Spectrophotometric analysis of a penny

8 August 2016

Once we have gathered that information, we will use the results in the other two experiments. The purpose of the second experiment is to obtain a calibration curve based on different concentrations and absorptions. The purpose of the last experiment is to determine the copper content in a penny. Pennies minted in the United States no longer contain pure copper metal. This change was due to the fact that the cost of copper metal required to produce a penny was higher than the face value of the penny. Pennies now consist of a copper coating on a core that contains primarily zinc metal.

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The percentage of copper in a penny is now about 2. 5% We will also be dealing with complex ions. Complex ions are formed by the bonding of a metal atom to two or more ligands by coordinate covalent bonds. A ligand is a neutral molecule attached to the central metal ion in a complex ion. In the simplest term, complex ion has a metal ion at its center with a number of other molecules or ions surrounding it. Many of these are highly colored due to their ability to absorb light in the visible region of the electromagnetic spectrum. In the first experiment, we will first dissolve a penny in a concentrated solution of nitric acid, HNO3.

In aqueous solution, most of the first-row transition metals form octahedral complex ions with water as their ligands as shown for copper: Cu(s) + 4 HNO3(aq) + 4 H2O(l) > Cu(H2O)62+(aq) + 2 NO2(g) + 2 NO3-(aq) Once our penny has been dissolved, we will then convert the aquated copper complex ions to their tetraamine complex ions : Cu(H2O)62+(aq) + 4 NH3(aq) > Cu(NH3)42+(aq) + 6 H2O(l) We will then be allowed to detect the presence of Cu(NH3)42+ ion by its characteristic deep-blue color, and you can measure its intensity with a spectrophotometer as a way to determine the percentage of copper in a penny.

Since copper is a transitional metal, it has colored compounds due their partially filled d orbitals. To better explain how the blue color is emitted, a white light hits a transition metal compound and a certain color of radiation is absorbed. The absorption of energy excites the electrons and splits up the electrons in the d orbitals. This is called the splitting of the d orbitals. As we learned in a previous lab called electromagnetic spectrum, the electrons are unstable at their excited state so they must come back down to their original state.

When the electrons go back to their original state, they release the energy absorbed as light. Since copper absorbs yellow radiation, the color it emits is commentary to the radiation color. In this case, since blue is opposite from yellow on the color wheel, blue will be the color visible to the human eye. To do this part of the experiment, we must construct a calibration curve that relates the measured absorbance, A, to known concentrations of Cu(NH3)42+ ion using the Beer-Lambert Law: A = a? b? c

The Beer–Lambert law, also known as Beer’s law or the Lambert–Beer law or the Beer–Lambert–Bouguer law (named after August Beer, Johann Heinrich Lambert, and Pierre Bouguer) relates the absorption of light to the properties of the material through which the light is travelling. It is the linear relationship between absorbance and concentration of an absorber of electromagnetic radiation. The law states that there is a logarithmic dependence between the transmission of light through a substance and the product of the absorption coefficient of the substance, and the distance the light travels through the material.

In simplest terms, Beer’s Law is a physical law stating that the quantity of light absorbed by a substance dissolved in a non-absorbing solvent is directly proportional to the concentration of the substance and the path length of the light. In order to get a calibration curve, you will need to plot your points and get the best fit line. This will give you a slope. The slope is the molar absorptivity constant because it is related to absorbance and concentration. When you plot five concentrations and absorbance readings, you can determine the molar absorptivity constant.

The calibration curve will allow us to solve for x using the given Beer’s Law formula. In this experiment, we will be using percent yields instead of theoretical yields. A percent yield is the end results you get after preforming an experiment. The method to find the percent error is to subtract the theoretical yield from the experimental yield and divide your answer by the theoretical yield and multiply by 100%. This will be used to calculate how much copper was in the penny we are dissolving. % error = | your result – accepted value | x 100 % accepted value

Procedure: As stated in the lab manual 1. Obtain a penny minted, and RECORD the physical properties – mass, thickness, diameter, year minted, city minted (D or P), appearance of penny) – in your data table. Place your penny in a 100 mL or 150 mL beaker and label the beaker; and obtain a watch glass to rest on top of the beaker. 2. In the fume hood you will measure out 20 mL of 8 M HNO3 in a graduated cylinder and add it to the beaker containing the penny. Cover the beaker with the watch glass.

The reaction of the copper and zinc metals in the penny with HNO3 is quite vigorous so you will not need to stir the reaction. RECORD your observations of the reaction, and allow the reaction to go to completion in the fume hood. 3. Fill a 100 mL volumetric flask approximately half way with distilled water. In the fume hood, add your HNO3 – penny solution. Add distilled water to the mark, cover the flask and invert several times to mix. 4. Obtain a clean 10 mL volumetric flask, and using a 1 mL volumetric pipet, transfer 0. 80 mL of the penny solution. 5. In the fume hood, add about 0.8 mL of 15 M NH3 carefully from the buret until the light-blue precipitate that initially forms dissolves and a deep-blue solution results. Then, fill the flask to the mark with distilled water, cover and invert several times to mix. 6. Using the same procedure you did for the standard solutions, measure the absorbance of the solution. 7. Repeat steps 16 through 19 two more times. These results from each trial should be very close to each other. Record these values in your data table. 8. Discard the remaining solutions in the labeled waste container in the fume hood.

