Question

Postlab Assignments . (a) Using your own standard curve, the line equation you generated and Beer's law equation Equation 1) calculate the molar absorpivity (e) of blue dye. For all your measurem cuvette used is 1.00 cm. Use this as the value of b in Equation 1 ents, the size of (b) Using Beer's law equation again, what does a high value of molar absorptivity tell you in relation to on and absorbance of species in solution? What does a very low molar absorptivity tell you? 2. The absorbance (A) of light is related to the amount of light transmitted through the equation: AlogT or T 10 A The standard curve below is generated after analyzing a series of standard solutions of FD&C Yellow dye at a certain wavelength. (a) If a commercial "vitamin" water sample containing yellow dye has a percent transmittance, % T, of 43, what is the molar concentration of the yellow dye in the vitamin water? (Hint: You have to use equation 4 first, then Beer's law (equation 1)] (b) If another commercial drink has a transmittance of 73% at the same wavelength, how much yellow dye does it contain? Show one sample calculation below:

Please help I am completely confused, I included the 2 graphs needed for this post lab. Thank you!

Answer 1

#1. a.

Absorption coefficient, E = Slope/Path length Given (from the graph, Blue Dye) minus Slope = 86923 M^minus 1 Path length = 1.00 cm # Putting the values in above expression E = 86923 M^minus 1/1.00 cm = 86923 M^minus 1cm^minus 1 Absorption coefficient, E = Slope/Path length Given (from the graph, Yellow Dye) - Slope = 25600 M^minus 1 Path length = 1.00 cm # Putting the values in above expression - E = 25600 M^minus 1/1.00 cm = 25600 M^minus 1 cm^minus 1

#b. Beer-Lambert’s Law, A = e C L             - equation 1,              

where,

                       A = Absorbance

                       e = molar absorptivity at specified wavelength (M^-1cm^-1)

                        L = path length (in cm)

                        C = Molar concentration of the solute

# Note that the absorbance (A) of a solution is directly proportional to the molar absorptivity (epsilon, e) of the solute. Also, e remains constant for the specified molecule under specified value of temperature and experimental conditions.

Since the question mentions higher value of e, we must consider the comparison of two different solutes/ molecules with different value of e.

# Note that the absorbance (A) of a solution is directly proportional to the molar absorptivity (epsilon, e) of the solute. So, a higher value of e indicates higher absorbance of the sample when compared to another solute of same concentration but with lower value of e.

# Re-arranging equation 1-

            e = A / (C L) – note that concentration (C) is inversely proportional to e.

So, a higher value of e indicates lower concentration of the solution in the specified solution when compared to a solution of another solute with same absorbance value. That is, when two solutions of different solutes (obviously with different value of e) have same absorbance value, the solution of solute with higher value of e has a lower concentration.

#2. Note that the question relates to Graph 2- FD&C Yellow Dye

#a. Absorbance, A = - log T = - log (43 %) = -log (0.43) = 0.3665

Given, trendline equation is: Y = 25600x + 0. In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation, 1 absorbance unit (= 1Y) is equal to 25600 units of x-axis. Unit of x-axis values is M.

Now, putting the value of A in trendline equation-

            0.3665 = 25600x M^-1

            Or, x = 0.3665 / 25600 M^-1 = 1.43 x 10^-5 M

Hence, [FD&Y] in given solution = 1.43 x 10^-5 M

#b. Absorbance, A = - log T = - log (73%) = -log (0.73) = 0.1367

Now, putting the value of A in trendline equation-

            0.1367 = 25600x M^-1

            Or, x = 0.1367 / 25600 M^-1= 1.43 x 10^-5 M

Hence, [FD&Y] in given solution = 5.47 x 10^-7 M