ACP Home | Organic Chemistry IOrganic Chemistry II  | Lecture | LaboratoryGeneral Chemistry | Laboratory Syllabus 

Separation & Purification of Components of an Analgesic Tablet
Experiment Description & Background

In this two-week experiment, students work individually to separate the three active constituents, aspirin (ASA), acetominophen (ACE) and caffeine (CAF), of an analgesic tablet using extraction.  The constituents are purified, melting point determinations done, the percent recovery (by weight) of each constituent determined, and the efficiency of the extraction evaluated by TLC analysis.

Ionizable Organic Functional Groups:  Acids and Bases
Table 1 lists some common ionizable organic functional groups (i.e., organic acids and organic bases).  Organic acids tend to donate protons (Bronsted-Lowry definition), while organic bases tend to accept protons.  Figure 1 shows the reaction schemes of the ionization processes for a phenol (acid) and an amine (base).  Note that the reaction scheme to illustrate the ionization process is always written to show the acid (or conjugate acid) on the left, donating the proton to generate the base (or conjugate base), drawn to the right of the arrows.
Organic Acids
pKa range
Organic Bases
pKa Range
(of conjugate acid)
Carboxylic acids  4-6   Amines 10-13
Phenols 8-10   Imines 10-13
Secondary imides 14-16      
Sulfonamides 7-10      

Table 1:  Some Ionizable Organic Functional Groups

Figure 1:  Ionization Schemes for a Phenol (acid) and an Amine (base)

The three components of the analgesic tablet that are separated in this experiment each have ionizable functional groups.  Aspirin contains an acidic carboxylic acid, acetaminophen contains an acidic phenol, and caffeine contains a basic amine.  Ionization schemes for these more complex structures are illustrated in Figure 2.  Note that the acid (or conjugate acid) is always on the left, and the base (or conjugate base) is on the right for each ionizable functional group.

Figure 2:  Ionization Schemes for Aspirin, Acetaminophen and Caffeine

back to top

The Hendersohn-Hasselbach Equation:  pKa  and pH
There are two factors that determine the extent of ionization of an ionizable functional group;  the pKa of the functional group and the pH of the environment.  The pKa of an ionizable functional group is an intrinsic property of the functional group that is a quantitative measure of the functional group's acid strength, i.e, how likely it is for the functional group to donate a proton (H+).  For acidic compounds, the pKa value refers to how likely it is for the acid to donate a proton.  For basic functional groups, the pKa refers to the acid strength of the base's conjugate acid.  Each functional group of a molecule with multiple ionizable functional groups has its own distinct pKa.  The pKa of a specific functional group is fixed and cannot be changed.  pka values of functional groups in specific organic compounds and drugs can be found in the Dictionary of Organic Compounds, CRC Handbook of Data on Organic Compounds and the appendix of a medicinal chemistry textbook like Foye's principles of Medicinal Chemistry.  The pH of the environment refers to the aqueous solution (or aqueous body fluid) where the molecule containing the ionizable functional group(s) is located.  The pH of the environment can be changed.
pH  =  pKa + log base/acid

Figure 3: 
Hendersohn-Hasselbach Equation
Ionized forms of organic acids and ionized forms of organic bases are generally soluble in aqueous environments (i.e., ionized forms are very polar and soluble in very polar solvents like water).  Organic acids and bases can be ionized in aqueous environments of the appropriate pH.  The Hendersohn-Hasselbach equation, given in Figure 3,  relates the pKa of an ionizable functional group,  the pH of its environment, and the ratio of ionized to unionized forms of the functional group. The Hendersohn-Hasselbach equation can be used to calculate exactly how much of a particular acid or base is ionized (relative to the total amount present) in an aqueous solution of a specific pH. 

