Archives for March 2010

Purpose

To measure and determine the relationship between a magnetic field generated by a line of current and a radial distance from a conductor, and to measure and determine the relationship between a magnetic field at the center of a coil and the number of turns in a coil.

Hypothesis

As the distance from the center of the conductor increases, the magnetic field strength will decrease in accordance with the equation B = μ_oI / 2πR. As the number of turns in the coil increases, the magnetic field strength will increase, in accordance with the equation B = Nμ_oI / 2R.

Labeled Diagrams

See attached sheet.

Data

Part 1

Current = 7 A

Part 2

Current = 7 A

Diameter of Cylindrical support = 0.130 m

Graphs

Part 1

Part 2

Questions

Part 1

1. Theory states that the magnetic field produced by a long straight current-carrying wire decreases in strength as you get further from the wire. The exact dependence of the magnetic field strength B on radial distance from the wire R is B = μ_oI / 2πR where μ_o is the permeability of free space and has a value of 4π x 10^-7 Tm/A. For your data from Part 1, plot a graph in Graphical Analysis of B vs. 1/R. Your graph should have a linear trend. Perform a Linear Fit on your graph. What is the value of the slope?

The slope is 9.2 x 10^-7 Tm.

2. Calculate your value of μ_oI / 2π. How does it compare to the slope of your graph? What is the percent difference?

The value is 1.4 x10^-6 Tm, which is larger in value in comparison to the slope. The percent difference is 41%.

Part 2

3. Theory states that the magnetic field produced by a circular loop of current- carrying wire increases in strength as the number of turns of wire is increased. The exact dependence of the magnetic field strength B on the number of turns N is B = Nμ_oI/ 2R where R is the radius of the loop of wire. For your data from Part 2, plot a graph in Graphical Analysis of B vs. N. Your graph should have a linear trend. Perform a Linear Fit on your graph. What is the value of the slope?

The graph of the slope is 6.53 x 10^-5 T/turn.

4. Calculate your value of μ_oI / 2R. How does it compare to the slope of your graph? What is the percent difference?

The value is 6.7 x 10^-5 T/turn, which is very similar yet slightly large in value than the slope. The percent difference is 2.6%.

Conclusion

During part one of the experiment, magnetic field strength was measured as a function of radial distance from a conductor. First, a piece of polar graph paper with concentric circles starting at a diameter of 0.5 cm increasing in increments of 0.5 cm to 10 cm was punched through a rigid aluminum conductor at the center of the concentric circles. The paper was placed on the plastic table of the apparatus and was aligned using a compass so that the parallel lines on the sheet were pointing north. The paper was then secured to the apparatus using tape. The high amperage DC power supply was connected in series with a high power resistor and the aluminum wire at the side and on top of the apparatus. The magnetic field sensor was zeroed and the DC power supply was set to 7 A. With the current on and kept constant, the magnetic field strength was recorded at each circle on the polar graph paper by holding the sensor in line with the parallel lines on the sheet and so that the white dot on the sensor was on the left and at a 90^o angle with the parallel lines pointing north. The magnetic field strength was recorded using Vernier Lab Pro. This value was recorded along with the respective radius.

Graphical Analysis was used to plot B vs. 1/R and perform a linear trend. The resulting slope was 9.2 x 10^-7 Tm and the correlation was 0.9734. It was somewhat difficult to get accurate readings at the smaller radii, which negatively affected this correlation. The slope from this plot in comparison to the value of μ_oI / 2π, 1.4 x10^-6 Tm, yielded a percent difference of 41%. The data followed the expected trend of decreasing in magnetic field strength as the radius increased. Possible sources of error include the difficulty of aligning the sensor perfectly along the radius of each circle, and it was also a challenge to get an accurate reading from the sensor as the readings kept jumping around. Additionally, if the sheet was not perfectly aligned northward, there would be interference from the earth’s magnetic pull which would have affected the readings.

