BASIC
HIGH PERFORMANCE LIQUID CHROMATOGRAPHY
ANALYSIS OF PHARMACEUTICAL COMPOUNDS
TROUGH INTERNAL STANDARD CALIBRATION METHOD IN HLPC
The main purpose of this
experiment is to quantify the active ingredients in commercial analgesic
tablets Excedrin, using internal standard calibration method.
Introduction:
An internal standard is a
known amount of a compound, different from analyte, that is added to the
unknown. Signal from analyte is compared with signal from the internal
standard to find out how much analyte is present.
The reasons for to use that is because internal standards are
especially useful for analyses in which the quantity of sample analyzed or the
instrument response varies slightly from run to run for reasons that are
difficult to control. For example, liquid flow rates that vary by a few
percent in a HPLC experiment could change the detector response. A calibration
curve is only accurate for the one set of conditions under which is obtained.
However, the relative response of the detector to the analyte and standard
increases by 8.4% because of a change in solvent flow rate, signal from the
analyte usually increases by 8.4% also. As long as the concentration of
standard is known, the correct concentration of analyte can be derived.
Internal standards are widely used in HPLC because the small quantity of
sample solution injected into the HPLC is not very reproducible in some
experiments.
Internal standards are also desirable when samples loss can occur
during sample preparation steps prior to analysis. If a known quantity of
standard is added to the unknown prior to any manipulation, the ratio of
standard to analyte remains constant because the same fraction of each is lost
in any operation.
The required features of an internal standard are sample chemically
inert, similar chemical structure from the analyte, the internal standard
should elute out faster, in about 3 minutes. It needs to add the same amount
of same compound to all calibration and samples. The internal standard must
have resolution and detectability and mimics analytes in pretreatment steps.
In this experiment the internal standard that follow the above
criteria is nicotinic acid, which was used as internal standard.
The method of calibration is concern with the calibration curve,
which shows the response of a chemical analysis to known quantities (standard
solutions) of analyte. When there is a linear response, the corrected
analytical signal is proportional to the quantity of analyte.
The mobile phase in HPLC refers to the solvent being continuously
applied to the column, or stationary phase. The mobile phase acts as a carrier
for the sample solution. A sample solution is injected into the mobile phase
of an assay through the injector port. As a sample solution flows through a
column with the mobile phase, the components of that solution migrate
according to the non-covalent interactions of the compound with the column.
The chemical interactions of the mobile phase and sample, with the column,
determine the degree of migration and separation of components contained in
the sample. For example, those samples which have stronger interactions with
the mobile phase than with the stationary phase will elute from the column
faster, and thus have a shorter retention time, while the reverse is also true.
The mobile phase can be altered in order to manipulate the interactions of the
sample and the stationary phase. So, the mobile phase can be optimizing
plotting the capacitor factor (k’) versus different concentration of the
same mobile phase and from the resulting graph we can choose the best mobile
phase. Such kind of graph is known as Relative Resolution Map (RRM).
The active ingredients of sample are acetaminophen (325mg), caffeine (65mg)
and the internal standard used was nicotinic acid.
Materials and System Set-up:
Column: C18, 15 cm x 4.6 mm, 5 ?m.
Detection: UV 254 nm
Excedrin: Label Claim per tablet: AI1 – acetaminophen 325mg, AI2 –
caffeine 65mg.
Internal Standards: Nicotinic acid.
Standards: acetaminophen and caffeine.
Injection Volume: 20 ?l
Mobile Phase: to be determined, using “RRM vs % organic” graph to
select a suitable mobile phase.
Flow rate: 1.5 ml/min.
Detector Sens: 1.0 AUFS.
Procedure:
Mobile Phase Optimization:
The mobile phase optimization is done by plotting
in a graph the capacity factor (k’) versus different concentrations of
mobile phase and observe the separation of internal standard (IS) and the
active ingredients (AI1 and AI2).
Preparation of Standard Solution:
Stock of Internal Standard solution: Transfer about
500.0 mg of Nicotinic Acid, accurately weighed, into a 100 ml volumetric flask,
half fill with M.P, sonicate for 3 minutes then dilute to volume with M.P.
