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Study Problems

Experiment 1:

Absorption Spectroscopy

> 1. (a) A 2 x 10-3 M solution of a compound X has an absorbance of .15 at 550 nm. Another solution of unknown concentration of the same compound has an absorbance of .06 at the same wavelength. What is the concentration of this solution? (the same cuvette was used for the two measurements).

> Solution:

> c1 = concentration of first solution = 2 x 10-3 M

> c2 = unknown concentration = ?

> A1 = Absorbance of first solution = .15 at 550 nm

> A2 = Absorbance of unknown solution = .06 at 550 nm

> l1 = l2 = cuvette width (path length) = remains constant

> A = ecl (Beer-Lambert's Law)

> = e l

> e is constant for a given wavelength and choice of solute and solvent.

> l is unchanged for both Absorbance measurements (it is the width of the cuvette).

> Therefore, el = constant

> Now we can set up the following proportion.

> c2 = .0008 M

> b. Knowing the cuvette had a path length of 1 cm, what is the extinction coefficient , e, of compound X?

> Solution:

> A = ecl

> e = 75 M-1 cm-1

> 2. With the data provided below, find the extinction coefficient, e, and the absorbance of a

> 7.5 x 10-3 M solution of [Fe(SCN)]2+. The cuvette has a path length of 1 cm.

> Absorbance (460 nm) | Concentration of [Fe(SCN)]2+.
> .4 | 10.0 x 10-3 M
> .2 | 5.0 x 10-3 M

> Solution:

> A = ecl

> 4 = e(10.0 x 10-3M)(1cm)

> e = 40M-1cm-1

> Now that we know the extinction coefficient, plug it into Beer-Lambert's Law in order to calculate the absorbance for a solution of the desired concentration.

> A = ecl

> A = (40M-1cm-1)(7.5x10-3M)(1.00cm)

> A = .3

> 3. An archeologist wanted to determine the content of nickel (Ni) in an ancient coin. She dissolved the .324±.001 g coin in about 10 mL of nitric acid, and after adding an excess of 8-hydroxyquinolinol (HQL), she diluted the solution with distilled water to a final volume of 250.0±.2mL. This final solution showed an absorbance of .628±.001 at 365 nm, the wavelength of maximal absorbance of the Ni (HQL) complex, when measured in a 1.00.01 cm cuvette. Knowing that the extinction coefficient for the Ni (HQL) complex at 365 nm is 2500 M-1 cm -1, and that the atomic mass of nickel is 58.7 g / mol, calculate the percent content of Ni in the coin. Express the previous result with the absolute error and the appropriate number of significant figures.

> Solution:

> A = ecl (Beer-Lambert's Law, again)

> c=

> where A = ecl

> A = .628±.001

> e = 2500 M-1cm-1

> l = 1.00±.01 cm

> c=

> c= .0002512M

> Error calculation

> Let's treat the extinction coefficient as a constant without error, since we were provided no error.

> relative error (c) = rel err (A) + rel err (l)

> rel err (c) =

> rel err (c) = .0015924 + .01

> rel err (c) = .011592

> We need to express our answers with absolute errors generally.

> We define relative error as follows:

> rel err in c =

> Therefore, if we multiply the relative error by c (which includes our constant, e), we should obtain the absolute error in c.

> abs err (c) = 3 x 10-6 M

> c = .000251±.000003M

> Try not to round off to a large degree until the end when reporting the answer to an appropriate number of significant figures. Why? "Rounding off errors" (errors due to rounding off too soon)

> Now we use the concentration to find the weight of nickel in the coin.

> molarity (M)= = unit of concentration

> .000251M =

> .00006275 mol =

> wt (Ni) = .00366875

> Error associated with the previous calculation is the following:

> We'll rewrite the equation, treating the atomic weight as a constant. (Note: Atomic weights are determined experimentally and have error, but again, for our purposes, it's not critical.)

