PRACTICAL CHROMATOGRAPHY (HPLCUV) REPORT

REPORT PRACTICAL CHROMATOGRAPHY Student name Du Ngoc Thao Nguyen ID 18247112 Abstract Determination of paracetamol, caffeine and aspirin in pharmaceutical product has been completed by reversed phase.

REPORT PRACTICAL CHROMATOGRAPHY Student name: Du Ngoc Thao Nguyen ID: 18247112 Abstract Determination of paracetamol, caffeine and aspirin in pharmaceutical product has been completed by reversed phase chromatography In this experiment, students have learned how to prepare standard and working solutions, operate HPLCUV system to separate and quantify components of Panadol Extra product By doing this, students can survey and find out which condition, such as pH range or solvent ratio, is the best to have good separation and shorten retention time of analytes Furthermore, calibration curve and standard addition are applied to calculate the amount of paracetamol and caffeine in pill of Panadol Extra product Experimental results showed that basic substances as paracetamol and caffeine have weaker affinity with stationary phase than aspirin at pH and pH while at pH 7, they interact with stationary phase more strongly Choosing MeOH:buffer (45:55) as optimal solvent ratio of mobile phase helps to incease selectivity and shorten analytical time In addition, by applying calibration curve, we determined that the quantity of paracetamol and caffeine in pill are 426.3 ± 43.2 mg and 46.6 ± 6.7 mg, respectively By using standard addition method, we calculated the recovery of paracetamol which was good of sample S1 and S2, while sample S3 have the recovery is more than 100% that can be caused in diluting procedure and contaminated injector Key words: chrmotography, paracetamol, caffeine, aspirin, reversed phase, HPLC, calibration curve, standard addition INTRODUCTION Chromatography is one of the most common methods used to separate substances from each other The principle of this method is based on the interaction between analytes and the stationary phase (which stays in place in column) or the mobile phase (which moves through the column) Depending on the purpose of the analysis and mechanism of separation, chromatography is divided into many different types The partition chromatography is widely used because it can analyze compounds ranging from non-polar to highly polar to ionic compounds having moderate molecular weight The partition chromatography includes normal phase chromatography (NPC) and reversed phase chromatography (RPC) In NPC, the stationary phase is more polar than the mobile phase This stationary phase has strong affinity with polar substances In RPC, it is a term to indicate that the stationary phase is less polar than the mobile phase.Most organic compounds which have long- chain carbon can be analyzed in detail with this method In this experiment, we use RPC to determine components in pharmaceutical products The stationary phase is silica which has long-chain C18 alkyl groups attached on The mobile phase has stronger polarity Because of the properties of stationary phase and mobile phase, substances which are less polar will be retained in the column longer based on the “likes dissolve likes” principle The polarity of analytes depends on many factors, including pH values Weak acids and bases will change its form to ions when pH changes because of its pKas values When acids or bases are converted into ions, the polarity will increase As a result, they will be eluted faster by mobile phase Therefore, pH controlling in chromatography is essential to maintain the desired form of analytes and the speed of elution For quantitative analyses in chromatography, we can use calibration curve and standard addition methods A calibration curve shows the response of an analytical method to known quantities of analyte[1] By constructing calibration curve, we can interpolate on linear range of the graph to determine the quantity of the unknown In standard addition, known quantities of analyte are added to the unknown From the increase in signal, we deduce how much analyte was in the original unknown[1] This method plays an important role in removing the effect of matrix on the sample High-performance liquid chromatography (HPLC) is a type of liquid chromatography, which is used extensively by the pharmaceutical industry.It is characterised by the use of a finely divided stationary phase in a column[2] Instead of the solvent flowing under the pressure of gravity, the