You are analyzing riboflavin in wheat flour by fluorescence spectroscopy (?ex =

Question

You are analyzing riboflavin in wheat flour by fluorescence spectroscopy (?ex = 450 nm, ?em =530 nm). The extraction procedure is fairly complex. You obtain a 100g sample of the flour. You then weigh out 4.542 g of the flour, add it to a 250 mL volumetric flask, and dilute to volume with 1 N boiling HCl. After thoroughly mixing this solution, you combine 2 mL with 7 mL 1 N HCl. You then freeze-dry 4 mL of the diluted solution, and redissolve the residue to a total volume of 1.5 mL. You combine 0.5 mL of this solution with 0.5 mL ddH2O and measure the fluorescence. The following fluorescence values are obtained for the sample and standards. Prepare a blanked standard curve, and determine the riboflavin concentration in the flour (ppm) (10 points)

Fluorescence (?ex=450 nm, ?em=530 nm).

Sample

1.706

Riboflavin (20 mg/L)

0.822

Riboflavin (10 mg/L)

0.368

Riboflavin (5 mg/L)

0.121

Riboflavin (2 mg/L)

0.004

Riboflavin (1 mg/L)

1.706

Riboflavin (0 mg/L)

0.476

Flour

Fluorescence (?ex=450 nm, ?em=530 nm).

Sample

1.706

Riboflavin (20 mg/L)

0.822

Riboflavin (10 mg/L)

0.368

Riboflavin (5 mg/L)

0.121

Riboflavin (2 mg/L)

0.004

Riboflavin (1 mg/L)

1.706

Riboflavin (0 mg/L)

0.476

Flour

Explanation / Answer

On the basis of given data, plot x as conc of riboflavin and y as sample to obatin a curve y = 0.088x - 0.0575

so, Riboflavin(mg/L) in flour =0.476+0.0575/0.088=6.0625mg/L

The amount of riboflavin in the flour sample will be: 6.0625mgL

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