CRYSTALLIZATION

ABSTRACT
Crystallization is the process of solid crystals formation precipitating from a solution. It is also a chemical
solid-liquid separation and purification technique, in which mass transfer of a solute from the liquid solution to a
pure crystalline phase occurs.
The most important points to be considered are: (1) selection of the solvent, the best solvent or solvent-
pair being the one in which the substance to be purified dissolves readily at the boiling point but is only slightly
soluble when cooled; (2) preparation of the solution which involves separation of the impurities by decoloration
and filtration; (3) collection of the crystals and repurification.
Crystallization can be hastened by seeding, agitation, scratching the sides of the container and freezing.
INTRODUCTION
Organic compounds synthesized in the laboratory or isolated from natural sources are often
contaminated with impurities. Recrystallization is a method for removing impurities from organic compounds that
are solid at room temperature. This method relies on the observation that the solubility of a compound in a
solvent generally increases with temperature.
Conversely, the solubility of a compound will decrease as a solution cools until the solution becomes
saturated, and crystals form. Recrystallization can produce very pure compounds. When a warm solution of a
compound cools and reaches the saturation point, small seed crystals of the compound form in the solution.
Slowly, additional molecules of the compound attach to the seed crystal and the crystals grow.
Since molecules in the crystals have a greater affinity for other molecules of the same type than they do
for any impurities, the process of crystal formation gives rise to relatively pure crystals. The impurities originally
present in the compound are left in solution.
In this laboratory you will have the opportunity to carry out the recrystallization of two organic compounds:
benzoic acid and naphthalene. The first experiment will give you practice in the technique as you recrystallize
benzoic acid from water. In the second experiment, your goal will be to purify an impure sample of naphthalene
by recrystallization. You will then assess its purity by taking a melting point.
PRE-LAB ASSIGNMENT:
1. What is the purpose of crystallization in an organic chemistry procedure?
 The main use of crystallization in the organic chemistry laboratory is for purification of impure
solids: either reagents that have degraded over time, or impure solid products from a chemical
reaction.

2. Explain why a Buchner funnel is used to isolate the final crystallized product instead of stem
funnel.
 A Hirsch funnel is used to isolate final crystallized product instead of a stem funnel because this
type of filtration proceeds much more quickly than allowing the solvent to drain through the filter
by force of gravity.

3. Give three criteria for a good recrystallization solvent.

a. The recrystallization solvent should NOT dissolve the substance to be purified at room
temperature, but it should dissolve it well at the solvent’s boiling point
b. The solvent should dissolve soluble impurities well at room temperature.
c. The solvent should not dissolve insoluble impurities even at the solvent’s boiling point. These
insoluble impurities can then be removed by gravity filtration.
TABLE OF CONSTANTS AND OBSERVATION:
Reagent Molecular mass Melting pt.(oC) Melting pt. (Observed) Mass of dried crystals

Benzoic acid _122.12 g/mol_ 122.3 oC ___ __121 oC____ __1.9 g__

Computation:

% Recovery =_mass of the dried crystals_ x 100%

Mass of crude benzoic acid
% Recovery =_1.9 g _ x 100%
2g
% Recovery = 95%
APPLICATION:
1. What is the function of the activated charcoal?
 Activated charcoal is sometimes used to remove small amounts of colored impurities from
solution.
2. Why do you have to filter the mixture while it is still hot?
 A hot filtration is generally used in some crystallization, when a solid contains impurities that are
insoluble in the crystallization solvent. It is also necessary in crystallization when charcoal is used
to remove highly colored impurities from a solid, as charcoal is so fine that it cannot be removed
by decanting.
3. Why did we choose water as the solvent for this experiment and not another non polar organic
solvent which could readily dissolve the crystals of benzoic acid?
 Benzoic acid can be separated from ionic solids because the materials have different solubility’s
in water. At room temperature, benzoic acid is not soluble in water, whereas ionic solids.
4. Are the crystals of benzoic acid obtained in the experiment pure? Why or why not?
 No, it is often the case that this basic form of re-crystallization is not sufficient to remove all of the
impurities--for instance, those impurities that are insoluble in the solvent used for re-
crystallization.
5. How does crystallization process applied in the sugar factories?
 Crystallization is the process after evaporation. During evaporation the clarified sugar cane juice
is boiled in evaporators which remove most of the water leaking a thick syrup. Then in the
crystallization process the syrup is boiled at low temperatures under partial vacuum and some
seedings are added which causes the development and growth of sugar crystals and the outcome
is massecuite (raw sugar crystals mixed with molasses). The sugar crystals and molasses are
then separated in centrifugal.
QUALITATIVE DETECTION OF ORGANIC COMPUNDS
DISCUSSION OF RESULT:
I.

