Mayonnaise production method

Objective:

1. To determine the type of food emulsion obtained by varying the mixture sequence

2. To produce the home made mayonnaise

Introduction:

Mayonnaise is a creamy, pale yellow, and mild-flavored food product which is frequently used in preparation of salads, sandwiches, and many other food products. Although the production of mayonnaise consisting of relatively few ingredients and processing steps, but successful formulation and processing are required an understanding of the role of each ingredient and the critical processing steps in order to create the delicate structure of mayonnaise. Mayonnaise is a unique emulsion which contains the mixture of water and oil. The major component in the mayonnaise is oil which dispersed throughout the lesser amount of continuous aqueous phase (water). The structure of mayonnaise is easily disrupted because of this unusual relationship between the oil and aqueous phase. Integration of processing and colloid chemistry is essential to understanding the formation and stabilization of the mayonnaise.

In this experiment, the mayonnaise is made by using 75% of oil, 8% of fresh egg yolk, 1% of mustard powder, 1.5% of salt, 1.3% of distilled water, and 13.2% of vinegar in producing mayonnaise. In a colloidal system, the minor component (dispersed phase) is usually being dispersed throughout the dispersion medium. However, the major component in mayonnaise (oil) is forced to be fine droplets to disperse throughout the lesser amount of continuous aqueous phase (dispersed medium). Mayonnaise is known as an oil-in-water (o/w) emulsion. When the emulsion of mayonnaise breaks, the water (dispersed phase) does not dispersed in oil (dispersed medium) and hence this allows the phenomenon of creaming formation to occur.

In order to maintain the stability of emulsion in mayonnaise, emulsifier acts as the most important role to stabilize the unique emulsion since the high amount of oil in water does not favor the formation of o/w emulsion. An emulsifier works at the surface of two immiscible liquids and tends to reduce to interfacial tension between two different liquids by reinforcing the in contact surface between them. The larger the ratio of surface area to volume of oil, the smaller the oil droplets dispersed in the water. Finer dispersion of oil droplets is required more emulsifier to surround them and thus this can maintain the stability of emulsion in the system. If emulsifier is not added, the two immiscible liquids will quickly separate after they mixed. Thus, emulsifiers are liaisons between the two liquids and serve to stabilize the mixture. In producing the mayonnaise, artificial emulsifier is not allowed. So, the source of emulsifier is normally obtained from egg yolk for stabilization of mayonnaise. The egg must be totally dissolved in the water before the addition of oil begins so that to achieve more efficient emulsion. Lecithin is a low molecular mass surfactant that can be found in egg yolk which acts as a hydrophilic effective emulsifier in oil-in-water emulsion.

Emulsion activity of emulsifier is based on its molecular structure. There is a hydrophobic part with a good non-aqueous solubility and hydrophilic part that is soluble in water. The hydrophobic part of the molecule is generally a long chain alkyl residue while the hydrophilic part of the molecule consists of a dissociable group. In a system containing two immiscible liquids such as oil and water, the emulsifier is located in the interface between two liquids which tends to reduce the interfacial tension of each liquid. The alkyl residues are solubilized in the oil droplets whereas the negatively charge of end group projected to the water. It maintains the stability of emulsion by involving the double electrostatic layer. The adsorption of negatively charge end group on the surface of oil globules causes the formation of a negative layer around them. Eventually, the oppositely charge particles will tend to approach to the particular negative layer and hence forming the second layer. This is known as double electrical layer. The electrical double layer creates repulsive forces greater than the attractive forces between the oil droplets. Hence, it can maintain the oil droplets from combining to each other in the emulsion.

Scattering light is one of the characteristics of colloidal system. Emulsion always exhibits the similar characteristic with colloidal suspension which is scattering the light. The oil-in-water emulsion present as a cloudy and turbid because of the existence of oil droplets dispersed in the aqueous phase. This is because the many phase interfaces between the two liquids scatter the light which passes through the emulsion system. The basic colour of emulsion is white. The Tyndall effect will scatter the light and distort the colour to blue if the emulsion is dilute. In a concentrated emulsion, the colour of emulsion system will scatter light and then distort the colour to yellow.

Several ways have been proposed to determine type of emulsion formed. The drop-dilution method can be used to determine the type of emulsion. To a small portion of the emulsion, add some water and stir slightly. If the water blends with the emulsion, it is an oil-in-water emulsion, but if oil blends with the outside phase it is a water-in-oil emulsion.

Another method of determining the type of emulsion is to use Sudan III, red dyes soluble in the oil but not in the water. A small portion of the finely powdered dye is dusted over the surface of the emulsion. If oil is the external phase the color gradually spreads throughout the emulsion. But if water is the external phase the color does not spread but is confined to the oil with which it comes in contact on the surface.

The microscope may be used to determine the type of emulsion formed. If the oil is dyed red, a red field with clear globules indicates a water-in-oil emulsion; red globules in a clear field show an oil-in-water emulsion. Sometimes a multiple emulsion is obtained, for example, a dispersed phase dispersed within a dispersed phase. The only means of identifying a multiple emulsion is by using the microscope.

Apparatus: beaker, measuring cylinder, glass rod

Materials: oil, egg yolk, salt, mustard powder, distilled water, vinegar

Procedure:

Three manufacturing procedures were investigated in order to illustrate the importance of the manufacturing procedure on emulsion stability.

Ingredients	%	Weight or volume
Oil	75.0	75 g
Egg yolk	8.0	8 g
Salt	1.5	1.5 g
Mustard powder	0.1	0.1 g
Distilled water	1.3	1.3 ml
Vinegar, distilled 5% acetic acid	13.2	13.2 ml
Total	100.0	100 g

In this formula, 30% of the aqueous ingredients equals to 99g water. In all the cases, make a paste of the mustard in a little water before using.

Procedure A

All the ingredients were mixed well in a beaker and stirred by using glass rod for 30 minutes.

Procedure B

1. The mustard powder, salt, water, vinegar, and egg yolk were mixed in a beaker.

2. The oil was slowly added into the mixture while stirring. Stir for 30 minutes.

Procedure C

1. Egg yolk was added to the beaker and blended thoroughly.

2. In a separate beaker, the mustard powder, 1.3g of water, 3.2g of vinegar, and salt were blended.

3. The mixture was stirred until all the salt dissolved.

4. The mixture was added into the egg yolk and was stirred for 5 minutes.

5. At this point, the oil was slowly added while stirring.

a. First 5 minutes, 10-15% of oil was added. The first addition should be small and gradual. Wait about 1 minute between additions.

b. During the next 15 minutes, 50% of oil was added.

c. During last 5 minutes, the remaining of oil was added.

