"Preparation of iron oxide nanoparticles (NPs)"

 

Preparation of iron oxide nanoparticles (NPs)

Theory:

The co-precipitation method involves the precipitation of iron hydroxide from iron salts in the presence of a base. In this case, the FeCl₃ and FeSO₄ salts are used to provide Fe⁺² ions for the reaction. By gradually adding NH₄OH as the base, the pH of the solution increases, leading to the formation of iron hydroxide. The iron hydroxide precipitate can then be further processed to obtain iron oxide nanoparticles.

Chemicals Required:

• Iron (III) chloride (FeCl₃) solution, 0.2 M
• Iron (II) sulfate (FeSO₄) solution, 0.1 M
• Ammonium hydroxide (NH₄OH) solution
• Solvent: Water or appropriate organic solvent

Materials Required:

1. Reaction vessel or flask
2. Magnetic stirrer or mechanical stirrer
3. pH meter
4. Centrifuge
5. Drying equipment (e.g., vacuum oven or desiccator)

Chemical Equation:

Procedure:

• Prepare a reaction vessel and add the FeCl₃ solution (50 ml, 0.2 M) and the FeSO₄ solution (50 ml, 0.1 M) to it.
• Start stirring the mixture using a magnetic or mechanical stirrer for 15 minutes to ensure proper mixing.
• Gradually add NH₄OH dropwise to the reaction mixture while monitoring the pH using a pH meter.
• Continue stirring the mixture for an additional 20 minutes or until the pH reaches 11. The pH adjustment to 11 promotes the formation of iron hydroxide precipitates.
• Once the reaction time is complete, stop the stirring and allow the precipitate to settle.
• Separate the precipitate from the solution using a centrifuge. Discard the supernatant.
• Wash the obtained precipitate several times with water or a suitable solvent to remove any impurities or unreacted reagents.
• Dry the washed precipitate using a vacuum oven or desiccator to remove any residual moisture.
• The resulting dried material will likely consist of iron oxide nanoparticles, which can be characterized using various techniques such as X-ray diffraction (XRD), transmission electron microscopy (TEM), or Fourier-transform infrared spectroscopy (FTIR).

Alternating method for synthesis of Iron Oxide Nanoparticles by using ammonia:

Theory:

Iron oxide nanoparticles are commonly synthesized through a precipitation method using iron chloride and iron sulfate solutions. The addition of ammonia helps in the formation of iron hydroxide, which is subsequently transformed into iron oxide nanoparticles through heat treatment or aging.

Formation of iron hydroxide:

FeCl₂ + 2NaOH → Fe(OH)₂ + 2NaCl

FeSO₄ + 2NaOH → Fe(OH)₂ + Na₂SO₄

Conversion of iron hydroxide to iron oxide:

2Fe(OH) ₂ → Fe₂O₃ + 2H₂O

Overall reaction:

2FeCl₂ + 4NaOH → Fe₂O₃ + 4NaCl + 2H₂O

2FeSO₄ + 4NaOH → Fe₂O₃ + 2Na₂SO₄  + 2H₂O

Procedure:

1. Prepare a 0.2 M solution of iron chloride (FeCl₂) by dissolving the appropriate amount of iron chloride in 50 ml of distilled water.
2. Prepare a 0.1 M solution of iron sulfate (FeSO₄) by dissolving the appropriate amount of iron sulfate in 50 ml of distilled water.
3. Mix both solutions (0.2 M FeCl₂ and 0.1 M FeSO₄) in a glass beaker.
4. Stir the mixture for approximately 15 minutes to ensure thorough mixing. Apparatus required: Magnetic stirrer, stir bar.
5. Add 40 ml of ammonia (NH₃) to the mixture while continuing to stir.
6. Continue stirring the mixture until the formation of iron oxide nanoparticles.

Note:

The stirring time required for nanoparticle formation may vary depending on the desired size and properties of the nanoparticles. Further heat treatment or aging may be necessary to complete the transformation.

Calculations:

Molar mass of FeCl₃ = 55.845 g/mol + 106.359 g/mol = 162.204 g/mol

To prepare a 0.2 M solution of FeCl₃ in 50 ml water:

Molarity (M) = moles of solute / volume of solution (L)

0.2 M = moles of FeCl₃ / 0.05 L (50 ml = 0.05 L)

Moles of FeCl₃ = 0.2 M x 0.05 L = 0.01 moles

Mass of FeCl₃ = Moles of FeCl₃ x Molar mass of FeCl₃

Mass of FeCl₃ = 0.01 moles x 162.204 g/mol

Therefore, the mass of FeCl₃ required is 1.62204 grams.