Method of What evidence of C and H

Detection did you observe in each
test?
A. Tests for COLOR OF THE FLAME INTENSITY PRESENCE OF
Carbon and CARBONACEOUS
Hydrogen MATERIAL

Low intensity Presence of black

1. Ignition of : Yellow flame residue
Toluene

Isopropyl alcohol Blue flame High Intensity Presence of white

residue

INTERPRETATION:
1. IGNITION OF TOLUENE
All hydrocarbons will burn in the presence of oxygen (in the air). This reaction is called combustion, and the
products of this reaction are water and carbon dioxide gas.
Benzoic acid and benzaldehyde are produced commercially by partial oxidation of toluene with oxygen. The
major aromatic exhaust products from toluene combustion are benzene, ethylbenzene, styrene,
benzaldehyde, and unburned toluene.
Toluene has high molecular weight, aromatic hydrocarbon with double bonds.

Toluene benzoic acid

Incomplete combustion occurs when there isn’t enough oxygen to allow the fuel to react completely with the
oxygen to produce carbon dioxide, water and undesirable products are also formed. Carbon monoxide (CO) is
an example of an undesirable product of incomplete combustion. This compound is toxic.
2. IGNITION OF ISOPROPYL ALCOHOL
In complete combustion, the reactant burns in oxygen, producing a limited number of products. When
a hydrocarbon burns in oxygen, the reaction will only yield carbon dioxide and water. When elements are burned,
the products are primarily the most common oxides
C3H7OH + 5O2 3CO2 + 4H2O
Isopropyl alcohol
 2. Heating of: Describe the crystal before Observation after heating
heating
oxalic acid White fine crystals It turned into black liquid

benzoic acid White cubic crystals

Turned to liquid and vaporize

White crystal
sucrose It turn into brownish black sticky
substance

INTERPRETATION:
1. HEATING OF OXALIC ACID
Oxalic acid crystals is a crystalline hydrate, (H2C2O4 . 2H2O). On heating, the water of crystallization is lost first
and begins to sublime at about 150 - 160° C whilst on heating to a still higher temperature it partially decomposes
into carbon dioxide and formic acid, or into carbon dioxide, carbon monoxide and water.
H2C2O4 · 2H20 heat H2C204 + H2O
Oxalic Acid Dihydrate
H2C204 + O2 CO2 + H2O + CO
Anhydrous Oxalic acid
2. Heating of Benzoic Acid
Benzoic acid is an organic aromatic carboxylic acid, when heated to decompose, it emits acrid smoke and
irritating fumes. It’s melting point is 121-123 C because of these it turned into liquid and vaporized immediately.
3. Heating of Sucrose
Sucrose (C12H22O11)
C12H22O11 + O2 CO2 + H2O
Sucrose is a disaccharide sugar; it is made up of two monosaccharide sugar units. In the case of sucrose, the
two units are glucose and fructose. When you heat sucrose, the water will be lost first. The crystalline structure
of sucrose breaks down and the molecules decompose into glucose and fructose and then lose water and then
become isomers and polymerize to form caramel, a red-orange colored solid at room temperature.
Heating of sucrose gently to the right temperature, it will become a caramel. Through a process
called caramelization the sucrose breaks down and reforms different sugars giving you the distinct flavor and
color of caramel.
 3. Heating a Observation Element Identified
mixture of Formation of white smoke and Presence of Carbon Dioxide
CuO- bubbles
Acetanilide Formation of white solution The calcium hydroxide(limewater) reacts
with the carbon dioxide forming calcium
carbonate (white precipitate)
Presence of droplets of water on Presence of hydrogen in the form of H2O
the mouth

