Question

Determining aluminum content in candy: To determine the aluminum content in sour apple candy, a 5.12 g piece is mineralized by heating, dissolved in 5.00 mL concentrated sulfuric acid and then diluted to the mark in a 100.00-mL volumetric flask. To 25.00 mL volumetric flasks are added a 5.00 mL aliquot of the dissolved candy sample. To each flask, 250.0 ppm aluminum standard was added in the volumes listed below. After dilution to volume, the solution is analyzed for aluminum by graphite furnace atomic absorption spectroscopy at 309 nm. The results are shown below. Report the aluminum content of the candy. Assume all solutions prepared assume the density of water Vol of added standard (mL) Signal (Abs) 0 0.100 0.200 0.300 0.214 0.386 0.554 0.728

Answer 1

Ans. Preparation of final aliquot (whose Abs is recorded)- Aliquot 2

            Volume of Al standard taken, V1 = 0.100 mL

            Concertation of standard Al soln. C1 = 250 ppm

            Final volume made upto = 25.0 mL

Now, Using C1V1 = C2V2              - equation 1

C1= Concentration, and V1= volume of initial solution 1         ; Standard Al soln.

C2= Concentration, and V2 = Volume of final solution 2         ; Final aliquot

Final [Al] in aliquot, C2 = (C1V1) /V2 = (250.0 ppm x 0.100 mL) / 25.0 mL = 1.000 ppm

Calculations for other aliquots are done in excel.

# Plot the graph using [Al] of aliquots on X-axis and their respective Abs on y-axis.

The trendline equation for graph is y = 0.171x + 0.214

            Where, c = y-intercept         , m = slope

# [Al in un-spiked aliquot]

Un-spiked aliquot (aliquot 1) contains 5.0 mL of original digested & diluted sample but no standard Al solution.

From the linear graph equation-

[Al] in un-spiked original Candy solution = (y-intercept / slope) concentration units

= (0.214 / 0.171) ppm

= 1.2515 ppm

# [Al] in original diluted digested sample:

For simplicity, lets label the 100.0 mL solution, that made prepared after diluting the 5.0 mL of mineral digest of the candy, as original diluted sample.

25.0 mL (V1) with 1.2525 ppm Al (C1) of us-spiked aliquot is prepared from 5.0 mL (V2) of original diluted sample.

Using equation 1,

Concentration of Al in original diluted sample, C2 = (1.2515 ppm x 25.0 mL) / 5.0 mL

                                                            = 6.2575 ppm

# Total Al in original diluted sample

Total Al in original 100.0 mL diluted sample =

[Al] in original diluted sample x Volume of original digested sample in mL

= (6.2575 ppm) x 100.0 mL                                     ; [1 ppm = 1 mg/ 1000 mL]

= (6.2575 mg/ 1000 mL) x 100.0 mL

= 0.62575 mg

# The 100.0 mL original diluted sample is prepared by digesting 5.12 g candy in H₂SO₄. So, total Al content of the original diluted sample is equal to the total Al content of the candy.

So,

Al content in Candy = Mass of Al in mg / Mass of candy in gram

                                                = 0.62575 mg Al / 5.12 g candy

                                                = 0.12 mg Al / g candy