Similarly, for FeSO₄:

Molar mass of FeSO₄ = 55.845 g/mol + 32.06 g/mol + 63.996 g/mol = 151.901 g/mol

To prepare a 0.1 M solution of FeSO₄ in 50 ml water:

Molarity (M) = moles of solute / volume of solution (L)

0.1 M = moles of FeSO₄ / 0.05 L (50 ml = 0.05 L)

Moles of FeSO₄ = 0.1 M x 0.05 L = 0.005 moles

Safety Precautions:

• Wear appropriate personal protective equipment, such as gloves and safety goggles, when handling chemicals.
• Work in a well-ventilated area or under a fume hood to avoid inhalation of fumes.
• Handle ammonia with care, as it is a corrosive substance. Avoid direct contact with skin and eyes.
• Follow proper waste disposal protocols for chemicals used in the synthesis.

"Preparation of p-aminoazobenzene (4-phenylazoaniline) from Aniline"

 

Preparation of p-aminoazobenzene (4-phenylazoaniline) from Aniline

The preparation of p-aminoazobenzene (4-phenylazoaniline) involves two main steps: the synthesis of diazoamino benzene (4-diazoanilinobenzene) and its subsequent conversion to p-aminoazobenzene. Here's a step-by-step outline of the process

Step 1:

Synthesis of Diazoamino Benzene (4-diazoanilinobenzene)

Theory: The synthesis of diazoamino benzene involves the conversion of aniline to its diazonium salt, followed by the reaction of the diazonium salt with an excess of aniline to form the diazoamino compound.

Chemicals Required:

• Aniline (C₆H₅NH₂)
• Concentrated hydrochloric acid (HCl)
• Sodium nitrite (NaNO₂)
• Ice-cold water

Materials Required:

1. Reaction flask
2. Ice bath
3. Stirring apparatus
4. Filtration setup
5. Drying oven

Chemical Reaction:

Mechanism:

Procedure:

• Start by setting up an ice bath and placing a reaction flask in it to keep it cold throughout the process.
• In the reaction flask, add 10 mL of concentrated hydrochloric acid (HCl) and cool the solution in the ice bath.
• In a separate container, dissolve 6.8 g of sodium nitrite (NaNO₂) in 30 mL of water to prepare a sodium nitrite solution.
• Slowly add the sodium nitrite solution dropwise to the cold hydrochloric acid solution in the reaction flask, while stirring continuously. Maintain the temperature below 5°C.
• In another container, prepare a solution of 7.2 g of aniline in 30 mL of water.
• Slowly add the aniline solution to the reaction flask containing the diazonium salt solution, while stirring vigorously.
• Continue stirring the reaction mixture for about 30 minutes to ensure complete reaction.
• Pour the reaction mixture into a large volume of ice-cold water to precipitate the diazoamino benzene.
• Collect the precipitate by filtration and wash it with water to remove impurities.
• Dry the precipitate in a drying oven at a suitable temperature.
• Weigh the dried diazoamino benzene product and record its mass for further calculations.

Calculations:

Aniline (C₆H₅NH₂) = 7.2 g

Sodium nitrite (NaNO₂) = 6.8 g

First, we need to determine the limiting reagent by comparing the moles of aniline and sodium nitrite used:

Molar mass of aniline (C₆H₅NH₂) = 93.13 g/mol

Molar mass of sodium nitrite (NaNO₂) = 69.01 g/mol

Number of moles of aniline: mass / molar mass

Moles = 7.2 g / 93.13 g/mol = 0.0774 mol

Number of moles of sodium nitrite = 6.8 g / 69.01 g/mol = 0.0984 mol

The stoichiometry of the reaction between aniline and sodium nitrite is 1:1. Since the number of moles of aniline is lower than the number of moles of sodium nitrite, aniline is the limiting reagent.

To calculate the theoretical yield of diazoamino benzene, we'll use the moles of aniline and the molar mass of diazoamino benzene:

Molar mass of diazoamino benzene = 181.2 g/mol

Theoretical yield = Moles of limiting reagent × Molar mass of diazoamino benzene

Theoretical yield = 0.0774 mol × 181.2 g/mol

Theoretical yield = 14.04 g

Therefore, the theoretical yield of diazoamino benzene, based on the given quantities, is approximately 14.04 grams.