The presence of carbon and hydrogen, in an organic compound is detected by heating the given compound with
Cupric Oxide (CUO), when carbon is present it will be oxidized to form Carbon dioxide and Hydrogen is oxidized
to water (H2O). It will undergo to complete combustion. Carbon dioxide turns lime water milky while water
condenses on the cooler parts of the test tube.
B. Test How did you
for recognize the
Halogen presence of the
halide?

1. Belstein The CCl4 emits The color is The appearance of green flame is due to the
Test green flame . given to the formation of volatile cupric halides indicates the
flame by the presence of halogens in the organic compounds
vapor of copper
halide formed. The Beilstein test is very sensitive; halogen-
containing impurities may give misleading
results.

2. AgNO3 Formation of
Test white precipitate The precipitates are insoluble silver halides:
silver chloride, It forms a AgCl, white insoluble
precipitate

3. Fusion Formation of
of an white precipitate
Organic A white precipitate soluble in ammonia and
cpd. insoluble in diluted Nitric Acid indicates the
presence of chlorine.

NaCl + AgNO3 AgCl + NaNO3

Also known
In the halogen tests, formation of a voluminous
precipitate on addition of silver nitrate indicates
that a halogen is present, and the color of the
precipitate (a silver halide) may suggest which
halogen: white for chlorine
the precipitate is silver chloride
INTERPRETATION:
1. BELSTEIN TEST
 The Belstein test is very sensitive; halogen-containing impurities may give misleading results. The
appearance of green flame is due to the formation of volatile cupric halides indicates the presence
of halogens in the organic compounds
2. AgNO3 Test
 The precipitates are insoluble silver halides: silver chloride, it forms a AgCl, white insoluble
precipitate
3. Fusion of an Organic cpd.
 A white precipitate soluble in ammonia and insoluble in diluted Nitric Acid indicates the
presence of chlorine.

NaCl + AgNO3 AgCl + NaNO3

In the halogen tests, formation of a voluminous precipitate on addition of silver nitrate indicates that a halogen is
present, and the color of the precipitate (a silver halide) may suggest which halogen: white for chlorine.
C. How did you recognize Also known as Soda-
Demonstration the presence of lime test
of Nitrogen in nitrogen?
an Organic
Compound

INTERPRETATION:
A smell of ammonia indicated the presence of nitrogen and it turns the litmus paper to blue means the solution
is basic

