Absorption Spectroscopy

Topics: Spectroscopy, Absorption spectroscopy, Atom Pages: 3 (565 words) Published: February 2, 2006
Abstract: This experiment explores the technique of absorption spectroscopy. The procedures deal with the wavelengths and absorption of dyes in a sample of grape Kool-aid. The use of Beer's Law helps to determine values of absorption. Introduction:

This experiment demonstrates another technique used in the analysis molecules with light. The study of light absorbed my molecules is known as absorption spectroscopy. This is very easily the opposite of emission spectroscopy because it occurs when an electron absorbs a photon and moves to a greater energy level. Absorption is thought to be an even more useful technique than emission, because emission requires an atom or molecule specific to its species. The educational objectives of the experiment include using a spectrophotometer, as well as understanding absorption spectroscopy as a quantitative technique, Beer's law, and quantitative dilutions.

Spectroscopy provides a technique for making direct measurements of numbers of atoms and molecules because of the absorption and, emission of light as a nuclear phenomenon. Because the amount of light absorbed or emitted is directly proportional to the number of atoms or molecules. Raw Data and Observations: Please see attached.


Red #40: absorbance vs. concentration

Blue # 1: absorbance vs. concentration

•% Composition (Grape):
% Composition = (mass of Red #40 dye in sample / total mass of sample) (100%)
= (4.068 x10^-4 g / 0.1019 g) (100%)
= 3.992 x10^-3 (100%)
= 0.399%

•% Composition (Grape):
% Composition = (mass of Blue #1 dye in sample / total mass of sample) (100%)
= (7.92 x10^-6 g / 0.1019 g) (100%)
= 8.07 x10^-7 (100%)
= 8.07 x10^-5%

Sample Calculations:

•Concentration of standard solution (Red #40):
Stock: 4.03 x10^-5 M
Dilution 1:4 = Stock (1/4)
= 4.03 x10^-5 M (1/4)
= 1.0075 x10^-5 M

•Concentration of dye...
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