Percentage of water in a hydrate- Copper(II) sulphate pentahydrate (CuSO4 ·5 H2O)

 Percentage of water in a hydrate- Copper(II) sulphate pentahydrate (CuSO4 ·5 H2O)

Aim / Objective:

To determine the percentage of water in a hydrate

Introduction: 

Many pure substances combine with water in a fixed mole ratio to yield compounds called hydrates. For example, copper sulphate combines with water to form crystalline, CuSO4. 5H2O, which is a stable compound at normal atmospheric conditions. All pure samples of this hydrate show the same percentage of water by analysis. Thus, this hydrated compound obeys the law of constant composition. Upon heating a sample of such a hydrate, it may lose all its water of hydration and revert to the anhydrous salt. Substances which have adsorbed water on the surface do not show constant composition and therefore are not hydrates. An example of this would be common table salt, NaCl, which becomes very sticky on humid, summer days. In these cases, the percentage of water is not constant for all samples of a particular compound, and the water is not chemically bonded as part of the crystal structure.

Other examples of hydrates are Nickel (II) sulfate hexahydrate – (NiSO4 ·6 H2O), lithium perchlorate trihydrate ( LiClO4 . 3H2), aluminum potassium sulfate dodecahydrate – (AlK(SO4)2 ·12 H2O) and magnesium carbonate pentahydrate – (MgCO3 ·5 H2O.)

In this experiment the percentage of water in a hydrate will be determined in an known hydrate. Water is removed from the hydrate by heating an accurately weighed hydrate sample until the residue has reached a constant weight. The percentage of water in the sample is calculated by using the weight of water lost and the initial hydrate sample weight multiplied by 100.

Materials/ Apparatus:

Porcelain crucible and cover, clay triangle, tripod stand, Bunsen burner, flint, tongs,1.000g  Copper(II) sulphate pentahydrate (CuSO4 ·5 H2O(s) , watch glass

 Method / Procedure:

  1. Clean and dry a porcelain crucible and cover.
  2. Place the empty crucible on a covered crucible on a clay triangle supported by a ring on a ring stand.
  3. Heat the crucible and cover in the hottest flame of the Bunsen burner for 5 minutes. Ensure that a dull red glow is observed on the crucible and cover.
  4. Cool the crucible and cover to room temperature for approximately 15 minutes.
  5. Using crucible tongs transfer the crucible and cover to a watch glass and weigh them to the nearest 0.001g.
  6. Add 1g of the 1.000g CuSO4 ·5 H2O(s) to the crucible and weigh the covered crucible to the nearest 0.001g.
  7. Place the covered crucible on the clay triangle with the cover slightly opened.
  8. Heat the crucible gently for a few minutes. Continue to heat for 15 minutes.
  9. Then allow the crucible to cool on the triangle after removing the flame until it reaches room temperature.
  10. Transfer it to the watch glass and weigh the covered crucible to the nearest 0.001g.
  11. Reheat the crucible and contents for about 5minutes. Cool, and then weigh again.
  12. Repeat this heating, cooling and weighing sequence for a total of two readings .
  13. Tabulate results, and complete calculations for the percentage of water in the copper(II) sulphate.

Suggested Results: 

Table showing the results of the heating and cooling of Copper(II) sulphate pentahydrate (CuSO4 ·5 H2O) to find the percentage hydrate in water

Table showing the results of the heating and cooling of Copper(II) sulphate pentahydrate (CuSO4 ·5 H2O) to find the percentage hydrate in water

Use the results from your experiment. Your teacher may request that you use a evaporating dish instead of a crucible and cover which would may the results above slightly different but the concept is the same.

percentage hydrate snippet

Calculations/Answers/Discussion:

  1. Write the formula for the reaction

>>>CuSO4 ·5 H2O(s)  +  HEAT =  CuSO4 (s)  +  5 H2O (g)

2. Calculate the experimental measurement of the percent hydration:

  •       Mass of hydrate before heating =  1.000g
  •       Mass of hydrate after heating = 0.6400g
  •       Difference- mass of water lost = 0.3600g