CH4N2O + NaOH NH3 + Na2CO3

Urea ammonia

Simple primary amides can be decomposed by boiling with alkali and thereby evolving ammonia.
EXTRACTION
Discussion
Extraction uses two immiscible phases to separate a solute from one phase into the other. The distribution
of a solute between two phases is an equilibrium condition described by partition theory
Liquid-liquid extraction also known as solvent extraction and partitioning, is a method to separate
compounds based on their relative solubility’s in two different immiscible liquids, usually water and an organic
solvent. It is an extraction of a substance from one liquid phase into another liquid phase. Liquid-liquid extraction
is a basic technique in a chemical lab where it is performed using a separatory funnel.
Immiscible solvents are those mixtures of liquids which are insoluble in each other, as, for example, oil
and water. Such solvents will form layers with the lighter liquid on top and the heavier liquid on the bottom.
The layers can be separated by pouring off or decanting the supernatant liquid, or through the use of a
separatory funnel. The former technique is satisfactory where the supernatant liquid need not be separated
quantitatively from the underlying solvent. In the use of separatory funnel the lower liquid is drawn off through
the funnel stem by manipulation of the stopcock, and the upper layer is poured off through the top of the funnel.
It is not good technique to draw off the upper layer through the funnel stem.
In order that intimate contact can be affected between the two liquids when it is desired to extract from
one layer into another layer, it is essential to shake the two solutions vigorously, and for this purpose, a stopper
should be inserted in the top of the funnel. Some caution should be exercised in this process, especially if low-
boiling solvents are being used. As considerable pressure may by produced by evaporation of the low boiling
solvent, it is suggested that the stoppered funnel with the liquid be held in an inverted position, with the stem
pointing upward, and the stopcock opened to release any pressure within the container. This procedure should
be repeated several times when extracting with low-boiling solvents. This avoids the spattering of the liquid which
may take place on opening the upper stopper if there is too much pressure within the funnel.
Water is usually one of the solvents used in an extraction process, and the other solvent is usually an
organic liquid of a non-polar (non-ionic) character. The organic or non-polar liquid such as n-butyl alcohol, may
have a slight polar property as indicated by its solubility in water. It is very slightly soluble in water, however, and
its efficiency in use can be improved considerably by the addition of a small amount of an ionizable salt, such as
sodium chloride to the water layer. The increase in the polar layer of this solution will cause a decrease in the
solubility of a non-polar compound. This is called the “salting out” process.
The partition coefficient generally refers to the concentration ratio of un-ionized species of compound,
whereas the distribution coefficient refers to the concentration ratio of all species of the compound (ionized plus
un-ionized).
In the chemical and pharmaceutical sciences, both phases usually are solvents. Most commonly, one of
the solvents is water, while the second is hydrophobic, such as 1-octanol. Hence the partition coefficient
measures how hydrophilic ("water-loving") or hydrophobic ("water-fearing") a chemical substance is. Partition
coefficients are useful in estimating the distribution of drugs within the body. Hydrophobic drugs with
high octanol-water partition coefficients are mainly distributed to hydrophobic areas such as lipid bilayers of
cells. Conversely, hydrophilic drugs (low octanol/water partition coefficients) are found primarily in aqueous
regions such as blood serum.
The process of salting out is the separation of an organic phase from an aqueous phase by the addition of salt.
The addition of an organic salt into a mixture of water and a water-miscible organic solvent causes a separation
of the solvent from the mixture and the formation of a two phase system. Sometimes this phenomenon is referred
to a “salt-induced phase separation. Observations of this “salting-out” phenomenon were made for a number of
water-miscible organic compound such as acetone, methanol, ethanol and acetronitrile.
 Volume of n-butyl alcohol 30 mL
Volume of water 150 mL
Volume of n-butyl alcohol recovered ___ mL
Volume of n-butyl alcohol recovered after salting out process ___ mL
1ST Extraction = 19 ml of n-butyl alcohol
2nd Extraction = 9.2 of n-butyl alcohol

III. Calculate the % recovery

% Recovery = volume of the recovered n-butyl alcohol x 100

Original Volume of the n-butyl alcohol

% Recovery of n-butyl = 1st volume n-butyl alcohol + 2nd vol x 100

Original Volume of the n-butyl alcohol

% Recovery = 19 ml + 9.2 ml x 100

30 ml
% Recovery of n-butyl alc = 94

ALCOHOLS
Abstract
Alcohols are compounds containing the hydroxyl group attached to the carbon of an alkyl group which
may be represented by the general formula R-OH. These compounds can be considered as alkane that have
one hydrogen atom replaced by an -OH group, the functional group of alcohols.
Primary, secondary and tertiary alcohols react differently. In the removal of the H atom from the -OH
bond, the rate of reaction is dependent on the relative polarization of the -OH group. When the reaction involves
the replacement of the -OH group, the reactivity of the alpha carbon depends on the stearic hindrance caused
by the substituents.