How to Use the Hendersohn-Hasselbach Equation
Suppose you want to know how much the functional groups of the drug shown in Figure 4 are ionized in a 0.05M solution of NaOH.  How can the Hendersohn-Hasselbach equation be used to determine the extent of ionization?  Numerous pieces of information must be gathered before the Hendersohn-Hasselbach equation can be used.
Figure 4:  Structure of isoproterenol, a bronchodilator
The ionizable functional groups of the molecule must be identified  and the pKa values for each of the ionizable functional groups must be found.

The pH of the environment must be determined (i.e., What is the pH of a 0.05M solution of NaOH?)

The ionization schemes for each ionizable functional groups must be drawn.

The drug in Figure 4 contains a phenol, a secondary alcohol, and a tertiary amine.  Only the phenol and tertiary amine are ionizable functional groups.  The pKa values for each of these functional groups were found in the appendix of "The Textbook of Organic, Medicinal and Pharmaceutical Chemistry, 10th edition".

The pH of a 0.05M solution of NaOH can be calculated.  NaOH is a strong base, meaning it is essentially 100% ionized.  The molarity of a solution of NaOH is the same as the molarity of HO- ions.  Thus, for a 0.05M solution of NaOH, the concentration of HO- ions is 0.05M.  Two equations are used to calculate the pH of this solution.  The first is to determine the pOH, then using the pOH, determine the pH.

The pOH is the negative log of the HO- ion concentration, which in this case is 0.05M.

pOH = -log [HO-] 
pOH = -log 0.05 
pOH = 1.3

The pH can be determined from the pOH using the equation:

pH  +  pOH  =  14
pH  +  1.3  =  14
pH  =  12.7

The ionization schemes for each of the two ionizable functional groups are shown in Figure 5.  Notice that for the acid (phenol), the unionized form is on the left and the ionized form is on the right, but for the amine, the ionized form is on the left and the unionized form is on the right.

Figure 5:  Ionization Schemes for Ionizable Functional Groups of a Drug Molecule

back to top

Now the Hendersohn-Hasselbach equation can be used to determine how much of each of these ionizable functional groups are ionized in a solution of 0.05M NaOH.  The equation must be applied to each functional group individually.  These calculations predict that the major form of the drug at a pH = 12.7 would look like the structure depicted in Figure 6.


Figure 6:  Major form of the drug at pH = 12.7

pH  = pKa  +  log base/acid

For acidic functional groups, since the conjugate base is ionized and the acid is unionized, the equation could also be written:
pH  =  pKa  +  log ionized/unionized  or
pH  =  pKa  +  log I/U

12.7  =  9.9  + log I/U
2.8  =  log I/U
631  =  I/U

So the ratio of ionized (I) to the unionized (U)  form of the phenol is 631.  This means there is much more of the ionized form of this functional group present in an environment where the pH = 12.7.

pH  = pKa  +  log base/acid

For basic functional groups, since the  base is unionized and the conjugate acid is ionized, the equation could also be written:
pH  =  pKa  +  log unionized/ionized  or
pH  =  pKa  +  log U/I
(Note that the U and I are reversed from the equation for an acid)

12.7  =  8.7  +  log U/I
4  =  log U/I
10,000  =  U/I

So the ratio of unionized (U) to the ionized (I)  form of the phenol is 10,000.  This means there is much more of the unionized form of this functional group present in an environment where the pH = 12.7.

It is common that the amount of the ionized form of a functional group is expressed as a percentage. The percent ionization of a functional group at a given pH can also be calculated using two equations.