Stock Solution of Acetaminophen and Caffeine:
Transfer about 65 mg of Acetaminophen, 13 mg of Caffeine, (in
duplicate), accuately weighed, into a 100 ml individual volumetric flask
(V.F.), add 0.5 ml Internal STD Stock Solution, half fill with M.P. then
sonicate for 3 minutes or until dissolve, dilute to volume with M.P.
Working: Pipette 1.0 ml of each of STD Stock Solution (containing
Acetaminophen, Caffeine and Internal STD) into a 10 ml V.F., dilute to volume
with mobile phase. Filter a portion through 0.45 ?m syringer filter into vial.
Inject working standard solution (STD-1) and STD-2 for 5 and 2 times
respectively.
Preparation of Samples:
It was weighted 10 tablets
for analysis and the average weight per tablet was 493mg. The 10 tablets,
which formed the composite was pulverized in a mortar by a pestle. It was used
the composite because one tablet can have a wrong amount of active product.
After that it was weighted out one tablet equivalent.
The solvent used to dissolve the sample was a solution of
acetonitrile…and the dissolution is accomplished by USP.
Procedure for the internal standard was as follow:
From a stock solution of 5mg/ml of Nicotinic Acid, it was taken 0.5
ml, which was added in volumetric flasks of 100 ml that contained the samples
to be analyzed. The final concentration of the internal standard was
0.025mg/ml.The amount of Internal Standard must comply with the USP. The
internal standard was added at preparation of the stock solution.
Preparation of the Stock Solution of the
Sample:
It was added 65.0 mg of
Acetaminophen in a volumetric flask of 100 ml. It was added 13.0 mg of
Caffeine in a volumetric flask of 100 ml, the volumetric flask was sonicated
for 3 minutes in half fill of mobile phase. It was added 0.5 ml of internal
standard, which it has a concentration of 5mg/ml. After that the
volumetric flask was completed to the mark with mobile phase.
Consequently, the final concentration AI1 – Acetaminophen is 0.65mg/ml and
AI2 – Caffeine is 0.13 mg/ml.
Preparation of the Working Solution of
the Sample:
It was pipette 1.0 ml of each standard stock
solution containing acetaminophen, caffeine and internal standard in 10 ml
volumetric flask. It was diluted with the mobile phase to the mark of 10 ml
volumetric flask and the solution was filtered through 0.45 ?m syringer filter
into vial. Consequently the final concentration of AI1 is 0.065mg/ml and AI2
is 0.013 mg/ml that is 65?g and 13?g respectively.
Preparation of Standard Solutions:
It was pipette 6.5 ml stock APAP and 1.3 ml stock
caffeine (2mg/ml) into a 200 ml V.F. and add 10 ml Nicotinic Acid making it to
(0.5 mg/ml) with water. Tranfer 1 ml to transfer 1 ml to 10 ml volumetric
flask and dilute with water. The final concentration for AI1 is 6.5?g and AI2
is 1.3?g. and the nicotinic acid 0.25?g.
Quantitative Analysis by
HPLC:
For quantitative analysis we assume that the area
(or the height) of our peak in the chromatogram is proportional to the amount
of substance that produced the peak. In the simplest method we measure areas
or heights, which are then normalized (this means that each area or height is
expressed as a percentage of the total). The normalized hieghts or areas give
the composition of our mixture.
There are two problems with this approach, which are:
a- We have to be sure that we have counted all the components, i.e. that
each component appears as a separate peak on the chromatogram. Components may
co-elute, or be retained on the column, or may elute without being detected.
b- We are assuming that we get the same detector response for equal amounts
of each component.
Results:
(A) Raw Data:
STANDARD CALIBRATION
Table # 1.