6. Gradually add the remaining vinegar and stir for 5 minutes.

Take the observation on the three different mixtures on viscosity and dilution test.

Results:

Mixture	Procedure A	Procedure B	Procedure C
Viscosity	Low viscous	Moderately viscous	Highly viscous
Dilution test	Water in oil (W/O)	Water in oil (W/O)	Oil in water (O/W)

Suspension-sedimentation

Objective:

1. To obtain a good formulation in controlling the rate of flocculation and sedimentation

2. To study the formation of flocculation

3. To study the relationship between flocculation and sedimentation

Introduction:

In colloid chemistry, a heterogeneous mixture contains the solid particles with a diameter more than 1μm which are large enough to settle out from the liquid through sedimentation is known as suspension. Suspension is a coarse dispersion in which the insoluble particles dispersed in a liquid medium and it will settle out of the liquid after left for some time. Suspension shows difference with a solution in which the solute does not exist in the solid form and the solvent and solute are homogeneously mixed. Suspension is thermodynamically unstable due to the properties of suspension is always changing with time as they will undergo sedimentation. Suspension may be flocculated or deflocculated.

Sedimentation is a process of the particles settle down to the bottom of the solution. This is due to the instability of the particles in the system. The factors that affect the sedimentation are particles size, densities of dispersed phase and dispersed medium, temperature of liquids and the viscosity of the liquid. In a colloidal system, the particles are apart from each other due to their double layer charge. In suspension, the particles also possess the particular surface charges, but the attractive force in between the suspending particles are greater (after flocculating agent was added) than the repulsive force causes the particles to bind together and form larger particles. Usually, the forces acting on the particles that cause sedimentation occurs due to the gravity. The larger particles will sink to the bottom through sedimentation process which will accumulate in the bottom and the particles stack on the sediment. The sedimentation rate can be calculated by using the formula as shown in the following:

The second factor is the densities between the dispersed phase and dispersed medium. Generally, the density of dispersed phase (particles) is greater than the dispersed medium, however in certain cases particle density is less than dispersed phase, so suspended particle floats is difficult to distribute uniformly in the system. If density of the dispersed phase and dispersion medium are equal, the rate of settling becomes zero hence sedimentation does not occur. Temperature and viscosity of liquid are highly in related in which the temperature might affect the viscosity of the liquid. If the liquid is highly viscous, the sedimentation rate of the particles will be slower compared to the less viscous liquid. This is because when the terminal velocity decreases, the dispersed phase will settle at lower rate which may remain dispersed for longer time. The viscosity is known as the resistance of a substance to motion under an applied force.

Flocculation is a process of destabilizing the suspending particles in suspension by adding in certain chemicals called flocculating agent. In industry, this process is purposely used to cause the particles suspended in solution to aggregate into clumps or masses that then sink or else it can be removed easily by filtering. The formation of floc is due to the decrease in Zeta potential (in slipping plane) between the particles until Van der Waals forces predominate. Usually, the larger particles are known as floccules. The floc may float on the surface of liquid instead of sediment at the bottom if the density of floc is less dense than liquid’s density. This is known as creaming. The flocculating agent are includes neutral electrolyte (eg: KCl, NaCl), surfactant, and polymers. When the neutral electrolyte added into the suspension, it will affect the electrical barrier between the particles and hence the Zeta potential could be altered. For surfactant, the cationic and anionic surfactant could bring about the flocculation of suspended particles. However, optimum concentration is necessary because these compounds acting as the wetting agents to achieve dispersion. Surfactant decreases the surface energy by reducing the interfacial tension between the particles which attracted by van der Waals force and the solvent-particles’ interfacial tension. Besides, the addition of polymeric flocculating agents can induce the floc formation. The polymers possess long chain in their structure. The long chain will adsorb on the surface of the particles and the remaining part projecting out into the dispersed medium. The formations of bridging between these later portions tend to cause the formation of floc.

The flocculation process is reversible if the interaction forces between the floc are not strong. Floccules form when weak van der Waals forces are holding the particles together. However, the floccules are easily to re-suspend by shaking. If the floccules trap solvent during sedimentation, they are easily to be broken apart. Otherwise, the flocculation is considered as non reversible if the particles settle into a tighter aggregate by the strong interaction force without trapped solvent. Most of the flocculants are multivalent cation and anion which are able to neutralize the different charges (positive charge or negative charge) of the particles suspended in liquid. After overcome the surface charges of the suspending particles, the particles can easily to form floccules since their barrier have been minimized. Other than that, the deflocculation to form may be occurred once the magnitude of Zeta potential of the particles achieve sufficiently positive or negative (usually more than +30mV or less than -30mV). Thus, the phenomenon of flocculation and deflocculation depends on zeta potential carried by particles.

Apparatus: boiling tube, stopwatch, test tube rack

Materials: potassium dihydrogen phosphate, purified water, bismuth subnitrate, methylcellulose, parafilm

Procedure:

1. 40ml of flocculating reagent (potassium dihydrogen phosphate) was prepared which will be used in four of five suspensions.

2. Five 25ml graduated cylinder were assembled and were numbered consecutively. 2g of bismuth subnitrate was added to each boiling tube.

3. Sufficient water was added to make the 25ml of suspension in the 1^st boiling tube.

4. 10ml of anionic flocculating reagent solution was added into the 2^nd boiling tube and enough water was added to make 25ml.

5. 10ml of anionic flocculating reagent solution was added into the 3^rd boiling tube and 0.5% methylcellulose was added to make 25ml.

6. 10ml of anionic flocculating reagent solution was added into the 4^th boiling tube and 1.0% methylcellulose was added to make 25ml.

7. 10ml of purified water was added into the 5^th boiling tube and sufficient 0.5% methylcellulose was added to make 25ml.

8. All the boiling tubes were covered with parafilm.

9. Each of them was inverted for several time to mix tehm well. The height of suspension were measured and recorded after 15minutes, 60minutes, and 120minutes.