From the Hendersohn-Hasselbach equation for the phenol, we know that

631  =  I/U
631U  =  I

We also know that the sum of the ionized and unionized forms of the functional group must equal 100% or:

100%  =  I  +  U

If 631U is substituted into this equation for I (from equation above):

100%  =  631U  +  U
100%  =  632U
0.16%  =  U
100%  -  0.16  =  I
99.84%  = I
The phenol is 99.84% ionized at pH = 12.7

From the Hendersohn-Hasselbach equation for the amine, we know that

10,000  =  U/I
10,000I  =  U

We also know that the sum of the ionized and unionized forms of the functional group must equal 100% or:

100%  =  I  +  U

If 10,000I is substituted into this equation for U (from equation above):

100%  =  I  +  10,000I
100%  =  10,001I
0.01%  =  I
100%  -  0.01  =  U
99.99%  = U
The amine is 0.01% ionized at pH = 12.7

back to top

Ionization and Solubility
Solubility of an organic molecule in a particular solvent is dependent on how closely the polarity of the molecule "matches' the polarity of the solvent.  Molecules with polarities very similar to the solvent will tend to be very soluble.  Polar molecules tend to be soluble in polar solvents, while non-polar molecules tend to be soluble in non-polar solvents. This is the basis of the simple rule "like dissolves like"
Solubility Rule 
Polar molecules are soluble in polar solvents.
Non-polar molecules are soluble in non-polar solvents. 
Ionization of functional groups within a molecules  makes the molecule more polar, and more soluble in polar solvents.  The increase in polarity that occurs as a result of ionization is due to the formal charges (and thus charge separation) of the conjugate acid or conjugate base forms of the ionizable functional groups. A molecule with ionizable functional groups will tend to be more soluble in a polar solvent when it is ionized and less soluble in polar solvents when it is unionized. Conversely, unionized forms of a molecule will tend to be more soluble in non-polar solvents and and less soluble in polar solvents.

Acid or Base Extraction
Acid or base extraction is an experimental technique that exploits the different solubility a molecule has depending on whether its functional groups are ionized or unionized.  During an extraction, a molecule is "partitioned" between two solvents that are not soluble in each other.  Typically the two solvents used for an extraction have very different polarities such as water (polar) and hexane (non-polar) and will form two distinct layers when combined.  As the molecule is "mixed" between the two solvents, it ultimately dissolves in the solvent that best matches its polarity.  A molecule in its unionized from would tend to dissolve in hexane, while a molecule in its ionized form would dissolve in water.

An acid or base extraction can be used to separate a molecule with ionizable functional groups from contaminants or other compounds.  For example, a base extraction could be used to separate the product of the reaction shown in Figure 8 from unreacted starting material.  The product contains an ionizable functional group (carboxylic acid) and the starting material does not.  Under the experimental conditions used, the reaction only yields 65% of the product, leaving ~35% of the starting material intact.  After the reaction is over, the product must be separated away from the unreacted starting material.

To use base extraction for separation of the product from the staring material, the reaction mixture, containing both the alcohol and the carboxylic acid, are dissolved in methylene chloride, a medium polarity solvent that is not water soluble.  The solution is added to the separatory funnel (with the stopcock closed).  An equal volume of aqueous base (at a pH sufficient to ionize the carboxylic acid) such as 0.05M NaOH is added to the separatory funnel.  Two distinct layers form in the funnel since the two solvents are not soluble in each other. (Their polarities are too different).  The two layers are then "mixed" by shaking the funnel allowing the reaction mixture to be partitioned between the two solvents (Figure 9).  Shaking the separatory funnel in this manner brings the basic aqueous layer into contact with the carboxylic acid, causing it to ionize and become more polar.  After the partitioning, each component of the mixture will end up dissolved in the solvent it is most soluble in.  The medium polarity alcohol will remain dissolved in the methylene chloride, while the ionized carboxylic acid product will move to the more polar basic aqueous layer.  The two layers separate out again after the funnel is returned to the ring stand and allowed to sit undisturbed.  The bottom layer can be drained out into a clean beaker by opening the stopcock and closing it again just at the boundary between the to layers.  A second beaker can then be used to collect the top layer in a similar manner.