AI1
Acetaminophen AI2
Caffeine Internal Standard
Nicotinic Acid
Injection # TR Area TR Area TR Area
1 0.873 1467.95 1.319 34.44 0.593 140.07
2 0.873 1470.51 1.318 32.76 0.592 140.84
3 0.869 1647.58 1.318 37.30 0.638 169.03
4 0.870 1652.80 1.314 27.30 0.640 168.50
5 0.866 1651.41 1.311 26.34 0.683 167.15
MEAN 0.870 1578.05 1.316 31.63 0.629 157.12
CV 0.003 99.36 0.003 4.69 0.038 15.23
Individual Std. - - - - 0.666 469.69
- - 0.826 2927.50 - -
0.827 74.09 - - - -
(B) Calculated Results and Discussion:
I am not able to write about the
results on optimization of mobile phase composition, RRM plot and discuss
whether it provides a good prediction for the composition you chose because we
only used the mobile phase at 29% of ACN.
Table # 2. Calculate Ratio of
AI1/IS, AI2/IS from Different Injections.
AI1/IS
AI2/IS
Injection #
1 10.480 0.246
2 10.441 0.233
3 9.747 0.221
4 9.809 0.162
5 9.880 0.158
MEAN 10.071 0.204
CV 0.358 0.041
The coefficient of variation (CV)
of a set of measurements is the standard deviation divided by the mean: CV = s
/ x. Usually the coefficient of variation is expressed as a percentage of the
mean: CV (%) = 100 X s / x. The smaller the coefficient of variation, the more
precise is a set of measurements. Therefore the ration AI2/IS is more precise
than AI1/IS and in addition is less than 2% that is the requirement from USP.
The active ingredients in the working solution of the Excedrin extract is
as follow.
Acetaminophen:
% Active Sample = 10.460 x 0.5mg/1ml x 25ml/500ml x
1ml/10ml x 500ml/325mg x 10ml/1ml x 100 = 40.23%
Caffeine:
% Active Sample = 0.2395 x 0.5mg/1ml x 25ml/500ml x
1ml/10ml x 500ml/65mg x 10 ml/1ml x 100 = 4.61%
The active ingredients in the stock solution of the Excedrin Extract.
Acetaminophen:
% Active Sample = 9.812 x 0.5mg/1ml x 0.5ml/100ml x
1ml/10ml x 100ml/65mg x 10ml/1ml x 100 = 3.77%
Caffeine:
% Active Sample = 0.180 x 0.5mg/1ml x 0.5ml/100ml x
1ml/10ml x 100ml/13.0mg x10ml/1ml x 100 = 0.35%
The Active Ingredient per Tablet:
Acetaminophen:
% Active Sample = 10.460 x 0.5 mg/1ml x 25ml/500ml x
1ml/10ml x500ml/325mg x 10ml/1ml x 100 x 493mg/495mg = 40.06%
Caffeine:
% Active Sample = 0.2395 x 0.5mg/1ml x 25ml/500ml x
1ml/10ml x 500ml/65mg x 10ml/1ml x 100 x 493mg/495mg = 4.59%
Acetaminophen:
40.06%
40.06 mg to 100 mg
In 495 mg of sample we got 198.3 mg of active
product. Once the average weight of the tablet is 493mg, consequently we have
197.5mg in average of active product.
% Label Claim = 197.5mg/325mg x 100
= 60.77%
Caffeine:
4.59%
4.59 mg to 100 mg
In 495 mg of sample we got 22.72 mg of active product. Once the average
weight of the tablet is 493 mg, consequently we have 22.63 mg in average of
active product.
% Label Claim = 22.63mg/65mg x100 =
34.81%
.
A huge source of error in many steps was done.
It is difficult to figure out at first glance the kind of mistake was done. So
many students were conducted this experiment, each one in charge of one part
of the experiment. Could be an error in the preparation of the reactive or in
the dilution. How knows that a contamination was the main source of error,
perhaps the glass were not clean correctly.
Many people working any thing could be done. The one that write
this believe that some dilution, or a huge mistake in preparation of the
solution happens because each person take care one step in the experiment.
Then it was difficult to follow the same thought. Therefore in each step from
this experiment could be done a mistake.
The solution for that dilemma was to repeat the experiment again.
However, each step, even the calculation needed to be conduct for the same
person. One believes that if a good result is necessary, even the glass need
to clean for the same person. In that experiment is fundamental to start by
taking care of the clean of all material to avoid surprises. A good night of
sleeping and try the experiment again would be the solution.
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