Results:

Table 1 The height of sedimentation in each boiling tube over the time

Number of boiling tubes	Mixture in the boiling tubes	0min	15mins	60mins	120mins
First	2g bismuth nitrate + water	6.5cm	0.3cm	0.3cm	0.5cm
Second	2g of bismuth subnitrate+10ml of flocculating agent+water	6.5cm	1.2cm	1.5cm	1.6cm
Third	2g of bismuth subnitrate+10ml of flocculating agent+0.5% methylcellulose solution	6.5cm	2.0cm	1.2cm	1.6cm
Fourth	2g of bismuth subnitrate+10ml of flocculating agent+1% methylcellulose solution	6.5cm	1.4cm	1.5cm	1.3cm
Fifth	2g of bismuth subnitrate+10ml of purified water+0.5% methylcellulose solution	6.5cm	0.0cm	0.0cm	0.0cm

Table 2 The height of suspension in each boiling tube over the time

Number of boiling tubes	Mixture in the boiling tubes	0min	15mins	60mins	120mins
First	2g bismuth nitrate + water	0.0cm	6.2cm	6.2cm	6.0cm
Second	2g of bismuth subnitrate+10ml of flocculating agent+water	0.0cm	5.3cm	5.0cm	4.9cm
Third	2g of bismuth subnitrate+10ml of flocculating agent+0.5% methylcellulose solution	0.0cm	4.5cm	5.3cm	4.9cm
Fourth	2g of bismuth subnitrate+10ml of flocculating agent+1% methylcellulose solution	0.0cm	5.1cm	5.0cm	5.2cm
Fifth	2g of bismuth subnitrate+10ml of purified water+0.5% methylcellulose solution	0.0cm	6.5cm	6.5cm	6.5cm

Precaution steps:

1. Do not shake or move the boiling tubes consist of suspension.

2. Make sure the content in each boiling tube is mixed well.

3. Make sure all the boiling tube are properly covered by parafilm

Verification of the Henderson-Hasseblach Equation Using Buffer Solution

Objectives:

1. To study the validity of Henderson-Hasseblach equation to a buffer system

2. To determine the buffer capacity of a buffer solution

Introduction:

Buffers are solutions that maintain a relatively constant pH when an acid or a base is added into a buffer solution. They protect other molecules in solution from the effects of the added acid or base. Buffer solution contains either a weak acid (HA) and its conjugate base (A^-) or a weak base (B) and its conjugate acid (BH⁺). They are extremely important for the proper functioning of biological systems such as blood, which is a natural example of buffer solution.

An acid is defined as a substance that can dissociate in water to produce hydrogen ions, H⁺ while a base is a substance that can dissolve in water to produce hydroxide ions, OH^-. The acid is categorized into two groups which are strong acid and weak acid. Strong acid is the acid that dissolves completely in water to produce a lot of H⁺ ions whereas weak acid is the acid that ionizes partially in water to produce less H⁺ ions.

An important characteristic of weak acids and weak bases is their ability to form buffer systems. The action of a buffer is caused by the presence of both the weak acid and its anion or the weak base and its cation. These are known as conjugate acid-base pair. If strong base is added into a weak base, it will dissociate into OH^- and conjugate acid. The conjugate acid produced will neutralize the base added and hence control it pH. On the other hand, if string acid is added, the weak base will react with the strong acid by neutralizing it. So, the pH of the solution can be maintained. This is same applied by the acid buffer system with H⁺ ion and its conjugate base.

Consider a sodium acetate-acetic acid buffer system in which containing the same amounts of acetic acid (CH₃COOH) and its conjugate base, acetate ion (CH₃COO^-). When the addition of a strong acid hydronium ion (H₃O⁺) to the solution. The acetate ion (base form) in the buffer reacts with the hydornium ion to neutralize it.

In the same way, the addition of the strong base provide hydroxide ion, OH^- to the solution, then the acetic acid will reacts with this ion and neutralize it.

The amount of strong acid or base can be added to a given volume of a buffer system without a significant change in pH in +/- 1.0 unit. This is known as the buffer capacity of a buffer system.

The strength of acid is depends on the degree of H⁺ ionization in the water. For base, its strength is depends on the degree of OH^- ionization in the water.

The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. The equilibrium constant for acid dissociation is

The equilibrium constant for base dissociation is

In order to obtain Henderson Hasselbalch equation, rearrange the following equation:

Buffer capacity of a buffered solution is defined as in terms of the amount of protons or hydroxide ions it can absorb without a significant change in buffer. A buffer with a large capacity contains large concentration of the buffering components and thus can absorb a relatively large amount of protons or hydroxide ions and show little pH change. The pH of a buffer system is determined by the ratio [A^-]/[HA]. The capacity of a buffer system is determined by the magnitude of [A^-] and [HA]. Buffer capacity is also depends on the concentration of the solution. The greater the concentration of a weak acid and its conjugate base, the greater the buffer capacity. Buffer capacity may expressed as

A buffer solution has the maximum buffer capacity when the ratio or = 1 for acidic buffer and basic buffer respectively. In such a solution pH = pK_a for acidic buffers and pOH= pK_b for basic buffers. A solution can act as buffer only if the ratio of salt to acid or base is between 0.1 and 10. Thus, the pH of an acidic buffer can have the range from pK_a - 1 or pK_a + 1. Similarly, a basic buffer can act as buffer in the pOH range from pK_b – 1 to pK_b + 1.

Materials: standard buffer solution (pH= 4.00), 1 M aqueous sodium hydroxide, 1 M aqueous acetic acid

Apparatus: burettes, beaker (100cm³), pH meter

Procedure:

1. The pH meter is calibrated by standard buffer solution, pH=4.00.

2. 50cm³ of the 1 M acetic acid and 1 M sodium hydroxide are pipetted into a 100cm³ beaker.