Experimental Measurement of percent hydrate

  • (0.3600g/1.000g) x 100=36%

3. Calculate the theoretical percentage hydration from the formula.

percentage hydrate snippet2

 

 

 

 

 

 

 

4.Using the theoretical value and the experimental values calculate the percent error

percentage hydrate snippet3

 

 

 

 

 

 

Source of Error/ Limitations/ Assumptions: 

- Allowing the crucible to cool to room temperature before weighting as if not cooled then convection currents will lower the mass and resulting in incorrect results

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Effect of Concentration on the rate of reaction of magnesium ribbon in hydrochloric acid

Effect of concentration on rate of reaction of magnesium ribbon in hydrochloric acid

Aim / Objective: 

To investigate the effect of concentration on rate of reaction of magnesium ribbon in hydrochloric acid.

Abstract:  

The rate of reaction was determined by measuring the time required for a given amount of magnesium metal to be consumed by Hydrochloric acid (HCl) solution of varying concentrations.

Introduction: 

The rate of a chemical reaction is the time required for a given quantity of reactant(s) to be changed to product(s). The unit of time may be seconds, minutes, hours, days or years.

The rate is affected by several factors, some of which are listed as follows:

(1) Nature of the reactants, i.e., one metal may react vigorously with acid while another does not react.

(2) The particle size of the reactants, i.e., a lump of coal burns slowly but powdered coal may explode.

( 3 ) Temperature increases in general increase the rate of reaction, i.e., a 2O°C rise in temperature doubles the reaction rate.

( 4 ) Catalysts affect the rate by using or allowing a different pathway for the reaction to follow.

(5) Concentration affects the rate of reaction, i.e., if the concentration of one of the reactants is doubled and is an integral part of the reaction then rate increases appropriately.

Some reactions are fast and other reactions are slow. The rate of a specific reaction can be found only by experiment.

Apparatus/Materials:

Magnesium ribbon , ruler, scissors, analytical balance, sandpaper, hydrochloric acid, measuring cylinder, graduated cylinder, distilled water, glass stirring rod.

Method / Procedure:

  1. Clean a 25cm length magnesium ribbon lightly by using sandpaper to remove the surface oxide layer. Cut the clean ribbon into five equal pieces of 6cm using a ruler.
  2. Weigh the five pieces together and determine the mass of one piece assuming all six are the same.
  3. Make up 30mL of each of the following five solutions of hydrochloric acid (HCl) with water: 2.0M, 1.5M, 1.0M, 0.5M and 0.25M. To do this, calculate the volumes of 3.0 mol L-1 stock hydrochloric acid solution and water that must be mixed using the dilution equation: (number mL stock HCL solution) x [HCL]= (3.0mL diluted solution) x [HCL] diluted
  4. Measure the calculated volume of stock hydrochloric acid solution using 10mL graduated cylinder. Pour this acid into a 50-mL graduated cylinder and then dilute to the 30-mL mark by carefully adding water from a bottle.
  5. Make up the other for solutions in a similar way. Pour each diluted solution into a 50-mL beaker.
  6. Drop a piece of magnesium into the 2.0 mol L-1 acid solution and start timing the reaction. Stir gently at first using the glass stirring rod to make sure the metal does not stick to the sides of the beaker.
  7.  Measure the time elapsed when the reaction stops and record the time.
  8. Repeat the procedure with the other four acid solutions.
  9. **Your instructor may require you to plot a graph

Suggested Results:

Table showing the effect of varying concentration of HCl on rate of reaction of Magnesium ribbon

Table showing the effect of varying concentration of HCl on rate of reaction of Magnesium ribbon

Discussion:

Write a balanced equation for the reaction

>>> Mg(s) + 2HCl(aq)      =         MgCl2(aq)   + H2(g)

 

Write the ionic equation for the reaction

>>>Mg(s) + 2H+(aq)           =        Mg2+(aq)  + H2(g)

 

Identify the variables in the experiment

  • The manipulated variables were the HCl and the water
  • The responding variable was time
  • The controlled variable was the magnesium ribbon

Based on your experimental data, make a general statement about the effect of concentration of reactants on time and reaction rate

>>>Concentration affects the rate of reaction. Therefore over time as the concentration of HCl increased then the rate of the reaction also increased.

