An environmental chemist working for the Environmental Protection Agency (EPA) w

Question

An environmental chemist working for the Environmental Protection Agency (EPA) was directed to collect razor clams from a heavily-contaminated river superfund site and analyze them for cadmium (Cd2*) content using graphite furnace atomic absorption spectrometry (GFAAS). The chemist freeze-dried the clams and processed them in a blender resulting in 50 g of homogenized dry weight. A representative 52.00 mg sample was taken from the 50 g dry material and dissolved in 100.00 mL of 0.1 M HCl to create a sample solution. Using the method of standard additions, the chemist prepared 5 standard solutions in 100.00-mL volumetric flasks, each containing 5.00-mL aliquots of the sample solution. Varying amounts of a 85.0 ppb (Ag L-1) cadmium standard were added to each of the flasks, which were then brought to volume with 0.1 M HCI. The solutions were then examined for their Cd2 content using GFAAS, resulting in the following absorbance data. Sample Vol. (mL) Cde Standard Vol. (mL) cd2 Standard Mass (Ag) Absorbance 0.000 0.080 5.00 0.00 5.00 2.50 0.213 0.119 0.425 0.163 5.00 5.00 0.638 0.200 5.00 7.50 10.00 0.850 0.241 5.00 Determine the amount of Cd2 per gram of dry clam weight. Express your final result as weight percent (grams of Cd2 per gram of dry clam multiplied by 100). Additionally, find the absolute uncertainty (standard deviation) and the 95% confidence interval of the measurement. A list of Student's t values can be found here. Report the absolute uncertainty and 95% confidence level with two significant figures. Number Number Scroll down to o lo view the entire wt t uncertainty question

Explanation / Answer

concentration of standard in 100 ml solution,

moles of Cd2+ in standard = moalrity x volume = 8 x 10^-5 g/L x 0.0025 L = 2 x 10^-7 mols

Molarity of Cd2+ standard in total volume = 2 x 10^-7/(0.1 + 0.0025) = 1.95 x 10^-6 M

gave an absorbance of 0.119

blank without standard absorbance = 0.080

absorbance due to standard = 0.119 - 0.080 = 0.039

1.95 x 10^-6 M of standard gave an absorbance of 0.039, so concentration of Cd2+ in sample

= 1.95 x 10^-6 x 0.08/0.039 = 4 x 10^-6 M

mass of Cd2+ in 100 ml solution = 4 x 10^-6 x 112.414 = 4.5 x 10^-4 g

This is in 54 mg sample = 0.054 g

So, in 50 g Cd2+ concentration = 4.5 x 10^-4 x 50/0.054 = 0.417 g = 417 mg

Thus, the mg of Cd2+ per gram of clam = 417/50 = 8.34 mg Cd2+ / g clam

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