solvent flows at high pressures up to 400 atm, increasing the flow rate of mobile phase solvents HPLC works based on the basic principle: separate components of a mixture based on the difference in affinity between analytes with the stationary phase and the mobile phase The purpose of this experiment is to study and operate a high-performance liquid chromatography system with UV detector (HPLC-UV), find a suitable gradient program as well as pH to separate the components of Panadol Extra product In addtion, we use calibration curve and standard addtion methods to determine the amount of caffeine, paracetamol and aspirin in one pill EQUIPMENTS, CHEMICALS AND PREPARATION PROCEDURE 2.1 Equipments HPLC-UV Shimadzu Corporation equipped with C18 column (150mm × 4.6mm, 5.0μm particles), 0.45μm cellulose acetate membrane, solvent filtration system, pH meter, micropipette 100μL, 200μL and 1000μL, 10mL and 25mL volumetric flasks, centrifugal tubes, analytical balances, vials, ultrasonic bath 2.2 Chemicals 10 pills of Panadol Extra product, concentrated acetic acid solution (to prepare buffer pH 3.0 ± 0.2), methanol (MeOH) solution, doubled-distilled water, stock solutions consisting of pure 1000ppm of each caffeine, paracetamol and aspirin 2.3 Calculation 2.3.1 pH buffer: We prepare 0.2% (v/v) acetic acid solution to make a pH buffer %V = The general information of compounds: White powder, odourless[3] Boiling point: 178oC[3] Melting point: 238oC[3] [3] (1) We need to prepare 600mL 0,2% acid acetic solution, according to equation (1): 0.2% = Kow as log pOW: -0.07 Caffeine Vsolute × 100 Vsolution Vacetic acid Vsolution ×100% = Vacetic acid 600mL ×100% Vacetic acid 1.2 mL pKa = 14[3] The volume of concentrated acetic acid need to be taken is 1.2mL White powder, odourless[4] Boiling point: > 500oC[4] Melting point: 170oC[4] Paracetamol Kow as log pOW: 0.49[4] pKa = 9.38[4] White powder, odourless[5] Boiling point: 140oC[4] Melting point: 135oC[4] Aspirin Kow as log pOW: 1.24[4] pKa = 3.5[4] 2.3.2 Standard solution preparation for qualitative analysis: From stock solutions, we prepare single standard solutions in 1mL-vial for qualitative analysis with 10ppm of each C1V1 = C2 V2 (2) With C1: concentration of stock solution, C2: concentration of single standard solution, V1: volume of stock solution need to be taken, V2: volume of single standard solution From equation (2), we calculate the volume of stock solution needed to be taken is 10μL of it Using micropipet for this procedure, we have single standard solutions Because the concentration of stock solutions are large, we need to prepare mixtures of 3-substance intermediate standard solution with the concentration is 50ppm for caffeine and paracetamol, 100ppm for aspirin in 1mL-vial Calculating from equation (2), the volume of stock solution needs to be taken is: Vcaffeine = Vparacetamol = 50μL Vaspirin = 100μL Total volume in vial is 200μL Therefore, we need to add 800μL MeOH:buffer (50:50, v/v) solvent to reach 1mL solution Working solution for chromatography: From the intermediate standard solution, we prepare vials with different concentration Table The concentration of mixture standard solutions and the volume need to be taken from immediate standard solution Vial Ccaf (ppm) 10 15 20 Cpar (ppm) 10 15 20 Casp (ppm) 10 20 30 40 Vintermediate mixture (mL) 0.02 0.1 0.2 0.3 0.4 Vmobile (mL) 0.98 0.9 0.8 0.7 0.6 phase By calculating from equation (2), we have the volume of intermediate mixture solution needed to be taken for each vial For example: In vial 1, total concentration is 4ppm (1ppm caffeine, 1ppm paracetamol, 2ppm aspirin) In intermediate mixture, total concentration is 200pm We calculate the volume of intermediate mixture need to be taken: Vintermidiate mixture = ppm×1mL = 0.02 mL 200 ppm Therefore, the volume of mobile phase is 0.98mL Similar calculation for the remaining vials, we have the results in Table 2.3.3 Standard preparation: 2.3.4 Vintermidiate mixture = C working ×Vvial Cintermidiate mixture (3) 2.3.5 For quantitation of caffeine and paracetamol in the product: According to the manufacturer, there are 500mg paracetamol and 65mg caffeine in pill In this experiment, because we use 5mg of the medicine, it is estimated that there are 3.57mg paracetamol and 0.46mg caffeine 5mg of the medicine will be dissolved in 25mL of MeOH:buffer (95:5, v/v) solvent to form a solution which has 142.8ppm paracetamol and 18.4ppm caffeine Because this