3. The solution is mixed thoroughly and pH of the solution is measured.

4. The pH measurement is repeated for further 1 cm³ addition each up to a total of 5cm³ of 1 M sodium hydroxide.

5. The pH of the solution is measured for further 5cm³ additions each up to a total of 45cm³ of 1 M sodium hydroxide.

6. The pH measurement is repeated for further 1 cm³ addition each up to a total of 50cm³ of 1 M sodium hydroxide.

Result and calculation:

Table 1 The average pH value of buffer solution

Volume of aqueous 1 M CH3COOH (X) ,cm3	Volume of aqueous 1 M NaOH (Y), cm3	pH	[ Salt ] / [ Acid ] = y/(x-y)	Log10 ( [ Salt ] / [ Acid ]
50	1	2.97	0.0204	-1.690
50	2	3.26	0.0417	-1.380
50	3	3.45	0.0638	-1.195
50	4	3.58	0.0870	-1.060
50	5	3.69	0.1111	-0.954
50	10	4.06	0.2500	-0.602
50	15	4.32	0.4286	-0.368
50	20	4.54	0.6667	-0.176
50	25	4.74	1.0000	0.000
50	30	4.95	1.5000	0.176
50	35	5.21	2.3333	0.368
50	40	5.65	4.0000	0.602
50	45	10.03	9.0000	0.954
50	46	11.76	11.5000	1.061
50	47	12.03	15.6667	1.195
50	48	12.19	24.0000	1.380
50	49	12.29	49.0000	1.690
50	50	12.37	-	-

Turning point at pH

= (5.65+10.03)/ 2

= 7.84

Synthesis of Metal Acetylacetonates

Objectives:
1. To prepare tris(acetylacetonato)manganese(III), Mn(acac)₃ {also called tris(2,4-pentanedionato)manganese(III)}
2. To identify and draw the structure of Mn(acac)₃
3. To find out the magnetic moment of Mn(acac)₃
Intorduction:
A coordination complex or metal complex is a structure which consisting of a central transition metal is bonded to one or more as ligands (neutral or ionic). Ligand is defined as the molecule or ion that is attached directly to the central metal atom or ion through dative covalent bond which both bonding electrons are contributed by ligand. The metal is Lewis acid which is being an electron pair acceptor while the same or different types of ligands are being the Lewis bases which donate their electron pair to central metal. The bond between metal and a ligand is a Lewis acid-base interaction.
When a ligand is bounded to a central metal through a single donor atom, then the ligand is known as unidentate (the ligand binds to the metal through a single point of attachment as it has one tooth). For a bidentate ligand, it can use two atoms donor to bind to a metal. Larger ligands may contain more than one capable of coordinating bonds which bonded to a single central metal. These kind of ligands are known as polydentate. When a bi- or polydentate ligand uses two or more donor atoms to bind to a single metal ion, it is said to form a chelate complex. Such complexes tend to be more stable than similar complexes containing unidentate ligands due to chelating effect.
The coordination number of a metal complex is determined directly from the number of ligand that can be accepted by a transition metal. In order to obtain the oxidation state of the metal in the complex to be synthesized, this may need prior oxidation or reduction of the starting material. The metal salt, manganese(II) chloride tetrahydrate is the starting material in this experiment. In the presence of a base, the acetylacetone is readily to lose a proton to form the acetylacetone anion, acac^- through monodeprotonation of acetylacetone. Hydrogen atoms on the carbon atoms that are adjacent to carbonyl (C=O) groups are relatively acidic. In the isomerism, the acetylacetonate ion has three resonance forms. This ligand is a bidentate which used its two oxygen atoms to coordinate by donating lone pair electrons towards central metal. The diagram 1 below shows the structure of acetylacetonate ion.

Diagram 1 Structure of acetylacetonate ion

In this experiment, the manganese complex is a good example of octahedral complex which containing three bidentate ligands. The metal ion is electrically neutral because it is carrying a 3+ charge and each ligand carried -1 charge. The complex is more soluble in organic solvent than water due to its neutral properties. Manganese exhibits an unusually wide range of oxidation state which includes +7 to -1, but in aqueous solution the +2 oxidation state is the most common. The three bidentate ligands are pack very efficiently around the trivalent ions of transition metal. As a result, the Mn²⁺ can be easily to be oxidized to form Mn(acac)₃ in the presence of acetylacetonate ions. Oxidizing agent (potassium permanganate solution) is used to oxidize the four equivalent of Mn²⁺ to become Mn³⁺ as shown in the equation below:

MnO₄^- + 4Mn²⁺ + 8H⁺ à 5Mn³⁺ +4H₂O

The formation of Mn³⁺ ions are used to react with acetylacetonate ions to produce Mn(acac)₃. At the same time, the deprotonation of acetylacetone will produce a large amount of H⁺. In order to maintain the acidity of the solution, the sodium acetate is added to neutralize the acid released which proton reacts with acetate ion to form weak acetic acid.

7H⁺ + 7CH₃COO^- à7CH₃COOH

Transition metals have variable oxidation state. In their complexes, the number of outer valence electrons varies with the oxidation state. Since the electrons spin and generate a magnetic field, the magnetic properties of transition metals in their complexes are of great interest in determining the oxidation state, electron configuration and so on. The magnetic moment of the metal is determined indirectly from the magnetic susceptibility. The unit measurement for magnetic moment is Bohr magneton, μB.
Apparatus:
Conical flask
Measuring cylinder
Glass rod
Hot-plate stirrer and magnetic bar
Thermometer
Beaker
Filter funnel
Suction filter
Materials:
Manganese(II)chloride tetrahydrate
Sodium acetate trihydrate
Acetylacetone
Potassium permanganate solution
Acetone
Procedure:
1) 5g (0.025 mol) of manganese(II) chloride and 1.3g (0.0095 mol) sodium acetate trihydrate were dissolved in 200ml of distilled water. The stirred solution was added slowly with 21cm³ (~0.20mol) of acetylacetone(check whether its purity is 99%, density 0.975g cm^-1; record the actual purity and density of the acetylacetone).
2) The resultant two phase system was treated with potassium permanganate solution (1g in 50cm³).
3) After a few minutes, small amounts of sodium acetate solution were added with stirring (13g or 0.095 mol of sodium acetate trihydrate).
4) The solution is heated to about 60°C with continuous stirring for 30 minutes.
5) The resultant solution was cooled in the ice-cold water and then the solid complex formed was filter by suction filtration. The complex was washed with acetone and was sucked until dry.
Result and calculation:
Weight of manganese(II) chloride tetrahydrate, MnCl₂.4H₂O = 5.0024g
Weight of first protion of sodium acetate trihydrate, Na(acac)₃ = 1.3038g
Weight of potassium permanganate, KMnO₄ = 1.0091g
Weight of second portion of sodium acetate tirhydrate, Na(acac)₃ in 50cm³ distilled water = 13.0120g
Weight of filter paper = 0.7962g
Weight of filter paper + weight of complex = 4.9926g
Weight of complex = 4.1964g
Purity of acetylacetone = 100%
Density of acetylacetone = 0.975g/cm³
Mass of acetylacetone = density x volume used
= 0.975g/cm³ x 21cm³
= 20.475g
Molecular mass of acetylacetone = [(12.01x5) + (16x2) + (1.008x8)] g/mol
= 100.11 g/mol
Number of mole of acetylacetone used = Mass/ Molecular mass
= 20.475g / 100.11g mol^-1
= 0.2045 mol
Molecular mass of manganese(II) chloride tetrahydrate
= [54.94 + 35.45(2) + 4 (18.016)] g/mol = 197.90 g/mol
Number of mole of manganese(II)chloride tetrahydrate = 5.0024g/ 197.90 g mol^-1
= 0.0253 mol
Mass of Mn(acac)₃ = 4.1964g
Molecular mass of Mn(acac)₃
= [54.94 + 3(12.01x5 + 16x2 + 1.008x7)] g/mol= 352.258 g/mol
Number of mole of Mn(acac)₃ = 4.1964g/ 352.258g mol^-1
= 0.0119 mol
No of mole of KMnO₄ = 1.0091g/ 158.0339 g mol^-1
= 0.0064 mol
4Mn²⁺ + MnO₄^- + 15Hacac à 4H₂O + 7H⁺ + 5Mn(acac)₃
Theoretical mass of Mn(acac)₃ = 5 x 0.0064 mol x 352.2617g/mol
= 11.2724g
Percentage yield of Mn(acac)₃ = 4.1964g/11.2724g x 100%
= 37.23%

