An environmental chemist working for the Environmental Protection Agency (EPA) w
ID: 548135 • Letter: A
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 their Cd2+ content using graphite furnace atomic absorption spectrometry (GFAAS). The chemist dried the clams at 95° C overnight and ground them in a scientific blender, resulting in approximately 50 g of homogenized dry weight. A representative 97.19 mg sample was taken from the approximately 50 g of dry material and dissolved in 100.0 mL of 0.1 M HCI to create a sample solution. Using the met prepared five standard solutions in 100.0-mL volumetric flasks, each containing 5.00-mL aliquots of the sample solution. Varying amounts of a 95.0 ppb (g/L) Cd2+ standard were added to each of the flasks which were then brought to volume with 0.1 M HCI. The Cd2*content of the solutions was then analyzed using GFAAS, resulting in the following absorbance data f standard additions, the chemist Sample Vol. (mL) Cd2 Standard Vol. (mL) Absorbance 5.00 5.00 5.00 5.00 5.00 0.00 2.50 5.00 7.50 10.00 0.080 0.119 0.163 0.200 0.241 Determine the amount of Cd2 per gram of dry clam. Express your final result as milligrams of Cd2 per gram of dry clam. Number 279.62 mg Cd g clamExplanation / Answer
Cd2+ conc., originated from std.sol 87 ppb, in 100 mL flasks
Absorbance
(corrected for errors)
Measured Absorbance – Absorbance in the first flask (without std. additions)
0 ppb
0.080
0.000
2.17 ppb
0.120
0.040
4.35 ppb
0.160
0.080
6.52 ppb
0.200
0.120
8.70 ppb
0.240
0.160
First solve the concentration measurement in the solutions prepared in 100 mL flasks.
The calibration is linear and the slope is
0.160 A.units / 8.70 ppb = 0.0184 a.u/ppb
The absorbance for the unknown itself is 0.080. Then the concentration of the unknown is
0.080 a.u. / 0.0184 a.u/ppb = 4.35 ppb in 100 mL flask (for AAS measurement).
In the 5 mL sample aliquot
Cd2+ conc. = 4.35 ppb x 100 mL/5 mL= 95 ppb = 95 x 10-9 g/mL
In 61 mg sample, the content of Cd2+ is
95 x 10-9 g/mL x 100 mL = 9.50x10-3 mg Cd
61 mg sample (dry clam) contains …….9.50 x10-3 mg Cd
1000 mg (1g)……………………………….x
X = 155.74 x10-3 mg Cd /g dry clam
Rounded result:
0.156 mg Cd /g dry clam
Cd2+ conc., originated from std.sol 87 ppb, in 100 mL flasks
Absorbance
(corrected for errors)
Measured Absorbance – Absorbance in the first flask (without std. additions)
0 ppb
0.080
0.000
2.17 ppb
0.120
0.040
4.35 ppb
0.160
0.080
6.52 ppb
0.200
0.120
8.70 ppb
0.240
0.160