Category: Chemistry

Introduction to NMR Spectroscopy

best-of-chemistry-lNMR is an acronym of Nuclear Magnetic Resonance. You may be familiar with the concept due to MRI, Magnetic resonance imagining. NMR Spectroscopy is used in modern chemistry to determine the structure of unknown species. It can describe the situations in which particular type of atom in a molecule is thus their structure.

Theory behind it.

  • Nuclei have spin like electrons
    • 1/2 – H1, C13, N15, F19, Si29
    • 0 – C12, O16
    • 1 – H2, N14
    • 3/2 – B11, Cl35
  • Any charged particle, i.e. the nuclei with spin, will generate a magnetic field. When an external magnetic field is applied to the system, the charged particles may align along the magnetic field or against it.
  • Alignment along the field (I=+1/2, a hydrogen) will be favored and will be lower in energy. But, Alignment against the field (I=-1/2 b hydrogen) will also exist despite the excess of a hydrogen. Their energy difference is very small; thus, excess will not be big
  • When radio wave is given, thus energy, the alignment will change direction so that b hydrogen becomes excess. When giving wave stops, the spins will orientate themselves back to equilibrium and release energy in forms of heat.
  • This phenomenon of nuclei absorbing energy from a particular frequency of wave is called resonance.
  • To yield a greater difference between the energy when aligned and when not aligned, the strength of the magnetic field has to be used. V=γB/2π thus E=h γB/2π. Since γ varies for all NMR active nuclei, the interference of a different nucleus in a particular NMR is very rare.
  • Superconducting magnets
  • Short pulse of radio frequency shot at the sample, as it is cooling to its equilibrium, the computer takes the data as intensity vs. time. Fourier Transform à Intensity vs. resonance frequency

Taking data from and Analyzing NMR graphs

  • Dissolve about 10mg of the compound in about 1mL of deuterated solvent or known ppm value such as water or DMSO and add 0.05% TMS.
  • Integral
    • Indicates the number of protons there are in a molecule that share the same environment.
  • Chemical shift
    • When a particular magnetic field (B) is given to a charged object, the charged objects too will create induced magnetic field (Bi). Thus, Bnet of each hydrogen will be different for hydrogens at different position for they will induce different fields. Also, since Bnet felt by most hydrogens are greater than the applied B due to electromagnetic surroundings, the molecules are deshielded and lies to the left of TMS.
    • TMS or tetra methyl silane are given 0ppm.
      • Alkanes
        • They usually appear around 1 because no electronegative group Deshields the molecule.
      • Electronegative group
        • Electron withdrawing groups will lessen the electron density, deshielding the molecule. This will lie further downfield
      • Allylic hydrogens
        • Carbon carbon double or triple bonds are electron withdrawing group, which Deshields the molecule
      • Vinylic hydrogens
        • Hydrogens attached directly to double bond is more deshielded
      • Delocalization
        • Since carbonylic acid groups and aldehyde group delocalizes the electron density of H, H is deshielded
      • Aromaticity
        • They always lie between 6.5 and 8 ppm
        • Ring current through the ring is created. Since the outside hydrogens receive magnetic field of both aromatic ring and Bo it feels more deshielded.
      • Attached to Oxygen or Nitrogen
        • They vary greatly on the nature of the solvent such as pH.
        • If pH is low, the proton in alcohol group will fall off.
      • Spin-spin coupling and Coupling Constant (J)
        • Hydrogens too have magnetic fields. Thus, Bnet experienced by a hydrogen will be affected by the hydrogen next to it. There are a lot of orientations of spin that the adjacent hydrogens may have. Let’s take an example of ethyl iodide. The hydrogens attached to the carbon with iodine will have four possibilities of different electron configurations: up, up, up, down, down, up, down, down arrangements. However, since up down and down up configurations are the same thing. Three peaks in total will appear with the integral ratio of 1:2:1 which in total will equal 2. Let’s look at the methyl group. There are 8 possible arrangements of spin. However, since six of the eight divide into three and three, only 4 peaks will appear with the integral ratio being 1:3:3:1 where the total will equal 3. As you may have noticed, the total number of peaks are proportional to the number of adjacent hydrogens. This is called the n+1 rule. Also, the integral ratios are given in Example in pg 729
        • Coupling constant accounts for the distance between the peaks. When peaks form from coupling, the distance between the two peaks are J Hz. This frequency value can be divided by the carrier frequency to obtain the ppm value. The coupling constants are varied by the bond distance between the affecting hydrogens and the affected hydrogen. For example, if the two hydrogens are three bond length away, meaning they are on adjacent carbons, the J value is between 6 to 8. Since J value for long range coupling is negligible, and the chance of two different protons being attached to a single carbon is highly improbable, only the coupling done over adjacent carbon will affect the NMR readings. The coupling constant also varies according to the angle that two hydrogens make. Dihedral angle is the angle between the two hydrogens when looked through newman projection. Karplus curve shows that J is highest when theta is 0 and 180 and lowest when theta is 90.
        • Decoupling is a technique which reduces the unnecessary coupling of hydrogens. If a C=C double bond exists, the vinylic hydrogen can couple with additional groups attached on the other end of the vinyl via long range coupling. Thus, NMR spectrometer apply two radio frequency wave, one which follows the routine and one which has the same frequency as the resonance frequency of the excess hydrogens. This allows rapid transition between spin states cancelling the coupling effect.

C13 NMR

C13 also has a nuclei spin of +_1/2. However, since C13 isotope rarely exists, it is not visible in the H1 NMR.

The second technique is called DEPT (distortionless enhancement with polarization transfer) measures the number of CH, CH2, CH3, and CR4 groups.

Today, NMR spectroscopy has advanced so much that now, using MRIs, (Magnetic Resonance imaging), images of nuclear magnetic resonance can be seen rather than peaks of resonance in a particular molecule. Yet, despite the advanced technology of spectrometer and structure determination, we have to be able to intuitively figure out the structure of experimentally obtained molecules by their physical and chemical properties.

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Chemical Equilibrium: Le Châtelier’s Principle

Chemical Equilibrium: Le Châtelier’s Principle

  1. Introduction

Equilibrium is a state where the forward rate of the reaction is equal to the reverse rate. At this state, chemical reaction is not observed making it look as if the reaction has stopped.

However, the equilibrium can be easily changed when there is any stress in the system. The equilibria’s response to these stresses is explained by Le Châtelier’s principle: if equilibria are disturbed by an outside factor, the system will react in a direction to adjust to the new environment, reestablishing the equilibrium. However, the principle does not only state that the equilibrium will shift but it also predicts how it will change. Knowing the factors that disturb the equilibrium helps figuring out whether the equilibrium will move. The factors that are specified in Le Châtelier’s principle are pressure, concentration of the reactant or product, and temperature.

The equilibrium constant (Kc) and the reaction quotient (Qc) are also important in finding which direction the equilibrium will shift to when the system is upset. The equilibrium constant is the concentration of product divided by the concentration of the reactant raised to their orders accordingly, and the constant dependant on temperature when the system is in equilibrium (Kc=K([A]a[B]b)/([C]c[D]d) when cC+dD←→aA+bB). The reaction quotient has the equal way of calculating as the equilibrium constant. However, the reaction quotient can be calculated at any point on the progression of the reaction(Qc= K([A]a[B]b)/([C]c[D]d) when cC+dDaA+bB).

The significance of the equilibrium constant and the reaction quotient is that they decide which direction the equilibrium will shift to according to their comparative values. The reaction quotient being bigger than the equilibrium constant indicates that the product’s concentration is bigger than the ones of the reactant’s shifting the equilibrium towards the left. In exact opposite, the reaction quotient being smaller than the equilibrium constant indicates that the reactant’s concentration is bigger than the ones of the product’s shifting the equilibrium towards the right.