Source of Error/ Limitations/ Assumptions:

  •  Inaccurate timing. Time may be lost during the experiment between the times taken to notice the cross has disappeared to the actual stopping of the watch.
  • Inaccurate measurement of reactants will affect the overall rate of reaction.
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Reactivity series of metals|Reaction with hydrochloric acid and sulphuric acid (Mg, Zn. Al, Fe, Pb, Cu)

Reactivity series of metals|Reaction with hydrochloric acid and sulphuric acid (Mg, Zn. Al, Fe, Pb, Cu)

Aim / Objective: 

To investigate the reaction of some metals: Mg, Zn. Al, Fe, Pb, Cu, with hydrochloric acid and sulphuric acid.

Apparatus/ Materials:

metals of magnesium (Mg) turnings, zinc (Zn) granules, aluminum (Al) turnings, iron (Fe) filings, lead (Pb) foil, copper (Cu), 6 test-tubes, test-tube rack, dilute hydrochloric acid (HCl), dilute sulphuric acid (H2SO4), splints, Bunsen burner and spatula.

Method / Procedure:

  1. Half fill a test- tube with dilute HCl.
  2. Add a spatula full of magnesium turnings to the acid, place cork stopper in the mouth of test-tube and observe for effervescence of gas.
  3. Record your observations. Heat gently under the Bunsen burner if no reaction is taking place or if it is too slow.
  4. Remove cork-stopper and place a lighted splint at the mouth of test-tube. Note the reaction.
  5. Repeat steps 1-4 with the remaining metals and dilute HCl. Then repeat the same procedure with dilute sulphric acid.
  6. Tabulate your results.

Suggested Results: 

reactivity table 1 reactivity table 2

 

Discussion:

reactivity series equations

reactivity series results

 

 

 

 

Source of Error/ Limitations/ Assumptions:  

  • Overheating of test-tube.
  • Allowing some gas to escape when HCl and H2SO4 is added
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Oxidizing and Reducing Agents

 Identification of oxidizing and reducing agents using solutions of acidified potassium dichromate (VI), hydrogen peroxide, acidified potassium iodide, acidified potassium manganate (VII), bleach

Aim / Objective: 

To identify oxidizing and reducing agents using solutions of acidified potassium dichromate (VI), hydrogen peroxide, acidified potassium iodide, acidified potassium manganate (VII), bleach.

Introduction: 

An oxidizing agent, or oxidant, is one that gains electrons and is reduced in a chemical reaction. They are also known as electron acceptors, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reduced. Some examples of oxidizing agents include potassium nitrate, halogens and nitric acid.

A reducing agent, or reductant, loses electrons and is oxidized in a reaction.  A reducing agent is usually in one of its lower possible oxidation states and is the electron donor. A reducing agent is oxidized because it loses electrons in the redox reaction. Examples of reducing agents include formic acid, sulfite compounds and earth metals.

Materials/ Apparatus:

Hydrogen peroxide (H2O2), acidified iron (II) Sulphate (FeSO4), acidified potassium iodide (KI), acidified potassium dichromate(VI) (K2Cr2O7) acidified potassium manganate (VII) (KMnO4), sulphuric acid (H2SO4), bleach (NaClO), 7 test tubes, test tube rack, glass stirring rod, metal boiling tube holder, droppers, sodium hydroxide (NaOH).

Method / Procedure:

  1. Label test tubes 1-7 and place in test tube rack
  2. Half fill test tube 1 with acidified KI and use a dropper to add acidified potassium dichromate (VI) (K2Cr2O7) to the solution. Note color change.
  3. Half fill test tube 2 with acidified KI and use a dropper to add acidified potassium manganate (VII) (KMnO4) to the solution. Note color change.
  4. Half fill test tube 3 with acidified KI and use a dropper to add NaClO to the solution. Note color change.
  5. To test tube 4 half fill with hydrogen peroxide (H2O2) and use a dropper to add acidified KMnO4 to the solution. Note color change.
  6. Half fill test tube 5 with KI and use a dropper to add H2O2 to the solution. Note color change.
  7. Half fill test tube 6 with H2O2 and use a dropper to add acidified K2Cr2O7 to the solution. Note color change.
  8. To the final test tube half fill with FeSO4 and use a dropper to add acidified K2Cr2O7 to the test tube. Add sodium hydroxide (NaOH) to the test tube and note color change.
  9. Tabulate results

Suggested Results:

 

oxidizing and reducing agents 1 oxidizing and reducing agents 2 oxidizing and reducing agents 3 oxidizing and reducing agents 4

 

Hope this one helps, remember to keep reading and studying as the weeks go by and your exams will be great.