Learning Outcomes:
At the end of this activity, the student shall be able to:
 Characterize the different types of alcohols as to their solubility, volatility and boiling point.
 Distinguish the structure of an alcohol as to whether primary, secondary or tertiary as far as reactivity is
concerned
 Infer from the structure of alcohol molecule about the extent of oxidizability
 Write the equation in each chemical reaction
DISCUSSION:

A. Solubility of alcohols
Types of alcohol
a. Primary alcohol – ethyl alcohol, n-butyl alcohol
b. Secondary alcohol – Isopropyl alcohol
c. Tertiary alcohol – tert-butyl alcohol

1. Which of the alcohols are soluble? All except tert-butyl alcohol

2. Least or Slightly soluble? N- butyl alcohol
3. What is the trend in the solubility of alcohols?
Ethanol > tert-butyl > isopropyl> n-butyl alcohol
 As the HC chain increases, solubility decreases. This is due to the increase in stearic
hindrance of the bulky group (long chain of the non-polar HC) thus, solubility decreases.
 For the isomeric alcohols, alcohols with more branching are more soluble. This is due to the
surface area of the non-polar portion of the alcohol decreases, solubility increases.
Structures:

Isopropyl alcohol

n-butyl alcohol

4. Describe the volatility and boiling point of: Ethyl alcohol, Isopropyl alcohol, Tertiary butyl
alcohol and n-butyl alcohol.

Volatility and Boiling point of alcohols

Alcohols with low mol. wt. are more volatile and have lower boiling point.
 The hydroxyl groups in alcohol molecules are responsible for hydrogen bonding between
the alcohol molecules. As greater energy is required to overcome these strong intermolecular forces,
the melting points and boiling points of alcohols are higher than those of alkanes with a corresponding
chain length.

B. Lucas Test
1. Which of them reacted immediately after mixing? TERT-BUTYL ALCOHOL
2. Which of them reacted after 30 minutes? ISOPROPYL ALCOHOL
3. Which of them did not react at all? ETHYL ALCOHOL AND N-PROPYL ALCOHOL
4. Describe the trend in the rate of reactivity of alcohols in Lucas test?
TERTIARY >SECONDARY>PRIMARY ALCOHOL

Lucas test is a solution consisting of anhydrous zinc chloride in concentrated hydrochloric acid. This test is
used to differentiate and categorize primary, secondary and tertiary alcohols.
The mixture of zinc chloride and concentrated hydrochloric acid is called Lucas reagent. It reacts with primary,
secondary and tertiary alcohols at different rates. This reagent forms a cloudiness on reacting with alcohols.
Tertiary alcohols react immediately and give cloudiness, secondary alcohols reacts slowly and gives cloudiness
after 5 to 10 minutes and there is no reaction with primary alcohols.
The chemical reactions are given below.

Note:
Cloudiness appears immediately → Tertiary alcohols
Cloudiness appears within five to ten minutes → Secondary alcohols
Cloudiness appears only on heating → Primary alcohols
 C. Oxidation

1. Which of the alcohols is oxidized readily? Ethyl alcohol and n-butyl alcohol
2. What is the color of the solution? Green solution
3. Which of them did not react at all? Tert-butyl alcohol
4. What is the trend in the oxidizability of alcohols? Reactivity primary > secondary > tertiary(no
reaction)

Oxidizing agent is K2Cr2O7 potassium dichromate. These are used along with H2SO4, H2O
 1o alcohol → Carboxylic acid
 2o alcohol → Ketone
 3o alcohol → No reaction

Primary alcohol can be oxidized to either aldehydes or carboxylic acids.

Alcohol is first oxidized to an aldehyde which is then oxidized to carboxylic acid.
secondary alcohols can be oxidized to give ketones.
Tertiary alcohols, in contrast, cannot be oxidized without breaking the molecule's C–C bonds.