 

Step 2: Conversion of Diazoamino Benzene to p-Aminoazobenzene

Theory:

The conversion of diazoamino benzene to p-aminoazobenzene involves the reaction of diazoamino benzene with aniline under specific conditions to form the desired product.

Chemicals Required:

1. Diazoamino benzene (from Step 1)
2. Aniline (C₆H₅NH₂)
3. Sodium hydroxide (NaOH)
4. Ethanol (C₂H₅OH)

Materials Required:

1. Reaction flask
2. Ice bath
3. Stirring apparatus
4. Filtration setup
5. Drying oven

Chemical Reaction:

Mechanism:

Procedure:

• Start by setting up an ice bath and placing a reaction flask in it to keep it cold throughout the process.
• In the reaction flask, add the diazoamino benzene obtained from Step 1.
• In a separate container, prepare a solution of 4 g of aniline in 30 mL of ethanol.
• Slowly add the aniline solution to the reaction flask containing diazoamino benzene, while stirring continuously.
• Add a solution of 5 g of sodium hydroxide (NaOH) in 30 mL of water dropwise to the reaction mixture while maintaining the temperature below 10°C.
• Continue stirring the reaction mixture for about 30 minutes.
• Pour the reaction mixture into ice-cold water to precipitate the p-aminoazobenzene.
• Collect the precipitate by filtration and wash it with water to remove impurities.
• Dry the precipitate in a drying oven at a suitable temperature.
• Weigh the dried p-aminoazobenzene product and record its mass for further calculations

Calculations:

Theoretical yield of diazoamino benzene = 14.04 g

The stoichiometry of the reaction between diazoamino benzene and aniline to form p-aminoazobenzene is 1:1. Therefore, the moles of p-aminoazobenzene formed will be equal to the moles of diazoamino benzene used.

Molar mass of p-aminoazobenzene = 197.23 g/mol

Theoretical yield of p-aminoazobenzene = Moles of diazoamino benzene × Molar mass of p-aminoazobenzene

Moles of diazoamino benzene = Theoretical yield of diazoamino benzene / Molar mass of diazoamino benzene Moles of diazoamino benzene

Moles of diazoamino benzene = 14.04 g / 181.2 g/mol = 0.0774 mol

Theoretical yield of p-aminoazobenzene = 0.0774 mol × 197.23 g/mol

Theoretical yield of p-aminoazobenzene = 15.26 g

Therefore, the theoretical yield of p-aminoazobenzene, based on the given quantities, is approximately 15.26 grams.

"Preparation of Acetanilide"

 

"Preparation of Acetanilide"

Theory:

Acetanilide is an organic compound that can be synthesized by acetylation of aniline. Acetylation refers to the process of introducing an acetyl group (-COCH3) to a compound. In this case, acetic anhydride will be used as the acetylating agent to react with aniline, resulting in the formation of acetanilide.

Materials Required:

• Acetic anhydride (C₄H₆O₃): Acetylating agent used to introduce the acetyl group.
• Aniline (C₆H₇N): Amine compound that undergoes acetylation to form acetanilide.
• Zinc dust (Zn): Catalyst used to promote the reaction.
• Glacial acetic acid (CH₃COOH): Solvent and reaction medium.
• Distilled water (H₂O): Used for washing and purification steps.

Chemicals Required:

1. Acetic anhydride
2. Aniline
3. Zinc dust
4. Glacial acetic acid
5. Distilled water

Chemical Equation:

Mechanism:

Procedure:

• Set up a round-bottom flask equipped with a reflux condenser and a magnetic stirrer.
• Add 10 mL of aniline to the flask.
• Slowly add 20 mL of acetic anhydride to the flask while stirring.
• Add a small amount (a few grams) of zinc dust to the reaction mixture as a catalyst.
• Heat the mixture gently under reflux for about 30 minutes. The reaction is exothermic, so monitor the temperature and adjust the heat accordingly to maintain refluxing conditions.
• After 30 minutes, remove the heat source and allow the reaction mixture to cool to room temperature.
• Carefully add the reaction mixture to a beaker containing ice-cold water while stirring vigorously. This will cause the acetanilide to precipitate out of the solution.
• Filter the precipitated acetanilide using a Buchner funnel or a simple gravity filtration setup.
• Wash the collected solid with cold distilled water to remove impurities. Repeat this washing process a few times.
• Finally, dry the acetanilide by spreading it out on a watch glass or filter paper in a well-ventilated area until it is completely dry.
• Weigh and record the yield of acetanilide obtained.
  
