Question

Given the following data for the fluorescence of Vitamin B2 (Riboflavin) found in energy drinks, plot a calibration curve in Microsoft Excel. Show the equation for the best fit line and the R² value.

A 25.00 mL sample of a new energy drink was diluted to 100.0 mL of DI water and analyzed at 530 nm. The fluorescence intensity measured 0.3013. Calculate the mass of Vitamin B2 in a sixteen ounce serving of the new energy drink. (MM Riboflavin = 376.4 g/mol). (1 ounce = 29.57 mL)

Concentration (M)                  Fluorescence at 530 nm

0                                              0.007

2.00 x 10^-4                               0.117

4.00 x 10^-4                               0.209

6.00 x 10^-4                               0.326

8.00 x 10^-4                               0.448

1.00 x 10^-3                               0.537

Answer 1

Ans.

The trendline equation for the graph “y = 537.14x + 0.0054” is in form of y = mx + c. where, c = y-intercept , m = slope.

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation y = 537.14x + 0.0054 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 537.14 units on X-axis (concentration) plus 0.0054

Given, absorbance of unknown, y = 0.3013

Putting y = 0.3013 in trendline equation-

            0.3013 = 537.14x + 0.0054

            Or, 537.14x = 0.3013 – 0.0054

            Or, x = 0.2959 / 537.14 = 5.5088 x 10^-4

Hence, [Riboflavin] in the diluted unknown sample = 5.5088 x 10^-4 M

# Preparation of diluted unknown sample:

25.0 mL original sample is diluted upto 100.0 mL with distilled water.

We have, [Riboflavin] in 100.0 mL diluted sample = 5.5088 x 10^-4 M

Now, using    (C1V1) original drink = (C2V2) diluted sample

            Or, (C1 x 25.00 mL) = (5.5088 x 10^-4 M x 100.0 mL)

            Or, C1 = (5.5088 x 10^-4 M x 100.0 mL) / 25.0 mL = 2.2035 x 10^-3 M

Hence, [Riboflavin] in the original drink/sample = 2.2035 x 10^-3 M

# Total volume of drink = 16 oz = 16 x (29.57 mL) = 473.12 mL

Moles of riboflavin in 1 serving original drink = [Riboflavin] x Volume of serving in liters

                                                = (2.2035 x 10^-3 M) x 0.47312L                  ; [1 mL = 0.001 L]

                                                = 1.0425 x 10^-3 mol

Now,

Mass of riboflavin in 1 serving original drink = Moles x Molar mass

                                                = 1.0425 x 10^-3 mol x (376.4 g/ mol)

                                                = 0.392 g