During part two of the experiment, magnetic field strength was measured as a function of the number of turns in a wire. First, the power supply was turned off and the center aluminum wire was removed along with the top table of the tangential galvanometer. The magnetic field sensor was inserted into the side of the cylindrical support by screwing it into the side of the top table support using a threaded plastic bushing. The white dot of the sensor was aligned to be in the center of the cylindrical support with the white dot facing upward. The sensor was zeroed and the power supply was turn on to 7 A. A piece of flexible wire was wound around the outside of this support to complete one full turn. The magnetic field sensor measured B using “Events with Entry” on Lab Pro. The magnetic field strength was measured for six full turns of the wire. The diameter of the cylindrical support was also measured, being 0.013 m.

Graphical Analysis was used to plot B vs N and perform a linear trend. The resulting slope was 6.53 x 10^-5 T/turn and the correlation was 0.9943. The data followed the expected trend of increasing in magnetic field strength as the number of turns increased. The slope from this plot in comparison to the value of μ_oI / 2R, 6.7 x 10^-5 T/turn, yielded a percent difference of only 2.6%. Possible error could have resulted from the earth’s magnetic field. Additionally, the wire was not perfectly wrapped around the base, which would throw the readings off as it would not have been an exact turn of the wire. Finally, it was difficult to align the sensor perfectly in the middle of the support, so again this would contribute to some error in the readings.

Equations

Percent Difference = |E2 – E1| / [ (E1 + E2) / 2 ] x 100%

B x R = μ_oI / 2π

B / N = μ_oI / 2R

Purpose

To investigate the current flow and voltages in series and parallel circuits, and also to use Ohm’s law to calculate equivalent resistances of series and parallel circuits.

Hypothesis

The calculated equivalent resistances for the series circuits will abide by the equation R_eq = R₁ + R₂ and for the parallel circuits the value will be similar to 1/R_eq = 1/R₁ + 1/R₂. The current flow is expected to be uniform throughout the series circuits, but will be stronger through the smaller resistor in the parallel circuits. The voltages across each resistor should add up to V_TOT in the series circuits, and the voltages should be uniform throughout the parallel circuits.

Labeled Diagrams

See attached sheet.

Data

Part 1: Series Circuits

Power supply: 2.5 V

Part 2: Parallel Circuits

Power supply: 2.5 V

Part 3: Currents

Power supply: 5 V

Graphs

Part 3: Currents

Trial 1:

Average current through 10 Ω: 0.0655 A

Average current through 51 Ω: 0.0656 A

Trial 2:

Average current through 68 Ω: 0.0648 A

Average current through 51 Ω: 0.0925 A

Questions

Part 1:

1. Examine the results of Part 1. What is the relationship between the three voltage readings: V₁, V₂, and V_TOT?

V₁ plus V₂ is about equal to V_TOT.

2. Using the measurements you have made above and your knowledge of Ohm’s law, calculate the equivalent resistance (R_eq) of the circuit for each of the three series circuits you tested.

See data table.

3. Study the equivalent resistance readings for the series circuits. For each of the three series circuits, compare the experimental results with the resistance calculated using the rule for calculating equivalent resistance outlined in the Theory section. In evaluating your results, consider the tolerance of each resistor by using the minimum and maximum values in your calculations.

The range for the 10 Ω resistor is 9.5 Ω to 10.5 Ω. This gives a theoretical R_eq range of 19.0 Ω to 21.0 Ω for trial 1. The calculated value of 19.0 Ω falls just on the edge of this range. For trial 2, the 10 Ω resistor has the same range and the 51 Ω resistor has a range of 48.45 Ω to 53.55 Ω. This gives a theoretical R_eq range of 57.95 Ω to 64.05 Ω. The calculated value of 58.2 Ω again falls just on the edge of that range. For trial 3, the theoretical R_eq range is 96.9 Ω to 107.1 Ω, and the calculated R_eq value of 106 yet again falls within that range.