  
  

Calculations:

Amount of Aniline = 10ml

Molar mass of Aniline = 93 g/mol

Density of Aniline = 1.022 g/mol

Mass of Aniline = 10 x 1.02 = 9.784g

Moles of Aniline =  m/M

                    9.784÷93g/mol = 0.1052 moles

Moles of Aniline = 0.1052 moles

Amount of Acetic anhydride = 20ml

Molar mass of Acetic anhydride = 102 g/mol

Density of Acetic anhydride = 1.08 g/mol

Mass of Acetic anhydride = 20 x 1.08 = 18.51g

Moles of Acetic anhydride = m/M

                       18.51g÷102g/mol = 0.1815 moles

Moles of Acetic anhydride = 0.1815 moles

From chemical equation;

1eq of Aniline = 1eq of Acetic anhydride = 1eq of Acetanilide

so,

Molecular mass of Acetanilide = 135 g/moles

Moles of Acetanilide = 0.105 moles

Mass of Acetanilide = n×M

                                (0.105mol)(135 g/mol) = 7.77g

Theoretical yield = 7.77 g

Actual Yield = 7 g

Percentage Yield = (7 g / 7.77 g) × 100

 Percentage Yield = 90.01%

Precautions:

The reaction should be performed with caution as acetic anhydride is a corrosive and toxic compound. It is also important to wear appropriate safety equipment, such as gloves and goggles, and work in a well-ventilated area.

"Synthesis of Copper Oxide Nanoparticles by using precipitation method"

 

"Synthesis of Copper Oxide Nanoparticles by using precipitation method"

Theory:

Copper oxide nanoparticles can be synthesized by the precipitation method, where copper ions react with a precipitating agent (in this case, sodium hydroxide) to form copper oxide nanoparticles.

Materials Required:

1. Copper sulfate (CuSO₄)
2. Sodium hydroxide (NaOH)
3. Distilled water
4. Volumetric flask (100 mL)
5. pH paper
6. Hot plate/stirrer
7. Centrifuge
8. Oven

Chemicals Required:

• Copper sulfate (CuSO₄)
• Sodium hydroxide (NaOH)
• Distilled water

Chemical Reaction:

Procedure:

1. Weigh 1.596 g of copper sulfate (CuSO₄) and transfer it into a 100 mL volumetric flask.
2. Add a small amount of distilled water to dissolve the copper sulfate. Then, fill the flask up to the mark with distilled water.
3. Place the volumetric flask on a hot plate/stirrer and stir the solution.
4. Heat the solution to a temperature between 50 to 60°C, ensuring complete dissolution of copper sulfate.
5. Add 100 mL of sodium hydroxide (NaOH) solution dropwise into the copper sulfate solution while stirring continuously.
6. Check the pH of the solution using pH paper. Continue adding NaOH solution dropwise until the pH reaches 14, which indicates high alkalinity.
7. Observe the color of the solution turning dark blue after the addition of NaOH. This color change indicates the formation of copper oxide nanoparticles.
8. Maintain the stirring and temperature for two hours to allow for proper nanoparticle formation.
9. Allow the solution to settle overnight, allowing the copper oxide nanoparticles to precipitate.
10. Carefully discard the supernatant solution and perform centrifugation to separate the copper oxide nanoparticles.
11. Dry the separated nanoparticles in an oven at 60°C for two days.
12. Once dried, the copper oxide nanoparticles are ready for further characterization and use.

 

Calculations:

Mass of copper sulfate taken = 1.596 g

Volume of the volumetric flask = 100 mL

Molar mass of CuSO₄ = (63.55 + 32.07 + (4 x 16.00)) = 159.61 g/mol

Moles of CuSO₄ = (mass of CuSO₄) / (molar mass of CuSO₄)

Moles of CuSO₄ = 1.596 g / 159.61 g/mol = 0.01 mol

Calculate the volume of water required to make a 0.1 M solution

Moles = Molarity x Volume (in liters)

0.01 mol = 0.1 M x Volume (in liters)

Volume (in liters) = 0.01 mol / 0.1 M = 0.1 L

Convert the volume to milliliters

Volume (in milliliters) = 0.1 L x 1000 mL/L = 100 mL

Therefore, to prepare a 0.1 M CuSO₄ solution using 1.596 g of CuSO₄, dissolve the given mass of CuSO₄ in distilled water and dilute it to a final volume of 100 mL.