Polymers Identification

Objectives:

1. To identity the types of polymer by using water

2. To identity the types of polymer by using copper wire

3. To identity the types of polymer by using alcohol

4. To identity the types of polymer by using acetone

5. To identity the types of polymer by using oil

6. To identity the types of polymer by using heat

Introduction:

Polymers are long chain organic molecules that are assembled from many identical and smaller molecules called monomers. Polymerization of is the process to assemble all the monomers together in order to form a huge and complicated molecule.

 

Monomer of high density polyethylene

 

Monomer of polyvinyl chloride

 

Polymer is containing many units of monomer in its long chain. When there is a lot of monomer joining together chemically, polymer is formed, which shown in the picture below:

Polymer of high density polyethylene

 

Generally, we can use the symbol below to show the structure of polymer instead of drawing the whole structure of polymer. Besides, we are not able to draw the whole structure because the number of monomer is usually over hundreds or even thousands of units.

The above picture refers to the simple structure in which it shows the whole structure of polymer or we named it as molecular formula as well.

The polymers shown in the picture above are the most common polymer that we always see and use in daily life. These includes polyethylene terephthalate (PETE), high density polyethylene (HDPE), polyvinyl chloride (PVC), low density polyethylene (LDPE), polypropylene (PP) and polystyrene (PS). These polymers are classified as class 1, followed by class 2, class 3 and so on. Identification of polymer by simply observing its appearance is difficult to differentiate them. However, we can identify them by using water, copper wire, alcohol, acetone, oil and heat.

Lassaigne's Test or Sodium Fusion test is also can be used to identify some of the polymers. For example, the presence of nitrogen, N in polyamides, and the presence of chlorine, Cl in PVC can be identified by using sodium fusion test. Besides, IR also can be used to identify the type of polymer via their major functional groups. The carbonyl stretches, O-H stretches, aromatic bends, etc could clearly show the identity of a polymer. Despite these test can be applied in identifying polymers, but today we only focus on the simplest test.

In this experiment, we will use some known plastic material and carry out some tests to identify the types of polymer. The flow chart below shows the overall tests on polymer identification.

We will use another three unknown sample of plastic to perform the test on them in order to identify the unknown samples.

Material:

Samples of polyethylene terephthalate (PETE), high density polyethylene (HDPE), polyvinyl chloride (PVC), low density polyethylene (LDPE), polypropylene (PP) and polystyrene (PS), three unknown samples of plastic, isopropyl alcohol, corn oil, copper wire, acetone

Apparatus:

Test tube, test tube rack, stopper, Bunsen burner, beaker, forceps, gauze wire

Procedure:

A) Water test

1. Pour 10ml of water into a test tube.

2. Place one of the samples into the water and stir the water by using a glass rod.

3. Observe whether the sample sink or float on the water surface.

4. Remove the sample from water and dry it for later use.

5. Repeat steps 1 to 4 by using each of the samples.

6. Take the samples that sank in the water for copper wire test while save the floated sample for alcohol test.

B) Copper wire test

1. Obtain a piece of copper wire about 6cm long and insert it into a small cork.

2. Burn the copper wire by using a Bunsen burner and heat until it is red hot.

3. Remove the copper wire from flame and touch on the sample of plastic. A small amount of sample should melt.

4. Place the wire copper with sample into the flame. A luminous flame is observed.

5. Repeat steps 1 to 4 by using the samples which sank in the water.

C) Acetone test

1. Prepare 20ml of acetone into a beaker under the fume hood.

2. By using a forceps, place the sample of plastic into the acetone for 15 seconds.

3. Remove the sample and press it firmly.

4. Scrape the sample to observe if the outer layer has softened.

5. Repeat steps 1 to 4 on the sample that gave yellow-orange flame.

D) Heat test

1. Prepare 100ml of water in beaker and heat it until boil.

2. By using a forceps, place the sample of plastic into the acetone for 15 seconds.

3. Remove the sample and press it firmly.

4. Scrape the sample to observe if the outer layer has softened.

5. Repeat steps 1 to 4 on the sample that shown negative result in acetone test.

E) Isopropyl alcohol test

1. Pour 10ml of isopropyl alcohol into a test tube.

2. Place one of the samples samples that floated on the water surface into the test tube and stir the water.

3. Observe whether the sample sink or float on the surface.

4. Remove the sample from water and dry it for later use.

5. Repeat steps 1 to 4 by using the samples that floated on the alcohol surface.

F) Oil test

1. Pour 10ml of oil into a test tube.

2. Place samples that floated on the alcohol surface into the test tube and stir the sample.

3. Observe whether the sample sink or float on the surface.

4. Repeat steps 1 to 4 on the unknown samples

Precaution steps:

1. Carry out the test under a fume hood.

2. Avoid the isopropyl alcohol and acetone from any sources of flame.

3. Hold the wire copper by using a forceps after heated.

4. Equip with personal protective equipment such as glove and mask.

Qualitative Analysis of Organic Compounds (Sodium fusion Test or Lassaigne Test)

Objective:

1. To carry out Lassaigne test in order to determine the elements (N,S halogens) present in the unknowns.

2. To identify the elements present in compounds and their colouration.

Introduction:

Qualitative analysis is always applied as a first step in identifying a compound when a new compound is readily prepared or isolated from some natural source. In an organic compound, elements carbon, hydrogen and oxygen are assumed to be present commonly. Nitrogen, sulphur and halogens (chlorine, bromine and iodine) may also present in the organic compound. The identification of elements in a given compound is a type of qualitative analysis since the experiment is dealing with the composition of a unknown compound. This experiment must be handled very carefully as further the analysis of the organic compound is according to the element present in it. Generally, the traditional technique is only can be applied to inorganic ions in aqueous solution.