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Qualitative Analysis on Compound X in order to identify cations and anions present

Qualitative Analysis on Compound X in order to identify cations and anions present

Aim / Objective: 

To perform qualitative analysis tests on compound X in order to identify cations and anions present.

Introduction: 

Qualitative analysis is a technique that is used to separate and detect cations and anions in a sample. Anions are atoms or groups of atoms that have gained an electron or electrons. The atoms that form ions most easily are the Group 17 or VII atoms, also called halides: Fluorine (F), Chlorine (Cl), Bromine (Br) and Iodine (I). These form anions with a -1 charge. Oxygen (O), Sulphur (S), Nitrogen (N) and Phosphorus (P) also form anions. Most anions are composed from multiple atoms, and are called polyatomic ions.

Cations are atoms that have lost an electron to become positively charged. For example: Sodium (Na)  has one valence electron, one electron in its outer energy level, so it tends to lose one electron, and to become an ion with a +1 charge

Materials/ Apparatus:

Bunsen burner, red litmus paper, blue litmus paper, distilled water, test tube rack, test tubes, glass stirring rod, metal test tube holder, dilute nitric acid (HNO3), sodium hydroxide (NaOH), potassium iodide (KI), barium chloride (BaCl2), ammonia solution, lead nitrate (Pb(NO3)2), silver nitrate (AgNO3), dilute hydrochloric acid (HCl), splint, dropper, compound X.

Method / Procedure:

  1. Heat solid compound X in a dry test tube.
  2. Test the gas evolved with red and blue litmus paper and a glowing splint.
  3. Record the observations.
  4. To a solution of compound X in a test tube add distilled water and dilute HNO3, then add AgNO3 and note observations.
  5. In a test tube mix solution X with distilled water, then add dilute HNO3 and BaCl2 to the mixture. Record the observations.
  6. In a test tube mix solution X with distilled water and heat the test tube moderately. Add dilute HNO3 and Pb(NO3) to the heated mixture. Record the observations.
  7. Mix a small portion of solution X with distilled water in a test tube. Use a dropper to add NaOH in excess to the mixture and heat test tube moderately. Record your observations.
  8. Mix a small portion of solution X with distilled water in a test tube. Add excess ammonia to the mixture and note the results.
  9.  Half fill a test tube with solution X and add KI(aq). Record the observations.
  10. Half fill a test tube with solution X, add NaOH(aq) and heat the test tube moderately. Dip the glass stirring rod with the mixture in a beaker of HCl. Test the gas evolved with litmus paper. Record the observations.
  11. Tabulate results

Suggested Results: 

qualitative analysis 1 qualitative analysis 2

 

Conclusion:

  1. The cations present In compound X were Al3+ and NH4+
  2. The anions present were SO42- and Cl-

 

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Separating a mixture of Oil and Water

Separating a mixture of Oil and Water using a Separating Funnel Aim / Objective: To separate a mixture of cooking oil and water. Abstract: In this experiment, an immiscible mixture of oil and water was separated using a separating funnel. Materials/ Apparatus: measuring cylinder, retort stand, beaker, clamp, separating funnel, conical flask, cooking oil and …

Continue reading ‘Separating a mixture of Oil and Water’ »

Solubility and Electrical Conductivity of molecular and ionic solids | Sugar| Sodium Chloride| Iodine|

Solubility and Electrical Conductivity of molecular and ionic solids | Sugar| Sodium Chloride| Iodine|

Aim / Objective:

To investigate the solubility and electrical conductivity of molecular and ionic solids.

