In the Amazon rainforest, a scientist is analyzing a chunk of bark to determine
ID: 1046309 • Letter: I
Question
In the Amazon rainforest, a scientist is analyzing a chunk of bark to determine if a particular substance is present. He was able to make a solution of the substance by dissolving 0.10 mmol of the substance in ethanol (total solution volume 500 mL). The scientist measured an absorbance of 0.578 au for the solution while just ethanol had an absorbance of 0.002 au. He then crushed up the bark sample (1.25 mmol) and made a solution by dissolving the bark in 50 mL ethanol. If he records an absorbance of 0.153 au, what is the mole percent of the unknown susbtance in the bark? Assume the scientist used a 1.00 mL cuvette and all absorbance in the bark sample was from the substance of interest.
Explanation / Answer
0.2095 %
According to Beer-Lambert Law
A = ecl; A = the absorbance, e = the molar absorption coefficient, l = path length of light (width of cuvette)
In the present case, since substance in standard and sample is same, and cuvette is also same,
So the absorbance of two solutions can be directly compared,
i.e. A1/A2 = C1/C2
Concentration of standard = 0.10 mmol, in 500 mL = 0.10x1000/500 = 0.2 mM
Absorbance of standard = 0.578 au
Corrected absorbance with blank = Absorbance of standard - absorbance of solvent = 0.578 - 0.002 = 0.576 au
Absorbance of the unknown sample = 0.153 au
Corrected absorbance = 0.153 - 0.002 = 0.151 au
A1/A2 = C1/C2
0.576 au / 0.151 au = 0.2 mM / x mM
x mM = 0.2 x 0.151/0.576 = 0.0524 mM
The amount of bark sample used = 1.25 mmol in 50 mL = 1.25 x 1000/50 = 25 mM
Mole percentage of the unknown substance in bark = 0.0524 mM x100 / 25 mM = 0.2095 %