Equilibrium

I. Purpose

To study how stress affects a system’s equilibrium and to determine the concentration of the complex ion in solutions of unknown concentration as well as the equilibrium constant.

  1. Procedure

LeChatelier’s Principle

Part 1

Solid sodium chloride was poured into a test tube and the test tube was then filled ¾ full of distilled water.  The solution was then decanted into a second test tube and Cl- ions in the form of concentrated HCl were added.

Part 2

A small test tube was filled about half full of distilled water.  Several drops of bromthymol blue indicator solution and 5 drops of 0.1 molar hydrochloric acid were added, and the resulting color was noted.  0.1 molar sodium hydroxide was then added drop by drop with stirring until no further color changed occurred. The color was then noted.  The attempt was then made to add the right about of acid to the test tube to cause the solution to be green in color after stirring.

Part 3

About 25 mL of 0.0020 molar KSCN solution was poured into a beaker.  25 mL of distilled water and 5 drops of 0.20 molar Fe(NO3)3 solution were then added and swirled.  The color of the KSCN and Fe(NO3)3 solutions were noted, as well as the color of the resulting complex ion. Equal amounts of the solution were poured from the beaker into four test tubes.  The solution in the first test tube was the reference solution.  To the second test tube 2-3 crystals of KSCN were added and the results were described.  To the third test tube 6 drops of Fe(NO3)3 solution were added, and the solution was stirred and the results noted.  To the fourth test tube small crystals of Na2HPO4 were added a few at a time and stirred into the solution; the results were noted.

Part 4

About 10 mL of ethanol was measured in to a beaker.  Solid cobalt (II) chloride was examined, with its color and formula noted, then a small amount was dissolved in the beaker of ethanol.  About 2 mL of the alcoholic cobalt solution was added into each of 3 small test tubes.  To the first test tube, 3 drops of distilled water were added, one drop at a time with stirring.  3 drops of distilled water were added to the other 2 test tubes and the effect was noted.  Using the first test tube as a control, 5 drops of 12 molar HCl was added one drop at a time with stirring and the result was noted.  To the third test tube a few crystals of sodium chloride were added and stirred.  The remainder of alcoholic cobalt solution from the beaker was put into a fourth test tube.  10 drops of 0.1 molar silver nitrate solution were added one drop at a time, and the color of the solution was noted.  A sealed Beral pipet containing some of the alcoholic cobalt chloride-water system was obtained and the large end of the pipet was immersed in some hot water and the color change was noted.  The Beral pipet was then chilled in an ice bath to see if the color change in the previous step was reversible.

Equilibrium Constant of FeSCN

Part 1                                                           

Using burets or pipets, 2.0, 3.0, 4.0, 5.0, and 6.0 mL of 2.0 X 10-3 M KSCN  in 0.50 M  HNO3 was measured into 100 mL volumetric flasks and diluted to 100 mL with 0.20 M Fe(NO3)3 in 0.50 M HNO3 and mixed well.  The concentration of FeSCN in each flask was then calculated, assuming the all of the SCN had reacted.  The absorbance of each of the standard solutions were measured at 445 nm, using distilled water as the reference in the spectrophotometer.  The molar concentration of FeSCN was then plotted against absorbance and the best fitting line was drawn through the data points.

Part 2

Using a buret or pipet, the following quantities were measured into 5 test tubes.  The solutions were mixed well with a glass rod and the absorbance of each was measured at 445 nm using distilled water as a reference.

Solution

2.0 X 10-3 M Fe(NO3)3 in 0.50 M HNO3

2.0X10-3 M KSCN in 0.50 M HNO3

0.50 M HNO3

1

5.0 mL

1.0 mL

4.0 mL

2

5.0 mL

2.0 mL

3.0 mL

3

5.0 mL

3.0 mL

2.0 mL

4

5.0 mL

4.0 mL

1.0 mL

5

5.0 mL

5.0 mL

0.0 mL

 II. Data

LeChatelier’s Principle

Part 1

A white solid precipitated at the top of the test tube.

Part 2

The result of step 1 is a yellow solution.  The number of drops of acid needed to create a lime green color is 40 drops.

Part 3

KSCN is a colorless solution.  Fe(NO3)3 is a purple solid.  The second test tube was blood red, the third test rube was a burnt orange color, the fourth test tube was clear, and the resulting complex ion was an orange color.

 Part 4

Step 8 caused the solutions to become more purple.  Step 9 caused the solution to become bright blue.  When placed in an ice bath, the solution turned pink.  When placed in a hot water bath, the solution reversed back to blue.

Equilibrium Constant of FeSCN
Part 1

Solution

mL 2.0*10-3 M KSCN in 0.50 M HNO3

% Transmittance

Absorbance

Concentration FeSCN, M

1

2.0

78.9

0.103

4.0 X 10-5

2

3.0

62.1

0.207

6.0 X 10-5

3

4.0

52.1

0.283

8.0 X 10-5

4

5.0

45.5

0.342

1.0 X 10-4

5

6.0

40.0

0.398

1.2 X 10-4

                                                Part 2

Solution

mL 2*10-3 Fe3+ in 0.50 M HNO3

ML 2*10-3 SCN- in 0.50 M HNO3

% Transmittance

Absorbance

Initial moles Fe3+

1

5.0

1.0

97.5

0.009

1.0*10-5

2

5.0

2.0

94.1

0.024

1.0*10-5

3

5.0

3.0

91.0

0.039

1.0*10-5

4

5.0

4.0

87.3

0.052

1.0*10-5

5

5.0

5.0

85.9

0.065

1.0*10-5

Solution

Initial moles SCN-

Concentration FeSCN2+, M

Moles FeSCN2+ at Equilibrium

Moles Fe3+ at Equilibrium

Moles HSCN at Equilibrium

1

2*10-5

2.7*10-6

2.7*10-7

9.7*10-6

2.0*10-5

2

4*10-5

7.1*10-6

7.1*10-7

9.3*10-6

4.0*10-5

3

6*10-5

1.2*10-5

1.2*10-6

8.8*10-6

5.9*10-5

4

8*10-5

1.5*10-5

1.5*10-6

8.5*10-6

7.9*10-5

5

1*10-4

1.9*10-5

1.9*10-6

9.8*10-5

9.8*10-5

 

 

Solution

Concentration HSCN at Equilibrium, M

Concentration Fe3+ at Equilibrium, M

Concentration H+ at Equilibrium, M

Equilibrium Constant Kc

1

2.0*10-4

9.7*10-5

0.50

70

2

4.0*10-4

9.3*10-5

0.50

95

3

5.9*10-4

8.8*10-5

0.50

116

4

7.9*10-4

8.5*10-5

0.50

112

5

9.8*10-4

9.8*10-5

0.50

99

 

 Average Kc = 98

Calibration Curve for Absorbance

 IV. Calculations

 

Equilibrium Constant of FeSCN

(See Data section for complete list of final calculation answers)

Part 1

Concentration of FeSCN

Equilibrium Constant

Concentration Fe at Equilibrium

9.7*10-6 / 0.1 L = 9.7*10-5 M Fe

Equilibrium Constant

Equilibrium Constant

V. Conclusions and Error Analysis

LeChatelier’s Principle

 

In this experiment the effect of stress on a system’s equilibrium was observed.  As more HCl was added in the second part, the equilibrium shifted to move the reaction in the reverse direction. In the third part, the color of the solution shifted as more of either KSCN or Fe(NO3)3 were added, shifting the equilibrium.  In the fourth part, the effect of temperature on equilibrium was observed.  As the solution sat in the ice bath, it turned pink because equilibrium shifted in the reverse direction because it had a ΔH of 50 kJ/mol.  As the solution warmed back up in the hot water bath, the reaction shifted again to favor the products.  In the first part of the experiment, the effect of a change in concentration was observed.  The HCl that was added reacted with the sodium chloride to form more solid sodium chloride on the top of the solution do to over-saturation.

 

Equilibrium Constant of FeSCN
The average Kc obtained was     . From the data it can be concluded that the concentration of FeSCN is inversely proportional to the absorbance and directly proportional to the amount of SCn used as a reactant.