Part 2:

4. Using the measurements you have made above and your knowledge of Ohm’s law, calculate the equivalent resistance (R_eq) of the circuit for each of the three parallel circuits you tested.

See data table.

5. Study the equivalent resistance readings for the parallel circuits. Do your results verify what is expected for R_eq from the Theory section?

Yes, the calculated equivalent resistances have about the same value as 1/R_eq = 1/R₁ + 1/R₂.

6. Examine the results of Part 2. What do you notice about the relationship between the three voltage readings V₁, V₂, and V_TOT in parallel circuits?

V₁, V₂, and V_TOT are all about equal.

Part 3:

7. What did you discover about the current flow in a series circuit in Part 3?

The current flow through each resistor is equal.

8. What did you discover about the current flow in a parallel circuit in Part 3?

The current flow through each resistor differs.

9. If the two measured currents in your parallel circuit were not the same, which resistor had the large current going through it? Why?

The smaller resistor had the largest current going through it because current prefers to go through the path of least resistance.

Conclusion

During part 1 of the experiment….procedure (what was done), results, expectations and sources of error.

During part 2 of the experiment….procedure (what was done), results, expectations and sources of error.

During part 3 of the experiment…procedure (what was done), results, expectations and sources of error.

Equations

R = V / I

R_eq = R₁ + R₂ + R₃ + …

1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + …

V_TOT = V₁ + V₂ + V₃ + …

Synthesis and Determination of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃

Abstract

The reaction of mesitylene with Mo(CO)₆ under reflux yields [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ in low percent yield (around 1%). ¹H NMR of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ shows singlets at δ 2.25 and 5.23 with absorption ratios of 9:3, respectively. ¹H NMR of mesitylene shows singlets at δ 2.25 and 6.78, also with absorption ratios of 9:3, respectively. This suggests addition of the metal complex to mesitylene causes downfield shifting of the signal for protons attached directly to the ring as they are unshielded from the backbonding of carbonyl groups. The IR spectrum of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ shows a strong antisymmetric C-O stretch at 1852 cm^-1 and a medium symmetric C-O stretch at 1942 cm^-1 with peak areas of 64.462 cm^-1 and 9.111 cm^-1 respectively. The calculated OC-Mo-CO bond angle is 108.32°.

Introduction

The reaction of mesitylene with Mo(CO)₆ under reflux yields [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃. The reaction specifically takes place in the following manner:

Scheme 1

This is compound of interest because it a metal-arene complex and can be considered to be an octahedral rather than tetrahedral complex. This is because the OC-Mo-CO bond angles are close to 90° instead of the expect 109.5° for tetrahedrals.¹ In order to determine the structure of said substance from its ¹H NMR, the peaks must be compared to the same spectrum of mesitylene for indication of identical methyl group peaks and downfield shifting a peak indicative of protons attached directly to the ring. The IR spectrum of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ can be analyzed for peaks indicative of symmetrical and antisymmetrical carbonyl stretching, whose areas can be used to calculate the bond angle between the carbonyl groups attached the to metal.

Experimental

All syntheses were carried out in nitrogen and the reagents and solvents were purchased from commercial sources and used as received unless otherwise noted. The synthesis of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ (1) was based on reports published previously.¹

[1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ (1). Mo(CO)₆ (2.083 g, 7.92 mmol) and mesitylene (10 mL, 72 mmol) were added subsequently to a 100 mL 3 neck round-bottom flask along with a small magnetic stir bar. A sand bath was constructed and set to 50% power. A greased sidearm, stopcock, and 30 cm cold water condenser were attached to the round-bottom flask. A greased gas inlet was then attached to the condenser and connected to a bubbler. The condenser was not connected to a cold water source; it was used only to allow air to circulate. The sidearm was connected to a nitrogen source, and the system was allowed to degas for 5 minutes. The system was then put on the sand bath and the stir bar was spun at a moderate speed via a magnetic stirring instrument.