Aim of the experiment is to define Le Châtelier’s principle according to the factors that upset the equilibrium: concentration of reactants and product, and Temperature; to observe the equilibrium using sparingly soluble salts; and to explain the direction of the shifts and their reasons.

2. Materials

    1. Apparatus
  • Chemicals in a Medicine Dropper
  • Test Tubes
  • 50mL Graduated Cylinder
  • 10mL Beaker
  • Clamp Stand, Clamp
  • Heating Plate
  • 100mL Beaker
    1. Chemical
  • 0.1M CuSO4
  • 0.1M NiCl2
  • 1.0M CoCl2
  • 0.1M KI
  • 0.1M Na2CO3
  • 0.1M AgNO3
  • 6M HNO3
  • 0.1M HCl
  • 1M HCl
  • 12M HCl
  • 15M NH3

3. Procedures

    1. Changes in Reactant or Product Concentration
      1. Copper and Nickel Ion
        1. Place 1mL of 0.1M CuSO4 in a 10mL beaker and record the color
        2. Add 15M NH3 dropwise until color change occur and the solution is clear
      • Solids may occur (Cu(OH)2)
        1. Record observations
        2. Mix the solution by tickling it while adding NH3
        3. Add 1M HCl dropwise and mix the solution until the solution is clear
        4. Record observation
        5. Repeat the procedure using 0.1M NiCl2 in place of CuSO4 and record observation
      1. Cobalt Ions
        1. Place about 0.5mL (10 drops) of 1M CoCl2 in a clean 10mL beaker
        2. Record color
        3. Drop 12M HCl to the test tube until a distinct color change occurs
        4. Record observation
        5. Slowly add water to the test tube while mixing
        6. Record observation
    1. Equilibria Involving Sparingly Soluble Salt
        1. To 0.5mL (10 drops) of 1.0M Na2CO3 in a 10mL beaker
        2. Add 1 drop of 0.1M AgNO3
        3. Record observation
        4. Add 6M HNO3 dropwise until you observe a change in appearance
        5. Record observation
        6. To the above solution, add 0.1M HCl dropwise
        7. Record observation
        8. While mixing, add 15M NH3 dropwise until chemical reaction occurs
        9. Add 6M HNO3 dropwise until chemical change is no more change.
        10. Record observation
        11. Continue to add 0.1M KI dropwise until you see evident chemical change
        12. Observe reaction
        13. Dispose the waste accordingly
    2. Effect of Temperature on Equilibria
        1. Heat 75mL of water to boil in a 100mL beaker on a heating plate
        2. Place about 1mL of 1.0M CoCl2 in a small test tube in the boiling content using a clamp stand
        3. Compare the color of the cool cobalt solution to that of the hot solution
        4. Record observation

4. Data

    1. Changes in Reactant or Product Concentration
      1. Copper and Nickel Ions
        1. CuSo4 (aq) – Light Blue
        2. [Cu(NH3)4]2+ (aq) – Dark Blue
        3. After HCl reaction – Extremely Clear, almost transparent but slightly light bluish
        4. NiCl2 (aq) – Light Green, turquoise
        5. [Ni(NH3)6]2+ (aq) – Pale violet
        6. After HCl reaction – Light Blue
      2. Cobalt Ions
        1. CoCl2 (aq) – Rose pink, slightly red
        2. After HCl reaction – Dark Blue, Gas evolve, Temperature increase
        3. After H2O reaction – Rose pink, same as the color before the HCl reaction
    2. Equilibria Involving Sparingly Solution Salts
        1. Na2CO3 (aq) – Transparent, slightly pinkish
        2. After adding AgNO3 (aq) – White precipitate formed
        3. After HNO3 reaction – White precipitate seems to dissolve, pale white
        4. After HCl reaction – Solution turned white, precipitate form
        5. After NH3 (aq) reaction – The solution turned transparent, gas evolved, extremely small amount of precipitate left
        6. After HNO3 reaction – Turned again into milky white, precipitate reappear
        7. After NH3 reaction – Turned transparent, however no gas evolved this time, precipitate disappear
        8. After adding KI – The solution turned milky green, white precipitates remained
    3. The Effect of Temperature on Equilibria
        1. Cool CoCl2 – Rose Pink
        2. Hot [CoCl4]2- – Very dark blackish rose pink

5. Analysis

    1. Changes in Reactant
      1. Copper and Nickel Ions
        1. [Cu(H2O)4]2+(aq) + 4NH3 (aq) ←→ [Cu(NH3)4]2+(aq) + 4H2O (l)
        2. [Ni(H2O)6]2+(aq) + 6NH3 (aq) ←→ [Ni(NH3)6]2+(aq) + 6H2O (l)
      • Addition of NH3 made the equilibrium shift right
          1. H+(aq) + NH3 (aq) ←→ NH4+
      • Reaction with HCl made the equilibrium shift left
      1. Cobalt Ions
        1. [Co(H2O)6]2+(aq) + 4Cl (aq) ←→ [CoCl4]2-(aq) + 6H2O (l)
      • Reaction with HCl made the equilibrium shift right
      • Reaction with H2O made the equilibrium shift left
    1. Equibria Involving Sparingly Soluble Salt
        1. 2Ag+(aq) + CO32-(aq) ←→ Ag2CO3 (s)
          1. 2H+(aq) + CO32-(aq) ←→ H2CO3 (aq)
          2. H2CO3 (aq) CO2 (aq) + H2O (l)
      • Reaction with HNO3 made the equilibrium shift left
        1. Ag+(aq) + Cl(aq) ←→ AgCl (s)
      • Reaction with HCl creates a new precipitate of AgCl and moves the equilibrium to the left
          1. Ag+(aq) + 2NH3 (aq) ←→ [Ag(NH3)2]+(aq)
      • Reaction with NH3 shifts the equilibrium to the left
          1. H+(aq) + NH3 (aq) ←→ NH4+(aq)
      • Reaction with HNO3 moves the equilibrium to the right
          1. I(aq) + Ag+(aq) ←→ AgI (s)
      • Reaction with the iodide ion creates a new precipitate of AgI and shifts the equilibrium to the left
    1. The Effect of Temperature on Equilibria
        1. CoCl2+ΔH Δ [CoCl4]2-
      • When the solution is heated, the equilibrium shifts to the right
      • When the solution is cooled, the equilibrium shifts to the left

6. Discussion

[Cu(H2O)4]2+(aq) + 4NH3 (aq) ←→ [Cu(NH3)4]2+(aq) + 4H2O (l)

[Ni(H2O)6]2+(aq) + 6NH3 (aq) ←→ [Ni(NH3)6]2+(aq) + 6H2O (l)

For above reaction NH3 reacts with both Cu2+ and Ni2+ ions to create [Cu(NH3)4]2+(aq) and [Ni(NH3)6]2+(aq) respectively. Thus, by adding NH3 to the solution, the equilibrium shifted right since while adding NH3 the concentration NH3 increases allowing more NH3 to react with the ions to create [Cu(NH3)4]2+(aq) and [Ni(NH3)6]2+(aq).

H+(aq) + NH3 (aq) ←→ NH4+

However, when HCl is added to neutralize NH3, the concentration of NH3 decreases, shifting the equilibrium left, thus creating less [Cu(NH3)4]2+(aq) and [Ni(NH3)6]2+(aq).