 

Based on the obtained R2 value of     it can be inferred that only a small amount of error was obtained up to this point.  However, some explanations for the small amount of error that did occur are not setting the features of the Spec 20 correctly, contamination of the chemicals, and aging of the chemicals.  If the chemical were aged or contaminated, this would affect the molarity and therefore the absorption.  If the settings on the Spec 20 were incorrectly adjusted, this would affect the absorption number obtained.  If an error was made in the calculation for the concentration of FeSCN using the data from the graph, this would affect the calculated number of moles of FeSCN, and therefore affect the calculations of the number of moles and concentrations of the Fe and SCN, as well as the Kc value.

 

Ways to prevent these types of error are to double check the Spec 20 settings before use and to use only fresh chemicals to avoid aging and impurities.

 

Because 100. mL of solution was mixed in a volumetric flask after being measured in a graduated cylinder, some of the solution may have been retained in the cylinder and was therefore unable to react with the other chemicals in the solution, which would affect the absorbency.  Because such small quantities of solution were used in part 2, any error in measurement would greatly affect the absorption.

 

Some ways to avoid this type of error are to be as accurate as possible when measuring and making allowances in calculations for lost or spilled solutions.

VI. Discussion of Theory  

 

Equilibrium occurs when the forward reaction and the reverse reaction are acting at the same rate.  In the first experiment, the effects of temperature, saturated solutions, and the addition of more of either reactants or products were observed.  As stress is applied to a system, the system responds accordingly in order to maintain a state of equilibrium.  For example, if an endothermic reaction were cooled, the reaction would shift in favor of the reactants in order to relieve the stress, as demonstrated in the following equation:

form 1

 

In the second experiment, the equilibrium constant was determined for a reaction taking place in solutions of unknown concentration.  The average calculated value of the equilibrium constant (98) is relatively high, which indicates that the reaction’s equilibrium favors the products side of the reaction as shown in the following equation:

form 2

A spectrophotometer is an electronic device that measures the absorbance and percent transmittance of a solution by comparing the light waves accepted by the standard solution to the control solution (in this case distilled water).  The standard solutions were obtained by diluting particular amounts of a solution of known molarity in order to obtain a percent transmittance.  The molarity of the diluted solution is then obtained by using the formula M1V1 = M2V2, which is

then plotted against the absorbance to produce a calibration curve graph.  This graph is then used to calculate the number of moles and concentrations of the FeSCN, Fe, and SCN solutions, as well as the Kc values. Three significant figures including one estimated digit can be obtained using the spectrophotometer.  The major source of error that stems from using the apparatus is that it can be unreliable if the absorption is measured using the device rather than calculating it from the percent transmittance.  Other experiments in which a spectrophotometer would be useful are those which are designed to determine a chemical substance’s identity by using the color transmitted by its electrons.

 

spectrophotometer

Throughout the experiment, 0.5 molar nitric acid was used in all solutions in order to maintain a constant molarity of H+.  If a higher concentration of nitric acid had been used, this would have caused the Kc value to increase due to the fact that the hydrogen was a product, and therefore a numerator in the Kc calculation.

Determining a Reaction Intermediate

By: Jaime Rodriguez

Background:

In this experiment, the concentrations of KMnO4 in each test tube were considered similar if their respective colors were similar. This step is what made it possible to calculate the amount of hydrogen peroxide in the original solution of melted ice. Unfortunately, the human eye is not keen to accurately detect the change in color intensity; therefore the colors in both test tubes may not have been very similar at all. Besides this method, there is another, more accurate, method that could determine when all of the hydrogen peroxide was consumed in the reaction. With the use of a ph monitor, the total consumption of the hydrogen peroxide could be determined. ( ph of H2O2 Solutions). In the reaction, the addition of the hydrogen peroxide turns the solution from basic to acidic. After all of the hydrogen has been consumed, the solution will return to a base. If the shifts in ph are monitored, it can accurately be determines when all of the hydrogen peroxide has been consumed.

Procedure:

No changes were made to the procedure. It was followed as it was written.

Data and Results:

detecting a reaction intermediateDetecting a reaction intermediate 2

Table of results:

Table of results

The number of moles of excess MnO4- present was calculated by taking the concentration of KMnO4 from the test tube with deionized water, and multiplying it by the total volume of solution in the test tube with melted ice.

6.87*10^-7M * .00552L

=3.8*10^-8 moles

The number of moles of MnO4- which reacted was found simply by subtracting the number of moles in excess by the number of moles added.

1.008*10^-7 moles  – 3.8*10^-8 moles

=6.28*10^-8 moles

The moles of H2O2 which reacted was found by using the stoichiometric ratio of number of moles of H2O2  to number of moles of MnO4- .

6.28*10^-8 moles of MnO4- * (5 moles of H2O2/2 moles of MnO4- )

=1.57*10^-7 moles

Finally, the concentration of H2O2  in the original melted ice solution was calculated by dividing the number of moles of H2O2  which reacted by the volume of melted ice.

1.57*10^-7 moles/.00472L

=3.33*10^-5 M

The reaction between H2O2  and MnO4- might proceed more slowly at the end because by this time, almost all of the H2O2  is being consumed in the reaction. The concentration of hydrogen peroxide that was measured was 3.33*10^-5 M

To calculate the free energy of the reactions for hydrogen gas reacting with gaseous water to form gaseous hydrogen peroxide, and gaseous hydrogen peroxide reacting with hydrogen gas to form water, were found by using the Gibbs free energy equation.

The equation states that,  delta G=delta H-delta S(T). To find delta S and H, the values of H and S for each of the reactants and products were found. Then, the sum of the products was subtracted by the sum of the reactants. Once this was done, the values were put in to the Gibbs free energy equation, and the temperature used was 198.15 K, room temperature. The free energy values were -105.5kJ, and -351.5kJ, respectively. The free energy values for these two equations should be equivalent to the free energy associated with the combustion of hydrogen, since both are intermediate reactions, with hydrogen peroxide being the reaction intermediate. Below is the corresponding potential energy diagram.

reaction coordinate

Reflection

The experiment was easy, relatively quick, and yielded good quality data. Some of the calculations got confusing, but they were eventually figured out. There were really no issues with the experiment, and it was overall relatively enjoyable.

References:

pH of H2O2 solutions? | H2O2.com – US Peroxide – Technologies for Clean Environment.” US Peroxide – Technologies for a Clean Environment. N.p., n.d. Web. 22 Apr. 2013. <http://www.h2o2.com/faqs/FaqDetail.aspx?fId=26>.

 

Diffraction Patterns from Polystyrene Beads and from Breath Figures

By: Ludmila Novikova

Abstract:

Before X-ray diffraction of crystals was applied to determine the arrangement of atoms in 1913, little was known about the geometric and stereochemical shape of molecules and about the science of crystallography in general. In fact, the atomic theory of matter was not completely accepted because there was no theory available to explain the existence of compounds of the same composition (Flack, 372). X-ray crystallography presented us with methods that would allow us to strike a chosen crystal with a beam of X-rays so that we could measure and interpret the diffraction patterns that would be produced from the crystal. The calculations of the angle of diffraction, wavelength paths and distances between the atoms in the crystal are useful in determining the positions of atoms in a crystal, their chemical bonds, electron density, and much more.  The resulting information is then used to analyze the structure and function of bio molecules. We can already get a sense of how important and advantageous the study of crystallography is, and one of the primary objectives in this chemistry project was to study diffraction patterns of crystals to better understand diffraction phenomena.  Throughout the semester, I worked with small sizes of polystyrene beads to analyze, compare, and interpret the outcomes of the scattering and diffraction patterns given out by the crystallites. The size of the beads had to small because they would create diffraction patterns that could have defined and distant spacing lengths between “points” of the crystal. These polystyrene beads ranged from .5mm to 3 mm in diameter and they were diluted in different mixtures of water, ethanol, glycerol, cationic surfactant, and even with a small amount of potassium chloride solids. They were then delivered onto glass slides, covered with glass cover slips so that then they could be placed into an apparatus that held the slide in place. A helium-neon laser was then used to shoot a beam of emitted photons of the same wavelength (632.8nm) so that diffraction patterns could be produced. It should be noted that the monochromatic light had to pass through a magnifying lens that focused in on one sample of the microscope slide. The goal in this crystallography project was to create a monolayer of polystyrene beads on the glass slides to see hexagonal packing. Measurements of the angle of diffraction and distances between a set of “points” could then be carried out and the calculations could then be applied to make an analogous method for finding protein structures. Towards the end of this project, it was also found that diffraction patterns could also be given out by breath figures. The condensed tiny droplets of water formed an epitaxial film on the glass slides and this observation turned out to be useful because breath figures are smaller than .1mm.  Overall, much was learned about the study of crystals, spectroscopy, diffraction gratings, and breath figures to come to the conclusion of how X-ray crystallography can help us study internal structure of crystalline materials.