During the oxidation, the orange dichromate ion is reduced to the green Cr 3+ ion. This can be used to detect
alcohols.
 D. Esterification
1. Which of the two alcohol samples has first developed the pleasant odor? Ethyl alcohol
2. What is the new substance formed? Ethyl acetate – pleasant odor
Carboxylic acids react with alcohols forming a fruit smelling ester. The reaction between an alcohol and a
carboxylic acid is called esterification. This reaction is a slow reaction catalyzed by concentrated sulphuric
acid.
The chemical reaction is given below.
R-OH + R-COOH → R-COOR + H2O
CH3OH + CH3-COOH → CH3-COOCH3 + H2O
Note: A sweet smell indicates the presence of alcoholic group.

E. Iodoform Test
Note the color of the precipitate. Yellow precipitate
1. What is the new substance formed? CHI3 – Iodoform

This test is given by secondary alcohols, ketones and acetaldehyde. First the compound is heated with sodium
hydroxide solution and iodine. A formation of yellow precipitate of iodoform shows the presence of alcohol.
The chemical reactions are given below.
CH3-CH(OH)-CH3 + I2 + 2NaOH → CH3-CO-CH3 + 2NaI + 2H2O
CH3-CO-CH3 + 3I2 + 4NaOH → CHI3(Iodoform) + CH3COONa + 3NaI + 3H2O
Note: The formation of yellow precipitate shows the presence of alcohol, acetaldehyde or methyl ketones.
QUIZ ANSWERS:

 Crystallization can be hastened by the following except

ANSWER: HEATING

 What is a good recrystallization solvent?

ANSWER: THE SOLVENT SHOULD DISSOLVE SOLUBLE IMPURITIES WELL AT ROOM
TEMPERATURE

 A method used for removing impurities from organic compounds that are solid at room
temperature.
ANSWER: RECRYSTALLIZATION

 A 5 g of crude Benzoic Acid was recrystallized and 72% of crystals was recovered. How many
grams of dried crystals was formed?
ANSWER: 3.6g

 Buchner funnel is better to used to isolate the crystal through the filter by
ANSWER: GRAVITY

 What is the formula of benzoic acid?

ANSWER: C6H5COOH

 A 3 g of crude Benzoic Acid was recrystallized and formed a 2.2g dried crystal of benzoic acid
What is the % recovery of the crystals.
ANSWER: 73.33%

 What is the melting point of pure benzoic acid?

ANSWER: INSOLUBLE

 What is the molecular mass of benzoic acid.

ANSWER: 122 g/mol

 Separation of impurities can be made by

ANSWER: DECOLORATION AND FILTRATION

 What do you call the raw sugar crystals mixed with molasses
ANSWER: MASSECUITE

 Used to removed small amount of impurities from solution.

ANSWER: ACTIVATED CHARCOAL

 What is the chemical name of limewater?

ANSWER: CALCIUM HYDROXIDE

 What is the product produce in heating toluene.

ANSWER: BENZOIC ACID

 Test used to determine the presence of halogen.

ANSWER: BELSTEIN TEST

 Visual evidence of chlorine ions in organic compound.

ANSWER: GREEN FLAME

 The product produce when oxalic acid dihydrate was heated.

ANSWER: ANHYDROUS OXALIC ACID

 Visual evidence of formation of silver chloride in silver nitrate test

ANSWER: WHITE PRECIPITATE

 The visual evidence in the presence of nitrogen in organic compound.

ANSWER: RED LITMUS PAPER TURNS TO BLUE

 In heating of CuO and acetanilide, it produces a moist on the mouth of the tube. What does it
indicate?
ANSWER: PRESENCE OF HYDROGEN

 The visual evidence of heating of benzoic acid.

ANSWER: IT LIQUIFY AND VAPORIZE

 The color of the flame when isopropyl alcohol was ignited.

ANSWER: BLUE FLAME

 The color of the flame in ignition of toluene

ANSWER: YELLOW FLAME

 Benzoic acid is what type of organic compound?

ANSWER: AROMATIC CARBOXYLIC ACID

 Suscrose is what type of sugar.

ANSWER: DISACCHARIDE

 The ignition of isopropyl alcohol is an example of complete combustion.

ANSWER: TRUE

 The process when hydrocarbons burns in the presence of oxygen in air.

ANSWER: COMBUSTION