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 1 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 3 and pH 5 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 1 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.

1 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.

2 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 3 buffer.
%V =

The general information of 3 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 3 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 3 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

5 vials with different concentration.
Table 1. The concentration of 5 mixture standard
solutions and the volume need to be taken from
immediate standard solution
Vial

1

2

3

4

5

Ccaf (ppm)

1

5

10

15

20

Cpar (ppm)


1

5

10

15

20

Casp (ppm)

2

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 =


4 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 1.

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 1 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 3 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 3 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 5

vials with the volume need to be taken according to
Table 1.
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 1 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 1 vial and fill up to the
1mL mark with solvent.

caffeine and paracetamol
Weight 10 pills of Panadol Extra product
respectively.

3 EXPERIMENTS, RESULTS AND
COMMENTS

Table 2. Mass of 10 pills Panadol Extra
Pill no.


Mass (mg)

Pill no.

Mass (mg)

1

696.6

6

695.5

2

685.2

7

707.3

3

687.8

8

688.8


4

690.7

9

691.7

684.1
10
Average mass (mg)

682.2
691.0

5

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.
2

Blank
Par

Table 3. 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
1
0.5
0
200

250


300

350

Wavelength (nm)

Figure 1. Spectroscopy of paracetamol, caffeine
and aspirin
From the spectroscopy of 3 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 2 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
0.0

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 min


Figure 3. Chromatogram of 3 compounds when
the solvent ratio is 45:55 with vial 4
Comment:
The distance between the first 2 peaks is larger than
when using 50:50 solvent ratio. The analytical time
is too long but we have successfully completed in
separation of 3 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

min

Figure 2. Chromatogram of 3 compounds when
the solvent ratio is 50:50 with vial 5
Comment:
The result is that the first 2 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 min

Figure 4. 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 3 (black

line). The peak of aspirin appearing at t = 9.20 (blue
line) matches with the last peak of mix 3. Thus, the
remaining peak in mix 3 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 5. Chromatogram at (a) pH 5, (b) pH 7 (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 3. 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 3. The appropriate
explanation for this elution is similar to at pH 3
because buffer pH 5 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 3.
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 2

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 3 (figure
4) and pH 5 (figure 6a) show that the peaks are
separated almost clearly but pH 5 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 7 (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 3 to protect (Si-O) groups of column from
collapse and conducting the survey at pH 3, 5 and 7
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 5 concentration
levels, corresponding to vial 1 to vial 5.
mV

Detector A Ch1:273nm

200

150

100


50

0
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 min

Figure 6. Chromatogram of caffeine in vial 1 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

min

Figure 7. Chromatogram of caffeine in vial 2 at
273nm

Peak area


0

mV
150

Detector A Ch1:273nm

125

y = 92226x + 4627.7
R² = 0.9983

0

5
10
15
Concentration (ppm)

100

20

75

Figure 11. Calibration curve of caffeine

50
25

0
2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

min

Figure 8. Chromatogram of caffeine in vial 3 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 5
concentration levels, corresponding to vial 1 to vial
5.
mV
150 Detector A Ch2:234nm

50
25

125
0

100
0.0

1.0

2.0

3.0

4.0

5.0


6.0

7.0

8.0

9.0

10.0

min

/3.005/38021

75

Figure 9. Chromatogram of caffeine in vial 4 at
273nm

RT9.260/9.260/1768688

1.0

RT3.248/3.248/1246982

0.0

50
25
0


RT3.983/3.983/1825230

200
150

0.0

2.0

3.0

mV

Detector A Ch2:234nm

200

50
0

150

1.0

2.0

3.0

4.0


4.0

5.0

6.0

7.0

8.0

9.0

10.0

min

Figure 12. Chromatogram of paracetamol in vial 1
at 234nm

100

0.0

1.0

5.0

6.0


7.0

8.0

9.0

10.0 min

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 5 at
273nm

50

0

Table 4. Retention time and peak area of caffeine in
5 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 min


Figure 13. Chromatogram of paracetamol in vial 2
at 234nm

Peak area

1

1

3.987

70126

2

5

3.989

496634

3

10

3.990

916055

4


15

3.955

1418619

5

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
0.0

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 3
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

0
-10
0.0

1.0

2.0

3.0


4.0

5.0

6.0

7.0

8.0

9.0

min

Figure 15. Chromatogram of paracetamol in vial 4
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

min

Figure 16. Chromatogram of paracetamol in vial 5
at 234nm
Table 5. Retention time and peak area of
paracetamol in 5 vials
Vial


CPar (ppm)

Using the optimal chromatographic conditions in
3.2, (MeOH:buffer : 45:55), we construct the
calibration curve for aspirin with 5 concentration
levels, corresponding to vial 1 to vial 5. Because the
Shimadzu HPLC-UV system can choose only 2
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 6. Retention time and peak area of aspirin in
5 vials

-2.5

0.0

The calibration curve of aspirin:

Vial

CAsp (ppm)

tR (min)

Peak area


1

2

9.190

142934

2

10

9.232

934708

3

20

9.260

1768688

4

30

9.027


2695911

5

40

9.254

3501378

Construct the calibration curve of aspirin:

tR (min)

Peak area

4.00E+06

1

3.243

98662

3.50E+06

2

5


3.247

663038

3.00E+06

3

10

3.248

1246982

2.50E+06

4

15

3.208

1849971

5

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

0

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

1.50E+06

3.6 Quantitation of caffein and paracetamol in
1 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
0

5


10

15

Concentration (ppm)

Figure 17. Calibration curve of paracetamol

20


Table 7. 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 1 pill, which
is calculated from sample M1. Calculate the same
for sample M2 and M3, we have the table below:
Table 8. 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 1 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 ) 2

(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 1 pill or in 691.0 mg of
the medicine:

coefficients (a and b):
Sb = Sre

Sa = Sre

n

nC i − (Ci ) 2
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 1
+ + 2
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 9. 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 9 times.
Uncertainty of paracetamol of the actual pill:
2

u par

u10 pills

m par / pill


2

uunknown
munknown

m1 pill
2

u par

uvolumetric flask

C par

2

(17)

Vvolumetric flask

umicropipet

2

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 1 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

5

0.4) 10

(x

Sx

u

xi ) 2

n 1


Sx

t0.95; n

426.3

2

u par

43.2 mg

Comment:
(14)
(15)

n

426.3

mpar /1 pill

Calculate uncertainty of weighing:

2

Vmicropipet solvent

b


4

2

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

6

3

(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
0
-25
0.0

43.63
7

in 1 pill (mg)
Average (mg/pill)

51.02
5

45.18
1

0.5


1.0

1.5

2.0

2.5

3.0

3.5

4.5

5.0

5.5 min

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

0.0

(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
0

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 5 times, 70μL of solvent
once and 100μL of solvent 4 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

2

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 1 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 3 standard addition
samples.
H

m par add
m par add / weight

(21)

100%

For example: Sample S1:
m par / medicine S 1

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

4

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 3 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 3 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 3 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 2 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 1 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,
/>Accessed 12 March, 2021.
[4] National Center for Biotechnology Information.
"PubChem Compound Summary for CID 1983,
Acetaminophen" PubChem,
/>phen. Accessed 12 March, 2021.

[5] National Center for Biotechnology Information.
"PubChem Compound Summary for CID 2244,
Aspirin" PubChem,
/>Accessed 12 March, 2021.