That we find a crystal or a poppy beautiful means that we are less alone, that we are more deeply inserted into existence than the course of a single life would lead us to believe.”    ~John Berger

Introduction:

Let us start off with the definition of a crystal. A crystal is a crystalline solid whose atoms are arranged in a repeating pattern. At a nano scale, the crystals’ complex arrays of atoms rarely have a perfect internal structure, and most often the crystals have misalignment of the arrangement of these atoms.

In theory, if we take a crystal (using high tech equipment) and we want to determine its crystal structure, we have to apply mathematical equations to the diffraction patterns it gives off when X-rays are shot at it. In X-ray crystallography, the X-rays are waves of electromagnetic radiation that are scattered by atoms. The X-rays essentially produce a diffraction pattern because their wavelength is typically the same order of magnitude as the spacing between the diffracting planes and the spacing between the points of the atoms. These patterns can be seen on some type of slide that shows the image of the diffraction pattern.

The diffraction pattern is produced because the electromagnetic waves from the X-rays interfere with each other constructively and destructively which results in the image we see when the X-ray light interacts with the crystallites.

When we finally determine the crystal structure of the crystal we’re studying, we can correlate its structure to the structure of bio molecules such as proteins. The applications of X-ray crystallography can help us better understand the structure of the proteins, and since “structure” of proteins is related to their “function”, we may also learn more about the function of the various proteins we might be studying.

Background Chemistry:

Crystal Structure Determination: A crystal’s structure may be determined if we take a sample of a small crystallite and measure the variations of the intensity of radiation passing through a portion of the crystal. We must then find the angle of diffraction between the diffracting planes and we must measure the path lengths of the rays.

There are many definitions for different terms when we’re discussing various locations on a crystal. Below is a list of the basic definitions applied to all crystals.

crystal_cove
  1. Lattice: An infinite array of points in space. The points have identical surroundings to all other points.
  2. Crystal Structure: The periodic arrangement of atoms in the crystal.
  3. Unit Cell: The smallest component of the crystal. Many unit cells compromise the crystal.
  4. Asymmetric Unit: Fraction of a unit cell

As for the mathematical equations that are applied to the diffraction patterns, most of them branch off from Bragg’s Law of diffraction. The Bragg’s Law of diffraction gives the angles for coherent and incoherent scattering from a crystal lattice.

Bragg’s Law of Diffraction: nλ=2dsinθ

It is for this law that we can confirm the existence of real particles at the atomic scale. Why? Because the tiny particles we’re observing scatter light. And although white light (400-700nm) consists of waves that have different intensities, the objects that scatter this light have different scattering patters. This is important to note because tiny objects usually do not give significant scattering of the light, but if we use a beam of X-rays, which have a wavelength of 10-35nm, we’ll then be able to see the scattering pattern from those tiny molecules. This is true because the size of the tiny objects correlate to the wavelength of the X-rays. Low wavelength (like X-ray wavelength) correlates to high energy. We can conclude that a lot of energy is proportional to high frequency.

Diffraction is the interaction of radiation with matter. In this project, the “matter” refers to polystyrene crystallites and the “radiation” refers to monochromatic light from a laser.

Polystyrene Beads: Beads of aromatic polymers. Polystyrene is commonly used in the industry of plastics. It is also solid at room temperature and the chemical formula is C8H8.

polystyrene

In this project, we are using monochromatic light because its wavelength is 632.8nm, it has a narrow frequency, and the energy levels are not strong enough to melt the polystyrene beads. Basically, if the wavelength is increased into the range of white light, we will not see any diffraction points. We must keep the wavelength 1.33 X 10-6  or lower to be able to see diffraction points. When the laser light is shot out at a crystal, the crystallites emit and absorb energy as they jump from an excited energy state to a ground state. The emitted and absorbed radiation occurs because of the interaction between the beads and the monochromatic light.

When many polystyrene beads are atop of one another, and monochromatic light is shot at the beads, the scattering pattern presents us with circular diffraction, called powder patters. These powder patterns are circular because the polystyrene beads are the same size and they’re giving out diffraction patterns in all directions. When there is a monolayer of polystyrene beads, you get a diffraction pattern. If you take a picture of this diffraction pattern, you can find the angle of diffraction using your knowledge of the wavelength of the laser, and your knowledge of the size of the polystyrene beads, and your measurements of the spacing between the “points” on the diffraction pattern.

One of the problems in X-ray crystallography, when determining crystal structure, is the issue of observing a crystal that is too small. It takes time to grow the perfect crystal and once you’ve finally chosen the crystal you’d like to study, you must keep in mind that when the crystal is placed on the goniometer head (X-ray crystallographic device), the crystal is rotated in the X-ray beam, so that it gives off diffraction patterns that reflect the radiation at different levels of absorption. This absorption depends on the path lengths of the X-rays through the crystal and it changes as the crystal is oriented. The mathematics is done through a computer programed system, but factors such as these are crucial in order to understand how the crystal structure correlated to the structure of various bimolecular chemicals.

To avoid crystallography problems in our experimentation, you must make sure to keep the different sizes of the polystyrene beads away from each other because they might mix in with each other and you will never know because the beads are tiny.

 

 

 

 

Pre-Laboratory Quiz:

1)      What is a crystal?

A crystal is a crystalline solid whose atoms are arranged in a repeating pattern.

2)      What is diffraction?

Diffraction is the interaction of radiation with matter

3)      Why do we need to use monochromatic light when observing 1mm polystyrene beads?

We are using monochromatic light because its wavelength is 632.8nm, it has a narrow frequency, and the energy levels are not strong enough to melt the polystyrene beads

4)      What would happen if we used white light to observe the diffraction patters from the polystyrene beads?

We would not see any diffraction patterns because the wavelength of the white light would be too long and it would not correlate to the size of the beads.

5)      What is the name of the X-ray crystallographic device that is used to determine the crystals’ structure?

Goniometer

6)      Name one problem that scientist might encounter when working with crystals?

The crystal might be too small to have its structure determined. It takes time to grow enough of the crystal for it to be observed and analyzed.

7)      What is the Bragg’s law of diffraction?

nλ=2dsinθ

8)      What are powder patters?

Powder patterns are circular diffraction patterns because the polystyrene beads (for example) are the same size and they’re giving out diffraction patterns in all directions.

9)      What is the chemical formula for polystyrene?

C8H8

10)    What is one benefit from determining the crystal structure of the crystal being studied?

We can use the crystal structure to understand the structure of proteins.

 

Laboratory Experiments: Flowchart of the Experiments

 

Section A: Powder Patterns

Section B: Diffraction Grating

Section C: .5mm Polystyrene Diffraction Pattern

Section D: 3mm Polystyrene Diffraction Pattern

Section E: Breath Figures

Section A: Powder Patterns

Goal: To recognize powder patterns and understand how and why they form.

Powder patterns are patterns of circular diffraction given out by many crystallites that are superimposed on each other, clumped together, and atop of one another.

Procedure:

1)      Set up a helium-neon laser

2)      Tape a magnifying lens to the opening of the laser

3)      Stack 3 1X96 well-trays atop of one another 10 cm in from of the laser opening.

4)      Place 4 straws into the top tray so that they can hold the glass slide. (Look at picture below)

5)      Set up a white slide 3 feet away from the apparatus to look at the diffraction patterns.

6)      Deliver one drop of .5mm polystyrene beads onto the glass slide and spread the beads with a brush.

7)      Let the mixture dry.

8)      Take the slide and place it between the 2 straws on each side.

9)      Look at the slide to observe the scattering pattern.

Apparatus Diagram:

Apparatus Diagram

The apparatus in the middle holds the slides.

The pattern should appear to have circular concentric rings around the undiffracted beam (the most intense beam) This pattern is called a powder pattern, and you should be able to see it because the drop of the polystyrene beads from the hydrophobic solvent were not diluted in water. If the polystyrene beads were highly diluted, you would see hexagonal packing in the diffraction pattern.