[Co(H2O)6]2+(aq) + 4Cl (aq) ←→ [CoCl4]2-(aq) + 6H2O (l)

Similar to above reaction, adding more HCl, the concentration of Cl(aq) increases shifting the equilibrium right. Through this reaction, more [CoCl4]2-(aq) is being created. However, when more water(H2O) is added, the concentration of H2O increases shifting the equilibrium left, creating less [CoCl4]2-(aq)

2Ag+(aq) + CO32-(aq) ←→ Ag2CO3 (s) …………….(1)

In reaction (1), CO32- reacts with Ag+ to create Ag2CO3 (s). The addition of Na2CO3 shifts the equilibrium to the right.

2H+(aq) + CO32-(aq) ←→ H2CO3 (aq)

H2CO3 (aq) CO2 (aq) + H2O (l)

In the above reaction, H+ reacts with CO32- to form H2CO3 that quickly decomposes to CO2 and water. Thus, addition of HNO3 reduced the concentration of CO32- ions shifting the equilibrium (1) left. Decrease in production of Ag2CO3 (s) is the result of this reaction. This process is observed through decrease in precipitates, Ag2CO3.

Ag+(aq) + Cl(aq) ←→ AgCl (s) …………………….(2)

From equilibrium (1), Cl from HCl reacts with Ag+ to create AgCl(aq). This decreases the concentration of Ag+(aq) shifting the equilibrium left. However, if excess HCl is added until chemical reaction has settled, the reaction has changed to equilibrium (2). This creates a new precipitate of AgCl(s).

Ag+(aq) + 2NH3 (aq) ←→ [Ag(NH3)2]+(aq)…………(3)

NH3 reacts with the silver ion to form [Ag(NH3)2]+(aq) which is a soluble substance. Thus, addition of NH3 reduces the concentration of Ag+(aq) ion shifting the equilibrium (2) left. This leads to less production of AgCl (s) so that the precipitates seem to disappear.

H+(aq) + NH3 (aq) ←→ NH4+(aq)

However, H+ react with NH3 to form NH4+(aq) which does not take part in reaction. This reduces the concentration of NH3 in equilibrium (3) which eventually increases the concentration of Ag+ in equilibrium (2). This shifts the equilibrium (2) right creating more AgCl (s) precipitate.

Process of adding NH3 and HNO3 – in this context H+ – is repeated several times. For all those times the equilibrium shifts for the same reason as explained above.

I(aq) + Ag+(aq) ←→ AgI (s)……………………….(4)

The iodide ion, I, reacts with Ag+ to form AgI(s) precipitate. This decreases the concentration of Ag+ in equilibrium (2) shifting the equilibrium left. However, if excess KI is added to the solution, the reaction completely changes to equilibrium (4).

CoCl2+ΔH Δ [CoCl4]2-

The process of CoCl2←→ [CoCl4]2- is an endothermic reaction. Thus, when heated, heat is given into the system. Thus, letting more heat get absorbed into the system shifts the equilibrium to the right.

On the other hand, if the reaction is cooled down, heat is taken out of the system. Thus, not letting heat take part in reaction shifts the equilibrium to the left.

Thus overall, all the reaction’s shift in equilibrium is explained by Le Châtelier’s principle. The change in concentration was the main reason of shifts in equilibrium. Furthermore, the change in heat demonstrated in the last reaction was also another factor that shifts equilibrium. Thus, Le Châtelier’s principle which explains the shift in equilibrium due to change in concentration and heat is well demonstrated throughout the experiment.

However, some errors that might be present on the process of the experiment are insufficient amount of substances (or even heat) that did not lead the experiment to completion and lack of observation.

One error observed was in the process of observing the color change of CoCl2+ΔH Δ [CoCl4]2- reaction. According to the text book, the color change in CoCl2+ΔH Δ [CoCl4]2- should have been from rose pink to blue. However, the color change observed was from rose pink to very dark blackish pink. One reason to explain this error is that, the amount of heat exerted in the reaction was insufficient to completely shift the equilibrium to its end but made it stop in the middle of shifting the equilibrium.

Another error that was observed was in the process of adding HNO3 to 2Ag+(aq) + CO32-(aq) ←→ Ag2CO3 (s). According to the process and its analysis, CO2 gas is supposed to bubble when HNO3 is added. However, there were no such bubbles observed. Two reasons that explain this error might be that, first; there wasn’t enough amount of solution present for any bubble to be observed and second; the amount of CO2 produced was too less that most of CO2 being produced dissolved in water before being able to form gas. The amount added to the 10mL beaker was 0.5mL of Na2CO3 which equals to about 10 drops, and a drop of AgNO3 which approximates to about 0.05mL. In total, only 0.55mL of solution was present in a 10mL beaker. This might have led to having too much surface area for bubble to be seen thus no significant bubble was observed.

Another error that was observed was that in process of adding highly concentrated acids and bases, gases evolved when it was not supposed to: when adding 12M HCl and when adding 15M NH3. One way to explain this is that HCl and NH3, being a very strong acid and a very strong base, reacts readily with water to form H2 gas for HCl and and NH4+ for NH3.

In order to improve the experiment, more time would have helped observe more definite data. Using definite data, observations of formation of more justifiable substances would have been given leading to more clear shifts of equilibrium. More time in the last experiment would have provided chances for more CoCl2 to form [CoCl4]2- thus changing the color into definite blue. To add on, addition of time would have given opportunities to carry experiments out for more than one time solidifying the argument of equilibrium shifts.

Work Cited

Brown and LeMay the Central Science Lab Manual

Brown, Lemay, Bursten, Murphy, Woodward, Stoltzfus. 2015. Chemistry the Central Science. 13th edition. United States of America: Pearson

John J. Carroll. John D. Slupsky. Alan E. Mather. June 13, 1991. The Solubility of Carbon Dioxide in Water at Low Pressure. Ref. Data, Vol 20, No. 6: University of Alberta

Activity Series

Activity Series

Worked with Jake Tae, Jeremy Chang

  1. Introduction

Oxidation reduction reactions are reactions in which electrons are transferred from element to element. Furthermore, in order to express the extent that an element has been oxidized, oxidation number is used. Oxidation number, expressing the oxidation state of an atom, shows the total number of electrons which have been removed or added to the element. When an electron is removed from an element, the oxidation number increases and the increase of oxidation number means that the atom has lost electron which is called oxidation. Likewise, when an electron is added to an element and the oxidation state decreases, the process of gaining an electron is called reduction.

Reduction potential is another concept that illustrates the losing and gaining of electron. However, reduction potential not only show the element’s electron loose and gain but also shows the tendency of atoms to lose and gain electrons when it is exposed to change with an introduction of a new element. Thus, if an atom has higher reduction potential than another, the former is more likely to be reduced thus gaining electron from the other when they two react. In other words, the latter is more likely to be oxidized and donate electron to the other when reaction occurs between the two. However, the tendency of losing or gaining electron is not absolute but is dependent on one another. This is because the definition of reduction potential states that reduction potential do involve introduction of other element meaning that all measurements will also depend on the later introduced element too. Thus, experimented reduction potential of hydrogen is set to 0 so that other atoms’ reduction potential, when reacted with hydrogen, can be easily measured and denoted.

Furthermore, reduction potential as the tendency to gain electron does have a close connection to the electronegativity which is the tendency to gain electron. As it seems, since electronegativity is the tendency to gain electron and reduction potential is the exact same, they are proportional to each other which means that an electron with high electronegativity will have a high reduction potential.