After 5 minutes, the solution in the round-bottom flask was not boiling as outlined, so the sand bath was turned up to 70% power. The sand bath was turned up to 85% another 5 minutes later. A rigorous boil was achieved when the sand bath was set to 95% 5 minutes after that. It was then set to 85% power in efforts to obtain a less extreme boil. After a total of 0.33 h of reflux, the solution was taken off the sand bath and allowed to cool to room temperature. When the apparatus was removed from the sand bath, it was dropped and roughly more than 60% of the solution was lost. The remaining solution cooled to a blackish yellow color.

The following and final procedures took place in the presence of air. Once cool, the solution was washed with 15 mL of hexane via suction filtration in a 15 mL frit. The solution was then washed with another 5 mL of hexane. About 10 mL CH₂Cl₂ was added to the blackish yellow powder precipitate remaining in the frit. The powder was washed with 25 to 30 mL of hexane and vacuum dried. This powder was discarded and the collected yellowish washings were rotovapped for about 0.33 h to obtain the desired product. The product was vacuum filtered, as it would not completely dry under the rotovap. The resulting yellowish powder was determined to be 1 (0.028 g, 1.18% yield based on the amount of Mo(CO)₆ used). ¹H NMR (CH₂Cl₂): δ 2.25 (s, -CH₃), 5.23 (s, C-H). FTIR (ATR) ν_(C-O) 1852 cm^-1 (s, C-O linkage), ν_(C-O) 1942 cm^-1 (m, C-O linkage).

1,3,5-C₆H₃(CH₃)₃ (2). The ¹H NMR spectrum of 2 was extrapolated from the literature.^{1 1}H NMR (CHCl₃): δ 2.25 (s, -CH₃), 6.78 (s, C-H).

Results

The reaction of Mo(CO)₆ with mesitylene yielded 0.028 g of the product, [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃. This translated to 0.09328 mmol, and thus a 1.18% yield based on the amount of Mo(CO)₆ used, which was the limiting reagent in the reaction. Mo(CO)₆ reacted to form the product in a 1:1 ratio, and 7.92 mmol of Mo(CO)₆ was used to start, so that proportion was taken into account when calculating the percent yield. Proton NMR spectroscopy yielded a two peaks of interest. A peak found at δ 2.25 was indicative of methylhydrogens and a peak noted at δ 5.23 was suggestive of hydrogens attached directly to the aromatic ring.¹ These peaks were noted with relative intensities of 9 to 3, respectively. Peaks seen at δ 7.25 and 1.54 were attributed to solvent and hexane, respectively. The ¹H NMR spectrum of mesitylene in CHCl₃ showed absorptions at δ 2.25 and 6.78 with relative intensities of 9 to 3, respectively.¹ The peak at δ 2.25 hinted of methyl protons and the peak at δ 6.78 was suggestive of protons bonded directly the ring. The mass spectrum of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ showed a peak at m/z = 302.0, which is nearly equal to the molar mass of said substance.¹ The IR spectrum showed a strong peak around 1852 cm^-1 of area 64.462 cm^-1 indicative of antisymmetrical C-O stretching and a medium peak around 2942 cm^-1 of area 9.111 cm^-1 indicative of symmetrical C-O stretching. These areas were used to calculate a OC-Mo-CO bond angle of 108.32°.

Discussion

The percent yield of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ is very poor. Much of this quantitative shortcoming can be attributed to clumsiness, as much of the solution containing future product was lost when the reaction vessel was dropped. This not only resulted in a direct loss of solution, but also exposed the solution to air. The solution had been kept under nitrogen as to prevent decomposition of the products. This exposure to air undoubtedly had a contribution to the poor percent yield. The solution was also not heated as desired because it was difficult to control the sand bath. The solution was to be brought to a moderate boil, but could not be controlled to do so. It would not boil, then boiled rigorously a moment later. Attempts were made to subdue the boiling, but were unsuccessful. This overheating was probably not favorable for the reaction. More adept control of the sand bath would have resulted in a better yield of product.