 

 

Section B: Diffraction Grating

Goal: To apply Bragg’s law of diffraction to find the angle of diffraction from a diffraction grating that has 75000 grooves per cm.

A diffraction grating is a piece of a plastic slide with many tiny spaced slits in it that diffract laser beam light into an order of points that have different intensities.  (Refer to picture below)

diffraction grating

When we use diffraction gratings with a laser, we know that the points displayed are 2-dimentional. Because the points are 2-dimentional, we use the Bragg’s law of diffraction that is not 3-dimentional. This equation comes out to be nλ=dsinθ. The wavelength of the laser was 632.8nm, the distance between the grooves was 75000 grooves per cm, and the integer number was 1. If we plug this information into the equation, we can find the angle of diffraction.

Procedure:

1)      Place a piece of diffraction grating 10cm away from the laser.

2)      Mark the points of interaction between the laser light with the diffraction grating.

3)      The mathematical procedure of finding the angle of diffraction is written out below.

The calculations should show that there was constructive interference between the grooves. The grooves allowed the light to pass through and as the light passed through, it separated between the slits. The angle of diffraction also depends on the wavelength of light. From previous experimentation, we can also conclude that the relationship between the angle of diffraction and wavelength determines the color we see.

 

Question to consider: If there was destructive interference between the rays, would the equation of nλ=dsinθ be true? No, the nλ would not equal dsinθ because the rays would cancel each other out.

Section C: .5mm Polystyrene Diffraction Pattern

Goal: To create a monolayer of .5mm polystyrene beads on a glass slide and prove that the wavelength of the laser correlates to the size of the polystyrene beads.

 

.5mm is a very small size. From the introduction and background chemistry, we may recall that is preferable to work with small crystallites because they give out a more precise image of the diffraction patterns. The beads give out “points” onto the slide whose spacing can be measured easily because it is not clumped together. The smaller the beads are, the more distant the spacing is between the points is. This is what we would like to prove in this experiment and in the next one where we will be using 3mm polystyrene beads.

 

Procedure:

1)      Get out a clean microscope slide with a clean cover slip and set them aside.

2)      .55mm polystyrene beads need to diluted in lots of water and ethanol. To prepare the mixture, you must deliver 1 drop of the beads from a pulled micropipette into a well in the 1X24 well tray.

3)      You must then add 20 drops of water and 10 drops of ethanol into this tray and mix them all together.

4)      Suck up the solution using a clean micropipette and deliver 1-2 drops of this solution onto the microscope slide.

5)      Make a wet mount by placing the cover slip over the solution. Try to avoid any air bubbles.

6)      Place the slide into the apparatus that holds microscope slides.

7)      Find a monolayer on the slide

8)      Observe the diffraction patterns.

9)      What do you see? (If you see scattering patterns, this means that the beads are not diluted enough)

10)   Mark the spots on the slide. Keep the recorded markings on the slide for further comparison with the 3mm polystyrene bead diffraction patters.

To get a defined image from the polystyrene beads, it is also useful to cut out a hole in the slide (paper) for the undiffracted beam to go through. This will allow you to see the different intensities of the various points on the image.  Also, make sure to record your data immediately after you have found a clear diffraction pattern because from experimentation, it was noticed that the polystyrene beads were moved by the laser light and the beads were also melted. It is hard to avoid these problems because the magnifying glass directs a lot of energy onto a sample on the microscope slide.  High energy is what is responsible for moving and melting the beads.

 

 

 

Section D: 3mm Polystyrene Diffraction Pattern

Goal: To compare the diffraction patterns from the 3mm polystyrene beads to the diffraction patterns from .5mm polystyrene beads.

The diffraction patterns from 3mm polystyrene beads should come out to be smaller, more clumped (packed) together and the spacing between the points should be small as well. In the following experiment, we want to prove that the spacing between the crystallites can be interpreted knowing the size of the polystyrene beads. The primary objective here is to show that the bigger the polystyrene beads are, the harder it is to detect a diffraction pattern. If the beads are bigger than the wavelength of the monochromatic light, we will not be able to see the diffraction patterns at all.

Procedure:

1)      Repeat everything from Section C using 3mm polystyrene beads.

2)      What diffraction pattern do you see?

Compare the diffraction patterns from both experiments and explain what you would expect to see if we used .01mm polystyrene beads.

schematic representation of what the

Section E: Breath Figures

Breath figures are condensed tiny droplets of water formed that form an epitaxial film on the glass slides. They are useful in our research because breath figures range from .1mm to .01mm. In theory, we may obtain diffraction patterns that have distant spacing between the water molecules. In the following experiment, we will observe breath figures.

Procedure:

1)      Take a styrofoam cup and fill it with hot water of 65°C

2)      Place the top lid of a petri dish onto the opening of the cup and let it sit there until a cloud forms underneath the lid

3)      Place the lid back onto the bottom half of the petri dish

4)      Allow laser light to pass through a sample of the cloud

5)      Observe and record diffraction patterns.

The diffraction patterns should come out to be well defined because the size of the water droplets ranges from .1mm to .01mm.

To continue on with these experiments, we can observe diffraction patterns from various chemicals like the interaction between HCl and NH3. As a chemical front is formed off the surface of the NH3 drop, smoke will fill the petri dish and the NH4Cl particles will form small solid crystallites that will too give off diffraction patterns.

6)      Take a clean petri dish and place one drop of NH3 on the top and one drop of the HCl onto the bottom half of the petri dish

7)      Close petri dish

8)      Allow the smoke to fill the petri dish

9)      Shoot laser light through the petri dish

10)   Compare the diffraction patterns given off by HCl and NH3 and water molecules from water vapour.

An advantage of studying diffraction patterns from breath figures rather than from polystyrene beads is that it is cheaper. The price to gather the apparatus materials for polystyrene beads came out to be over a $100. Thanks to Dr. Thompson, I was able to study diffraction patterns through the use of these micron sized polystyrene beads.

 

 

 

 

 

Bibliography:

Works Cited

“Bragg’s Law.” Wikipedia, the Free Encyclopedia. Web. Feb. 2011. <http://en.wikipedia.org/wiki/Bragg’s_law>.

Clegg, William. Crystal Structure Determination. Oxford: Oxford UP, 1998. Print.

“Crystallography.” Wikipedia, the Free Encyclopedia. Web. 28 Jan. 2011. <http://en.wikipedia.org/wiki/Crystallography>.

“Epitaxy.” Wikipedia, the Free Encyclopedia. Web. 18 Apr. 2011. <http://en.wikipedia.org/wiki/Epitaxy>.

Feynman, Richard P. QED: the Strange Theory of Light and Matter. Princeton, NJ: Princeton UP, 1985. Print.

Flack, H. D. “Louis Pasteur’s Discovery of Molecular Chirality and Spontaneous Resolution in 1848, Together with a Complete Review of His Crystallographic and Chemical Work.” Web. Jan. 2011. <http://library.epfl.ch/en/periodicals/?recId=12869839>.

Taylor, Charles Alfred. Images: a Unified View of Diffraction and Image Formation with All Kinds of Radiation. London: Wykeham Publications, 1978. Print.

Thompson, Stephen. Chemtrek: Small-scale Experiments for General Chemistry. Englewood Cliffs, NJ.: Prentice Hall, 1989. Print.

Analysis of Florida Well-Water

By: Timothy Pemberton

Introduction:

For this project, the purpose was to experimentally determine the qualities of a sample of well water including ‘hardness’, amount of minerals, and they types of minerals present. The results obtained will then be compared to other lab groups who also conducted the same experiments on the same sample of water to reveal more conclusive evidence of the contaminants and quality of the water. Several methods will be used to test the purity and qualities of the water with the final goal being to determine if it is safe for human consumption. These methods include spectrophotometry, flame testing, conductivity testing, a calcium indication test, and a titration with Ethylenediaminetetraacetic acid (commonly known as EDTA) to determine the water’s ‘hardness’ or amount of minerals present (namely calcium and magnesium). The hypothesis for this experiment is that the water that will be tested is pure (or safe for human consumption) and is reasonably soft.

The flame test can indicate the minerals in the water, but only if they are part of a special set of minerals that burn different colors. These minerals include calcium, iron, and potassium—all minerals that have a high probability of being in the sample water.