When a new compound contains covalent bonding instead of ions, its molecule can be broken up into ions by controlled decomposition of the compound. The reagent used which cause the decomposition of the original unknown compound into the ions produced by the decomposition will reflect clearly those elements that were present initially in the compound. In this experiment, sodium fusion test (Lassaigne’s test) is used in elemental analysis of qualitative determination of elemental halogens, sulphur and nitrogen in a sample. Sodium is a very strong reducing agent that will able to break up the organic compounds carbon atom chain. It also will convert the atoms which are covalently bonded to the carbon chain to inorganic ions. The elements are detected by sodium fusion test. The organic compound is fused with metallic sodium to convert these elements into ionic mixture which dissolved in water and the filtrate is used to perform the tests.

The organic compound undergoes sodium fusion test which the carbon present in the particular compound is reduced partially to elemental carbon. The nitrogen present in the compound is reduced to cyanide ion, CN^- while the sulphur present is converted into the sulfide ion, S^2-. Any halogens (Cl, Br, I) that are present in the compound are reduced to the halides ions, Cl^-, Br^- and I^-. A precipitate of an iron/cyanide complex with the characteristic of dark blue colour (Prussian blue) will be formed if cyanide ion is present in the compound. If sodium is not heated well, very little amount of cyanide ion is produced during fusion test. So, a greenish solid may result. It is always helpful to repeat the sodium fusion test in order to confirm the presence of nitrogen if only the greenish solid is being produced.

Several techniques can be used to test the presence of sulfide ion in the unknown organic compounds. Hydrogen sulfide gas will be produced if the filtrate obtained from sodium fusion products is acidified. This can be noted by the formation of lead(II) sulphate precipitate is formed after the lead(II) ions introduced into the solution.

Besides, the sulfide ions can be tested by adding with sodium nitroprusside reagent. The formation of purple or violet colour shows the presence of sulfide ion. Silver ion reagent can be used to confirm the presence of halide ions in the compound. The formation of precipitate between halide ion and silver ion as shows below:

For the compound consists of more than one halogen, the mixed precipitate of silver halides can be observed as a result. However, it may be necessary to remove the cyanide ion and sulphide ion if they were present. Otherwise, these ions will precipitate with the silver ions to form precipitate which will mislead to the presence of halogens in the compound.

Apparatus:

Fusion tubes

Test tube

Evaporating dish

Gauze

Tongs

Materials

Sodium metal

Unknown organic compounds A,B,C

0.5M ferrous sulfate

0.5M ferric chloride

3M sulfuric acid

3M acetic acid

3M nitric acid

Sodium nitropusside

Lead acetate

0.1M silver nitrate

Procedure:

Part A: Sodium Fusion Test (Lassaigne)

1. 2-3 drops of unknown substance (for liquid) or 0.1g (for solid) was put into the fusion tube.

2. A piece of sodium was placed into the tube.

3. The tube was hold with a pairs of tong. The heating was gently to avoid spurting out of the sodium. When the sodium was molten, the compound was heated strongly until the end of the tube was red hot (continue heating for 1 minute).

4. The hot tube was plunged into 20ml of distilled water in a evaporating dish and covered with a gauze.

5. The tube was crushed with a pestle.

6. The mixture was filtered into a clean test tube. The colourless and clear filtrate was readily to be examined for the presence of nitrogen, sulphur, and halogens.

Part B: Test for nitrogen (Cyanide test)

1. To 1ml of filtrate, few drops of 0.5M ferrous sulfate was added and a dark green precipitate of iron(II) hydroxide is formed.

2. If sulphur is also present, precipitate will be black. More ferrous sulfate solution was added dropwise to precipitate all the sulphide ion until no more black precipitation is formed.

3. The mixture was heated to boiling with shaking, cooled and acidified with 3M sulfuric acid. If nitrogen is present, a blue colour (Prussian blue) appears immediately on addition of a trace of 0.5M ferric chloride solution.

Part C: Test for sulphur (sulphur test)

1. To 1ml of filtrate, a few drops of dilute solution of sodium nitroprusside were added. If sulphur ion is present, a deep purple colour will appear.

2. Alternatively, sulphide ion may be detected by precipitation as black lead(II) sulphide with lead(II) acetate solution which has been acidified by 3M acetic acid.

3. If halide is also present, white or yellow lead(II) sulphide precipitate may also formed.

Test for halogens

1. 1ml of filtrate was acidified with a few drops of 3M dilute nitric acid.

2. Silver nitrate solution was then added into the solution. Halides are indicated by the formation of a white or yellow precipitate.

Part E

The above procedure was repeated for unknown B and C. For each test, the observations were written down and the elements in the unknown were deduced.

Results:

Table 1 Observations of the unknown compounds on nitrogen test

Unknown compound	Observation on nitrogen test	Inference
A	i) Some dark green precipitate and a little amount of dark precipitate were formed after brownish orange FeSO4 is added. ii) The precipitate become greenish after more FeSO4 is added. iii) The colour of precipitate remained unchanged after heating. iv) The precipitate does not change after sulphuric acid is added. v) Blue precipitate is formed immediately after adding of FeCl3	Nitrogen is present.
B	i) The solution turns to yellow after brownish orange FeSO4 is added. ii) Brownish orange precipitate is formed after heated. iii) Colourless solution is formed after H2SO4 is added. iv) No change after FeCl3 is added.	Nitrogen is absent.
C	i) The solution turns to yellow after brownish orange FeSO4 is added. ii) Brownish orange precipitate is formed after heated. iii) Colourless solution is formed after H2SO4 is added. iv) No change after FeCl3 is added.	Nitrogen is absent.

