### Objective:

The purpose of this lab is to demonstrate that there is a linear relationship between the number of molecules that can absorb light present in a solution and the amount of light absorbed by a solution. This lab should prove that Beer’s law and the equation A=a x b x c, is a linear relationship. Procedure:

The only deviations in the lab procedure was that the stock solution was made before arrival to the lab with 0.570 g of KMnO4 in 0.500 L. The diluted solutions and the Spec 20 were used as directed in the lab manual. The same cuvette was used each time to eliminate error. Cuvettes are all made differently and have a difference in how they measure. If a new cuvette was used each time, the data would be slightly off due to the possibility of each cuvette having different characteristics which affect the measurements in the Spec 20. Data Calculations:

### To find the molarity of the stock solution:

*Note: Molarity is moles/ Liters so in the equation below the first half is finding the number of moles of KMnO4 and the second half is dividing the moles by the liters of the solution. Grams of KMnO4 x (1 mole / molar mass (158.04g)) / Liters of stock solution = molarity of Stock Solution 0.570g KMnO4 x (1 mole / 158.04g) / 0.500 Liters = 0.00721 M

### To find the molarity of solution #1:

*Note: To find the molarity of the first solution, use the molarity found for the stock solution. Since 5.00 mL of the stock solution was used to make solution 1, multiply the molarity of the stock solution by 5.00 mL to get the moles of solution 1. Once the moles of solution 1 have been found, divide that by the liters of water that were added to solution 1. The 0.10000 L comes from the 100 mL volumetric flask the solution was made in. mL of stock solution x (moles of stock solution / liter) / total liters of solution 1 (volumetric flask) = M of solution 1 5.00 mL stock solution x (0.00721 moles / 1000mL ) / 0.10000 L = 0.000361 M

### The Essay on Determine The Macromolecules Present In An Unknown Solution

... the presence of proteins in the solutions. The purpose of this lab was to successfully identify which macromolecules were ... determine the identity of the unknown solution. The macromolecules being tested in this lab were carbohydrates (monosaccharides and polysaccharides) ... The objective of this lab was to correctly identify which macromolecules the unknown solution was comprised of using various ...

### To find the molarity of solution 2:

*Note: To find the molarity of solution 2, follow the same steps for solution 1 except use 2.00 mL instead of 5.00 mL. The same steps are used due to solution 2 being diluted from the stock solution. mL of stock solution x (moles of stock solution / liter) / total Liters in solution 2(Volumetric flask) = M of solution 2 2.00 mL stock solution x (0.00721 moles / 1000 mL) / 0.10000L = 0.000144 M

### To find the molarity of solution 3:

*Note: To find the molarity of the third solution the same procedure is followed as finding the molarity of the first solution, except you will be using the molarity of the first solution since solution three was made using the first solution. mL of solution 1 x ( moles of solution 1 / 1 L) /total liters in solution 3( volumetric flask) =M of solution 3 50.00mL solution 1 x (0.00721 moles / 1000 mL) / 0.10000 = 0.000181 M

### To find the molarity of solution 4:

*Note: to find the molarity of the fourth solution follow the steps for finding the molarity of the third solution except use the molarity of solution 2 since solution 4 was made with 50.00 ml of solution 2. mL of solution 2 x ( moles of solution 2 / 1 L) /total liters in solution 4 (volumetric flask) = M of solution 4 50.00 mL solution 2 x (0.000144 moles / 1000 mL ) / 0.10000 L = 0.000072 M Table 1. The molar concentration, absorbance values, percent transmittance, average absorbance and transmittance values are shown in the table below. Solution #

Molar Concentration

Trial

Absorbance

% T

Average Absorbance

### The Essay on Solution Preparation

... concentration of FeSCN2+ for each solution? 9. Determine the corrected absorbance of the solutions. The corrected absorbance is the difference between the absorbance of the analyte solution ... the slope of the best fit curve, determine the molar absorptivity or the molar extinction coefficient, ε of FeSCN2+. The path length is ...

Average % T

1

0.00003605 M

1

0.821

15.1

0.814

15.3

2

0.811

15.4

3

0.811

15.5

2

0.0001442 M

1

0.324

47.4

0.325

47.3

2

0.326

47.2

3

0.324

47.4

3

0.0001805 M

1

0.388

40.9

0.402

39.6

2

0.406

39.2

3

0.413

38.7

4

0.000072 M

1

0.208

62

0.209

61.8

2

0.208

61.9

3

0.211

61.5

Figure 1. The figure below shows the absorbance vs. the molar concentration of KMnO4.

To find the extinction coefficient:

The extinction coefficient is found by A/bc = a. A/c is the slope of the line from figure 1. 3139.9/(mol/L) x 1.00 cm =a

a= 3139.9 L * mol-1 * cm-1

Discussion and Conclusion:

In this lab the equation of Beer’s law was proven to have a linear relationship. The purpose was to show that molar concentration and absorbance are proportional to each other. This was proved through diluting solutions and using a spec 20 to determine the absorbance values. The solutions were diluted to give different molar concentrations and each concentration was placed in the spec 20. After creating a scatter plot it was obvious to see as the molar concentration increases the absorbance increases. This is because there are more particles present at higher molar concentrations and therefore more light will be absorbed by the particles present.

There were many possible sources of error in this experiment. First, if one solution was diluted incorrectly all of the following solutions were diluted incorrectly since they came from the first incorrectly diluted solution. One of the solutions in the experiment could have been diluted wrong, causing all of the solutions to have incorrect dilutions and the calculated values, especially the extinction coefficient, to have incorrect values. Another source of error is that when diluting the solutions not all of the solution transfers were done exactly due to some of the solution being transferred was often left in the pipet.

### The Essay on Osmosis Of Potatoes In Different Sucrose Solutions

... the potato placed in a less sweet solution. Independent variable: concentration of sucrose, concentrations: pure water H20, 0.1 M, 0.5M, 1M ... . Materials used: forcebs, beaker, scale, filter, paper, potato sticks, sugar solutions, and water. Methods: 1. Take the forcebs and pick one ... liquid, the time the potato sticks have been in the solution, the type of filter paper used, the same scale used ...

The drops left in the pipet after the transfer could make a difference in the actual molar concentration of each solution. The last source of error occurs from not placing the cuvette in the spec 20 at the same orientation. Although the same cuvette was used each trial, some carelessness may have resulted in the cuvette not being placed in the same orientation each time. Because the sides of the cuvette may be different the readings from the spec 20 may be off. The conclusion of the lab is that Beer’s Law equation is indeed linear, and the absorbance is proportional to the molar concentration. If this lab were preformed again the stock solution should be placed in the spec 20 machine and the absorbance should also be found. The measurements from the stock solution could have provided even more evidence to the conclusion. Overall though the lab was very successful in determining the relationship of the equation in Beer’s Law.

### Questions:

2. A larger cuvette diameter will produce a higher absorbance value. The diameter of the cuvette is the path length, or b, in the equation A = a x b x c. The larger the path length, the higher the absorbance will be because you are multiplying a and c by a higher value. Also there is more particles present in a larger path length to absorb light. 3. To find the extinction coefficient the equation A/cb= a is used. A larger cuvette diameter, or path length, would result in a smaller extinction coefficient. The larger the number is on the bottom the smaller the value of the extinction coefficient. 4. Solution 4 probably has the greatest error because it was the last solution to be diluted. Any errors made in diluting a solution will carry through to the last solution diluted because the first solutions are used to dilute the latter solutions. For example if solution 1 is incorrectly diluted then solution 3 will be incorrectly diluted and then solution 4 will be incorrectly diluted.