Question

A student used a DCP solution (standardized to 9.93x10^-4 M) to analyze a sample of extract from 3.25 g of solid food. The titration required 21.49 mL of DCP. 84.50% of all the ascorbic acid in the food was collected in the extract. (MM Ascorbic Acid = 176.124 g/mol)

What mass of food (in grams) would be required to attain the RDA of ascorbic acid (60 mg)

Answer 1

1 mole of DCP oxidizes 1 mole of ascorbic acid

moles of DCP used = 9.93 x 10^-4 M x 21.49 ml = 0.02134 mmol

moles of ascorbic acid present = 0.02134 mmol

mass of ascorbic acid in food = 0.02134 mol x 10^-3 mol x 176.124 g/mol = 0.004 g

Only 84.50% ascorbic acid was collected

So, actual ascorbic acid present in 3.25 g food = 0.004 x 0.8450 = 3.4 mg

For 60 mg ascorbic acid be 84.50%, then 100% = 71.00 mg

To get 71 mg of ascorbic acid we would need = 71 mg x 3.25 g/4 mg = 57.69 g

So we would need 57.69 g (or 57.7 g) of solid food to get 60 mg of ascorbic acid