Introduction: 

Ionic solids are formed by ionic bonding where the particles in the solids are ions. These solids are crystals because of their structures. The regular structure of a crystal tells you that the ions are arranged in an orderly manner. This orderly arrangement is a crystal lattice.

Most substances that contain covalent bonds are also described as simple molecular substances, because they consist of separate molecules. Sugar is a molecular crystal which has a low melting and boiling point.

Sodium Chloride is an ionic compound which has a high melting and boiling point due to strong forces which exist between the ions.

Iodine, like sugar is described to be a molecular crystal for it consists of separate molecules.

Materials/ Apparatus:

Iodine and Sugar solid (molecular solids), sodium chloride (ionic solid), distilled water (polar solvent), tetrachloromethane (non-polar solvent), beaker, electrolytic cell, test tubes, measuring cylinder.

Method / Procedure:

  1. Label three test-tubes A, B and C. In test tube A put one crystal of iodine solid.
  2. In test-tube B, put a spatula full of sugar and finally in test-tube C put a spatula full of sodium chloride.
  3. Add 5cm3 of distilled water to the three test tubes. Record your observations.
  4. Repeat steps 1 and 2, then add 5cm3 of tetrachloromethane and record your observations.
  5. Tabulate results.
  6. Label three (3) beakers A, B and C. In beaker A, mix one crystal of iodine and 25cm3 of tetrachloromethane.
  7. In beaker B, mix three spatulas full of sugar and 25cm3 of water and in beaker C, mix three spatulas full of sodium chloride and 25cm3 of water.
  8. Place electrodes into the solutions and close the switch. Watch the ammeter for readings and record the registered reading to determine if the solution conducted an electric current.
  9. Tabulate results.

Suggested Results: 

Table showing the electrical conductivity of iodine, sugar and sodium chloride

Table showing the electrical conductivity of iodine, sugar and sodium chloride

Table showing the solubility of sugar, iodine and sodium chloride in water and solvent

Table showing the solubility of sugar, iodine and sodium chloride in water and solvent

Discussion/Answers:

1. State the reason(s) why sugar did not conduct electricity when dissolved in water

>>>This is due to the absence of ions, for when ions are set free they move to an oppositely charge electrode when a voltage is applied.

2. State the reason for the differences in solubility of sugar in water and tetrachloromethane

>>>The reason why sugar dissolves in water and not in tetrachloromethane is because sugar has polar hydroxyl groups and polar substances dissolve in polar solvent (water), whereas they do not dissolve in nonpolar organic solvent (tetrachloromethane).

3. Explain why sodium chloride was the only substance that conducts electricity

>>>The sodium chloride solutions conducts electricity because the ions are set free upon melting and these ions will furthermore move to an oppositely charge electrode when a voltage is applied.

4. State the reason for the differences in solubility of NaCl in water and tetrachloromethane

>>>The reason why sodium chloride dissolves in water and not in tetrachloromethane is because NaCl has polar hydroxyl groups and polar substances dissolve in polar solvent (water), whereas they do not dissolve in nonpolar organic solvent (tetrachloromethane).

5. Why did iodine not conduct electricity?

>>> Iodine did not conduct electricity when molten or dissolved due to the absence of ions from its structure.

6.  State the reason for the differences in solubility of iodine in water and tetrachloromethane

>>>Iodine consists of molecules and dissolves in tetrachloromethane and is insoluble in water. This is because tetrachloromethane is a non-polar organic solvent and water is a polar solvent, and with iodine being a non-polar molecular substance it will readily dissolve in solvents that are non-polar. This means essentially that the atoms of the element are identical and has the same attraction to the tetrachloromethane.

Source of Error/ Limitations/ Assumptions: 

-          Incorrect reading from the ammeter

-          Contamination of solids resulting in incorrect voltage readings

-          Parallax error

-          Incorrect measurement of solutions

-          Not dissolving solids completely

Remember to always consult your textbook for any other useful information that would really impress your teacher.

