Chromatography Lecture 4 PDF

Chromatography Lecture 4 Chromatograms – qualitative and quantitative analysis 1) To use chromatograms for quantitative analysis 2) To know how to choose the most appropriate form of analysis (e.g. internal normalisation vs external calibration vs the use of internal standards. Remin...

Chromatography Lecture 4 Chromatograms – qualitative and quantitative analysis 1) To use chromatograms for quantitative analysis 2) To know how to choose the most appropriate form of analysis (e.g. internal normalisation vs external calibration vs the use of internal standards. Reminder Qualitative analysis: Provided the chromatography is carried out on the same equipment, using the same conditions, the retention time tR of compounds will remain relatively constant for any given compound. This can be a useful tool in identification of analytes. Quantitative analysis: We will look at this separately for GC and HPLC as different techniques are often used. Mainly there are three methods for quantitative analysis in GC and HPLC: -internal normalisation -use of external std -use of internal std Quantitative analysis in GC – Internal Normalsation Internal normalisation: If an equal amount of any compound will give the same peak area on a chromatograph, internal normalisation can be used to find the percentage of each component in the mixture. e.g. % of component x = (area of component x peak / total area of all peaks) x 100 This is often not the case i.e. the same number of moles of different components give different detector responses. In such cases the detector response must be calculated from a set of standards (external calibration), or use of an internal standard must be used. External standard In this approach a series of samples each containing various known amounts of the test substance are subjected to GC under identical conditions. A calibration curve of response area vs. concentration is then constructed and from this the concentration of the test sample may be read from the GC response.  External standard Example question: An analyst wants to determine the concentration of aspirin in tablet extract. They prepared a range of known concentrations of aspirin in methanol (0-1% v/v), injected them into the GC and recorded the area of the aspirin peak in each resulting chromatogram. Using this data a calibration graph of aspirin concentration vs. peak area was plotted: Y = 0.5027x In the original mixture, the peak that was assigned to aspirin had a peak area of 0.35cm2. Use the calibration graph to calculate the concentration of aspirin in the mixture. Show how you worked this out. What answer did you get? A. 0.70 % v/v B. 0.69 % v/v C. 0.35% v/v D. 0.50 % v/v E. None of the above Questions may be slightly more complex when they involve dilutions An aspirin tablet was dissolved in 10 mL of water (pH7) (Solution A). 1 mL of this solution was further diluted by placing it in a 10 mL flask and diluting to the mark with water (to make Solution B). Solution B was analysed by HPLC, and the peak area of the chromatographic peak for aspirin was found to be 0.25 cm2. Calibration standards had been prepared previously with concentrations ranging from 0.0 %w/v to 2.0 %w/v. The calibration curve of the standards had a linier fit of y = 0.955x + 0.01. What is the concentration of aspirin in Solution A? a) 0.025 %w/v b) 25.1 %w/v c) 0.25 %w/v d) 2.51 %w/v e) None of the above Quantitative analysis in HPLC Analysis based on calibration with an external standard Weigh out analytical standard for the analyte and dissolve it in a precise volume of solvent  stock solution. Prepare suitable dilutions (hopefully to cover the expected range of concentration of your unknown solution). Run them on the HPLC and plot a calibration graph. Make up standard solution of your tablet/formulation and inject. Use calibration graph to determine the concentration of analyte in the mixture from its peak area. Example: Analysis of paracetamol tablets using a calibration curve H Tablets: Expected to contain 500 mg Paracetamol N O Mobile Phase: 0.05M acetic acid/acetonitrile (90:15) HO The slightly acidic mobile phase ensures that the phenol group of paracetamol does not ionise. Column: ODS 4.6mm x 150mm Dilution: 1.25mg/100ml of mobile phase (high dilution needed to bring it within range of UV detector) Method: 1. Weigh out 125 mg  10 mg of paracetamol standard and transfer to 250 mL volumetric flask made up to volume with acetic acid (0.05 M) [STOCK SOLUTION] Actual weight 126.1 mg (in 250 mL) = 50.44 mg/100 mL 2. Prepare a series of solutions from the stock solution containing 0.5, 1.0, 1.5, 2.0 and 2.5 mg/100 mL of paracetamol. 