Please help me solve this question Not sure where to start

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 80.83mg sample was taken from the approximately 50 g of dry material and dissolved in 100.0 mL of 0.1 M HCl to create a sample solution. Using the method of standard additions, the chemist prepared five standard solutions in 100.0 mL volumetric flasks, each containing 5.00 mL aliquots of the sample solution. Varying amounts of a 100 ppb (μg/L) Cd2+ standard were added to each of the flasks, which were then brought to volume with 0.1 M HCl. The Cd2+ content of the solutions was then analyzed using GFAAS, resulting in the absorbance data given in the table.

Sample Volume (mL)Cd2+Cd2+ Standard Volume (mL)Absorbance
5.00	0.00	0.080
5.00	2.50	0.119
5.00	5.00	0.163
5.00	7.50	0.200
5.00	10.00	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

Answer 1

Probably the easiest place to start is by looking at your table and finding the ppb for each internal standard (spike). The ppb will be equal to the mls used, i.e. 2.5 mls = 2.50 ppb final; 5.00 mls = 5.00 ppb final, etc.

0 ppb

2.50 ppb

5.00 ppb

7.50 ppb

10.00 pb

Next step is to construct a standard curve from the 0 ppb to 10.00 ppb added. So plot the ppb on the x-axis against the Absorbance values on the Y-axis. Either do a linear regression analysis or simply eye-ball the best straight line. If you have the linear regression line, plug in 0 for Y and solve for x, or divide the y-intercept by the slope. That will be the ppb in the sample used.

Then you will have to correct for dilution and aliquoting factors. If you can get to the point of obtaining the straight line and finding ppb from that graph, I'd be happy to work through the correction factors to get the answer in mg Cd²⁺ / g dry clam. But it's difficult to do without actual numbers.

If x = 4.9 ppb, then this was in a 5 ml sample from a 100 ml solution. Thus, the 100 ml solution is also 4.9 ppb. But this 100 ml solution came from diluting 5 mls of the original 100 mls extract. So that represents a 20-fold dilution. Thus, the original 100 ml (0.1L) extract has 20 x 4.9 ppb = 98 ppb. This is equal to 98 ug/L Cd²⁺.

So, total ug of Cd²⁺ in the original extract = 98 ug / L x 0.1 L = 9.8 ug Cd²⁺

This arises from 80.83 mg of clam tissue. Thus we have 9.8 ug Cd²⁺ / 80.83 mg tissue = 0.1212 ug / mg tissue

Converting this to mg Cd²⁺ / g clam tissue,

0.1212 ug Cd²⁺ = 1.21x10^-4 mg Cd²⁺

1 mg tissue = 1x10^-3 g tissue

[Cd²⁺] = 1.21x10^-4 mg Cd²⁺ / 1x10^-3 g tissue = 0.121 mg Cd²⁺ / g clam