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Rate of Diffusion of two gases – Ammonia (NH3) and Hydrochloric Acid (HCl)

Rate of Diffusion of two gases – Ammonia (NH3) and Hydrochloric Acid (HCl)

Aim / Objective:

To compare the rates of diffusion of two gases- Ammonia (NH3) and Hydrochloric Acid (HCl)

Introduction: 

The particles in a fluid are in continuous random motion. In light of this, gases and gaseous mixtures spread out to occupy any space available to them, eventually acquiring a uniformed composition. The process by which fluid mixtures kept a constant temperature becomes uniformed is known as diffusion.

In this experiment, a comparison on the rate of diffusion of ammonia and hydrogen chloride will be done.

Materials/ Apparatus:

Glass tubing, rubber bung, concentrated ammonia solution, concentrated hydrochloric acid, cotton and tape, stop-watch

Method / Procedure:

  1. Set up apparatus as shown below.
  2. Soak one cotton wool in the concentrated ammonia solution and a next piece in the concentrated hydrochloric acid.
  3. Insert both cotton wools simultaneously at points A (one end of the tube) and B (the other end of the tube) respectively and then quickly insert rubber bungs at both ends of the as shown in the diagram.
  4. Begin timing and record the time taken for the white ring to for in the glass tube.
  5. Record the following distances:
    •  From End A to the center of the white ring and label this distance as X
    •  From End B to the center of the white ring, and label this distance as X2
Diagram of apparatus showing the diffusion of ammonia and hydrochloric acid

Diagram of apparatus showing the diffusion of ammonia and hydrochloric acid

 

Suggested Results:

Diagram showing the reaction of  the diffusion of ammonia and hydrochloric acid

Diagram showing the reaction of the diffusion of ammonia and hydrochloric acid

  1. After the cotton wools were placed in the tube, it took approximately 13mins and 22 seconds for a reaction to start taking place.
  2. When the two gases reacted there was a white ring formed 2.5cm from the cotton wool with the HCl and 29.5cm from the cotton wool with the NH3
  3. The length of the tube was estimated to be 32cm and from the experiment, one can deduce that HCl is more dense than NH3 because the NH3 traveled a farther distance than the HCl in the same period of time , hence the white ring formed closer to the cotton with the HCl (Hydrogen Chloride)

Discussion/ Answers:

1. Define diffusion.

>>>> Diffusion is the process by which molecules or ions move from an area of high concentration to a region of low concentration along a concentration gradient.

2. Use the values of X1 and X2 and time (t1) to calculate the relative rates of movements of ammonia gas and hydrogen chloride under similar conditions.

>>> Rate = distance / time

Time = 13m 22s gives 802 seconds

(a) Distance travelled by HCl = 0.25m

   Rate of HCl = Distance/ Time = 0.25m/ 802 seconds

                                                         = 3.12 x10-4 ms-1

(b) Distance travelled by NH3 = 0.295m

Rate of NH3 = Distance/ Time = 0.295m/ 802 seconds

                                                         = 3.67 x10-4 ms-1

3. Calculate the relative molecular masses of ammonia Mr(NH3) and Mr(HCl)

>>>  Relative molecular mass of HCl =

H                                   Cl                      = HCl

1                                   35.5                   = 36.5

>>>Relative molecular mass of NH3 =

N                                   H3                      = NH3

14                                 (3×1)                   = 17

Explain why the rates of diffusion for the two gases are different.

>>> The rate of diffusion for the gases were different even though they were under the same conditions was a result of ammonia being less dense and moving at a further distance than the hydrogen chloride. The formula given for rate is distance/ time thus ammonia’s rate of diffusion higher than that of HCl as it moved a further distance.

In one short sentence explain why concentration affects rate of diffusion.

>>> The steeper the concentration gradient is the faster the rate of diffusion.

Write a balanced equation for the formation of the white ring.

>>> NH3(g)       +                HCl(g)                      =         NH4Cl

Ammonia    +      hydrogen chloride       =      ammonium chloride.

 

Sources of Error/ Limitations/ Assumptions:

  1.  Inaccuracy in timing due to strips not changing entirely and the time is recorded too soon.
  2. Inconsistency with volume of ammonia used.