However, unlike the reduction potential, activity series shows the tendency of an element to lose electrons when it is exposed to change with an introduction of a new element. Furthermore, activity series, unlike the reduction potential, does not have values but are series of metals in order of their tendency to react where the most reactive metals are on top and the least reactive at the bottom. Thus, if a metal is higher on the activity series than another, this means that the former metal wants to give away electron more readily thus be oxidized quicker.

In this experiment, six different metals will be tested with acid and their ions present on varieties of solutions to experimentally determine the tendency of oxidation and eventually create an experimented activity series of those six metals.

2. Materials

    1. Apparatus
      1. Small Test Tubes
      2. Test Tube Rack
    2. Chemicals
  1. 0.2M Ca(NO3)2
  2. 0.2M Zn(NO3)2
  3. 0.2M Mg(NO3)2
  4. 0.2M Fe(NO3)3
  5. 0.2M FeSO4
  6. 0.2M CuSO4
  7. 0.2M SnCl4
  8. 6M HCl
  9. 7 Small Pieces Each of Calcium, Magnesium, Zinc, Iron Wool, Tin, Copper

3. Procedure

    1. Reactions of Metal with Acid
      1. Add 0.5mL of dilute 6M HCl to 6 test tubes
      2. Add each piece of metal into each test tube
      3. Record any change
    2. Reactions of Metal with Solutions of Metal Ion
      1. Add 0.5mL of Ca(NO3)2 in 6 test tubes
      2. Add a small piece of calcium, copper, magnesium, iron wool, tin, and zinc
      3. Observe any color changes, evolution of gas and whether the metal dissolves or not
      4. Repeat the process with 6 other solutions and observe
    3. Reactive Activity Series
      1. Rank the metals in order of their reactivity judging from the most reactive to the least reactive

4. Data

    1. Reaction of Metals with Acid
      1. Ca
  • Observation: Extremely hot, visible gas evaporation, bubble eruption, still transparent.
      1. Cu
  • Observation: N.R.
      1. Mg
  • Observation: Hot temperature, rapid gas visibly erupts, bubble evolves, violent reaction
      1. Fe
  • Observation: Very slow reaction, bubbles however at very slow rate Gas evolves, Slowly rises up to the surface
      1. Sn
  • Observation: N.R.
      1. Zn
  • Observation: Bubble eruption. Is rapid reaction but not as much as calcium
    1. Reaction of Metals with Solution of Metal Ion
      1. Ca
  • Ca2+ N.R.
  • Cu2+ White Gas evolve, bubble eruption, becomes blueish green in color, blue precipitate
  • Fe3+ Bubble eruption, White precipitate, turquoise
  • Fe2+ Bubble erupts, visible white gas, color of rust, temperature increase
  • Mg2+ Bubble erupts, Gas evolution, violent reaction
  • Sn4+ Bubble, White gas
  • Zn2+ Hot, white bubble form, gas form, White precipitate
      1. Cu
  • Ca2+ N.R.
  • Cu2+ N.R.
  • Fe3+ N.R.
  • Fe2+ N.R.
  • Mg2+ N.R.
  • Sn4+ N.R.
  • Zn2+ N.R.
      1. Fe
  • Ca2+ N.R.
  • Cu2+ Solution turned murky yellow, the iron turned copper red, very slow reaction
  • Fe3+N.R.
  • Fe2+ N.R.
  • Mg2+ N.R.
  • Sn4+ N.R.
  • Zn2+ N.R.
      1. Mg
  • Ca2+ N.R.
  • Cu2+ Bubble erupts, liquid turns greenish yellow
  • Fe3+ slow bubble eruption, turned murky orange with a taste of white
  • Fe2+ Bubble erupts extremely slowly
  • Mg2+ N.R.
  • Sn4+ White Gas evolves, Solution turned yellowish translucent
  • Zn2+ solution turned murky
      1. Sn
  • Ca2+ N.R.
  • Cu2+ N.R.
  • Fe3+ N.R.
  • Fe2+ N.R.
  • Mg2+ N.R.
  • Sn4+ N.R.
  • Zn2+ N.R.
      1. Zn
  • Ca2+ N.R.
  • Cu2+ Liquid turned transparent, zinc turned black, little bubble eruption
  • Fe3+ Solution turned red
  • Fe2+ N.R.
  • Mg2+ N.R.
  • Sn4+ Bubble erupts
  • Zn2+ N.R.

5. Analysis

    1. Reactions of Metal with Acid
      1. Ca(s) + 2H+(aq) Ca2+(aq) + H2(g)
      2. Cu(s) + 2HCl(aq) Cu(s) + 2HCl(aq) : N.R.
      3. Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)
      4. Fe(s) + 2H+(aq) Fe2+(aq) + H2(g)
      5. Sn(s) + 4HCl(aq) Sn(s) + 4HCl(aq) : N.R.
      6. Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
    2. Reactions of Metal with Solutions of Metal Ion
  • The reactions with metal and the same ion are not written since they do not and should not react with each other when put together.
      1. Ca(NO3)2(aq)
        1. Cu(s) + Ca+2(aq) Cu(s) + Ca+2(aq) : N.R.
        2. Mg(s) + Ca+2(aq) Mg(s) + Ca+2(aq) : N.R.
        3. Fe(s) + Ca+2(aq) Fe(s) + Ca+2(aq) : N.R.
        4. Sn(s) + Ca+2(aq) Sn(s) + Ca+2(aq) : N.R.
        5. Zn(s) + Ca+2(aq) Zn(s) + Ca+2(aq) : N.R.
      2. CuSO4(aq)
        1. Ca(s) + Cu2+(aq) Cu(s) + Ca2+(aq)
        2. Mg(s) + Cu2+(aq) Cu(s) + Mg2+(aq)
        3. Fe(s) + Cu2+(aq) Cu(s) + Fe2+(aq)
        4. Sn(s) + Cu2+(aq) Sn(s) + Cu2+(aq) : N.R.
        5. Zn(s) + Cu2+(aq) Cu(s) + Ca2+(aq)
      3. FeSO4(aq)
        1. Ca(s) + Fe2+(aq) Fe(s) + Ca2+(aq)
        2. Cu(s) + Fe2+(aq) Cu(s) + Fe2+(aq) : N.R.
        3. Mg(s) + Fe2+ (aq) Fe(s)+ Mg2+(aq)
        4. Sn(s) + Fe2+(aq) Sn(s) + Fe2+(aq) : N.R.
        5. Zn(s) + Fe2+(aq) Fe(s)+ Zn2+(aq)
      4. Fe(NO3)3(aq)
        1. 3Ca(s) + 2Fe3+(aq) 2Fe(s) + 3Ca2+ (aq)
        2. Cu(s) + Fe3+ (aq) Cu(s) + Fe3+ (aq) : N.R.
        3. 3Mg(s) + 2Fe3+ (aq) 2Fe(s) + 3Mg2+ (aq)
        4. Sn(s) + Fe3+ (aq) Sn(s) + Fe3+ (aq) : N.R.
        5. Zn(s) + Fe3+ (aq) Zn(s) + Fe3+ (aq) : N.R.
      5. Mg(NO3)2(aq)
        1. Ca(s) + Mg2+ (aq) Mg(s) + Ca2+ (aq)
        2. Cu(s) + Mg2+ (aq) Cu(s) + Mg2+ (aq) : N.R.
        3. Fe(s) + Mg2+ (aq) Fe(s) + Mg2+ (aq) : N.R.
        4. Sn(s) + Mg2+ (aq) Sn(s) + Mg2+ (aq) : N.R.
        5. Zn(s) + Mg2+ (aq) Zn(s) + Mg2+ (aq) : N.R.
      6. SnCl4(aq)
        1. 2Ca(s) + Sn4+(aq) Sn(s) + 2Ca2+(aq)
        2. Cu(s) + Sn4+(aq) Cu(s) + Sn4+(aq) : N.R.
        3. 2Mg(s) + Sn4+(aq) Sn(s) + 2Mg2+(aq)
        4. 2Fe(s) + Sn4+(aq) Sn(s) + 2Fe2+(aq)
        5. 2Zn(s) + Sn4+(aq) Sn(s) + 2Zn2+(aq)
      7. Zn(NO3)2(aq)
        1. Ca(s) + Zn2+(aq) Zn(s) + Ca2+(aq)
        2. Cu(s) + Zn2+(aq) Cu(s) + Zn2+ (aq) : N.R.
        3. Mg(s) + Zn2+(aq) Zn(s) + Mg2+ (aq)
        4. Fe(s) + Zn2+(aq) Fe(s) + Zn2+ (aq) : N.R.
        5. Sn(s) + Zn2+(aq) Sn(s) + Zn2+ (aq) : N.R.
    1. Experimental Activity Series
  1. Ca Ca2+ + 2e
  2. Mg Mg2+ + 2e
  3. Fe Fe3+ + 3e
  4. Zn Zn2+ + 2e
  5. Fe Fe2+ + 2e
  6. Sn Sn4+ + 4e
  7. Cu Cu2+ + 2e

