using the same solvent system and adsorbent. Alternatively, control spots (using pure samples of individual compounds that may be present in the mixture) can be run on the same TLC plate - a direct comparison can be made to identify each chemical in the mixture.
Why do chemicals have different Rf values?
- As the solvent rises up the TLC plate, the mixture moves with it.
- The separation occurs as each chemical in the mixture experiences different attractions to the stationary phase.
- The attraction between a substance and the surface of the plate it called adsorption.
- The higher the relative adsorption of a substance, the slower it moves, and therefore the shorter the distance it travels up the plate.
[Note: a chemical might be very attracted to the mobile phase, while experiencing weak adsorption with the stationary phase - if this is the case, it might not be separated using TLC]
6.3.1 b) Explain what gas chromatography is, and how it works.
Gas chromatography (GC) is useful for separating and identifying volatile organic compounds present in a mixture. The stationary phase is either a solid or liquid coating on the coiled tube - this coating is usually a hydrocarbon with a high boiling point. The mobile phase is an inert carrier gas (e.g. helium, neon or nitrogen).
To carry out GC:
- A sample of the volatile mixture is injected into the gas chromatograph.
- The mobile gas carries the components through the coiled tube, where they either dissolve (in a liquid stationary phase) or adsorb (onto a solid stationary phase).
- They then evaporate back into the gas, and the whole process is repeated (with each chemical re-dissolving/re-adsorbing) as the mixture travels through the tube.
- The individual components are separated and the time takes for a substance to pass through the coiled tube and reach the detector is called the retention time.
- This produces a chart with absorption (y) against time (x), called a gas chromatogram.
6.3.1 b) i) Explain what retention time is, and outline why different chemicals have different retention times.
Each peak on a gas chromatogram corresponds to a substance with a particular retention time. The retention time is the time taken for each component to travel through the column and reach the detector. Retention times can be compared with known values (recorded using the same phases) to identify each component.
- The more soluble a component is in the liquid stationary phase, or the higher the relative adsorption of a substance is in the solid stationary phase, the slower it moves through the coiled tube.
- The compound that travels through the column the fastest, and reaches the detector first, has the lowest retention time.
[Note: a substance with a high boiling point will spend more time condensed as a liquid in the tube than as a gas. This means it will take longer to travel trough the tube than one with a lower boiling point]
6.3.1 b) ii) Outline how the concentration of a component in a worked out using the peak integration.
The area under each peak in a gas chromatogram is proportional to the relative amount of each substance in the original mixture.
However, the relative amount of a substance differs from its exact concentration - to actually work out the concentration of this component you need to create an external calibration curve:
1) Prepare standard solutions of different known concentrations of analyte - analyte is the individual compound being investigated.
2) One by one, inject your standard solutions into a GC apparatus - this is to obtain a gas chromatogram for each concentration.
3) Calculate the area under the peak of each gas chromatogram.
4) Plot a calibration curve of area (y) against concentration (x).
Now, given the original gas chromatogram of the mixture, calculate the area under the peak for the compound being investigated. Compare this area with the calibration curve to calculate the concentration of the compound in the mixture.
[Note: the area under a peak is also called the peak integration]
6.3.1 c) i) Outline the test for identifying an alkene.
To test for alkenes:
Add bromine water, drop-wise, to the sample, and shake the test tube well.
- In the presence of an alkene, the bromine water will decolourise.
6.3.1 c) ii) Outline the test for identifying a haloalkane.
To test for haloalkanes:
Add aqueous silver nitrate and ethanol to the sample. Plate the test tube in a 50°C water bath to warm it.
- In the presence of a choroalkane, a white precipitate of silver chloride would form.
- In the presence of a bromoalkane, a cream precipitate of silver bromide will form.
- In the presence of an iodoalkane, a yellow precipitate of silver iodide will form.
6.3.1 c) iii) Outline the test for identifying a phenol.
To test for phenols, there are two steps:
1) Add solid sodium hydroxide to the sample.
- This is a strong base, and should react with phenols to form a colourless solution of sodium salt (i.e. the solid will dissolve). Alternatively, universal indicator or a pH probe can also be used to identify the weak acidity of a phenol.
2) Add either a solid carbonate or a carbonate solution (e.g. aqueous sodium carbonate) to the sample.
- The carbonate ion, CO3(2-), only reacts with strong acids. In the presence of a weak acid like phenol, there should be no reaction, and no effervescence observed.
6.3.1 c) iii) Outline the test for identifying a carboxylic acid.
To test for carboxylic acids:
Add aqueous sodium carbonate to the sample:
- In the presence of a carboxylic acid, effervescence will be observed (and the carbon dioxide gas that is produced will turn limewater cloudy).
6.3.1 c) vi) Outline the test for identifying a carbonyl compound.
To test for carbonyl compounds:
Add Brady’s reagent (2,4-DNP) to the sample.
- In the presence of an aldehyde or a ketone, an orange precipitate is formed.
6.3.1 v) Outline the test for identifying an aldehyde.
To test for aldehydes:
1) Add Tollens’ reagent to the sample.
- In the presence of an aldehyde, a silver mirror is produced.
2) Add Fehling’s solution to the sample.
- In the presence of an aldehyde, the blue solution forms a dark red precipitate.
3) Add acidified potassium dichromate.
- In the presence of an aldehyde, the green solution turns orange.
6.3.1 v) Outline the test for identifying an alcohol.
To test for alcohols:
Add acidified potassium dichromate to the sample.
- In the presence of a primary or secondary alcohol, the green solution turns orange.
- In the presence of a tertiary alcohol, there is no colour change.