Table 2 Observations of the unknown compounds on sulphur test

Unknown compound	Observation on sulphur test	Inference
A	a) Sodium nitroprusside i) A deep purple solution is formed immediately. b) Acidified lead(II) acetate i) A black precipitate is formed.	Sulphur is present.
B	a) Sodium nitroprusside i) A pale yellow solution is formed. b) Acidified lead(II) acetate i) A white precipitate is formed.	Sulphur is absent.
C	a) Sodium nitroprusside i) A pale yellow solution is formed. b)Acidified lead(II) acetate i) A white precipitate is formed.	Sulphur is absent.

Table 3 Observations of the unknown compounds on halogens test

Unknown compound	Observation on halogens test	Inference
A	Formation of white precipitate after nitric acid and silver nitrate are added.	Halogen is present.
B	Formation of white precipitate after nitric acid and silver nitrate are added.	Halogen is present.
C	Formation of yellow precipitate after nitric acid and silver nitrate are added.	Halogen is present.

Discussion:

The organic compounds to be analyzed consist of basically of a chain of carbon atoms which various other atoms are attached. Since these elements are covalently bonded to the carbon chain, it is unable to dissolve in water to form cations and anions. However, sodium fusion test can be used to reduce those atoms that are covalently bonded to the carbon chain to inorganic soluble ions since sodium is a very strong reducing agent. In the Lassaigne’s test, the nitrogen can be reduced to form cyanide ions, CN^-:

For sulphur, it had been reduced to form sulfide ion, S^2- in Lassaigne’s test as shown in the following:

If both nitrogen and sulphur are present in the organic compound at the same time, then the chemical reaction below will take place in the test:

If halogens (Cl, Br, I) are present in the compound, the halogens will be reduced to form halide ions (Cl^-, Br^-, I^-) during the sodium fusion test.

The inorganic ions in aqueous solution could be easily observed after undergo certain tests which can indicates the presence of elements in the particular compounds.

In the cyanide test, the filtrate of compound A was added with ferrous sulfate, a dark green precipitate was formed. The formation of ferrous hydroxide was produced from the reaction between ferrous sulfate and sodium hydroxide.

The sodium hydroxide was formed by the reaction of unreacted sodium metal with water due to incomplete reaction of sodium fusion with compound A.

The FeSO₄ solution was added to confirm the presence of NaOH and to react completely with it in the filtrate. At the same time, a small amount of black precipitate was formed at the bottom but it was disappeared after more ferrous sulphate was added. The formation of black precipitate may be due to the ferrous sulphide exists in the mixture.

The equation below shows that the ferrous sulphate was reacted with the sodium cyanide to form sodium ferrocyanide as the main product.

The sulphuric acid and increase in temperature was used to increase the suitable medium for the formation of complex. As a result, ferric-ferrocyanide complex with the colour of Prussian blue was precipitated out after ferric chloride is added to oxidize the Fe²⁺ to become Fe³⁺. This Prussian blue precipitate indicates that the unknown A contains nitrogen in the compound.

Some of the Fe³⁺ was formed before the oxidation of ferric chloride. This might be due to the air oxidation of iron(II) ions in the mixture before the ferric chloride is added. For compounds B and C, a negative result is obtained which end up with colourless solution as results. Hence, these shown nitrogen are absent in the both organic compounds.

The reduced sulfide ion can be confirmed by using two different tests which were sodium nitroprusside test and lead(II) acetate test. For the first test, the appearance of deep purple solution shows the positive result. The formation of sodium sulphonitroprusside is a complex that was formed between the sodium nitroprusside and sodium sulphide.

In another test, the black precipitate will be formed if the sulphur is present in the compound. The formation of black precipitate shows a positive result for this test.

The compounds B and C did not consisting sulphur in their structure because they cannot give the positive result on both tests. A pale yellow solution was formed after sodium nitroprusside was introduced and the solution shows the colour of sodium nitroprusside. In the latter test, white precipitate was formed may be due to the precipitation of the lead (II) ions with the halide ions. Based on the observation, the organic compound A containing sulphur while compound B and C did not containing sulphur.

In the halogens test, if white or yellow precipitation takes place after silver nitrate was added into the filtrate from compound A, B and, C respectively is known as the positive result. If cyanide ion or sulfide ion present in the compound, the acidified solution must be heated until boiled in order to expel the hydrogen cyanide gas and hydrogen sulfide gas. This step was taken to avoid the cyanide ions and sulfide ions cause the error in the halogens test. The sodium halide formed during the sodium fusion test was reacted with the silver nitrate to form the insoluble silver halide as precipitate in the solution.

The white precipitate was formed in the filtrate from compound A while yellow precipitate were formed in the both filtrate from the compound B and C. This is mean that the halogen presents in the compound A probably is chlorine whereas the halogens exist in the compound B and C were possibly bromine or iodine.

Precaution steps:

1. Safety glasses should be worn in the laboratory at all the time.

2. Do not add too much of water into the sodium fusion products unless if has been established with certainly that all elemental sodium has been destroyed.

3. The fusion of unknown compound is carried out in the fume hood.

4. When heating the liquids, use a small flame instead of large and move the test tube quickly through the flame.

5. The heating process of unknown compound must carried out in fume hood.

6. Avoid the corrosive sulfuric, nitric and acetic acids touch on skin because it can be dangerous to skin.

The Use of Volumetric Flask, Burette and Pipette in Determining the Concentration of NaOH Solution

Objectives:

1. To carry out titration of a strong acid (HCl) with a strong base (NaOH).

2. To determine the end point of titration with the use of phenolphthalein as indicator.

3. To determine the concentration of base when the concentration of acid is known by doing calculations related to titration.

Introduction:

The main purpose of this experiment is to determine the unknown concentration of a known reactant. Volumetric analysis is a method of quantitative chemical analysis that always used in titration. The technique is carried out by using a reagent of a known concentration (standard solution) and volume to react with a solution where the concentration is unknown. There are various types of titration carried out for different purposes such as acid-base titration and redox titration. Within the acid-base titration, there are four types of reactions:

(1) titration involving a strong acid and a strong base

(2) titration involving a weak acid and a strong base

(3) titration involving a strong acid and a weak base

(4) titration involving a weak acid and a weak base

The titration carried out is the acid-base titration which is based on the neutralization reaction that occurs between an acid and a base to produce salt and water. The base is added slowly into the conical flask with acid until there are all exactly reacted. This is called the end point or also known as the equivalence point of the reaction.