3. Weigh and powder 20 tablets Weight of 20 tablets = 12.1891g 4. Weigh out powder containing ~125mg paracetamol. Transfer to a 250 mL volumetric flask and make up to volume with acetic acid (0.05 M). Weight of tablet powder taken = 150.5 mg Methods continue 5. Filter ca 50 mL of the solution into a conical flask. Transfer 25 mL aliquot of the filtrate into a 100 mL volumetric flask and adjust the volume to 100 mL with acetic acid (0.05 M) 6. Take 10 mL of the diluted extract and transfer to a further 100 mL volumetric flask and make up the volume with acetic acid (0.05 M). 7. Analyse the standards and the extracts using HPLC conditions described. Area of paracetamol peak from extracted tablets = 45205 Calculate the percentage of the stated content of paracetamol in the tablet powder analysed.  Methods continue Calabration curve of paracetamol standards The dilution steps to take into account were: 100000 Area of chromatographic - 25ml into 100ml (x4) 80000 - 10ml into 100ml (x10) 60000 Total x40 peak 40000 y = 35657x + 80.803 20000 R2 = 1 0 0 0.5 1 1.5 2 2.5 3 [Paracetamol] mg/100ml x (conc. of diluted extract) = 45205 – 80.8 = 1.266 mg/100ml 35656 Therefore, the concentration of paracetamol in undiluted tablet extract (i.e. the one where 150.5 mg of powder was put in 250 mL ) = 1.266 mg/100 mL x 40 = 50.64 mg/100 mL Since our original volume was 250 mL. So the amount of paracetamol in the 150.5mg of powder weighed out is = (50.64 mg/100) x 250 = 126.7 mg. Weight of 20 tablets = 12.1891g so Weight of one tablet = 0.6095g (609.5 mg) Stated content per tablet = 500mg In 150.5 mg of tablet powder there was 126.7mg paracetamol, so in 609.5 mg there would be (126.7mg/150.5mg) x 609.5mg = 513.1mg Percentage of stated content = (513.11mg/500mg) x 100 = 102.6%  External standard The weakness of this technique is: The chromatography must be performed under exactly the same conditions every time. Accurate volumes must be used. This can be difficult in GC. The use of an internal standard. How does it work? To overcome the problems with external calibration, we can use an internal standard (something added to the sample you want to analyse). This is known as “spiking” For example, if its difficult to inject accurate volumes in GC, the peak heights may vary, but the ratio between the peaks will remain constant. The use of an internal standard. How does it work? An internal standard must: be closely related to the assayed component  be completely resolved from both the assay component and all other components of the mixture  ideally have a retention time close to that of the compound under analysis  ideally be used at a concentration similar to that you expect to find in the sample under analysis The BP format for assays requiring internal standards The BP format for using both GC and HPLC chromatograms quantitatively requires three samples to be run: 1. A calibration standard containing approximately equal amounts of the pure standard (of the component/s to be measured in the unknown) and an internal standard (SOLUTION 1) 2. An extract from the sample containing no internal standard (to check for interference from the formulation matrix) (SOLUTION 2) (validation) 3. An Extract from the sample containing the same amount of internal standard as solution 1 (SOLUTION 3) (unknown).  Solution 1: Standard solution (Calabration solution) Detectors do not always respond equally to the same amounts of different compounds. We can work out a response factor (Rf) from a standard solution to tell us how much more/less the detector responds to the analyte compared to the Rf for standard solution (Rfs) = internal standard. Area of peak due to analyte Area of peak due to Internal Standard Solution 3: Unknown solution (Sample solution spiked with IS) A known concentration of internal standard (the same concentration used in solution 1) is added to the unknown mixture. The ratio between the area of the sample peak compared to the internal standard is known as the response factor (Rfu) for the unknown solution. Rf for unknown mixture (Rfu) = Area of peak due to analyte Area of peak due to Internal Standard Combining all this gives us an equation we can use to calculate the concentration of an analyte in a unknown mixture spiked with an internal standard: The Advantages The advantage of this method is since all reponses are treated as ratios to the internal standard, this technique: Compensates for the changes in chromatographic conditions (providing these changes effect the sample and internal standard equally Is unaffected by exact volume injected. The analysis of methyl testosterone in a tablet formulation using testosterone as an internal standard Solution 1: Calibration solution containing 0.04% w/v methyl testosterone and 0.043% w/v testosterone in ethanol. Solution 2: Dissolve tablet powder containing ~ 20mg of methyltestosterone in 50 mL ethanol. Weight of 5 tablets = 0.7496 g Stated content of methyltestosterone per tablet = 25 mg Weight of tablet powder taken for assay = 0.1713 g Solution 3: Dissolve tablet powder (0.1713 g) in exactly 50 mL ethanol containing exactly the same concentration of testosterone as solution 1 (0.043%). 