Beers Law Lab Essay Example for Free

Beers faithfulness Lab EssayObjectiveThe purpose of this science lab is to demonstrate that there is a linear human relationship between the outlet of molecules that can absorb light exemplify in a stem and the tot of light absorbed by a resultant. This lab should prove that Beers practice of law and the equation A=a x b x c, is a linear relationship. ProcedureThe only deviations in the lab procedure was that the pullulate firmness was made before arrival to the lab with 0.570 g of KMnO4 in 0. ergocalciferol L. The diluted resolvents and the Spec 20 were utilise as directed in the lab manual. The same(p)(p) cuvette was used each(prenominal) time to eliminate misplay. Cuvettes atomic number 18 altogether made differently and capture a end in how they measure. If a new cuvette was used each time, the data would be slightly off cod to the possibility of each cuvette having different characteristics which affect the measurements in the Spec 20. Data CalculationsT o find the thou of the telephone line solution* agate line 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 crinkle solution = molarity of Stock Solution 0.570g KMnO4 x (1 mole / 158.04g) / 0.500 Liters = 0.00721 MTo find the molarity of solution 1*Note To find the molarity of the first solution, use the molarity tack for the gillyflower 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. kilobyte0 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 flas k) = M of solution 1 5.00 mL stock solution x (0.00721 moles / 1000mL ) / 0.10000 L = 0.000361 MTo 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 MTo find the molarity of solution 3*Note To find the molarity of the thirdly solution the same procedure is followed as finding the molarity of the first solution, except you go forth 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 MTo fi nd 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 ConcentrationTrialAbsorbance% TAverage AbsorbanceAverage % T10.00003605 M10.82115.10.81415.320.81115.430.81115.520.0001442 M10.32447.40.32547.320.32647.230.32447.430.0001805 M10.38840.90.40239.620.40639.230.41338.740.000072 M10.208620.20961.820.20861.930.21161.5Figure 1. The figure below shows the absorbance vs. the molar concentration of KMnO4.To find the extinction coefficientThe 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 =aa= 3139.9 L * mol-1 * cm-1Discussion and ConclusionIn this lab the equation of Beers law was proven to have a linearrelationship. 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 deliberate different molar concentrations and each concentration was placed in the spec 20. After creating a circularize 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 offsets of error in this experiment. First, if one solution was diluted ill-judgedly all of the following solutions were diluted wrong since they came from the f irst 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 directs were done exactly due to whatsoever of the solution being transferred was often left in the pipet.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 finding of the lab is that Beers Law equation is indeed linear, and the absorban ce 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 boffo in determining the relationship of the equation in Beers Law.Questions2. 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 becauseyou 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. Solut ion 4 probably has the greatest error because it was the last solution to be diluted. Any errors made in diluting a solution will require through to the last solution diluted because the first solutions are used to dilute the latter solutions. For precedent if solution 1 is incorrectly diluted then solution 3 will be incorrectly diluted and then solution 4 will be incorrectly diluted.