The spectrophotometry test will help determine the types of contaminants found in the sample. The working theory behind conducting this test is that the contaminants in the sample will have differing wavelengths that correspond to contaminants. This is used by many laboratories. One of the largest uses for this type of identification is used for defining all the elements present in a gas; when a light is shined through it, the gas absorbs some of the wavelengths of light. This makes it possible to identify the components of the gas. The same idea will be used with our sample. Since pure water will not absorb any light, the contaminants alone will absorb light, making it possible to identify the particles in the water. Each different element has its own particular wavelength that it absorbs. As such, it will be possible to prove what contaminants are in the water.

Conductivity testing will help indicate the concentration of minerals that aid in the conduction of electricity through the sample. This will not determine the type of contaminants present, but rather a good approximation of the sample. All minerals, if dissolved into water, will raise the conductivity of the water. This means less resistance to the conduction of an electric current.

The calcium indication test will use calcium bentonite in conjunction with the conductivity test to definitively see if calcium is one of the contaminants present in the sample of water. The nH2O in the clay compound breaks free so that the clay compound is negatively charged. Because it is negatively charged, it acts almost as if it was a magnet—attracting all the positively charged ions around it, which includes calcium carbonate and potassium, among others.

Finally, the EDTA test will show the “hardness” of the water. This hardness is a measure of the calcium carbonate (CaCo3) and magnesium content.1 This water hardness does not affect the purity of the water, but it is a quality that can be important to consumers. As long as the water is not extremely hard, it will be safe to consume. The purpose of running this experiment is to prove that the water is above a certain standard as softer water is generally regarded as purer.

Procedure:

The procedure will be done as follows:

Gather materials for the experiment, and place approximately 250 mL deionized water into a flask. This water will serve as a waste container for used or contaminated samples. This container will be placed at the center of the table and can be emptied and refilled as often as needed.

pH Test. Take a piece of litmus paper and dip into the sample to determine the pH of the water. If the paper turns ‘cool’ colors (green/blue) then the sample is basic; if it turns a ‘hot’ color’ (yellow/red) it is acidic.

Separate the solid materials from the water in the sample. Take 30.0 mL of the sample and place into a 50 mL beaker. Place on top of a heating plate and place on the plate on the highest setting until all liquid has boiled out. Use the clamp to remove the beaker from the heating plate to prevent burning. Take the spatula and scrape all of the minerals out into another 50 mL beaker after cooling. These are the minerals present in the water.

Flame Test. Acquire the Bunsen burner and light it. Take the minerals made in Step 2 and place onto a wire loop. If the resulting minerals do not cohere to the loop, dip loop into a test tube with the sample water in it then into the powder and proceed. Hold the tip of the loop over the tip of the flame and observe the changes in color made by the substance. Different colors indicate differing materials. Repeat as many times as needed to

Spectrophotometry Test. Take the pure sample and place into a 50 mL beaker; transfer enough liquid to fill a cuvette up to the fill marker. Be sure to calibrate the spectrophotometer properly. Place the sample into the spectrophotometer and observe the peaks in absorbance for light between about 390 and 700 nm (the visible spectrum). Each material has a different wavelength, so this test will indicate what minerals are in the water, but not how much of them. If the readings are not pronounced enough to draw accurate conclusions, take the sample made in Step 2 and add a measured amount to 10.0 mL of the pure sample to increase the amount of contaminants in the water. Do this only after completing the flame test. This should deliver more robust results. Repeat two more times, making sure to wash the cuvette between each test.

Conductivity Test. Calibrate the conductivity meter. Use 30 mL of deionized water in a 50 mL beaker as a control and measure the conductivity of the water in Ohms. Be sure to not let the probe touch the bottom of the beaker, while still remaining fully submerged. Take 30 mL of the sample in another 50 mL beaker and properly place the conductivity probe into the sample and record the resistance of the sample in Ohms (Note: the resistance in Ohms should be larger for the deionized water than for the sample). Compare to a chart to see whether the sample water has few enough minerals for safe consumption. Repeat two more times, making sure to wash the containers thoroughly before each reading.

Calcium Indication Test. The purpose of this test is to experimentally prove if calcium is in the sample. Calcium bentonite is used to do this. Take 10.0 g of this substance and pour into 20.0 mL of the sample. Allow to sit for at least 30 minutes to allow the mixture to settle. If, after half an hour, the mixture is not completely settled, allow to remain at rest for 30 minutes more. After the mixture is settled, use the conductivity meter to measure the conductivity of the liquid at the top of the beaker. If the resistance is the same as deionized water, the calcium bentonite reacted with the calcium in the water and settled.

EDTA Titration. Gather 7.306 g of EDTA and pour into a 500 mL beaker with 250.0 mL of deionized water. This will create a 0.1 M mixture of the EDTA. Heat for approximately 5 minutes or until the mixture is fully dissolved into the water so that it is fully transparent. Be careful not to let the water boil out as this can affect the molarity of the final solution. After, place into an ice bath to quickly cool the mixture. During the heating and the cooling process, mix 10.0 mL of the sample water with 40 mL of deionized water in a 250 mL Erlenmeyer flask. Add eight drops of the Eriochrome Black T indicator and 6 drops of the weakly basic ammonium (NH4+). The result should be a pink-red color.2 Once the EDTA mixture has cooled, pour into the burette. Be sure to mark where the burette is filled up to; although it is possible to be filled up to anyplace to get good results, it must be in the marked area. The final volume left in the burette will be subtracted from the initial volume. Place the Erlenmeyer flask holding the sample and indicator under the burette containing the 0.1 M EDTA. Allow the EDTA to slowly drip into the sample until the indicator turns the mixture to a blue shade. Find the volume of 0.1 M EDTA that was allowed to drip; this number will be used in the final calculations. Repeat five more times.

After the experiment is complete, wash all equipment thoroughly with soap and clean work space.

Results:

pH Test

When the litmus paper was dipped in the sample water, the litmus turned a slightly green tint. This indicates the sample water is about neutral.

Boil Test

When the sample was boiled so that no water remained, the resulting precipitate left at the bottom of the beaker was grainy and white. It dissolved easily in water. It was adhering to the bottom of the beaker firmly, but could be somewhat easily scraped off.

Flame Test

The white minerals made in the boil test burned an orange or red-yellow color when placed in a Bunsen burner’s flame on a wire loop. A purple flame may have appeared, but only for an instant.

Spectrophotometry

maximum wavelengths of the sample water

*This is not from the first set of tests conducted. The first time the sample water was tested using the spectrophotometer, the results were inconclusive. The data obtained was too nominal to include as decisive data. The maximum wavelength was essentially zero for each point on the graph, and constantly fluctuating from the negative part of the graph to the positive. This data taken was thus thrown out as unusable.

Conductivity Test

deionized water

Calcium Indication Test

resistance of the sample water

EDTA Titration

EDTA Titration

Sample Math for Calculating the Parts Per Million of Calcium

Using the results from Sample 1 in Figure 4:

results

Using this method, parts per million can be calculated for all the tests conducted:

conducted titrating EDTA

Discussion:

Our results were overall conclusive, save for the flame test. Though it was clear that the flame burned a vivid red-orange color, it was brought up that there were possibly purple flashes visible in the flame. These bursts were either very brief for the entire group to see or nonexistent. It is clear through the flame test however (and the titration with EDTA) that calcium was present. After the water had boiled out and just the contaminants remained, the white powder that remained was uniform in look and texture. When the remaining minerals were dissolved in the sample water to enhance the quality of the spectrophotometry results, the minerals dissolved extremely easily; no stirring or agitation was necessary for them to completely dissolve into the sample water. The conductivity test was quite straightforward and went as expected. The sample had a much higher conductivity than the deionized water. At first, there were issues with the results from the spectrophotometry. At most points along the graph, the plot was jumping above and below the zero line—which gave questionable data concerning the maximum wavelength and its absorbance. The problem was averted by simply using the excess minerals from the flame test and adding to the sample for a higher dilution of the minerals in the sample. The amount of minerals did not matter for the spectrophotometer, as just what minerals were present mattered. As long as no outside contaminants were allowed into the sample, the sample remained the same for the purposes needed. The only thing that changed was the total concentration of the contaminants. The titrations in the first week went poorly. The basic ammonium buffer was not added to the sample, so the indicator did not work as expected. This problem was fixed in the second week of testing by adding the buffer, which yielded workable results. The conductivity test gave a result that the water was just below city water according to the following chart:

average resistance

This supports the hypothesis that the water is pure enough to drink. There are not enough contaminants to label the water as impure.