range of concentration does not fall within the linear regression of the calibration curve, we need to dilute to 10ppm for each components in 1mL-vial From equation (2), we calculate the volume needed to be taken from 25mL solution is 70μL for determination of paracetamol and 543μL for caffeine For standard addition, we add about 5mg of pure paracetamol to 5mg of the medicine and dissolve it in 25mL solvent Because the amount of paracetamol is doubled, the volume needed to be taken from this 25mL solution is 35μL 2.4 Preparation procedure: 2.4.1 pH buffer Use disposable pipette to deliver about 1.2mL of concentrated acetic acid solution to a 1L-flask Then add 600mL doubled-distilled water to the flask and mix carefully Perform pH measurement using by pH meter and the result is 3.09 The value is valid The buffer solution is filtered by 0.45μm cellulose acetate membrane to remove solids and residues to avoid clogging the column After that, ultrasonic cleaning is performed in 30 mins for degassing 2.4.2 Standard solution: Using 200mL of buffer after filtration to prepare standard solutions Add 50mL of MeOH solution and 50mL of buffer solution to 200mL-beaker, we have MeOH:buffer (50:50, v/v) solvent Using micropipette to take 10μL of each stock solution to vial for caffeine, paracetamol and aspirin, respectively Then add solvent to fill up to reach 1mL mark with M is unknown sample used for calibration curve, while S is for standard addition method 2.4.3 Then add 10mL of MeOH:buffer (95:5, v/v) solvent to stabilize the form of components and extract easier Intermediate mixture solution Using micropipette to take 50μL of caffeine, 50μL of paracetamol and 100μL of aspirin and deliver to one vial Then add 800μL of MeOH:buffer (50:50, v/v) solvent to this vial to have 1mL intermediate mixture solution 2.4.4 Standard solution for calibration curve From intermediate mixture solution, we dilute to vials with the volume need to be taken according to Table 2.4.5 Preparation of the medicine to determine Mix M samples well, then take it to ultrasonic bath for 15 minutes After that, deliver solution from centrifugal tubes to 25mL-volumetric flasks and fill up to the mark Use 0.45μm membrane to remove residues Then use micropipette to take 70μL to one vial (for determination of paracetamol) and 530μL to other vial (for determination of caffeine) and fill up to the 1mL mark with solvent For S samples, mix the mixture well and leave the samples for hour to the amount of addition penetrate the sample matrix After that, addition and samples will have same behaviour in solution Use micropipette to take 35μL to vial and fill up to the 1mL mark with solvent caffeine and paracetamol Weight 10 pills of Panadol Extra product respectively EXPERIMENTS, RESULTS AND COMMENTS Table Mass of 10 pills Panadol Extra Pill no Mass (mg) Pill no Mass (mg) 696.6 695.5 685.2 707.3 687.8 688.8 690.7 691.7 684.1 10 Average mass (mg) 682.2 691.0 After that, crush 10 pills into fine powder to form homogenized sample Weight 5mg powder in centrifugal tube for determination of caffeine and paracetamol by calibration curve and standard addition methods 3.1 Selecting the wavelength for UV detection Spectrophotometry of caffeine, paracetamol and aspirin is provided by the instructor The experiment is carried out using Shimadzu UV1800 spectrophotometer, the concentration of analyte is 10ppm of caffeine, paracetamol and aspirin which was dissolved in MeOH:buffer (50:50, v/v) solvent Scanning the wavelength from 200nm to 400nm in order to obtain the best sensitivities for all to operate HPLC-UV system Blank Par Table Mass of the medicine used for calibration curve and standard addition Sample Mass (mg) Addition (mg) M1 5.12 M2 5.20 M3 4.82 S1 4.96 3.7 S2 4.98 4.2 S3 4.94 3.5 Abs 1.5 0.5 200 250 300 350 Wavelength (nm) Figure Spectroscopy of paracetamol, caffeine and aspirin From the spectroscopy of compounds (Fig.1), the maximum wavelength of caffeine, paracetamol, aspirin are 273nm, 244nm and 296nm, respectively Because the Shimadzu HPLC-UV system can choose only wavelengths, we select 234nm and 273nm for analysis to ensure that we can see the signals clearly and there are no duplicates efficiently by decreasing MeOH solution and increasing the buffer The ratio has been changed to 45:55 mV Detector A Ch1:273nm 200 3.2 Optimizing the mobile phase gradient 150 In RPC, the stationary phase is silica which has long-chain C18 alkyl attached on The mobile