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Paper Chromatography Lab/Experiment of the coloured Substances in Leaves

Paper Chromatography Lab/Experiment of  the coloured Substances in Leaves

Aim / Objective:

To separate the colored substances in leaves using the process of chromatography

Abstract:

The method of separation used was paper chromatography and the pigments were separated based on rates moved by each pigment. The extract was also separated based on the solubility of the various components and was found to contain green chlorophyll and yellow carotenoids

Introduction:

Chromatography is a technique for separating and identifying mixtures of compounds based upon their different rates of adsorption. All types of chromatography employ two different immiscible phases in contact with each other namely the mobile phase and the stationary phase.

Normally, the stationary phase is a solid such as paper, starch, alumina, or silica and the mobile phase is a liquid such as water, common organic solvents such as ethanol, or solvent mixtures.

The basis for paper chromatography is the fact that porous paper, cellulose, has an enormous surface area to which molecules or ions of substances are attracted (adsorbed) and then released (desorbed) into the solvent as an aqueous solution passes over the paper.

Separation of components occur, that is, they will travel at different speeds in a moving solvent because the varying attractions between these components and the paper. This method of separation is known as partitioning.

The identity of the components can be deduced by comparing a chromatogram of the unknown mixture with chromatograms of mixtures with known composition (standards). An additional aid in the identification of substances is its Rf value. The Rf value of a compound is a characteristic of the compound and solvent used and serves to identify the constituents of a mixture

This can be done by calculation    Rf = Ds / Df

Where, Ds = distance traveled by a spot, and

Df = distance traveled by the solvent.

Apparatus/Materials:

3 Green leaves, mortar and pestle, ethanol, chromatography paper / filter paper, scissors, 10cm3 beaker and cover, glass cover, dropper, pencil

Method / Procedure:

  1. Use scissors to cut leaves and place in the mortar.
  2. Add purified sand and 3cm3 ethanol to the leaves and ground with pestle until green extract is obtained.
  3. Place mixture to one side and prepare chromatography paper.
  4. Fill the 10cm3 beaker with ethanol to about a height of 2cm.
  5. Use pencil to draw a light line 1cm from one end of the strip and label as “A”
  6. Draw a second line, 1cm from line A and label it as “B”. Line A will represent the level to which the strip should be submerged into the solvent, ethanol. Line B will represent the line to which the ink is applied, that is the Line of Origin.
  7. Using the leaf extract apply thin spots across Line B and allow it to dry.
  8. Place the chromatography paper in the beaker using the glass cover to hold it in place.
  9. The solvent was allowed to move up the paper. When the solvent front has almost reached the top of the paper, remove it and mark this line with a pencil.
  10.  Allow paper chromatogram to air dry.
  11. Measure the distance from line B to each separated dye in the leaf extract.
  12. Record and tabulate the distances measured.

Suggested Results:

Paper Chromatogram of the different substances present in leaves - chlorophyll and carotenoids

Paper Chromatogram of the different substances present in leaves – chlorophyll and carotenoids

Table showing the different substances present in leaves and the calculated Rf values

Table showing the different substances present in leaves and the calculated Rf values

 

Discussion / Answers:

1.Define adsorption:

>>> Adsorption is the process by which one substance is being attracted and held to the surface of another substance.

2.Distinguish between adsorption and absorption

>>> Adsorption is the process by which atoms, molecules or ions from a substances either a gas, dissolved solid or liquid adhere to the surface of an adsorbent.

Absorption is the process by which a fluid is dissolved by a solid (absorbent) or a    liquid.

3.Why was ethanol used as a solvent?

>>> Ethanol is used to decolorize leaves. The separation into constituent parts is dependent on differences in solubility in ethanol.

4.Explain why chlorophyll B moved the shortest distance

>>>The different pigments identified have different polarity and due to this separation in polarity, it is possible to separate. Xanthophyll is non-polar so will travel the highest distance, followed by chlorophyll A. Chlorophyll b is the most polar of the pigments and therefore will travel the shortest distance.

5.How could you ascertain if there are more pigments present in the leaf than the ones identified?

>>>The filter paper could be turned sideways and carry out another chromatography to determine if there are more pigments present in the leaf.

Source of Error/ Limitations/ Assumptions: 

  1. Handling the paper with wet hands allows for additional uncontrolled moisture to the paper which will interfere with separation of components.
  2. Incorrect drawing of lines and measurements

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