Question

A chemist performed a calibration method analysis to determine the concentration of vitamin c in a fruit sample. The standard vitamin c samples were prepared by mixing a complexing agent with the standard solution. The absorbance of the resulting solution was obtained. Once the absorbance of each standard solution was measured, he prepared the following calibration curve.y = 1.6421x -0.1113The unknown sample was prepared by taking 2.0 g of crushed fruit and extracting the concentrated juice. The extracted juice was added to 100.0 mL volumetric flask and the final volume was adjusted 100.0 mL with DI water. 10.0 mL of diluted fruit extra was pipetted into 50.0 mL volumetric flask and after adding 5.0 mL of complexing agent, the DI water was added until the final volume of the solution reaches 50.0 mL. This solution gave an absorbance of 0.461. I. Find the concentration of vitamin C in the original fruit sample. II. What is the total mass of vitamin C in the sample? III. Calculate the wt% of vitamin C in the fruit sample.

Answer 1

Answer:

I.  1.743mg/mL

II. 0.174g

III. 8.7 wt%

Explanation:

As absorbance is directly proportional to concentration. The calibration curve was:

Y = 1.6421X - 0.1113

Where Y is absorbance of sample and X its concentration in mg/mL

The diluted sample obtains an absorbance of 0.461, that means its concentration is:

0.461 = 1.6421X - 0.1113

0.3485mg/mL = X

I. The sample was diluted to 50.0mL from an aliquot of 10.0mL, thus, concentration in the original fruit sample was:

0.3485mg/mL × (50.0mL / 10.0mL) = 1.743mg/mL

II. The total mass of vitamin C in the sample (100.0mL) is:

100.0mL × (1.743mg/mL) = 174mg = 0.174g

III. As the fruit sample was of 2.0g, the wt% is:

0.174g / 2.0g = 8.7 wt%