phase has stronger polarity Because of properties of stationary phase and mobile phase, substances which are less polar will be retained on column longer based on the “likes dissolve likes” principle When the mobile phase has the larger ratio of more polar part to less polar solvent, compounds that is nonpolar will have longer retention time because they are retained on the column and have stronger interaction with the stationary phase In order to increase the power of elution, we have to increase the less polar solvent of the mobile phase, the analytes will be eluted more quickly In addition, chromatography requires good retention time, selectivity and efficiency Thus, surveying and choosing the appropriate solvent ratio to conduct separation are very important to achieve highest efficiency and shortest retention time Firstly, open Shimadzu HPLC-UV operation, use channel A to withdraw MeOH solution, channel B for buffer The flow rate is of 0.7 mL/min, an injection volume of 20μL and temperature of column of 40oC Conduct purge gas by MeOH solution and wait until the system is stable Control the ratio of MeOH to buffer: 50:50 100 50 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 Figure Chromatogram of compounds when the solvent ratio is 45:55 with vial Comment: The distance between the first peaks is larger than when using 50:50 solvent ratio The analytical time is too long but we have successfully completed in separation of compounds without overlapping, tailing and fronting The appropriate solvent ratio is MeOH:buffer 45:55, thus choosing it for the next steps 3.3 Qualitative of caffeine, paracetamol and aspirin Using mixture in vial 5, paracetamol and aspirin single standard solutions to determine which peaks are of which components uV(x10,000) 7.0 Red: paracetamol 10ppm 6.0 Blue: Aspirin 10ppm uV(x100,000) 3.0 5.0 2.5 4.0 2.0 3.0 1.5 2.0 1.0 1.0 0.5 0.0 Black: mixture 0.0 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Figure Chromatogram of compounds when the solvent ratio is 50:50 with vial Comment: The result is that the first peaks are not well separated (figure 2) In order to separate the first two substances, the second substance is required to appear more slowly Thus, increase the polarity of the mobile phase so that the second compound has less interaction with the solvent than the first analyte and is retained longer on the column Therefore, we have to change the solvent ratio to separate more 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 Figure Chromatogram of mix 3, paracetamol and aspirin standard solutions Result: The peak of paracetamol appearing at t = 3.24 (red line) matches with the first peak of mix (black line) The peak of aspirin appearing at t = 9.20 (blue line) matches with the last peak of mix Thus, the remaining peak in mix line is of caffeine Therefore, the order of appearance of the peaks is paracetamol, caffein and aspirin, respectively 3.4 The effect of pH on selectivity uV(x100,000) uV(x100,000) 5.5 2.25 Green: mixture Red: Aspirin 1.75 Blue: Paracetamol Black: Caffeine Red: mixture Blue: Caffeine 5.0 2.00 Black: Paracetamol 4.5 4.0 1.50 3.5 1.25 3.0 1.00 2.5 0.75 2.0 0.50 1.5 1.0 0.25 0.5 0.00 0.0 -0.25 3.0 3.5 4.0 4.5 5.0 5.5 6.0 m in 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 m in (a) (b) Figure Chromatogram at (a) pH 5, (b) pH (provided by the instructor) Comment: In HPLC, pH is a factor that can affect the elution of bases or acids If a solute is an acid or base, its charge depends on pH The neutral form is retained by reversed-phase columns (“like dissolves like”), whereas the ionized form is hydrophilic and so weakly retained [1] Caffeine and paracetamol are base, while aspirin is acid Use buffer pH 3, bases tend to be protonated in a moderate acidic environment according to equation B + H+ ⇆ BH+ to form ions which are more polar than neutral compounds and interact with the non-polar stationary phase more weakly Paracetamol with pKa 9.38 has hydroxyl group and amide group which are very polar because it is protonated in pH Therefore, it appears firstly Caffeine is less polar than paracetamol, thus appearing later Aspirin is almost neutral with pKa = 3.5 in pH 3, and although it has carboxyl and ester groups but both of these are attached at ortho position, it is easy to form hydrogen intramolecular bonds that makes it less polar Thus, aspirin has longest retention time At buffer PH 5, although the experimental conditions