0.5ul of each solution was injected into the GC on a RTX-1 column 15 m x 0.25 mm i.d x 0.25 mm film. Programmed 150o (1min) then 10o/min to 320oC (5 min). The chromatograms are shown;  Results: Solution 1: Peak area testosterone: 216268; Peak area methyltestosterone: 212992 Solution 3: Peak area testosterone: 191146; Peak area methyltestosterone: 269243. Calculation We know that: Cu = RFu x Cs RFs Step 1: Cs = 0.04 % w/v methyl testosterone in the standard solution 1 (We were given this value) Step 2: Calculate RFs (Response factor for standard solution 1) RFs=212992 = 0.9849 216268 Step 3: Calculate RFu (Response factor for unknown solution 3) RFs=269243 = 1.4086 191146  Putting them all into the equation gives: Cu = 1.4086 x 0.04%w/v = 0.0572%w/v = 0.0572 g in 100 mL 0.9849 = 0.0286 g in 50 mL (the volume added to the 0.1713 g of tablet powder) Therefore 0.1713 g (171.3 mg) of tablet powder contains 28.6 mg of methyl testosterone. The weight of 5 tablets is 0.7496 g, therefore one tablet would be on average 0.14992 g. If there is 28.6 mg of methyl testosterone in 0.1713 g of tablet powder, then there will be 25.0 mg in the average tablet. If you want to go through some more examples to help with your understanding, then look at the Chromatography chapters in Pharmaceutical analysis by Watson Applications of GC in Biomedical and Pharmaceutical analysis Measurement of drugs and their metabolites in biological fluids. The characterisation of some unformulated drugs, particularly with regard to detection of process impurities. Limit tests for solvent residues and other volatile impurities in drug substances Sometimes used for quantification of drugs in formulations, particularly if the drug lacks a chromophore Characterisation of some raw materials used in the synthesis of drug molecules. Characterisation of volatile oils, proprietary cough mixtures and tonics, and fatty acids in fixed oils. Strengths Capable of the same quantitative accuracy and precision as HPLC, particularly when used in conjunction with an internal standard. Much greater separating power than HPLC when used with capillary columns Readily automated Can be used to determine compounds which lack chromophores The mobile phase does not vary and does not require disposal. Helium is relatively cheap compared to organic solvents used in HPLC. Weaknesses Only thermally stable and volatile compounds can be analysed. The sample may require derivatisation to convert it to a volatile form Quantitative sample injection is more difficult than HPLC because of small volumes injected Aqueous solutions and salts cannot be injected into the instrument. Applications of HPLC in Biomedical and Pharmaceutical analysis  Measurement of drugs and their metabolites in biological fluids  Provides an accurate, precise and robust method for quantitative analysis of pharmaceutical products and is the industry standard method for this purpose.  Monitoring the stability of pure drug substances in formulations with quantitation of any degradation products  Determination of partition coefficients and pKa values of drugs and of drug protein binding. Strengths Easily controlled and precise sample introduction ensures quantitative precision. Good technology – advancing rapidly. Variety of columns detectors, stationary phases and mobile phases – great diversity and applications. Less risk of sample degradation as in GC – and can be used for non-volatile compounds which can not be run with GC. Can be readily automated. Weaknesses  Still require a reliable and inexpensive detector which can monitor compounds lacking a chromophore.  Large amount of organic solvent waste.  Drugs often have to be extracted from their formulations prior to analysis. Summary 1) Chromatographs can be uses qualitatively and quantitively 2) There are three different methods of Quantification 3) Internal normalisation – risky, but the simplest method 4) External Calibration - good when conditions of analysis are stable 5) The use of an internal standard – to account for things that may cause conditions to change each time you undertake chromatography  END This is the end of the Chromatography topic