Figure 8:  Oxidation of an alcohol to a carboxylic acid

Figure 9:  Shaking the separatory funnel
(adapted from Zubrick, p. 161)

How to Handle a Separatory Funnel
The separatory funnel is used to perform simple extractions. The separatory funnel has three distinct parts:  1) the stopper (at the top), 2) the vessel itself, and 3) the stopcock (at the bottom). When the separatory funnel is in use, it is generally positioned through a ring (of appropriate size) mounted to a sturdy ring stand (Figure 7)

Figure 7:  Separatory Funnel
(adapted from Landgrebe, p.124; taken fron General Chemistry I)

The solvents and solutions used in the extraction are added to the separatory funnel through the ground-glass joint at the top of the funnel.  Solvents and solutions should be added to the funnel using a glass funnel to avoid contaminating the joint.  (Contaminants will prevent a tight seal from forming).  The stopcock at the bottom of the funnel must be in the closed position (perpendicular to the stem or "drain").  When two immisible solvents (solvents insoluble in each other) are placed in the separatory funnel, two distinct layers should be visible.

Shaking the separatory funnel allows the two solvents to mix and provides the opportunity for components dissolved in one of the solvents to be transferred to the other solvent.  Proper technique for shaking the separatory funnel is illustrated in Figure 9.  The stopper should be placed in the top of the funnel before removing it from the ring and the stopcock should be in the closed position. The stopper must be held firmly in place by holding it with the index finger or palm of the hand.  The stem (out spout) must be directed into the back of the hood and the funnel should be positioned with the stem pointed upward at an angle of ~ 45degrees.  Mixing occurs by gently shaking the funnel for ~10-30 seconds, after which the stopcock should be opened, with the funnel tilted upward,  to vent any gases generated by the shaking process.  After venting the gases, the stopcock should be closed and the funnel returned to the ring.

The mixture in the funnel will be cloudy when the two immiscible solvents mix during shaking.  The two distinct solvent layers will separate out after the funnel is allowed to sit undisturbed in the ring for a few moments.  Sometimes emulsions form and the two layers do not separate cleanly.  emulsions are mixtures of two immiscible solvents where one solvent becomes encapsulated ("trapped") by the other solvent.  Techniques have been developed to help break emulsions that form during the extraction process and are summarized Table 2..  Be aware that breaking an emulsion is often tricky and requires a bit of finesse.  one or possibly all of the techniques listed in Table 3.2 may be necessary to break the emulsion.

Addition of small quantities (10-100mg) of salt (NaCl or equivalent) to the funnel, followed by gently shaking or swirling.   Adding too much salt to the separatory funnel may clog the stopcock, so add sparingly.
Addition of one or the other of solvents used in the extraction.
Addition of ethanol which has solubility in both aqueous and organic solvents.
Addition of small quantities of dilute aqueous acid or base (1-2ml).  Note that some components may react with acid or base so use this method as a last resort if the components are acid or base sensitive.

Table 2:  Techniques for breaking emulsions

In addition to emulsions, separation of the two solvents is often complicated by a "third layer" or the interface between the two solvents.  A large interface is likely just a small emulsion and can be handled using the techniques listed in table 3.2.  Usually the best technique to use for an intereface problem is to add on or the other of the two solvents used in the extraction.  The interface contains both solvents and likely all the components of the mixture.  If the interface is collected with a specific layer, that layer will be contaminated with all components of the mixture.  To ensure a completely clean separation of the two layers, it is sometimes nessary to collect the interface in a third beaker.

The two layers are separated after shaking by draining the lower layer from the bottom of the separatory funnel.  The stopper must be removed from the funnel prior to draining.  The bottom layer is drained into a clean, labeled beaker or flask by opening the stopcock and closing it quickly when the interface region reaches the bottom of the funnel.  A second beaker or flask is used to collect the remaining layer in the funnel.

back to top

Creating and Using Flowcharts for Experimental Procedures
A flowchart is a visual or graphic representation of the organization, sequence and content of a project.  Good flowcharts have a clearly defined beginning, logical sequencing of steps, and easily identified endpoints.  Flowcharts are used extensively in computer programming but are also effective tools for organizing and sequencing steps of an experimental procedure in the laboratory.  The laboratory flowchart is generally a summary of steps involved in an experimental procedure.  The main advantage of the laboratory flowchart is that it often provides a simpler format to use when conducting a laboratory experiment.  It also allows the user to see how various elements of the procedure fit together.  However, the flowchart generally does not provide significant details and assumes the user is familiar with the basic techniques of the experiment.