Calcium seemed to be the most reactive amongst all the metal judging from the fact that it showed the most violent reaction with HCl and its metal reacted with every single solution present which means that calcium likes to oxidize more readily compared to other metals

Magnesium seemed to be the next reactive metal since its reaction with HCl was as violent as with calcium however did not readily oxidize compared to the calcium ion.

Zinc was predicted to be the next reactive amongst the metals since it did react with both HCl and many other solutions, however it neither reacted as violently as the previous two did nor reacted with solutions containing calcium and magnesium which means that it does not want to oxidize compared to them.

Iron generally seemed like the next reactive metal judging from the fact that it slowly reacted with HCl and with solutions containing copper and tin. In this experiment, two different ions of iron took part which were Fe(II) and Fe(III), ferrous and ferric. However, they both showed different traits of observation. Fe2+ in solution did react with zinc which means that zinc oxidizes more readily compared to Fe2+. Thus Fe2+ is lower on the activity series compared to zinc. On the other hand, Fe3+ did not react when zinc metal was added which means that Fe3+ oxidizes more readily than zinc. This locates Fe3+ higher on the activity series compared to zinc. However since Fe2+ did not react with tin and copper, Fe2+ stays between zinc and tin, when since Fe3+ did react with both magnesium and calcium, it stays between magnesium and zinc in the activity series.

Both tin and copper did not show much trait of reaction in both HCl reaction and reaction with varieties of ions. They neither reacted with HCl nor did with each other’s ions. Thus distinguishing their levels on the activity series is very difficult. However, the fact that iron only reacted with copper and not tin shows that tin oxidizes more readily compared to copper. Thus, tin has been put after Fe2+ and copper following tin on the activity series.

6. Discussion

The experimental activity series show similar traits as the actual activity series. However the major difference between them might be that known activity series does not distinguish the ions of irons but just categorizes them into iron so that further details cannot be seen. Furthermore, the known activity series unlike the experimental series has leveled out many more metals along the series that with experiment, would have been too vague to be differentiated.

Connecting the activity series to the periodic table, as discussed before in the introduction section, electronegativity is closely connected to the reduction potential thus the activity series. Electronegativity which shows the tendency of an atom to gain an electron is indirectly proportional to the activity series which shows the tendency of an atom to lose an electron since the elements as they go up the series prefers to oxidize more readily thus lose an electron. This allows the going up of the activity series the exactly opposite periodic trend as the electronegativity. Thus, the periodic trend of the electronegativity, is inversed in order to illustrate the periodic trend of metals going up the activity series. Furthermore this is proven through the experimental data and the expanded activity series. They show that calcium, a metal right below magnesium in the periodic table is more reactive compared to it and iron on the left of copper on the same period is more reactive than copper. This shows that the general trend of going up the activity series is from right to the left and from top to bottom of the periodic table.

However, there are exceptions on this trend as electronegativity do too, such as the filled and half-filled orbital exceptions do apply to this rule as well. To elaborate on the exceptions, the elements whose valence electrons either completely fill an orbital such as zinc whose d orbitals are filled tend to show a slight dent in the trend. This is due to their decreased electronegativity at the state of filled and half-filled orbitals. Thus, zinc, according to the general rule of periodic trend should have been under copper, however, because of the exceptions of the trend, saying that zinc has lower electronegativity than copper, it can be easily concluded that zinc is more reactive than copper.

Even though the experiment did generally show successful results, some of the observations were not as clear as it should have been. This mistake might have been committed due to lack of time. The experiment was performed under great time constraints since about 50 different reactions should have been carried out and all their observations should have been recorded at about an hour and a half. Furthermore, the acknowledgement given by the lab instructor at the beginning of the lab also created anxiety amongst the observers. Additionally, most of the reactions took very long time to even show some traits of reaction. For example, the reaction between iron and HCl might have been ignored if the test tubes were disposed before the first bubble erupt. This shows that there were ample number of chances that many of the observations made above couldn’t have been observed and vice versa, that there are chances that the observers might not have observed long enough for the reaction to be visible thus missing on some reactions.

Furthermore, during the experiment there were confusions regarding whether some particles visible were precipitates or residues of leftover metals. Due to lack of conciseness, the experiment was not carried out in controlled environment but was carried out by adding large pieces of metals to a small amount of solutions. This led to the ions present in the solution being the limiting reagent thus allowing the metals to leftover. This created small particles of unused metals. On the other hand, many of the above reaction created precipitates. However, since many precipitates are of similar color to the original metal, it was extremely hard for the observers to distinguish whether a particle is a residue or a precipitate thus creating mistakes on observation.

In order to improve on the mistakes made above, more time allotted would have helped the experiment. Increase of time would have helped the observers carry out the experiment in a more relaxed environment allowing better observations. Furthermore, time increase would have allowed the reactions to go to their completion allowing more accurate observations and results. Additionally, accurate measurement and calculation of the reactants would have helped to see a complete reaction. And finally, pre calculations of the reaction would have helped to determine whether a precipitate might form in a certain reaction or not allowing more accurate observations.

Work Cited

Brown and LeMay the Central Science Lab Manual

Brown, Lemay, Bursten, Murphy, Woodward, Stoltzfus. 2015. Chemistry the Central Science. 13th edition. United States of America: Pearson

Extraction of Caffeine

Extraction of Caffeine from Coffee by liquid/liquid extraction, tested for purity with Thin Layer Chromatography

Worked with Jaeone

  1. Introduction

Coffee has become a major part of modern society. It has been consumed since the 15th century. Due to its property of causing sleeplessness and increase in concentration when absorbed it is consumed largely by students or businessmen who are required to focus for a long time. However, the trend is diverging in the recent years. Scientists discovered that caffeine is the substance in coffee that contributes to the decreasing of the drowsiness. Thus, the demand for caffeine than raw coffee has increased significantly. In this experiment, caffeine will be extracted from coffee to be tested for purity which can be further used for other benefits.