"Preparation of p bromo aniline from p bromo acetanilide"

 

"Preparation of  p bromo aniline from p bromo acetanilide"

Theory:

Para-bromoaniline is synthesized by the conversion of p-bromoacetanilide to p-bromoaniline through a hydrolysis reaction using sodium hydroxide (NaOH). The p-bromoacetanilide is reacted with NaOH, resulting in the replacement of the acetyl group with a hydroxyl group.

Chemicals Required:

1. 5 g p-Bromoacetanilide
2. Ethyl alcohol (as a solvent)
3. 5% Sodium hydroxide (NaOH) solution
4. 7 mL Concentrated hydrochloric acid (HCl)

Apparatus Required:

1. Round-bottom flask
2. Reflux condenser
3. Separatory funnel
4. Buchner funnel
5. Filter paper
6. Glass stirring rod
7. Glassware for measuring and transferring chemicals (graduated cylinders, pipettes, etc.)

Chemical Equation:

Mechanism:

Procedure:

• In a round-bottom flask, I added 5 g of p-bromoacetanilide.
• Then I fitted a reflux condenser to the flask.
• I added 7 mL of concentrated hydrochloric acid (HCl) to the flask.
• Heat the reaction mixture under reflux (boiling) for 1-2 hours. The reflux temperature should be around 70-80°C. During the reaction, the p-bromoacetanilide will react with HCl to form p-bromoacetanilide hydrochloride.
• After the reaction, Icooled the reaction mixture to room temperature.
• I transferred the reaction mixture to a separatory funnel and added water to the separatory funnel. I shaked the separatory funnel gently, allowing the layers to separate.
• I drained the lower aqueous layer (containing impurities) and discard it properly.
• I transferred the upper organic layer (containing p-bromoacetanilide) to a clean beaker.
• To the organic layer I added 5% sodium hydroxide (NaOH) solution dropwise until the pH is around 9-10. This will convert p-bromoacetanilide to p-bromoaniline.
• I stirred the mixture for some time to ensure complete conversion.
• I extracted the p-bromoaniline product by adding a suitable organic solvent (e.g., ethyl acetate or dichloromethane) to the beaker and shaked the mixture gently to ensure good extraction by using the above mentioned procedure of extarction
• I concentrated the organic solution using a rotary evaporator or by heating, once the volume was reduced, I cooled the concentrated solution in an ice bath. And collected the solid p-bromoaniline product by filtration using a Buchner funnel and filter paper.
• I rinsed the solid product with a cold solvent (e.g., cold water or ice-cold ethanol) to remove impurities.

Calculations:

Mass of p-bromoacetanilide = 5 g

Moles of p-bromoacetanilide = Mass / Molar mass

Moles of p-bromoacetanilide = 5 g / 214.1 g/mol

Moles of p-bromoacetanilide = 0.0234 mol

The stoichiometric ratio between p-bromoacetanilide and p-bromoaniline is 1:1. Therefore, the moles of p-bromoaniline formed will be the same as the moles of p-bromoacetanilide

Mass of p-bromoaniline = Moles of p-bromoaniline x Molar mass

Mass of p-bromoaniline = 0.0234 mol x 172.1 g/mol

Mass of p-bromoaniline = 4.026 g

The theoretical yield of p-bromoaniline is approximately 4.026 grams.

To calculate the percentage yield, we need the actual yield of p-bromoaniline.

Let's assume the actual yield of p-bromoaniline is 3.5 grams.

Percentage yield = (Actual yield / Theoretical yield) x 100 

Percentage yield = (3.5 g / 4.026 g) x 100 Percentage yield = 86.8%

Therefore, the theoretical yield of p-bromoaniline is approximately 4.026 grams, and the percentage yield is approximately 86.8%.