such as the solvent ratio may be different from at pH 3.0, the order of appearance of the peaks is similar to the order of pH The appropriate explanation for this elution is similar to at pH because buffer pH forms weakly acidic environment, basic compounds tend to be protonated to create ions such as paracetamol and caffeine Aspirin with pKa = 3.5 is dissociated to form its conjugated base according to the equation: HA ⇆ H+ + A-, which is more polar Therefore, it is eluted more quickly than at buffer pH At buffer pH 7, there is not the presence of aspirin in chromatogram The order of appearance of the peaks is paracetamol and caffeine, respectively The concentration of H+ decreases at pH 7, both of analytes tend to re-create the neutral form (Le Chatelier‘s principle) but paracetamol is more polar than caffein because of its structural formula Thus, paracetamol is weakly retained on column and appear firstly From the survey, chrmomatograms at pH (figure 4) and pH (figure 6a) show that the peaks are separated almost clearly but pH is more optimal because the peaks are separated better and retention time is shortest, while at pH 3.0, the peak of paracetamol is still close to the peak of caffeine and the analytical time is too long At pH (figure 6b), the order of elution is determined but the result is not evaluated so much because the absence of aspirin In conclusion, carrying out the experiment at pH to protect (Si-O) groups of column from collapse and conducting the survey at pH 3, and to consider the effect of pH on selectivity of chromatographic system on the analytes 3.5 Construct the calibration curve: 3.5.1 The calibration curve of caffeine: Using the optimal chromatographic conditions in 3.2, (MeOH:buffer : 45:55), we construct the calibration curve for caffeine with concentration levels, corresponding to vial to vial mV Detector A Ch1:273nm 200 150 100 50 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 Figure Chromatogram of caffeine in vial at 273nm mV 150 Construct the calibration curve of caffeine: Detector A Ch1:273nm 125 2.E+06 2.E+06 2.E+06 1.E+06 1.E+06 1.E+06 8.E+05 6.E+05 4.E+05 2.E+05 0.E+00 100 75 50 25 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Figure Chromatogram of caffeine in vial at 273nm Peak area mV 150 Detector A Ch1:273nm 125 y = 92226x + 4627.7 R² = 0.9983 10 15 Concentration (ppm) 100 20 75 Figure 11 Calibration curve of caffeine 50 25 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Figure Chromatogram of caffeine in vial at 273nm mV 150 Detector A Ch1:273nm 125 100 75 3.5.2 The calibration curve of paracetamol: Using the optimal chromatographic conditions in 3.2, (MeOH:buffer : 45:55), we construct the calibration curve for paracetamol with concentration levels, corresponding to vial to vial mV 150 Detector A Ch2:234nm 50 25 125 100 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 /3.005/38021 75 Figure Chromatogram of caffeine in vial at 273nm RT9.260/9.260/1768688 1.0 RT3.248/3.248/1246982 0.0 50 25 RT3.983/3.983/1825230 200 150 0.0 2.0 3.0 mV Detector A Ch2:234nm 200 50 150 1.0 2.0 3.0 4.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Figure 12 Chromatogram of paracetamol in vial at 234nm 100 0.0 1.0 5.0 6.0 7.0 8.0 9.0 10.0 RT9.027/9.027/2695911 250 RT3.208/3.208/1849971 mV 300 Detector A Ch1:273nm /2.954/68180 100 Figure 10 Chromatogram of caffeine in vial at 273nm 50 Table Retention time and peak area of caffeine in vials CCaf (ppm) tR (min) 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 Figure 13 Chromatogram of paracetamol in vial at 234nm Peak area 1 3.987 70126 3.989 496634 10 3.990 916055 15 3.955 1418619 20 3.983 1825230 RT3.243/3.243/2479400 mV 300 Detector A Ch2:234nm 250 200 150 /3.000/55464 100 RT9.254/9.254/3501378 Vial 0.0 50 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 m in Figure 14 Chromatogram of paracetamol in vial at 234nm 60 50 /3.000/91118 40 30 20 10 3.5.3 RT9.232/9.232/934708 70 RT3.247/3.247/663038 mV 80 Detector A Ch2:234nm -10 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 Figure 15 Chromatogram of paracetamol in vial at 234nm 10.0 7.5 5.0 RT9.190/9.190/142934 /2.998/75978 RT3.243/3.243/98662 mV 12.5 Detector A Ch2:234nm 2.5 0.0 -5.0 -7.5 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 Figure 16 Chromatogram of paracetamol in vial at 234nm Table Retention time and peak area of paracetamol in vials Vial CPar (ppm) Using the optimal chromatographic conditions in 3.2, (MeOH:buffer : 45:55), we construct the calibration curve for aspirin with concentration levels, corresponding to vial to vial Because the Shimadzu HPLC-UV system can choose only wavelength, we select 234nm and 273nm for analysis Therefore, when we scan at 234 nm, we can determine the signals of both aspirin and paracetamol and their area Chromatograms of paracetamol above contain signals of aspirin Table Retention time and peak area of aspirin in vials -2.5 0.0 The calibration curve of aspirin: Vial CAsp (ppm) tR (min) Peak area 9.190 142934 10 9.232 934708 20 9.260 1768688 30 9.027 2695911 40 9.254 3501378 Construct the calibration curve of aspirin: tR (min) Peak area 4.00E+06 3.243 98662 3.50E+06 3.247 663038 3.00E+06 10 3.248 1246982 2.50E+06 15 3.208 1849971 20 3.243 2479400 Construct the calibration curve of paracetamol: 2.00E+06 1.50E+06 1.00E+06 3.00E+06 5.00E+05 y = 123800x + 4848.1 R² = 0.9993 2.50E+06 0.00E+00 10 20 30 40 Figure 18 Calibration curve of(ppm) aspirin Concentration 2.00E+06 Peak area y = 88225x + 8933.6 R² = 0.9991 Peak area 1.50E+06 3.6 Quantitation of caffein and paracetamol in pill of Panadol Extra product 3.6.1 Calibration curve method: 1.00E+06 5.00E+05 Using HPLC-UV to analyze sample M1, M2 and M3 The result is showed in table below 0.00E+00 10 15 Concentration (ppm) Figure 17 Calibration curve of paracetamol 20 Table Retention time and peak area of paracetamol and caffeine in the unknown Par Sample Caf tR (min) Peak Peak area tR (min) M1 3.268 1045038 4.008 636806 M2 3.277 1144947 4.009 755386 M3 3.282 1060834 3.998 620819 area 3.6.1.1 Determine the quantity of paracetamol in each pill: m par/pill = m par/25mL ×691.0 mg 5.12 mg 3.001mg×691.0 mg m par/pill = 5.12 mg There are 405.0 mg of paracetamol in pill, which is calculated from sample M1 Calculate the same for sample M2 and M3, we have the table below: Table The results obtained from calculating from calibration curve for paracetamol Par M1 M2 M3 Mass of the unknown 5.12 5.20 4.82 CPar 8.402 9.209 8.530 (calibration curve) Equation of calibration line of paracetamol: (3) Mass of Par in with x: concentration of paracetamol (ppm), y: peak area m mg (weight) y = 123800x + 4848.1 Interpolate from the equation (3), the concentration of paracetamol in sample M1: 1045038 123800 x x 8.402 ppm 405.0 437.1 436.7 in pill (mg) 426.3 Calculate a and b of calibration line: b= (4) With C1: the concentration of paracetamol in 25mL solution, C2: the concentration of paracetamol in sample interpolated from calibration curve, V1: the volume is withdrawn from 25mL solution, V2: the volume of vial μg mL C i2 Ai − Ci Ci Ai nC i2 − (Ci ) (6) (7) Calculating residual variance: Ai2 − aAi − bAi Ci  S y2 n−2 (8) Calculating standard deviation for regression The mass of paracetamol in 25mL solution or in 5.12 mg of the medicine (in sample M1): mpar/25mL = Cpar/25mL × Vvolumetricflask a= nCi Ai − Ci Ai nCi2 − (Ci )2 Sre2 = 8.402 ppm 1000μL 120.03 ppm = 120.03 Mass of Par Equations for calculating of uncertainty: C1V1 = C2 V2 C1 3.001 3.289 3.046 Average (mg/pill) 4848.1 Because we dilute the sample from 25mL to 1mL in vial, we need to calculate the amount of paracetamol in 1mL solution with the volume withdrawn was 70μL C1 70μL 405.0 mg (5) μg × 25mL mL = 3000.1 μg =3.001mg m par/25mL = 120.03 The mass of paracetamol in pill or in 691.0 mg of the medicine: coefficients (a and b): Sb = Sre Sa = Sre n nC i − (Ci ) 2 Ci2 nCi2 − (Ci )2 (9) (10) Confidence interval a, b:  a = t0.95, f  Sa n (11)  b = t0.95, f  Sb (12) n Calculating uncertainty of the concentration of each components: ux = n( Ax − Ai )2 1 + + n m b [nCi2 − (Ci )2 ] Sre b (13) With a: the absolute value of the vertical intercept, b: the absolute value of the slope, n: the number of data points for the calibration line, m: the number of replicate measurements of the unknown, Ax : the mean value of peak are of the unknown, Ai : the mean value of Ax for the points on the calibration line, Ci: the concentration of points on the calibration line, Ai: the peak area of points on the calibration line According to the equation of calibration line of paracetamol: n = 5, m = 3, Ax =1083606, Ai = 1267611, we have the table below Table Calculating of uncertainty for calibration curve of paracetamol Table 10 Uncertainty of equipments used in diluting procedure uvolumetric flask (mL) 0.007 umicropipette/70μL (μL) 0.33 umicropipette/30μL (μL) 0.15 umicropipette/100μL(μL) 0.17 with σvolumetric