A process in which an organic compound is separated from a mixture of compounds is called extraction. Liquid/liquid extraction in specific is a method of extraction through using solvent that can dissolve only certain compounds, leaving the rest undissolved. In this process, the solvent that has dissolved the specific compound is called the extract. An important point in liquid/liquid extraction process is that the two solutions, the original mixture and the solvent, has to have a different polarity so that they are immiscible with each other. Immiscible by definition is that two solutions cannot mix and maintain homogeneity. Thus an immiscible solution to the original mixture that is capable of dissolving the desired compound while leaving the unwanted compounds is ideal in separating a compound from the mixture. In this experiment, dichloro-methane (also called methylene chloride) was used. Since dichloro-methane is non-polar compared to water and since like dissolves like, it can dissolve more caffeine compared to water while being immiscible with water. However, a problem with using dichloro-methane in this experiment is that it can also dissolve tannin, a non-polar polyphenolic compound that is also present in coffee, thus making hard to isolate caffeine only from coffee. However, Tannin can be converted to salt by deprotonating its hydroxyl (–OH) group to (-O) group thus making the molecule highly polar. To solve the problem, sodium carbonate was used to react with tannin in order to produce phenolic anion, a salt which does not dissolve in dichloro-methane due to high difference in polarity. Furthermore, even though it has been assumed that dichloro-methane and water are highly immiscible, there may be a slight chance of water being dissolved through a phenomenon called emulsion. Emulsion is a process in which immiscible solutions mix due to droplets of either of the solutions getting trapped in the other solution’s bubble. To purify caffeine as much as possible, anhydrous sodium sulfate is added. Anhydrous compounds are substances that lack water of crystallization, thus absorbing water to create crystals of itself. Thus, when enough of anhydrous sodium sulfate is added to the solution, the salt crystallizes by absorbing the water in the solution, leaving the solution free of water, isolating caffeine.

To test for the purity of the caffeine that has been extracted, chromatography was used. Chromatography is a method of separating mixture. It uses two substances that vary in their polarity in which one dissolves the mixture, called mobile phase and one that does not, called stationary phase. To be more precise, stationary phase also builds the structure for the mobile phase to travel. In performing a chromatography, mobile phase is allowed to travel through the stationary phase. As it travels, due to difference in polarity for all compounds in the mixture, the distance that each compounds travels will vary. Thin layer chromatography, or TLC for short, is a branch of chromatography where a layer of absorbance on a glass plate is the stationary phase and the liquid that is being absorbed by the absorbance on the glass plate is the mobile phase. In this experiment, a non-polar absorbance, silica coating (SiO2) and a polar mixture of 5% acetic acid and 95% ethyl acetate was used. Rf value is a mathematical way of representing the result of TLC. It is the value of the distance that the substance traveled divided by the total distance that the mobile phase traveled. Thin Layer Chromatography was used for the testing of the result of the extraction since it allows the comparisons of the distance that pure caffeine and the extracted material travels thus making it easier to distinguish if the extracted material is made of pure caffeine or is a mixture including other impurities.

  1. Materials
    1. Brewing of Coffee
      1. Roasted and grinded coffee bean
      2. 500mL Beaker
  • 500mL Erlenmeyer flask with side arm
  1. Erlenmeyer flask stopper
  2. Filter paper
  3. Pump
  • Rubber cable
  • Glass cylinder funnel
  1. Glass support base with sand core permeable top
  2. Metal clamp
    1. filtration_vacuum-filtration-appartus
  1. Liquid/liquid extraction
    1. 6g sodium carbonate
    2. 75ml methylene chloride
  • Spoonful anhydrous sodium sulfate
  1. 250mL Beaker
  2. Separating funnel
  3. Plastic stopper
  1. Thin Layer Chromatography
    1. TLC plate coated with SiO
    2. Ethanol
  • Acetic Acid
  1. Ethyl Acetate
  2. Hexane
  3. Pure Caffeine
  • 10mL Beaker
  • Hot Plate
  1. Glass capillary
  2. Glass chamber
  3. Solid Iodine
  1. Procedure
    1. Brewing of Coffee
      1. Set the vacuum filtration apparatus by connecting the sidearm of the Erlenmeyer Flask to the pump by the rubber pipe, applying the stopper, connecting it with the glass base, setting a filter paper on the base, placing the glass cylinder on the set and holding it firm with the metal clamp like the Figure 1.1.
      2. In 500mL beaker, put all the roasted coffee powder and pour water to make 300mL of coffee. Stir it thoroughly to brew as concentrated as possible.
  • Pour the coffee into the glass cylinder and turn the pump on. If in the process of filtering, coffee bubble erupts and is about to go above the level of sidearm in the Erlenmeyer Flask, turn the pump off to slow the filtering process. Turn the pump back on when the bubble subsides.
  1. Liquid/liquid Extraction
    1. To the 300mL of coffee solution in the Erlenmeyer Flask, add 6g of Sodium Carbonate (NaCO3). Swirl the solution until all sodium carbonate is dissolved
    2. Divide the 300mL of coffee to two separation funnel for each group, 150mL of coffee each.
  • To each separation funnel, add 36.5mL of dichloro-methane. Close the funnel entirely and swirl it vigorously for 1 or 2 minutes. Do not swirl it excessively or emulsion will form. In the process of doing so, open the tab to allow the gas to escape. Repeat the process several times
  1. Allow the mixture to settle and separate into two layers, the dark top layer and a clear dichloro-methane bottom layer. A middle layer can exist, but it should not be significant.
  2. Carefully open the tab to pour out the bottom layer into a clean beaker. Pour out the middle layer into another beaker and the bottom layer into a separate beaker.
  3. Pour the middle layer into the same separation funnel and repeat process iii. by adding only as much amount of dichloro-methane as the solution itself. By this process, middle layer may not exist. If it does not, collect the bottom layer with the beaker used to collect the bottom layer in process v. and the top layer with the beaker used to collect the top layer. The top layers can be discarded.
  • To the dichloro-methane solution, add a spoonful of anhydrous sodium sulfate and swirl it vigorously. The salt will form crystals. When crystallization is no more observed, filter out the crystals of hydrous sodium sulfate using a filter paper and a funnel.
  • Let the dichloro-methane evaporate, leaving pure caffeine. Dry the solid and weigh the mass.
  1. Thin Layer Chromatography
    1. Measure out about the same amount of pure caffeine as the extracted caffeine.
    2. Dissolve it in 8mL of ethanol solution in a 10mL beaker.
  • Use a capillary tube to let it dissolve. You may use the hot plate to heat the solution to speed the dissolving process.
  1. Using the glass cutter, cut TLC plates, into rectangular pieces that will fit in the glass chamber.
  2. On the coated side of the TLC plates, drip the ethanol-caffeine solution into a small dot. Let it dry and drip it again. Repeat the process three to four times. Dry the solution and mark the position it has been dripped.
  3. Make a stock solution of 5% acetic acid, 95% ethyl acetate.
  • Pour a little amount of stock solution into the glass chamber and put the TLC plate in so that the dripped point is not below the solvent level but has enough distance to travel.
  • When the stock solution has traveled till the very end and is about to reach the end, take out the TLC plate and mark the starting point, and the ending point of the traveled solvent.
  1. To a dry glass chamber, add 20 to 30 beads of solid iodine.
  2. Place the TLC plate in the iodine glass chamber and wait until the plate has turned yellowish shows other dots above the originally marked points.
  3. Take out the TLC plate and quickly mark the newly made dots. Calculate the Rf value of the substance.
  • Repeat from process ii. to process xi. with the extracted caffeine. Also record the data.
  • Compare the Rf of the two solutions to check what you have extracted is caffeine.
  1. Data
    1. Qualitative Observation
      1. Liquid/liquid Extraction Process

The aqueous layer of coffee rested on top of the organic layer of dichloro-methane and caffeine. The top layer was dark brown and the bottom layer was transparent with a yellowish hue. The middle layer was a lighter brown than the top layer and had an oily texture to it.