Conclusion & Relation to Research:

In conclusion, the sample water in ‘Sample B’ of well water definitively had calcium carbonate in it, and possibly potassium. The results from the spectrophotometry indicate that potassium was present in the sample water along with calcium and magnesium. These elements all have wavelengths found in the 520-572 nm wavelengths that we obtained with the spectrophotometer. According to our calculations, the water was extremely hard. With values well over 4000 mg/L, the data obtained is questionable according to the water hardness chart given by the US Geological Service which states that 0-60 mg/L is soft, 61-120 is moderately hard, 121-180 is hard, and >181 mg/L is very hard. Our smallest value was approximately 3920 mg/L. This value is over 2100% above the highest value for hard water. This fact throws the values obtained for the titration into serious doubt. The results from the conductivity test contradict these results, and will be preferred over the titration. Simplicity is the reason this is so. The conductivity test was simple, while the EDTA titration needed large amounts of math where errors could easily arise unnoticed. The experiment itself was not a complete loss however, as long as no poisonous or relatively unnoticed contaminants that are highly fatal were present, according to the combination of all the tests the water was completely safe for human consumption. The hypothesis was half proved right. The water is safe to drink, but was not very soft at all.

This experiment is important to those in charge of maintaining and testing the water supplies of major water distributers across the nation. The company who releases unhealthy water to the people would be in serious threat of lawsuits or jail time. Without scientists present to periodically check the purity levels of the water supply, water companies could be unknowingly selling and distributing a serious health concern. Some of these same methods used in this experiment are almost exactly those ran by the scientists that are in charge of testing water for its purity and mineral content. All water processing facilities put their product through a strenuous purification process that leaves good minerals in the water while removing the harmful contaminants. The city of Tampa’s water supply is checked over 1000 times per month.6 According to the City of Cincinnati’s Water Works’ website, the water they produce is checked for quality over 600 times per day.5 If one is not on city water or any that comes from a water plant, those who use well water can send their water in to a laboratory to be tested for contaminants and to see if the water is safe for consumption. These labs also use many of the same methods to check the purity of the water. If the water fails the tests, a report is sent back that the water is unsafe for using. These processes are important to millions of people each and every day. If a bad batch of water gets out to the public, an epidemic affecting all these of people would certainly strike. Thus, it is important that water goes through the treatment process and testing to ensure that the people will be safe when drinking water.

References:

  1. City of Glendale Water and Power. Water Hardness. Glendalewaterandpower.com (Web) accessed Mar 18, 2013.
  2. ACS Publications. Complexes of Eriochrome Black T with Calcium and Magnesium. Pubs.acs.org (Web) accessed Mar 19, 2013.
  3. US Geological Service. Water Hardness and Alkalinity. Water.usgs.gov/ (Web) accessed Mar 19, 2013.
  4. National Environmental Service Center. “What does ppm or ppb mean?”. (Web PDF) nesc.wvu.edu accessed Mar 19, 2013.
  5. City of Cincinnati. Water Quality. cincinnati-oh.gov/water (Web) accessed Mar 19, 2013.
  6. City of Tampa. Water Quality. tampagov.net/dept_Water/ (Web) accessed Mar 19, 2013.
  7. Myron L Company. Deionized Water. myronl.com (Web) accessed Mar 19, 2013.

Multistep Reaction Sequence: Benzaldehyde to Benzilic Acid

Seth Dingas* and Jakkrit Suriboot

ABSTRACT

Benzilic acid

Benzilic acid was synthesized through a multistep reaction from the starting material of benzaldehyde and through the formations of benzoin and benzil.  The first reaction produced benzoin by using the thiamine hydrochloride catalyst, followed by an oxidation reaction to produce benzil, and a rearrangement to synthesize benzilic acid.  By utilizing crystallization, pure solid products of each step were collected and analyzed through IR, NMR spectroscopy, and other physical properties.

Introduction
Multistep synthesis reactions involve many advantages and disadvantages.  Disadvantages include time-consuming experiments, error within intermediate steps, or the presence of side reactions.  Advantages imply the production of ideal, marketable end products, and the synthesis of compounds that otherwise could not be produced through a simple reaction.  Research has enhanced the sustainability, time efficiency, and design of multistep synthesis reactions to be utilized in many industrial situations.1

Different organic processes and characteristics were utilized in the multistep synthesis involved in this reaction.  Green chemistry was involved in the preparation of benzoin by the choice of catalysis, thiamine hydrochloride.  In a biochemical environment, thiamine acts as a coenzyme that proceeds as the chemical reagent.2  Regarding the second step, an oxidation reaction was involved utilizing a mild oxidizing reagent, nitric acid, in pyridine.  Finally, the third step of this reaction involved the compound benzil that has attracted many speculations throughout the century.  Through the interaction with other molecules, the rearrangement characteristics of benzil have been proven based on the intramolecular oxidation and reduction forces of gaining and losing electrons.3 Overall, the combination of the various organic characteristics and experiments allow for the success of a multistep synthesis reaction.

Reaction Mechanisms
Scheme 1 depicts the reaction between the catalyst thiamine hydrochloride and two equivalents of benzaldehyde.  Once a proton was removed from thiamine hydrochloride, forming ylide, it acted as a nucleophile that allowed for the addition of the carbonyl group of benzaldehyde. A proton is removed from the intermediate and the new alkene bond attacks the carbonyl group of the second benzaldehyde.  The ultimate products of ylide and benzoin are produced.  The ylide is the regenerated catalyst and performs the mechanism again.

 

The production of benzoin

As the second step of the multistep synthesis, the alcohol group of benzoin must be oxidized.  By utilizing the mild oxidizing agent of nitiric acid, benzoin was oxidized to produce benzil through the mechanism in scheme 2.

benzoin and nitric acid

The final mechanism, shown in scheme 3, involves the synthesis of the carboxylate salt intermediate, potassium benzilate, which drives the reaction to produce benzilic acid through workup.

The formation of benzilic acid

Results and Discussion For the first reaction, the presence of crystals after the combination of the ylide and benzaldehyde appeared pale yellow, solid but mushy.  After filtration, a total of 4.68 g of crude benzoin were collected.  Through recrystallization, a pure product of 2.07 g was collected, which produced a 44% yield.  This product produced a melting point of 129-132 °C.  This corresponds to the melting point of the crude product concluding that purification failed.  Purification could have been improved by adding more 95% ethanol to wash the crude product.  The final product of benzoin contained 13C NMR peaks at 199.2 ppm accounting for the carbonyl group and eight peaks in the range of 139.0-127.8 ppm representing the alkene bonds as well as the carbons of the aromatic rings.  Finally, a peak at 76.2 ppm represented the carbon with the alcohol group attached.  Regarding the 1H NMR spectra, four multiplet peaks appeared in the range of 7.79 and 7.14 ppm representing the hydrogens surrounding the aromatic rings.  A peak at 5.82 ppm accounted for the hydrogen attached to the carbon containing the alcohol group.  A peak at 3.92 ppm represented the hydrogen of the alcohol group.  An impurity of ethanol appeared at 4.42 ppm.  Finally, the IR spectra displayed a peak at 3403 cm-1 representing the C-H stretches, a peak at 3003 cm-1, accounting for the alcohol group, and a strong peak at 1761 cm-1 representing the carbonyl group.  Overall, the spectra confirmed the condensation of benzoin.

When benzoin was reacted with nitric acid, an orange/red color appeared.  When this mixture was heated and refluxed, a strong red color appeared in the reflux condenser, proving the presence of nitric gas.  A total of 1.91 g of purified benzil was produced from this reaction which contained an observed melting point of 89-92 °C and a 77% yield.  This appeared to be less than the ideal melting point of 95 °C, which could account for the lack of purity.  The 13C NMR produced a peak at 192.0 ppm representing the two carbonyl groups.  Four peaks appeared between 132.3 and 126.5 ppm accounting for the carbons within the aromatic ring and the alkene bonds.  The 1H NMR displayed three multiplet peaks at 7.86, 7.56, and 7.53 ppm representing the hydrogens around the aromatic ring that coupled with the surrounding hydrogens.  Finally, the IR spectrum produced a C-H stretch peak at 3010 cm-1 and a carbonyl peak at 1668    cm-1.  This data proved the success of the oxidation of benzoin to produce benzil.