"Preparation of Schiff’s Base from benzaldehyde"

 

Preparation of Schiff’s Base from benzaldehyde

Theory:

Schiff bases are a class of compounds that contain a carbon-nitrogen double bond with an imine functional group (R-N=CR'). They are typically synthesized by condensation reactions between a primary amine and an aldehyde or ketone. Schiff bases find applications in coordination chemistry, organic synthesis, metal ion detection, biological activities, dye chemistry, optoelectronic devices, corrosion inhibition, and chelation. They are versatile compounds with uses ranging from catalysis and drug development to sensing and materials science. In this case, we will be synthesizing a Schiff base by reacting aniline (primary amine) with benzaldehyde (aldehyde) in the presence of ethanol as a solvent.

Chemicals Required:

1. Aniline (C₆H₅NH₂)
2. Ethanol (C₂H₅OH)
3. Benzaldehyde (C₆H₅CHO)

Apparatus Required:

1. Round-bottom flask
2. Magnetic stirrer
3. Ice bath
4. Heating source (e.g., hot plate)
5. Filtration setup (filter funnel, filter paper)
6. Weighing balance
7. Glassware (beakers, measuring cylinders)

Chemical reaction:

Mechanism:

Procedure:

1. Measure 2 mL of aniline (C₆H₅NH₂) using a measuring cylinder and transfer it to a round-bottom flask.
2. Add 20 mL of ethanol (C₂H₅OH) to the round-bottom flask containing aniline. Stir the mixture until the aniline is completely dissolved.
3. In a separate container, measure 3.5 mL of benzaldehyde (C₆H₅CHO).
4. Place the round-bottom flask with the aniline-ethanol solution on a magnetic stirrer.
5. Add the benzaldehyde dropwise to the aniline-ethanol solution while stirring continuously. The reaction mixture should turn yellow as the Schiff base is formed.
6. After the addition is complete, continue stirring the reaction mixture at room temperature for 10 minutes to ensure complete reaction.
7. Remove the round-bottom flask from the heating source and allow it to cool to room temperature.
8. Set up a filtration setup with a filter funnel and filter paper. Transfer the reaction mixture to a separating funnel. Extract the Schiff base precipitate from it.
9. Dry the filtered Schiff base by placing it in an oven at a suitable temperature (e.g., 60°C) or by placing it in a desiccator until a constant weight is achieved.
10. Weigh the dried Schiff base to calculate the yield and determine its purity.

Observations:

Colour	pale yellow to dark brown
Crystal shape	can be diverse, including needle-like crystals, plate-like crystals, or irregular-shaped crystals
Melting point	2200C
Boiling Point	2480C

 

Calculations:

To calculate the theoretical yield, we need to determine the limiting reagent. The reactants are aniline and benzaldehyde. We will calculate the moles of each reactant and compare them to find the limiting reagent.

Molar mass of aniline (C₆H₅NH₂)= 93.13 g/mol

Molar mass of benzaldehyde (C₆H₅CHO): 106.12 g/mol

Moles of aniline = (volume in mL x density) / molar mass Moles of aniline = (2 mL x 1.021 g/mL) / 93.13 g/mol = 0.0219 mol

Moles of benzaldehyde = (volume in mL x density) / molar mass

Moles of benzaldehyde = (3.5 mL x 1.044 g/mL) / 106.12 g/mol = 0.0345 mol

Now, let's compare the moles of each reactant:

Moles of aniline= 0.0219 mol

Moles of benzaldehyde= 0.0345 mol

From the comparison, we can see that aniline is the limiting reagent because it has fewer moles than benzaldehyde. Therefore, we will use the moles of aniline to calculate the theoretical yield.

The reaction between aniline and benzaldehyde produces 1 mole of the Schiff base.

Molar mass of C₆H₅-CH=N-C₆H₅ = (Molar mass of C₆H₅) + (Molar mass of CH) + (Molar mass of N) + (Molar mass of C₆H₅)) = (12.01 g/mol x 6) + (1.01 g/mol) + (14.01 g/mol) + (12.01 g/mol x 6)

Molar mass of Schiff base = 152.19 g/mol

Theoretical yield = Moles of limiting reagent x molar mass of the Schiff base

Theoretical yield = 0.0219 mol x 152.19 g/mol

Theoretical yield = 3.34 g

Actual yield = 2.8 grams

Theoretical yield = 3.34 grams

Now we can calculate the percentage yield:

Percentage yield = (Actual yield / Theoretical yield) x 100

Percentage yield = (2.8 g / 3.34 g) x 100

Percentage yield = 83.83%

Therefore, the calculated percentage yield of the Schiff base would be approximately 83.83%.