The equivalence point of the neutralization reaction is the point at which both acid and base have been consumed and neither is in excess. In other words, the hydrogen ion and hydroxide ion concentrations are equal in the solution. In order to determine the equivalence point in a titration, acid-base indicators need to be added to the acid solution before the titration start. The end point of a titration is indicated when the indicator changes color. An indicator is a weak organic acid or base that has distinctly different colors in its non-ionized and ionized forms. This will be discussed further in discussion section. Different indicators show different colour changes at the same pH, therefore choosing of indicator for a particular titration depends on the acid and base used.

Hydrochloric acid (HCl) and sodium hydroxide (NaOH) is used as the reactants in this experiment. HCl is a monoprotic acid which dissociate to give out one H⁺ ion. Monoprotic acids have acid dissociation constant, K_a, which indicates the level of dissociation in water. Hydrochloric acid has large K_a value as it is a strong acid and dissociate completely in water. NaOH is a metallic base and ionic which composed of sodium cation and hydroxide anion. The hydroxide ion makes sodium hydroxide a strong base which reacts with acids to form water and corresponding salts.

Procedure:

1. Volumetric flask was cleaned and rinsed with distilled water.
2. 25ml of NaOH solution was transferred into the volumetric flask.
3. The volumetric flask was topped up to 250ml with distilled water and rotated several times.
4. Burette was cleaned with distilled water and rinsed with 5ml NaOH solution few times and was filled.
5. Pipette was cleaned and rinsed a few times with HCl.
6. 25ml of HCl solution was pipetted into 3 conical flasks.
7. 3 drops of phenolphthalein was added into HCl solution.
8. The initial and final readings of burette were recorded.

 

Result and calculations:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Discussion:

The titration of a strong acid and strong base in this experiment can be illustrated with a graph called a titration curve. It is a graph of pH versus volume of the solution titrated. The figure below represents the pH versus volume data of the titration curve for the HCl-NaOH titration. From the graph we may explain the chemical changes happen during titration and decide which indicators best to be used to determine the endpoint which matches the equivalence point of the neutralization.

Based on the graph, the pH has a low value at the beginning of the titration which shows the concentration of the HCl in conical flask. As the titration proceeds, the pH changes slowly until it reaches just before the equivalence point. At the equivalence point, the pH rises sharply by just adding only two drops of base. This is because when the equivalence point is reached, the number of moles of hydrogen ions and hydroxide ions is equal to each other; therefore a slight addition of base will result in a steep increase of pH. Beyond the equivalence point, the pH again rises only slowly. According to the graph, any acid-base indicator whose color changes in the pH range from about 4.0 to 10.0 is suitable for HCl-NaOH titration.

The acid-base indicator that is used for the titration of HCl-NaOH is phenolphthalein which is situated within the pH range of 8.3 to 10.0. Other indicators only have pH range within 1.0 to 8.8 which does not include the pH range beyond 9.0 as phenolphthalein where the pH range is until 10.0. Based on the interpretation of the graph, indicator whose color changes in the pH range from 4.0 to 10.0 is only suitable for the HCl-NaOH titration. Therefore phenolphthalein is chosen rather than the other indicators.

As we known an indicator is usually a weak organic acid or base that has distinctly different colour in its non-ionized and ionized forms, but what are the characteristics of an indicator that make let us determine the endpoint of a titration by showing different colours. Acid-base indicators exist in two forms, a weak acid represented as HIn and having one colour and its conjugate base represented as In- and having a different colour. The indicator does not affect the pH of the solution if only just a small amount of indicator is added to a solution. However, the ionization equilibrium of the indicator is affected by the concentration of H₃O⁺ in the solution. When in a solution, the acid ionizes to the following ions:

Based on the Le Châtelier’s principle, increasing [H₃O+] in the solution shifts the equilibrium to the left, increasing the amount of HIn and thus showing the acid colour. On the other hand, decrease in [H₃O+] in a solution shifts the equilibrium to the right, increasing the amount of In^- and hence displaying the base colour. In general, if 90% or more of an indicator is in the form HIn, the solution will show the acid colour. If 90% or more is in the form In^-, the solution takes on the base colour. If the concentrations of HIn and In^- are about equal, the indicator is in the process of changing from one form to the other and has an intermediate colour which is the mixture of acid and base colour.

Based on the results obtained, within the three titration carried out only Titration 3 is less than 3 and within the range. However, Titration 1 and 2 is more than 3 and is out of the range. The causes of the results to be out of range can be due to human errors. First of all, the NaOH solution could have been diluted as the burette used to fill the NaOH solution is rinse with distilled water and not with NaOH before use. This causes the concentration of NaOH to be slightly different from the standard solution that has been prepared. The same possibility does happen to the conical flask which is used to fill the HCl which will be titrated against NaOH where the flask only rinse with distilled water not with HCl. Thus, the concentration of HCl may be less than 0.01M. Besides that, the reading on the burette could have some minor error because during the recording of readings the meniscus shown on the burette is not clearly view. In order to correct the error, a white paper should be situated behind the burette in order to have a clearly view on the position of the meniscus so that a more accurate readings can be obtained.

Precaution steps:

Firstly, is the determination of the titration end point which is based on the colour changing of the indicator added to the conical flask. Confusion arises about when to stop the titration as the colour changes is difficult to be observed. Therefore, a white tile or a piece of white paper should be placed at the bottom of the conical flask so that the changes of colour can be easily seen. Next, NaOH solution will react quickly with the carbon dioxide in the air. Therefore NaOH should be cover when it is not in use. This is the reason the prepared standard solution of NaOH is closed with the cap. Lastly is the dilution or the preparation of the standard solution of NaOH. The standard solution is prepared in a 250mL volumetric flask by adding 5mL of NaOH and distilled water should be added to the graduated line in the flask. During the adding of distilled water, water could have overshoot the line and cause the concentration of the standard solution to be different from the expected concentration. Thus, use dropper to fill the water into flask when the meniscus level approaches to the graduated line of the flask to avoid the overshooting of distilled water during the preparation of standard solution.