Washing the product with excess amounts of hexane and CH₂Cl₂ also may have added to the loss of product. Excess washing would make it difficult to extract the product from the solution, as there would have been a relatively small amount of product compared to the amount of solution it was dissolved in. The solution could not be completely dried with the rotovap, which means there was an excess of hexane and/or mesitylene in the solution. Vacuum filtration then had to be used to collect the product, which was not ideal. Best case scenario, the rotovap would have completely dried the product and it would have been scraped out of the flask. Vacuum filtration gives a better chance for loss of product.

The product seemed pure as it produced clear ¹H NMR and IR spectra readings. The ¹H NMR spectrum shows a methyl peak at δ 2.25 and a C-H peak at δ 5.23, and the reagent in the reaction, mesitylene, also gives a peak at δ 2.25. This seems to confirm the structure and addition of the metal complex, as only the C-H peak was shifted downfield. The methyl protons are too shielded to be affected by the metal. The downfield shift is caused by backbonding of the carbonyls. IR spectroscopy revealed 3 C-O stretches, 2 of which were accounted for by a strong peak at 1852 cm^-1 and the 3^rd of which was accounted for by a medium peak at 1942 cm^-1. Two peaks were seen because of symmetrical and antisymmetrical stretching of the carbonyls.¹ The two antisymmetical modes have exactly identical absorption frequencies, and the symmetrical mode has a different absorption frequency than them, which means that a total of two peaks should be seen.¹ The areas of these peaks, 64.462 cm^-1 for antisymmetical stretch and 9.111 cm^-1 for symmetrical stretch, allowed for discovery of the bond angle between the CO ligands. The calculated angle was 108.32°. This seems to make sense as [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ is predicted to be a tetrahedral complex and the expected bond angle for tetrahedral complexes is 109.5°, but in reality [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ acts more like an octahedral complex and the bond angle should be slightly less than 90°.¹ This error could be due to excess solvent or impure an sample, which resulted in skewed IR spectrum readings and thus incorrect peak areas. The oxidation state and electron count of Mo in [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ are 0 and 6 electrons, so it is an 18 electron complex.

Conclusion

The main purpose of the experiment was to interpret the ¹H NMR and IR spectra of [1,3,5-C₆H₃(CH₃)₃]Mo(CO)₃ to confirm the product and to decipher the bond angle between the carbonyls. The ¹H NMR spectrum of the product showed peaks at δ 2.25 and 5.23, while the ¹H NMR spectrum of mesitylene gave peaks at δ 2.25 and 6.78, both in ratios of 9:3, respectively. This ratio seems to confirm methyl groups and single protons attached to the ring. The downfield shifting of the second peak is attributed to the coordination of mesitylene to the metal complex. The protons attached directly to the ring are affected by backbonding of the carbonyl groups. The methyl hydrogens are shielded, and thus are not affected by the metal complex. The IR spectrum yielded two peaks near 2000 cm^-1. These two peaks account for 3 C-O stretches, 2 of which are accounted for by a strong peak at 1852 cm^-1 and the 3^rd of which are accounted for by a medium peak at 1942 cm^-1. The strong peak accounts for antisymmetrical stretching of the carbonyls and the medium peak accounts for symmetrical stretching of the carbonyls. The areas of these peaks, 62.462 cm^-1 and 9.111 cm^-1 respectively, provide for a theoretical angle between the carbonyls of 108.32°. In reality, this angle should be only nearly 90°. The percent yield for the reaction was poor and could have improved with a more steady hand, more precise heating of the reagents, less exposure of the solution to air, and less solvent used for washing.

References

(1) Angelici, R. J.; Girolami, G. S.; Rachufuss T. B. Synthesis and Technique in Inorganic Chemistry: A Laboratory Manual; University Science Books: Sausilito, CA, 1999; pp 161-170.