For the final reaction, once benzil and aqueous potassium hydroxide were combined, the reaction turned from black to brown.  This intermediate step produced potassium benzilate.  After workup, a total of 0.41 g of the crystallized product were collected, which produced a melting point of 151-152 °C and a 17% yield.  The melting point corresponded to the known melting point of 150 °C. The 13C NMR spectra displayed a weak peak at 175.8 ppm, which accounted for the carbonyl group within the carboxylic acid.  Four peaks at 141.4, 128.3, 128.2, and 127.4 ppm represented the carbons within the aromatic rings.  Finally, a peak at 82.0 ppm represented the carbon attached to the alcohol group.  The 1H NMR spectrum produced a peak at 7.47 and 7.26 ppm representing the two groups of equivalent hydrogens attached to the aromatic rings.  A peak at 2.18 ppm represented the hydrogen of the alcohol group.  A peak did not appear at 12 ppm that would have represented the hydrogen of the carboxylic group, which means the reaction was not carried to completion.  In the IR spectrum, a hydroxyl peak appeared at 3399 cm-1.  A broad peak appeared at 2889 cm-1 representing the carboxylic acid functional group of compound.  Finally, a peak at 1718 cm-1 represented the carbonyl group and a peak at 1177 cm-1 accounted for the carbon-oxygen bond in both alcohol groups.  From this data and the low percent yield, the rearrangement of benzil was not achieved successfully.

 

Conclusion Through the multistep reaction, a 44% yield of benzoin, a 77% yield of benzil, and a 17% yield of benzilic acid were obtained.  The IR and 1H NMR data displayed peaks that allowed for the distinction and identification of the different products that led to the ultimate synthesis of benzilic acid.

 

Experimental

General: All reagents were provided by Texas A&M University Chemistry Department. 1H and 13C spectra were taken on a Mercury 300 MHz NMR spectrometer.  An IR spectrum was provided by PerkinElmer UATR Two Spectrophotometer.

 

Benzoin: Thaimine hydrochloride (1.52 g, 0.45 mmol), water (2mL) and 95% ethanol (15 mL) were combined in a 50-mL Erlenmeyer flask and swirled until dissolved and homogeneous.  Aqueous sodium hydroxide (4.5 mL) was added and swirled until the solution appeared pale yellow.  Finally, pure benzaldehyde (4.5 mL, 4.41 mmol) was added to the flask and the mixture was stored for two days.  The crystals that formed at room temperature were placed in an ice bath and then filtered under vacuum.  The crystals were washed with 5-mL portions of ice-cold water and left to dry.  To isolate the pure product, the crude material was crystallized with 95% ethanol (24 mL).  The pure product of benzoin showed the following physical characteristics: 2.07 g (44.3 % yield) mp: 129-132°C (lit: 135-135 °C).  1H NMR (CDCl3, 300 MHz) δ: 7.79 (d, J=1.5, 2H) 7.25 (m, 2H), 7.24 (m, 2H), 7.19 (m, 2H), 7.14 (m, 2H), 5.82 (d, J= 1.2, 1H), 3.92 (s, 1H) ppm.  13C NMR (CDCl3, 75Hz) δ: 199.2, 139.2, 139.1, 134.0, 129.2, 129.1, 128.7, 128.5, 127.8, 76.2 ppm.  IR 3403, 3003, 1761, 1203 cm-1.

 

Benzil: Benzoin (2.51g, 1.18 mmol), concentrated nitric acid (12 mL, 28.8 mmol), and a stir bar were placed into a 25-mL round-bottom flask with a water condenser and heated in a hot water bath at 70 °C for one hour.  After heating and magnetically stirring, the mixture was added to 40 mL of cool water and stirred until crystallized into a yellow solid.  Vacuum filtration was used to collect the crude product.  The pure product was collected through recrystallization by using 95% ethanol (20 mL).  The product displayed the following properties: 1.91 g (76.8 % yield) mp: 89-92 °C (lit: 95 °C).  1H NMR (CDCl3, 300 MHz) δ: 7.86 (m, 4H), 7.56 (m, 2H), 7.53 (m, 4H) ppm.  13C NMR (CDCl3, 75Hz) δ: 192.0, 132.3, 130.4, 127.3, 126.5 ppm.  IR 3010 (w), 1668 (s) cm-1.

Benzilic Acid: Benzil (2.10 g, 1.0 mmol), 95% ethanol (6 mL), and a boiling stone were added to a 25-mL round-bottom flask with a reflux condenser and heated until the solid benzil was dissolved.  Aqueous potassium hydroxide (5 mL, 18.2 mmol) was added dropwise to the flask and the mixture was boiled for 15 minutes.  The mixture was cooled, transferred to a beaker, and placed in an ice-water bath until crystallized.  The crystals were isolated through vacuum filtration and washed with 4-mL portions of cold 95% ethanol.  The solid was transferred to a 100-mL flask of hot water (60 mL) and mixed until completely dissolved.  Concentrated hydrochloric acid (1.3 mL) was added drop-wise until a permanent solid was present and a pH of 2 was maintained.  The solution was cooled in an ice bath and the crystals were filtered through vacuum filtration and washed with 2, 30-mL portions of ice-cold water.  The remaining crystals were identified by the following properties: 0.41 g (17.4% yield) mp: 151-152 °C (lit: 150 °C).  1H NMR (CDCl3, 300 MHz) δ: 7.47 (m, 6H), 7.26 (s, 4H), 2.18 (s, 1H) ppm.  13C NMR (CDCl3, 75Hz) δ: 175.8, 141.4, 128.3, 128.2, 127.4, 82.0 ppm.  IR 3399 (s), 2889 (s, b), 1718 (s), 1177 (s) cm-1.

Supporting information IR, 1H NMR and 13C NMR spectra of benzil, benzoin, and benzilic acid are attached.

 

1 Bruggink, A.; Schoevaart, R.; Kieboom, T. Org. Proc. Res. Dev., 2002, 7, 622-640.

2 Pavia, L; Lampman, G; Kriz, G; Engel, R. A Small Scale Approach to Organic Laboratory   Techniques, 2011, 266-269.

3 Lachman, A. J. Am. Chem. Soc., 1922, 44, 330-340.

 

The Formula, Synthesis, and Analysis of Alum

By: Veronica Smith

 

I. Purpose

Formula of a Hydrate:

      The purpose of this experiment is to determine the formula for the hydrate alum.

Synthesis of a Hydrate:

      The purpose of this experiment is to become familiar with the formation of hydrates and using sequential reactions.

Analysis of a Hydrate:

The purpose of this experiment is to verify that the compound formed in the previous experiment is alum.

II. Procedure

Formula of a Hydrate:

      After determining the mass of an empty evaporating dish and watch glass, approximately 2.00g of alum were added to the evaporating dish and massed.  The evaporating dish and watch glass were set up on a ring stand with wire gauze, and were heated over a low flame until no steam was observed for about two minutes.  The evaporating dish, watch glass, and product were allowed to cool for about 10 minutes, and then massed to determine the amount of product produced.  The procedure was repeated once more in order to verify the original results.

Synthesis of a Hydrate:

      Approximately 1g of aluminum foil was weighed to the nearest centigram, torn into small pieces, and placed into a 250mL beaker.  25mL of 3M potassium hydroxide solution was added slowly and was allowed to react until the foil was dissolved. Undissolved solids were removed and discarded through vacuum filtration.  The solution was allowed to cool then acidified slowly with constant stirring using 45mL of 3M sulfuric acid.  The solution was then boiled until the water evaporated to give a volume of about 50mL of solution.  The solution was cooled overnight.  After collecting the crystals by vacuum filtration, they were washed with 50mL of a 50% by volume water and ethanol mixture.  The crystals were then allowed to dry at room temperature, and then massed.  The theoretical yield of alum was then calculated, assuming that aluminum was the limiting reactant and that the foil was 100% aluminum, and was then used to calculate the percent yield.

