Watch this lesson online: https://edrolo.com.au/vce/subjects/chemistry/vce-chemistry/aos-1-chemical-analysis/chromatography-hplc-glc/column-chromatography/#watch CHEMISTRY Unit 3, Area of Study 1: Chemical Analysis Chromatography HPLC and GLC 1
Column chromatography (The basis for HPLC and GLC) In the simplest form, column chromatography is carried out in the laboratory in a glass column packed with a solid such as alumina, Al 2 O 3. 2
Separation occurss in a short column Stationary phase: very Iine solid particles tightly packed eg silica HPLC Mobile phase: liquid under high pressure to force its way through the solid 3
The Chromatogram R t - retention time - determines sample identity R t mau R t Area or height is proportional to the quantity of analyte. Injection time We do our best to make these slides comprehensive and 4 up-to-date, however there may be errors. 4
HPLC Applications HPLC is very sensitive! Chemical polystyrenes dyes phthalates Bioscience proteins peptides nucleotides Environmental Pharmaceuticals polyaromatic hydrocarbons Inorganic ions herbicides tetracyclines corticosteroids antidepressants barbiturates Clinical Consumer Products amino acids vitamins homocysteine lipids antioxidants sugars We do our best to make these slides comprehensive and 5 up-to-date, however there may be errors. 5
HPLC Sample exam question HPLC uses a non- polar stationary phase together with a polar mobile phase. Which of the following molecules will have the greatest retention time on such a column? (VCAA June 2009 Q 5) 6
Answer The answer is A Molecules B, C and D are all polar. To see this observe the OH and NH groups on the hydrocarbon chain of the molecules. A non- polar molecule will have the greatest attraction to the stationary phase, so will take the longest time to move along the column = greatest retention time. 7
GC Sample and the solvent it is dissolved in must be volatile as they are heated in an oven Mobile phase: Inert gas Stationary phase: porous beads coated with high bpt liquid 8
Most sensitive of all chromatography techniques Used for testing urine samples for drugs Compounds must be able to be vaporised without decomposing (usually low molecular mass <500 g mol - 1 samples) 9
Separa0on of mixture Sample passes along column with mobile phase Passes in and out of solution in stationary phase Component which is least soluble in stationary phase will be eluted Iirst (moves quickest along column) As components are eluted they are detected by Ilame ionisation detector. 10
Chromatogram A graph with peaks which correspond to components of mixture Retention time = time taken for component to pass through column 11
Quantitative use of GC Run a series of standards of known concentration under the same conditions as the sample. The area under the peak is proportional to the amount of the substance present. Draw a calibration graph to link area under peak to concentration Find sample concentration using graph. 12 12
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GC Sample exam questions 1. Analysis of the components of a mixture using gas- liquid chromatography normally involves measuring the A. R f values of components using a gas as the mobile phase B. R f values of components using a liquid as the mobile phase C. Retention times of components, using a gas as the mobile phase. D. Retention times of components, using a liquid as the mobile phase. (VCAA June 2007 Q71) 14 14
2. This diagram shows the gas chromatogram of a sample containing 4 straight chain alkanes. Which of the statements that follow are true? I. The boiling points of the compounds arranged from highest to lowest are Z>Y>X>W II. The retention times will stay the same if the temperature at which the chromatogram is recorded is increased and all other conditions remain constant. III. Hydrogen gas could have been used as a carrier gas to obtain this chromatogram. (VCAA June 2009 Q6) 15 15
Answers 1. C is the answer R f values are used for TLC and paper chromatography, while retention times are used for column- based chromatography. In GC, the mobile phase is a gas. 2. Statements I and III are correct Changing the temperature will change the retention time of all of the alkanes 16 16
QUESTION 1 A student used paper chromatography to separate two components, I and II, in a solution. A spot of the solution was initially placed at the origin. When the spot corresponding to component I (R f = 0.50) had advanced 4.0 cm, the spot corresponding to component II was 1.0 cm behind. The R f value of component II is closest to A. 0.68 B. 0.38 C. 0.25 D. 0.13 (VCAA Q6 June 2006) 17
QUESTION 1 (ANSWER) A student used paper chromatography to separate two components, I and II, in a solution. A spot of the solution was initially placed at the origin. When the spot corresponding to component I (R f = 0.50) had advanced 4.0 cm, the spot corresponding to component II was 1.0 cm behind. The R f value of component II is closest to A. 0.68 B. 0.38 C. 0.25 D. 0.13 The answer is alternative B First we need to calculate the distance that the solvent had travelled. Since R f We can calculate that the distance travelled by the solvent So now we can find the R f of component II which travelled only 3.0 cm = (1.0 cm behind component I) 3.0 8.0 = 8.0cm R f = = 0.375=0.38 (to 2 sig figs) (VCAA Q6 June 2006) 4.0 0.50 18
QUESTION 2 Chromatogram 1 was obtained by analysis of a sample of a mixture of two sugars, A and B, using high performance liquid chromatography (HPLC). Chromatogram 2 was obtained by analysing another sample of the same mixture by HPLC under different conditions. Consider the following changes, which could be made to the operating conditions for HPLC. I decreasing the pressure of the mobile phase II decreasing the temperature III using a less tightly packed column Which of the changes would be most likely to produce chromatogram 2? A. I only B. II only C. III only D. I and II only (VCAA Q5 June 2007) 19
QUESTION 2 (ANSWER) Chromatogram 1 was obtained by analysis of a sample of a mixture of two sugars, A and B, using high performance liquid chromatography (HPLC). Chromatogram 2 was obtained by analysing another sample of the same mixture by HPLC under different conditions. Consider the following changes, which could be made to the operating conditions for HPLC. I decreasing the pressure of the mobile phase II decreasing the temperature III using a less tightly packed column Which of the changes would be most likely to produce chromatogram 2? 20
A. I only B. II only C. III only D. I and II only The answer is alternative C (III only). The retention times in chromatogram 1 are greater than those in chromatogram 2, so changes which slow the progress of the mixture will result in chromatogram 1 and those that speed it up will result in chromatogram 2. Decreasing the pressure of the mobile phase will slow down the progress of the components through the column, so this will match chromatogram 1, not 2 Decreasing the temperature of the column will also slow down the progress of the components (chromatogram 1) Using a less tightly packed column will allow the components to move more quickly through the column so this will result in chromatogram 2. (VCAA Q5 June 2007) 21
QUESTION 3 Consider the following TLC plate of compounds X, Y and Z developed using a suitable mobile phase on a polar stationary phase. The R f value of the most polar component in this TLC separation is: A. 0.29 B. 0.62 C. 0.78 D. 0.80 (VCAA Q6 June 2010) 22
QUESTION 3 (ANSWER) Consider the following TLC plate of compounds X, Y and Z developed using a suitable mobile phase on a polar stationary phase. The R f value of the most polar component in this TLC separation is: A. 0.29 B. 0.62 C. 0.78 D. 0.80 The answer is alternative A The question states that the stationary phase is polar, so the most polar component will have the greatest attraction to the stationary phase and will take the longest to move along the TLC plate. It will have the smallest R f value. It is component X that travels only 0.80 cm (BUT this is not its R f ) R f = distance travelled by component/distance travelled by solvent = 0.8/2.75 =0.29 (VCAA Q6 June 2010) 23