Please I need HELP with the calculations for the Experiment: Determination of Mn
ID: 926563 • Letter: P
Question
Please I need HELP with the calculations for the Experiment: Determination of Mn and Fe in Dog Food by Atomic Absorption Spectrometry.
Use the mean Absorbance from the triplicate readings for each concentration obtained from the AA as the final Absorbance (A) value to construct the MOSA calibration curve – one for Mn and one for Fe. Give equation and R2 value for each. From these calibration curves find the ppm (microgram/mL) concentrations, in the final solution, of Fe and Mn originated in the food sample. Then convert these concentrations to their corresponding ppm concentrations in the original dry food sample.
I have calculated the Mn and Fe in concentrations in Fe in final solutions, but I am having trouble converting them into ppm concentrations in original dry food samples.
Solution:
From the Calibration curve of ppm of Mn vs absorbance: y = 0.0406x + 0.0179 R2 = 0.9983
Let y = 0 Then: x = -b/m = - 0.0179/0.0406 = 0.441 ppm = [Mn] in the final solution originated in the food sample.
While from the calibration curve of ppm of Fe vs absobance y = 0.0092x + 0.0137 R2 = 0.9976
Let y = 0 Then: x = -b/m = - 0.0137/0.0092 = 1.49 ppm = [Fe] in final solution originated in the food sample.
The dog food sample initial mass was 1.5015g.
Thank you
Explanation / Answer
You have not mentioned your experimental details.
But here is what you should do to calculate the concentation in original sample
The final solution concentration x volume = moles of Fe
you do the same for Mn = moles of Mn
Now the volume of intial solution in which the sample was dissolved must be given.
divide the moles of Mn and Fe you have with the volume of original solution
This will give you concentration of Fe and Mn in original sample.
Make sure you account for any unit change at all possible steps of calculation.