Analysis of a Hydrate

      Part 1:

Approximately 0.5g of dry alum were pulverized with a mortar and pestle, and then packed into a capillary tube to a depth of about 1 cm.  A rubber band was used to fasten the capillary tube to a thermometer with the alum level with the bulb of the thermometer.  A universal clamp and cork stopper were used to fasten the thermometer to a ring stand.  The bottom of the capillary tube and thermometer were immersed in a Thide-Dennis tube, which was filled with water to about 1 inch past the second arm.  The sample was heated rapidly in the beginning, but slowly as it approached the melting point in order to get an accurate reading.  The temperature at which the alum melts was recorded, and was compared to the published value.

Part 2:

After finding the mass of an evaporating dish and watch glass, approximately 2g of alum crystals were added to the evaporating dish.  The evaporating dish, alum, and watch glass were then placed on wire gauze held in a ring on a ring stand and heated very gently over a low blue flame.  After the vapor had appeared to be driven off, the alum was heated more strongly for about five minutes.  After cooling, the mass of the anhydrous alum and the water driven off were calculated and compared to the accepted value of alum.

Part 3:

After weighing about 1g of alum into a 250mL beaker, it was dissolved in about 50mL of distilled water.  After calculating the volume of 0.2M barium nitrate that would be needed to totally precipitate all of the sulfate ions present in the solution, twice that amount was added slowly while stirring.  The beaker was then covered with a watch glass and heated nearly to boiling, where it was kept for about 15 minutes.  The precipitate was then filtered using vacuum filtration and, along with the beaker, was washed several times with small quantities of distilled water.  The filter and precipitate was then transferred to an evaporating dish and allowed to dry in an oven.  After cooling, the precipitate was massed and the percent sulfate present in the alum was compared to the value calculated from its formula.

III. Data

Formula of a Hydrate:

Trial #

Mass of dish & glass

Mass of hydrate

Mass of product

Mass of anhydrous alum

Mass of H2O

Mole ratio

Alum: H2O

1

111.54g

2.00g

112.36g

0.82g

1.18g

1:20

2

111.49g

2.00g

112.61g

1.12g

0.88g

1:11

      Bubbles appeared on the anhydrous alum during heating in the second trial.

Synthesis of a Hydrate:

Trial #

Mass of empty container

Mass of container and crystals

Mass of crystals

Percent yield

1

11.04g

22.55g

11.51g

65.4%

2

10.99g

———-

———-

0%

 

      When the sulfuric acid was added to the solution, the beaker became very hot.  No more crystals were formed in the second trial.

Analysis of a Hydrate: 

Part 1:

Trial #

Sweating Point (oC)

Melting Point (oC)

Percent error

1

85.0

86.0

4.2%

2

85.0

86.0

4.2%

Part 2:

Trial #

Mass of dish & glass

Mass of hydrous alum

Mass of dish, glass, & product

Mass of anhydrous alum

Mass H2O

Mole ratio alum : H2O

1

142.54g

2.00g

143.65g

1.11g

0.89g

1:11

2

142.54g

2.01g

143.62g

1.08g

0.93g

1:12

Part 3:

Trial#

Mass of crucible

Mass of alum

Volume Ba(NO3)2 needed

Volume H2O

Mass of precipitate

Percent sulfate

Percent error

1

89.17g

1.00g

42.0mL

50.0mL

1.20g

49.4%

20.0%

2

89.17g

1.01g

42.6mL

50.0mL

1.17g

48.2%

17.1%

 

IV. Calculations

Formula of a Hydrate
Analysis of a Hydrate
Convert grams of alum to moles

 

V. Conclusions and Error Analysis

Formula of a Hydrate:

The data collected has very little precision, and it can therefore be inferred that error has most definitely occurred.  A likely source of error is that in the first trial, some of the anhydrous alum was lost, either by spillage or splattering during heating, making the mass of the anhydrous alum appear smaller, and after the appropriate calculations, making the mole ratio of alum to water greater in the water’s favor.

Some modifications that could be made to the procedure to help avoid this type of error are to make sure that the watch glass is securely on the evaporating dish, and to take extra precautions so as to avoid spillage.

Synthesis of a Hydrate:

In the first trial we obtained a percent yield of 65.4%, from which it can be inferred that mass was lost during the syntheses.  This could have happened during the crystal collection by vacuum filtration.  Some of the crystals could have gotten stuck in the filter, and therefore not accounted for in the final mass.

A modification that could be made to the procedure to avoid this type of error is to carefully remove all crystal pieces from the filter and to be careful of spillage.

During the second crystallization, no more alum crystals were obtained.  An explanation for this could be that all of one of the reactants was used up during the first trial, or that the temperature was not cold enough for the crystals to form.

A modification that could be made to prevent this is to keep the solution in the ice bath for a longer period of time.

Analysis of a Hydrate:

Part 1:

In both trials a melting point of 86.0oC was obtained, which is 4.2% error.  Although there was some error, the results are very precise, from which it can be inferred that the source of error was something that affected the entire sample of alum before the experiment was carried out.  The most likely source of error was the obtainment of impurities.  This would have affected the entire sample of alum, and therefore changed the melting point of both trials.

A way to prevent this type of error is to ensure that all objects that come into contact with the alum have been thoroughly cleaned and dried first.

Part 2:

In the first dehydration we obtained a mole ratio of 1 mole of alum to 11 moles of water.  Because we know that the true ratio is 1:12, it can be inferred that an error was made.  The most likely source of error is that not all of the water was driven off during the heating.  If some water remained in the product that was assumed to be anhydrous, then when the mole ratio was calculated, it would have seemed as if the moles of water were smaller.

A way to prevent this type of error would be to heat the alum thoroughly, either by heating more vigorously or for a longer period of time.

In the second dehydration we obtained a mole ratio of 1 mole of alum to 12 moles of water, which is the correct ratio, so it can be inferred that if an error was made, it was not a large enough error to affect the mole ratio.

VI. Discussion of Theory

Formula of a Hydrate:

A hydrate is a chemical compound that contains water it its structure.  To find the formula of a hydrate, the hydrous sample is heated until all water is gone from the compound.  By comparing the mass of the hydrous sample and the anhydrous sample, we are able to determine the mole ratio of product to water, which gives us the formula of the hydrate.  Assuming that the compound has been properly dehydrated and that no mass has been lost, the experiment is expected to yield fairly accurate results.

Synthesis of a Hydrate:

            Alum is a general name for the type of compound in which many combinations of an alkali metal and ammonium or a trivalent metal such as aluminum, iron, or chromium.  A synthesis reaction is a reaction in which two or more chemicals are combined to create a new compound or compounds.  The following sequential reactions take place in this experiment to synthesize alum (KAl(SO4)2 * 12H2O):

Synthesis of a Hydrate

This synthesis demonstrates the aluminum hydroxide’s ability to act amphoterically. Amphoterism describes a substance’s ability to act as both an acid and as a base.  This reaction also produces a complex covalent bond between the potassium, aluminum, sulfate, and water.  A complex covalent bond is a covalent bond in which both electrons are furnished by one atom or ion.

A 100% yield should not be expected from this reaction because aluminum is the limiting reactant, and a certain amount of aluminum is lost when the solid aluminum hydroxide is filtered out.   The crystals should be washed with a 50% ethanol and water by mass solution because they dissolve easily in pure water.  As the temperature decreases, the crystal’s solubility decreases, thus why they can be put into an ice bath after boiling in order to rapidly produce crystals.  The hydrated aluminum ion is in the shape of an octahedron, which is a eight-sided figure as shown below in figure 1.

octahedron

Analysis of a Hydrate:

There are three parts involved in the analysis of a hydrate: determining the melting point, determining the formula, and determining the percent sulfate.

A melting point is a usually a fairly reliable way to determine the identity of a substance, however, there are influences which could affect the results and therefore result in a wrong identification.  One of these influences is the obtainment of impurities into the sample being tested.

To find the formula of a hydrate, the hydrous sample is heated until all water is gone from the compound.  By comparing the mass of the hydrous sample and the anhydrous sample, we are able to determine the mole ratio of product to water, which gives us the formula of the hydrate.  However, before massing, the sample and container must be cooled because the large temperature difference affects the surrounding air and therefore can affect the density of the sample.

To find the percent sulfate, an amount of barium nitrate is allowed to react with the sample to precipitate all of the sulfate in the compound as barium sulfate.  The precipitated barium sulfate is filtered, dried, massed, and then compared to the theoretical value of sulfate in the sample.  A percent error can then be found which, if no large errors were made, should help to identify the sample.