When anhydrous sodium sulfate was added, the powder of sodium sulfate started to form spherical, white and non-transparent crystals. As the swirling continued for a longer time, the crystals broke down to smaller units of still spherical and solid shape.

The extracted caffeine was yellowish white compared to the pure caffeine which is purely white.  It was in powder form, stuck to the bottom of the beaker.

  1. Thin Layer Chromatography

The chromatography results of the pure caffeine contained only a single dot. However, the result of the extracted substance had two define dots with different Rf values.

  1. Thin Layer Chromatography Results
    1. Chromatography of pure caffeine
      • Distance from the origin to the dot
        • 5mm
      • Distance from the origin to the end point of the solvent travel
        • 5mm
  1. Chromatography of the extract
    • Distance from the origin to the first dot
      • 0mm
    • Distance from the origin to the second dot
      • 0mm
    • Distance from the origin to the end point of the solvent travel
      • 0mm
  1. Analysis
    1. Analysis of the Qualitative data

Caffeine has a pure white color. Thus, looking at the fact that dichloro-methane solution and the final, dry extract had a yellowish color shows that it would have contained impurities.

The fact that anhydrous sodium sulfate broke down to smaller units show how reaction of absorbing of water has gone to completion. This is because if reaction did not go to completion, and the solution still had water, the anhydrous sodium sulfate would keep forming bigger crystal. However, since it did not, it can be safely stated that the reaction has completed.

Chromatography result shows that the extracted substance had one more dot compare to the chromatographic result of the pure caffeine. This implies that the extracted substance contained impurity other than caffeine.

  1. Analysis of the Chromatographic Results and Calculation of Rf Values
    1. Calculation of Rf value
      1. Rf calculation
    2. Rf value of pure caffeine
      1. Rf value of pure caffeine
    3. Rf value of the Extracted Caffeine
      1. Rf value of extracted caffeine

Looking at the evaluation of the Rf values, it can be inferred that the second dot was caffeine and the first dot, impurity.

  1. Discussion

It can be easily inferred that this experiment has failed to isolate caffeine only. Firstly, caffeine is a single compound that does not decompose due to chromatography. Thus, in chromatography, pure caffeine should result in only one dot. However, the chromatographic result of the extracted substance turned out to have two dots. This shows that the experiment has failed to isolate caffeine only. This is further shown through the fact that the chromatography result of the extracted substance did not exactly match the chromatography result of the pure caffeine. As analyzed above, the first dot that appeared in TLC can be interpreted as the impurity since it does not match any of the Rf value. On the other hand, the second dot, whose Rf value matches the one of pure caffeine can be interpreted as the extracted caffeine.

To discuss about the impurity that has been measured, there seems to be two possibilities; water and tannin. Of course, impurities in beaker and several human mistakes would have created this mistake too. However, to be only reasonable and sound reason for mistake was discussed. Firstly, the fact that the reaction between anhydrous sodium sulfate and water has gone to completion, dealt with in analysis section, shows that it couldn’t have been water molecule left in dichloro-methane. On the other hand, tannin seems to be a likely candidate for the impurity. As explained in the introduction, tannin, is another stimulant commonly found in coffee. It is non-polar like caffeine, even more so than caffeine due to its high molecular mass, making itself easy to be dissolved in dichloro-methane in the process of liquid/liquid extraction. Furthermore, even though it has been tried to eliminate as much tannin as possible by deprotonating them through dissolving carbonate ion, if carbonate ion dissolved was not enough to eradicate all tannin, some tannin would have taken part in the reaction, getting dissolved in dichloro-methane and showing its appearance in chromatography. To explain the chromatographic results, the impurity, assumed to be tannin, has traveled less than caffeine along the non-polar stationary phase. It can be safely said that tannin is more so non-polar even compared to caffeine due to its polyphenolic structure and high molecular mass. This makes tannin more stable with non-polar stationary phase than caffeine is. On the other hand, caffeine is more stable with polar mobile phase. Thus, caffeine will travel further than tannin with polar mobile phase due to the principle of like-dissolves-like making tannin not travel further than caffeine.

This mistake as discussed above should have been created due to the lack of sodium carbonate in the coffee solution leaving non-polar tannin behind. Thus, to improve the experiment, more sodium carbonate should have been dissolved in the solution, even past its saturation point to be filtered. This would have completely deprotonated tannin leaving only caffeine to be the non-polar molecule in the coffee solution. This would have definitely isolated caffeine.

Work Cited

Nam, Y. E; Hill, M. P; Randall, J; Organic Chemistry with Vernier; Vernier Software and Technology, 2012

Ramalakshmi, K; Raghavan, B; Caffeine in Coffee: Its Removal. Why and How?. Crit. Rev. Food Sci. Nutr., 2010, 39, 441-456

Extraction of Caffeine from Tea. http://vlab.amrita.edu/?sub=3&brch=64&sim=169&cnt=1 (accessed April 18, 2016)

For the Actual Document click Extraction of Caffeine

Synthesis of Aspirin Lab

Synthesis of Aspirin

Worked with Grace Choi

  1. Introduction

In status quo, aspirin is a commonly bought, widely used over-the-counter drug. It has lots of functions: it can reduce fever, pain, swelling, soreness and redness. Historically, its first account was by a Greek physician called Hippocrates when he recorded the effects of the powder created from the willow tree at curing childbirth and fever. Then, its active ingredient, salicylic acid was first discovered by an English pastor, Reverend Edward Stone, when he administered powdered willow bark on patient and observed its effect on them. Friedrich Kolbe then successfully described both structure of salicylic acid and its method to synthesize it. Salicylic acid was used to cure headache and pain since then. However, for its high acidity, salicylic acid could not be prescribed for long. Felix Hoffman then synthesized acetylsalicylic acid to activate the effects when decreasing the irritation in body, and named it aspirin. (Szczeklik 1) Proved by history, aspirin has lots of pharmacological benefits for curing illness. In this experiment, aspirin was synthesized with salicylic acid and acetic anhydrate.

Aspirin is synthesized through a mechanism, now called aspirin synthesis mechanism. For the sake of time and efficiency, sulfuric acid was added as a catalyst to speed up the reaction. In the process of aspirin synthesis mechanism, the protons of the acid reacts with the acetic anhydrate to bond with both oxygen with double bond and in between the acetate groups. This will create positive charge on both the protonated oxygen. Thus, the oxygen will hog the electrons from the C=O double bond to move its positive charge to carbon. Salicylic acid is consisted of a phenol attached to a carboxylic acid group. The oxygen in phenol, being more electro negative than carbon will be slightly negative than its surrounding. This will attract the positive carbon creating a bond with it. To release the positive charge on oxygen, the deprotonated acid will take away H+. Then, the protonated oxygen in between the acetate group, having a positive charge, will hog electrons from the C-O bond that it has with the carbon nearer to the phenol group. This will break the bond between oxygen and carbon creating acetic acid. Due to this, now carbon has a positive charge which will attract the lone pairs of oxygen creating a double bond, yet again, creating positive charge on oxygen. However, the second deprotonated acid will take away H+ from the oxygen to minimize formal charge. The final product is the phenol group attached to a carboxylic group and an acetate group, aceticsalicylic acid, aspirin, and acetic acid. To visualize the mechanism, a figure below can be used.

Aspirin synthesis mechanism

Figure 1

When the synthesis was over and the crystals formed, water was added to destroy all unreacted acetic anhydride.