Creating a flowchart to summarize an experimental procedure requires familiarity with the experiment and all the techniques used to complete the experiment.  It also requires planning ahead on how to best organize various parts of the procedure, and thoroughly working out appropriate sequencing of steps.  Usually a flowchart can be built from a text-based procedure.

Computer scientists use standard symbols (and some custom symbols) for constructing flowcharts.   Simple laboratory flowcharts use boxes and arrows to indicate actions (in boxes) and sequencing of actions (arrows).   Flowcharts always start at the top and flow down, or begin at the left and flow right.  Most flowcharts have branch points where a single action leads to two or more additional actions.

The flowchart that summarizes the procedure for "The Separation and Purification of the Components of an Analgesic" experiment (Figure 3.10).  (Refer to the text procedure for the experiment to help follow through the description of  the flowchart).  The first step of the experiment, "grind four Excedrin tablets" is placed in a box at the top of the page.  (View sample flowchart).  The second step of the experiment is indicated in the flowchart by an arrow leading from this box to a second box stating the next action, "mix the ground tablets in 15ml CH2Cl2 and heat".  Note that details about flask size or exactly how to warm the solution are not included in the flowchart.  The second box leads to a third box with the action "vacuum filter the mixture".  A branch point results after completion of the third step of the procedure, since the process of filtration leads to both a solid and a solution (filtrate) upon which further actions must be performed.  A branch point is indicated in the flowchart using a 'T' or "Y", with arrows leading to two distinct boxes.  View the complete flowchart for the of "The Separation and Purification of the Components of an Analgesic" experiment.

Calculating Percent Recovery
Rarely can all of a specific component in a multi-component mixture be isolated completely.  The amount of material isolated from a multi-component mixture is often reported as a percent of the total amount of material present in the mixture.  This amount is referred to as the "percent recovery" of the material.  In the experiment "The Separation and Purification of the Components of an Analgesic", three individual components of a mixture (i.e., the tablet) are isolated.  The percent recovery for each component of the mixture is the amont of each component that is isolated in the experiment relative to the amount of that component in the four tablets used.

Two values must be known to calculate the percent recovery of a specific component in a mixture, the experimental mass (weight) of the isolated component, and the amount of the component reported to be in the starting mixture.  Each Excedrin tablet contains 250 mg of aspirin, 250 mg of acetaminophen and 65 mg of caffeine.  If four tablets are used in this experiment, then the total reported amount of aspirin and acetaminophen is 1000 mg each, and the total amont of caffeine is 260 mg.  the equation, given below can then be used to calculate the percent recovery for each component.
Be sure to use the same units for the experimental value and the reported value, i.e., mg.

      x mg of ASA isolated        X       100       =       % recovery

                                                                                        1000 mg ASA

      x mg of ACE isolated        X       100       =       % recovery

                                                                                      1000 mg ACE

       x mg of CAF isolated        X       100       =       % recovery

                                                                                        260 mg CAF

Feiser, L.F.; Williamson, K.L. Organic Experiments, 8th Edition, Houghton Mifflin Co.: New York, 1998.
Zubrick, J.W. The organic Chem Lab Survival Manual, 4th Edition, John Wiley & Sons: New York, 1997.
Landgrebe, J.A. Theory and ractice in the Organic Chemistry Laboratory, 4th edition, Brooks/Cole Publishing Co.: Pacific Grove, CA, 1993.