SPECTROPHOTOMETERY

SPECTROPHOTOMETERY

SPECTROPHOTOMETERY
AND PREPARATION OF SOLUTIONS


THINGS THAT WERE USED:
Volumetric flasks (4-100 mL, 1-250 mL, 1-500 mL); pipettes (5, 10, 15, 20 mL) w/bulb; spectrophotometer, sample tubes thereof; Kimwipes; weighing boat; wax pencil; bottle for distilled water.

THINGS THAT WERE TALKED ABOUT:

The colors of chemical substances are due to the absorption of visible light. The amount of light absorbed depends on the wavelength (and thusly "color") of the light and the nature (plus amount of substance) absorbing the light. When a substance absorbs light of certain amounts at certain "colors"; a graph can be made ploting amount of light absorbed versus the wavelength. Such a graph is called an Absorption Spectrum, and is unique to a substance much like fingerprints are to humans. The method of "fingerprinting" various substances is called Spectrophotometry and an Absorption Spectrophotometer is what is used to "take fingerprints" (actually measuring the light trapped by the substance at different "colors"). The importance of spectral analysis is twofold: the first significance is that compounds and their molecular structures can be identified by the light that they absorb, the second (and the focus of this study) is to determine the amount of a known compound in a sample (this is determined by the amount, rather than the type of wavelengths absorbed).

Erythrosin B, commonly used to color maraschino cherries, is the compound studied in this "color light" study. This carbon-based ionic compound has the chemical formula Na2C20H6O4I4 and a molar mass of 879.87 g and is a deep red when in solid form. Solutions being of different concentrations (that is, in the amount of dye they had in them) were prepared and then subjected to a light of a singular wavelength (one color and one color only) in the absorption spectrophotometer which measures the amount of light the sample absorbed.
(Molecular Formula gleaned from the figure of Erythosin B structural formula.)

When a number of samples of known concentration are subjected to the light blocking test and the amount of light they blocked is plotted on a graph against their respective concentration. How much dye an unknown has in it can be determined by seeing how much of that "magic light" it takes up and comparing it to those we already know. Details of the parts that make up this piece of equipment is given in the lab manual.

The standard unit for the amount of light absorbed is usually given in einsteins (named after the famous scientist, yes, but not implying an army of clones) , but instead will be measured by the decimal fraction of the light let pass through the sample(called I) when compared to a "blank" of distilled water; because d. water blocks no light (and hence let all light pass through) this "reference blank" will be refereed to as Io. The absorbance of light by the solution is defined by the following equation:

Absorbance= log10(Io/I)
Because spectral absorbency is dependent on the concentration "c" of the solution, the above equation can be restated according to the Beer-Lambert Law:
A=e'lc
where "l" is the length that the light photon must pass thru the sample and "e'" is a constant called the molar absorbtivity and concentration is in moles per liter. In this experiment "l" , being the diameter of the test tube, is also a constant.


WHAT LIGHT FROM YONDER TEST-TUBE BREAKS?


Solution A: c= 1.63 * 10-6 M A=0.139
Solution B: c= 3.25* 10-6 M A=0.270
Solution C: c= 4.875*10-6 M A=0.406
Solution D: c= 6.50 *10-6 M A=0.532

the Unknown (#01) A=0.192
the color of the unknown was a "lighter than thou" pink

%uF06C= 525 nm (and like a newlywed couple, it was not disturbed!)

Calculations: done graphically (see graph)

(For the record; the Baush & Lomb Spectronic 20 spectrophotometer was the kind we used.)

(not included here, text only!)


THE UNKNOWN KNOWS

Molarity of Unknown #01 solved for using the graph: 2.41 * 10-6 M
[M] of unknown using the equation A=e'c : 6.35 * 10-5 M


COMPARING THE RESULTS: This makes no sense! I can't under- stand how two completely different answers could have been obtained. Intuitively, I'd trust the graphed one because that is in-between the range for the known solutions (and right where I thought it would be) of 1.63*10-6 and 6.50*10-6; while 6.35*10-5 M isn't even close!


THINGS THAT WERE ANSWERED:

1) Using the equation A=e'c; calculate e' for
a) Solution "A": e'= 8.53 * 104 for c= 1.63*10-6 and A= 0.139
Sltion "B": e'= 8.31 * 104 for c= 3.25*10-6 and A= 0.270
Sltion "C": e'= 8.33 * 104 for c= 4.875*10-6 and A=0.406
Sltion "D": e'= 8.38 * 104 for c= 6.50*10-6 and A= 0.532
b) Using 8.39 * 104 for e' the concentration of solution "X" is
4.77 * 10-6 M (A=0.40)
c) Calculating e' for solution "X" using %uF06C= 534nm and c
calculated in part "b": e'= 6.30 * 104

2) Concentration of solution "Y" is 5.56 * 10-6 M

e' is the slope of the graphed line (q.v.) and for this experiment has the value of 1.22 * 10-5 (no, c is not "cubed")