Both salicylic acid and aspirin are not very polar. Thus, they are slightly soluble in water when being highly soluble in ether or other organic compounds. (Rainsford 77, 78) Thus, acetic anhydrous, being an ether and an organic solvent that can dissolve salicylic acid and aspirin easily, was used as the solvent in which the reaction took place in. However, due to their large molecular mass, salicylic acid and aspirin has limited solubility meaning it is still not highly soluble as common salts such as NaCl. Due to this property, the solubility of both salicylic acid and aspirin depends largely on the temperature of the solvent. Thus, cold water and ethanol was used to rinse the solution and the solvents in which salicylic acid and aspirin needed to be dissolved was placed in the warm bath to increase temperature.

Recrystallization is a process which increases purity of crystallized solid product. When aspirin crystals are collected, there are possibilities that impurities that are still soluble in polar solvent remain in solid form. Thus, the solid product was dissolved in alcohol, a polar solvent, with heat. Then, the crystals were collected again.

To check for impurities that might remain in the product, titration and back titration method was used. Titration is a laboratory procedure that is used to determine the concentration of an unknown solution using a solution of known concentration. It is done by slowly adding the acid or base solution of known concentration to the solution of unknown concentration to look for the point where all substances are neutralized. Similarly, back titration is a process that is used to ensure the process of titration. It is carried out by slowly adding the solution of opposite acidity of known concentration to the solution that is once titrated and is added with excess of the first titrant in order to look for the point where the solution is neutralized completely. Both salicylic acid and aspirin, containing a carboxylic acid group, is acidic in nature and they both, along with other acid impurities, may exist in the final product. However, aspirin out of all compounds not only can deprotonate using NaOH but also can use NaOH to turn itself back into salicylic acid through a slow process called hydrolysis as shown in figure 2. As shown in figure 2, aspirin reacts with NaOH in 1:1 ratio to produce salicylic acid. Thus, after the first titration in which all acidic compounds are deprotonated, more NaOH was added to react with aspirin to form hydrolyzed aspirin, salicylic acid. When all aspirin has reacted with NaOH to hydrolyze, HCl was added to react with excess NaOH. Therefore, at the point of neutralization, (the moles of excess NaOH added after the first titration) – (the moles of HCl added for back titration) equals to the moles of NaOH reacted with aspirin to produce salicylic acid. This value also equals to the amount of aspirin that existed in the product in the first place since aspirin did not hydrolyze during the first titration, just deprotonated. Therefore, to obtain the purity by percent, the following equation can be used.

Titration of Aspirin

Figure 2

  1. Apparatus
    1. Salicylic Acid
    2. 125mL Erlenmeyer Flask
  • Acetic Anhydride
  1. 16M Sulfuric acid
  2. 500mL beaker
  3. Heating plate
  • Stand
  • Clamp
  1. Pipet
  2. Cold water and ethanol prepared in fridge
  3. 250mL beaker
  • Vacuum Filtration set
    • 500mL Erlenmeyer flask with side arm
    • Erlenmeyer flask stopper
    • Filter paper
    • Pump
    • Rubber cable
    • Glass cylinder funnel
    • Glass support base with sand core permeable top
    • Metal clamp
  • Weighing boat
  • Phenolphthalein Indicator
  1. Distilled Water
  • 0.1M NaOH
  • 0.1M HCl
  • Buret
  • Magnetic Stirrer
  1. Procedure
    1. Synthesis
      1. Weigh 40 of salicylic acid in a 125mL Erlenmeyer Flask. Record the exact mass.
      2. Add 6mL of acetic anhydride to the flask
  • Add 5 drops of sulfuric acid to the flask, swirl gently, and place the flask in a beaker of boiling water. Clamp the flask to a stand and heat for 20 minutes. Stir with glass rid, or if not present, glass pipet. Solid must all dissolve
  1. Remove the flask from the water bath and cool in room temperature. It should crystalize.
    • If crystals grow, let them grow until it stops. Pour 40mL of cold water
    • If not, pour the solution into a 250mL beaker of 40mL of cold water, mix thoroughly.
      • Water will destroy unreacted acetic anhydride and insoluble aspirin will precipitate in solution
  1. Collect the crystal by vacuum filtration.
  2. Rinse the crystal with 10mL of cold water and 10mL of Ethanol, let it dry and weigh it.
  1. Recrystallization
    1. Dissolve your crude product in about 20mL ethyl alcohol in a 125mL Erlenmeyer flask. Warm the alcohol in a warm bath to speed up dissolution.
    2. If any solid remains, filter it.
  • Add 50mL of warm water to clear alcohol solution.
  1. Keep heating to dissolve it.
  2. Set the flask aside to cool.
  3. If crystals form, cool the flask by surrounding it with cold water.
  • Collect the crystals by vacuum filtration.
  • Allow the crystals to dry.
  1. Save the sample.
  1. Analysis by Titration
    1. Measure out 0.5g of the synthesized Aspirin.
    2. Dissolve it in 15mL of distilled water. Add 5 drops of Phenolphthalein Indicator.
  • Use a stock solution of 0.1M NaOH solution to titrate the aspirin solution
  1. When the indicator turned color, add twice as much of the NaOH used for the titration to the solution. Add as much of the NaOH as the Aspirin solution first acquired (15mL).
  2. Heat the solution in warm bath for 10 minutes.
  3. Back titrate the solution using 0.1M of HCl.
  • When the solution turned color, test for the accuracy by adding drops of NaOH and HCl to titrate and back and forth. If the titration was accurate, the solution will change color with a drop of NaOH or HCl.
  1. Result
    1. Synthesis
      1. Mass of salicyclic acid + weighing boat
        • 807
      2. Mass of weighing boat
        • 777
  • Mass of Salicyclic acid
    • 030g
  1. Mass of Weighing boat
    • 781g
  2. Mass of Aspirin and weighing boat
    • 629g
  3. Mass of Aspirin
    • 048g
  4. Recrystallization of Aspirin
    1. Mass of filter paper
      • 785g
    2. Mass of filter paper + Recrystallized Aspirin
      • 069g
  • Mass of recrystallized aspirin
    • 284g
  1. Analysis by titration
    1. Mass of Aspirin used for titration
      • 503g
    2. Volume of NaOH used for first titration
      • 24mL
  • Additional NaOH included
    • 24 + 15 (mL)
    • 39mL
  1. Total volume of NaOH used
    • 24 * 24 + 15 (mL)
    • 63mL
  2. Volume of HCl used for back titration
    • 5mL
  3. Analysis
    1. Purity test
      1. Mass of Aspirin first created
        • 048g
      2. Mass of Aspirin after recrystallization
        • 284g
  • Mass of impurity discarded by recrystallization
    • 764g
  1. Percent purity by titrationpurity calculation
  2. Test for purity
    • 39mL*0.1M*1L/1000mL – 15.5mL*0.1M*1L/1000mL = 0.00235moles = total moles of aspirin
    • 00235moles*180.16g/mol = 0.423g = total mass of aspirin that is present in product
    • (Total mass of product) – (Total mass of aspirin) = (Total mass of impurities)
      • 503g – 0.423g = 0.08g
    • (Total mole of NaOH used to titrate all acidic compound of solution) – (Total mole of NaOH used to hydrolyze aspirin) = (Total mole of NaOH used to titrate impurity)
      • 24mL*0.1M*1L/1000mL – 23.5mL*0.1M*1L/1000mL = 0.5mL*0.1M*1L/1000mL = 0.00005mole of NaOH used to titrate impurity = Total mole of impurity assuming 1:1 reaction ratio
    • 08g/0.00005mole = 1600g/mol = average molar mass of impurity assuming full reaction and 1:1 reaction ratio
  3. Discussion

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