Rinse the flasks with small portions of tap water and discard the rinses. Data: Part I Wavelength (nm) Absorption (A) Wavelength (nm) Absorption (A) 440 .45 580 1. 350 3 .100 600 1. 420 480 .226 620 1. 440 500 .400 640 1. 360 520 .628 660 1. 230 540 .925 680 1. 100 560 1. 170 700 .955 Wavelength with maximum absorbance on spectrophotometer (nm) 620 Part II Concentration of copper (II) nitrate solution (M) Wavelength setting on the spectrophotometer (nm) Flask mL 0. 40 M Cu2+ Distilled H2O (mL) Concentration Absorbance of Cu(NH3)42+ 1 .2 9 .004m .282 2 .4 9. 2 .008m.528 3 .6 9. 4 .012m .825 4 .8 9. 6 .016m 1. 090 5 1. 0 9. 8 .02m 1. 380 Properties of the penny Mass (g) Thickness (nm) Diameter (nm) 2. 502 g 1. 1 nm 19. 02 nm Year City 1984 Denver Trial Absorbance Average Absorbance 1 .106 .092 2 .079 3 .092 Appearance of reacted penny/ammonia solution: The appearance of the penny that reacted with the ammonia solution went from a brown copper color to a blue/green color. While it was dissolving, there brown smoke like gas forming at the top of the lid. The bottom of the flask also got very hot to the touch. Data Analysis: 1.

Plot absorbance (y-axis) versus the concentration of the Cu(NH3)42+ ion from part II. Add a best fit line (trend line) and determine the slope of the line. Print the graph in “portrait” orientation and attach as page 2. Show that the slope of the line is equal to the molar absorptivity, a, of the Cu(NH3)42+ ion at this wavelength. Include proper units. Slope: 65. 94 concentration/absorbance 2. From the average absorbance of your penny samples, and the line equation from your standard curve, calculate the concentration of the Cu(NH3)42+ ion in your sample from the penny.

The ammonia molecules attach slowly, and in between each attachment, there is a chemical equilibrium. The more ammonia is added, the more complex is formed, as the equilibrium is pushed to the product side. The blue color is the result of the complex absorbing light in the visible light spectrum, and having a concentration high enough for the eyes to detect. 7. If the atomic radius of a copper atom is 1. 28 x 10-8 cm, how many atoms thick is the copper coating on your penny? R = 1. 28 x 10^-8 cm (2) = d = 2. 56 x 10^-8 cm Thickness in atoms = 9. 11 x 10^-4 cm/ 2. 56 x 10^-8 = 35623. 8 atoms

Error Analysis: 8. What assumptions were made in the experiment and in the calculations? We assumed that the reaction would go to completion and that we would get a close to accurate calculation. 9. Research the actual percent copper in a penny. Calculate your percent error for the percent of copper in the penny. Comment on any significant sources of error. l1. 8-2. 5/2. 5l x 100%=28% error Conclusion: Up to mid-1982, cents were about 98% copper. Since our penny was made in 1984, it had a low percentage of copper contained in it. The rest was made out of tin and/ or zinc depending on the year of issue.

In the middle of 1982 the price of copper shot up so the mint changed to coin’s composition to pure zinc core with a plating of copper so that the color would be identical to prior issues. The copper plating makes up about 2. 5% of the coin’s weight. Since our copper came to 1. 9%, we concluded that the experiment was done correctly minus a few errors. Discussion: Our experiment went really well considering our copper percentage was close to that of the actual percentage. Although we had a 28% error, it could have been accounted for during the experiment.

There was a possibility that we did not measure our penny accurately or that we did not let the reaction go to 100% completion. Overall our data was sufficient and our results proved we performed the experiment correctly. Questions: N/A MSDS: Chemical Names: Nitric acid Formula: HNO3 Molecular Wt: 63. 01 Clear, colourless or yellowish liquid with an acrid, suffocating odour. Hygroscopic. Will not burn. During a fire, nitric acid decomposes with the release of corrosive nitrogen oxide gases. Closed containers may develop pressure on prolonged exposure to heat. STRONG OXIDIZER.

Contact with combustible and easily oxidizable materials may result in fire and/or explosion. Highly reactive. May react violently or explosively and/or ignite spontaneously with many organic and inorganic chemicals. Releases extremely flammable hydrogen gas on contact with many metals, particularly in powered form. Generates heat when mixed with water. Nitric acid poses a very serious inhalation hazard. Symptoms of exposure include dryness of the nose and throat, cough, chest pain, shortness of breath and difficulty breathing. Causes lung injury-effects may be delayed. CORROSIVE to the eyes, skin and respiratory tract. Causes severe burns.

May cause permanent eye injury or blindness and permanent scarring. Chemical Name: Ammonia, Anhydrous Synonyms: Ammonia Formula: NH3 MOL. WT. : 17. 03(NH3) Ammonia is an irritant and corrosive to the skin, eyes, respiratory tract and mucous membranes. May cause severe burns to the eyes, lungs and skin. Skin and respiratory related diseases could be aggravated by exposure. Additional References: Spectrophotometric Analysis of A Penny Lab http://www2. volstate. edu/chem/1110/CopperClad. htm Clark, J. (2003). An introduction to complex metal ions. Retrieved from http://www. chemguide. co. uk/inorganic/complexions/whatis. html

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