> wt of Ni = (AW)(conc.)(volume)

> rel err(wt) = rel err(conc.) + rel err (volume)

> rel err (wt) =

> rel err (wt) = .01195 + .0008 = .01275

> Multiply by the weight which include our constant (the atomic weight).

> .01275 x .00366875g = .00005g

> Therefore, .00367± .00005g

> Now we need to calculate what percent of the coin is nickel.

> %Ni = x 100%

> %Ni = x 100% = 1.13%

> Finally, let's calculate the error assoicated with our answer.

> rel err (%Ni) = rel err [wt(Ni)] + rel err [wt(coin)]

> rel err (%Ni) =

> rel err (%Ni) = .01362 + .003086 = .016706

> abs err (%Ni) = .02%

> Therefore, there is 1.13±.02% Ni in the coin.

> 4. If instead of setting the transmittance of the blank to 100% with water you used a more dilute solution of [Fe(SCN)]2+ than the one you want to measure, how would the value of the transmittance that you read from the spectrophotometer be? Higher or lower than the real?

> Answer:

> The absorption value you read from the spectrophotometer, when you use a dilute cuvette, instead of a "blank", will be lower than the real value. Your dilute solution will absorb some electromagnetic energy, so your real solution will seem to absorb less than before (because it is now being compared to a dilute solution instead of a blank). The spectrophotometer will register a higher transmittance reading also because A and T (Transmittance) are related through the familiar equation: A = - log T or A = log (1/T).

> 5. Given the concentrations of the stock solutions of Fe3+ and SCN-:

> [Fe3+] = .005M and [SCN-] = .005M

> find the concentration of [Fe(SCN)]2+ in

> (a) a solution prepared by mixing 7.5mL of each stock solution

> Let's restate molarity's definition and then calculate the number of moles of each stock solute with which we're dealing.

> molarity =

> .005M =

> moles Fe(III) = .0000375 mol

> We would have the same number of moles of SCN- since that solution has the same molarity and liters of solution. Therefore, when the two ions combine, Fe(III) and SCN-, we should produce .0000375mol [Fe(SCN)]2+ .

> Now we need to calculate the molarity of the resulting solution. How many liters is it?

> 7.5mL + 7.5mL = 15mL = .015L

> Therefore, c = = .0025 M

> (b) a dilution of .5mL of the [Fe(SCN)]2+ solution and 4.5mL of H2O

> c1v1 = c2v2

> This is the familiar dilution equation where

> c1 = initial concentration

> v1 = volume of initial concentrated solution

> c2 = final concentration (ie, dilute solution concentration)

> v2 = final volume of dilute solution.

> Plug into the equation as follows:

> (.0025M)(.5mL) = c2(5.0mL)

> c2 = .00025M

> c2 = .0003M

> It's OK to use mL although Molarity is defined with Liters. Why? The mL terms cancel here, so you are left with a unitless ratio of volumes that would be the same whether you started with your units in mL or L.

> 6. What is the purpose of finding the slope of the graph of Absorbance vs. concentration?

> Answer:

> The slope of Absorbance versus concentration gave you el. This can be seen by re-examining Beer-Lambert's Law.

> c =

> Since you know the pathlength (ie, cuvette width) was 1 cm, you can solve for the extinction coefficient, e. Then you can use this information in Beer-Lambert's Law to find the concentration of unknowns.

> 7. Why should the instrument be reset to %T = 100% (A= 0) with the blank each time you change the wavelength?

> Answer:

> The intensity of the light source and the sensitivity of the detector change with wavelength. The extinction coefficient also varies with wavelength. As we know from Beer-Lambert's Law, an increase or decrease in this important coefficient means a new value for A and additionally T.

> 8. Normally, calibration curves are plots of absorbance (y-axis) vs. concentration (x-axis). Why couldn't a plot of transmittance vs. concentration be used as easily?

> Answer:

> Once again we need to consult the germane formula: A = ecl = - log T. We see from this that concentration and transmittance are related to each exponentially and thus their relationship is less straightforward than that between A and c.