Im doing an experiment that is using Flames Atomic Absorptio

I\'m doing an experiment that is using Flames Atomic Absorption Spectrophotometry to find the Calcium and Magnesium Contents in Limestone, but I\'m unsure on how to find the Calibration sensitivity and then how to go about calculating the concentration of the element.

Table 1. Analysis of Ca and Mg Contents in Limestone by direct calibration Limestone sample code Mass of limestone (g) 3-941 60 Vol. of stock (mL) | 250.0 Mass conc. (g/mL) 4414.8 Calcium 398.4 100.0 17.58989735 10.04 2 0.398 1.1037 Limestone test solution Aliquot size of stock (NL) Vol. of test solution (mL Mass conc. (g/mL) Calibrant (ug/mL of Metal Magnesium 40.1 50.0 3.5404485 0.395 2 0.628 3 0.413 3 0.617 0.430 Calibrant absorbance Calib. sensitivity (au/ppm) Test sample absorbance Conc. of element (ug/mL Result and Statistics Conc. of element (ug/mL) 0.620 0.164 0.157 0.168 0.886 0.881 0.887 mean rsd mean rsd Conc. of oxide (ug/mL) % metal oxide in limestone

Solution

The calibration sensitivity is nothing but the sensitivity of the absorbance to the small change in analyte concentration and can be obtained by calibration curve where absorbance measured for stock solutions of known concentration can be plotted versus the concentration and slope of the curve is a calibration sensitivity.

In this case, calibrant absorbance is for three known concentration so from that you will get the calibration sensitivity.

From calibration sensitivity or calibration curve, you will get the concentration corresponding compound ( such as CaCO3 or MgO) in the test sample by using the corresponding absorbance.

From that, you need to calculate element concentration, the molecular weight of the corresponding compound and the atomic weight of each element you will get the percent weight of each element in one mol from that you calculate the concentration of the element.

for example, in CaCO3 the Molecular weight is 100.08 g

Ca atomic weight is 40.078 g

so the percent weight = (40.078 / 100.08) *100 = 40.05 %

that means out of the the total weight of CaCO3 40.05 percent is Calcium