flask, σmicropipette/70μL, σmicropipette/30μL, σmicropipette/100μL are 0.03, 1.4, 0.6, and 0.7, respectively Use micropipette 20-200μL to take 70μL of sample once, 30μL of solvent once and 100μL times Uncertainty of paracetamol of the actual pill: u par u10 pills m par / pill uunknown munknown m1 pill u par uvolumetric flask C par (17) Vvolumetric flask umicropipet umicropipet solvent Vmicropipet Using the results above, we calculate 123800 Sa 23727.3 → upar = 13.6 mg a 4848.1 εb 2767.1 Mass of paracetamol in pill: Sre 29412.4 εa 41854.8 mpar /1 pill Sb 1936.03 upar 0.2 ppm With t0.95; n-2 = 3.182 → y (12.4 0.3) 10 x (0.05 0.4) 10 (x Sx u xi ) n Sx t0.95; n 426.3 u par 43.2 mg Comment: (14) (15) n 426.3 mpar /1 pill Calculate uncertainty of weighing: Vmicropipet solvent b From equation (14) and (15), we calculate the uncertainty of weighing 10 pills: The quantity of paracetamol which is analyted and calculated is lower than the published amount of the manufacturer The result is valid because of the acceptable uncertainty (about 10.2%) 3.6.1.2 Determine the quantity of caffeine in each pill: u10 pills = 2.3 mg Calculating the same for weighing of unknown sample M1, M2 and M3 Equation of calibration line of caffeine: y = 92226x + 4627.7 (18) uunknown = 0.12 mg with x: concentration of caffeine (ppm), y: peak area Uncertainty of equipments: Using equation from (4) to (13), calculating the same with paracetamol, we have the results in table below for caffeine u equip equip (16) Table 11 The results obtained from calculating from calibration curve for caffeine Caf Mass of the unknown M1 M2 M3 5.12 5.20 4.82 3.6.2 Standard addition method Because the amount of caffeine used to add to the unknow is too small, we only carry out with paracetamol Ccaf RT3.276/3.276/1146669 mV 150 Detector A Ch2:234nm 125 8.140 6.681 (calibration curve) 100 /3.025/278815 6.855 75 50 Mass of Caf in 0.323 0.384 0.315 m mg (weight) 25 -25 0.0 43.63 in pill (mg) Average (mg/pill) 51.02 45.18 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.5 5.0 5.5 Figure 19 Chromatogram of standard addition sample S1 46.6 mV Detector A Ch2:234nm 150 125 100 /3.023/262716 Table 12 Calculating uncertainty of caffeine 4.0 RT3.274/3.274/1217273 Mass of Caf 75 50 b 92226 Sa 26729.4 25 a 4627.7 εb 3103.6 -25 Sre 33133.9 εa 38036.8 Sb 2181.0 ucaf 0.3 ppm 0.0 (9.2 0.3) 104 x (0.05 0.4) 105 1.0 1.5 2.0 2.5 3.0 mV 175 Detector A Ch2:234nm 150 125 100 /3.013/274605 Using equation (14) and (15) to calculate uncertainty of weighing: 75 50 u10 pills = 2.3 mg 3.5 4.0 4.5 5.0 5.5 m in RT3.264/3.264/1352063 → y 0.5 Figure 20 Chromatogram of standard addition sample S2 25 uunknown = 0.12 mg -25 0.0 Uncertainty of equipments are showed in Table 10 Use micropipette 20-200μL to take 30μL of sample once, 100μL of sample times, 70μL of solvent once and 100μL of solvent times Using equation (17) to calculate uncertainty of caffeine of the actual pill 0.5 46.6 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 m in Table 13 Retention time and peak area of paracetamol in standard addition sample Sample t0.95; n ucaf With t0.95; n-2 = 3.182 mcaf /1 pill 2.0 Figure 21 Chromatogram of standard addition sample S3 tR (min) Peak area S1 3.276 1146669 S2 3.274 1217273 S3 3.264 1352063 Mass of caffeine in pill: 426.3 1.5 Par → ucaf = 2.1 mg mcaf /1 pill 1.0 6.7 mg Comment: The quantity of caffeine which is analyted and calculated is lower than the published amount of the manufacturer The result is valid because of the acceptable uncertainty (about 14.4%) The quantity of paracetamol in the total mass of each S sample is calculated by using the same aforementioned equation from (3) to (5) We have the table below: Table 14 Calculating the mass of paracetamol in standard addition samples Standard addition (par) Total mass (mg) CPar (calibration curve) Mass of Par in total mass (mg) S1 S2 S3 8.66 9.18 8.44 9.223 9.793 10.882 6.588 6.995 7.773 Calculating the amount of paracetamol in inital sample weight for standard addition, it means that calculating the mass of paracetamol in 4.96 mg medicine for S1, 4.98 mg for S2 and 4.94 mg for S3 m par / medicine m par / pill mweight (19) m1 pill Calculating the actual quantity of pure paracetamol used to add in sample mpar add mpar / total mass m par / medicine (20) Calculating recovery of standard addition samples H m par add m par add / weight (21) 100% For example: Sample S1: m par / medicine S mpar add S1 H 426.3 mg 4.96 mg 691.0 mg 6.588 mg 3.528 mg 100% 3.7 mg 3.060 mg 3.060 mg CONCLUSIONS HPLC is considered a "powerful" analytical method because of its high sensitivity; easy to operate; can analyze compounds ranging from polar to nonpolar, ionic or non-charged compounds, proteins, polymers, Therefore, HPLC is widely used to evaluate purity and determine content molecules in the pharmaceutical samples In this experiment, determination of caffeine, paracetamol and aspirin in pharmaceutical product – Panadol Extra by reversed phase chromatography was completed Carrying out the experiment in pH environment is to know how pH affects the selectivity Paracetamol and caffeine are separated better without overlapping but the analytic time is too long if the mixture has more than components By choosing the appropriate solvent ratio (MeOH:buffer : 45:55), analytes are separated completely and retention time is shortened 3.528 mg 95.4% Use equation (19) to (21) to calculate similarly The results are showed in the table below Table 15 Calculating recovery by standard addition method Standard addition (par) mpar/medicine (mg) mpar add (mg) Recovery (%) be contaminated by remaining solvent or other substances Consider recovery of both samples, it is still not stable This leads to low accuracy and repeatability There are many influencing factors such as the amount of standard added each time is not the same in diluting procedure; the sample matrix contains more or less of the analyte needed; chemical or equipment could be contaminated with the analyte In addition, skills of injection and diluting may be not proficient S1 S2 S3 3.060 3.528 95.4 3.072 3.923 93.4 3.047 4.726 135.0 Comment: Recovery of the first samples are close to 100%, which indicates that addition and uknown samples have same behaviour in solution Recovery of sample S3 is greater than 100% and the reason for this is that in injection procedure, the sample may Using calibration curve method helps to determine the amount of caffeine and paracetamol in pill of Panadol Extra The quantity of these components analyted is lower than the published amount on the seal of the product Because of the high correlation coefficients of calibration curve, the results interpolated from this can be valid Standard addition method is useful to when the sample composition is unknown or complex and affects the analytical signal By using this method, we can calculate the recovery and survey what can affect the analytical procedure and skills of students in diluting and injection The results obtained are quite large uncertainty This error and uncertainty are partly due to the error of the tool, partly proving that the skill is not proficient, the improper manipulation of using micropipette leads to the wrong volume of solution Therefore, the concentration of diluted solution is not highly accurate, or the sample injection into the column has not been repeated in each injection This leads to low measurement repeatability and accuracy In addition, the solvent may evaporate during storage, resulting in changea in the concentration of analytes affecting the analysis REFERENCES [1] Daniel C Harris, Quantitative Chemical Analysis, W H Freeman and Co (2015) [2] Oona McPolin, An introduction to HPLC for Pharmaceutical Analysis, Mourne Training Services (2009) [3] National Center for Biotechnology Information "PubChem Compound Summary for CID 2519, Caffeine" PubChem, https://pubchem.ncbi.nlm.nih.gov/compound/Caffeine Accessed 12 March, 2021 [4] National Center for Biotechnology Information "PubChem Compound Summary for CID 1983, Acetaminophen" PubChem, https://pubchem.ncbi.nlm.nih.gov/compound/Acetamino phen Accessed 12 March, 2021 [5] National Center for Biotechnology Information "PubChem Compound Summary for CID 2244, Aspirin" PubChem, https://pubchem.ncbi.nlm.nih.gov/compound/Aspirin Accessed 12 March, 2021  ... role in removing the effect of matrix on the sample High-performance liquid chromatography (HPLC) is a type of liquid chromatography, which is used extensively by the pharmaceutical industry.It... the mobile phase The purpose of this experiment is to study and operate a high-performance liquid chromatography system with UV detector (HPLC-UV), find a suitable gradient program as well as pH... we need to add 800μL MeOH:buffer (50:50, v/v) solvent to reach 1mL solution Working solution for chromatography: From the intermediate standard solution, we prepare vials with different concentration