Question

 

Experiment 3: Quantitatively Determine the Food Dyes in an Orange Drink Mix

Prepare an standard 8 ounce serving of orange drink as follows: Placed a 250 mL beaker on the workbench & added 235mL water to it.

Added 5.0g of orange drink powder to the beaker.

Placed a cuvette on the workbench & added 1mL of the orange drink from the beaker to it; then added 9mL of water to it.

Set the spectrophotometer to Vis and the wavelength to the lambda max value for Yellow 5 dye. Placed in spectrophotometer, readings were:

Lambda max - Wavelength = 440     absorbance = 0.37

Set the spectrophotometer to Vis and the wavelength to the lambda max value for Red 40 dye. Placed in spectrophotometer, readings were:

Lambda max - Wavelength = 500     absorbance = 0.25

Given that the molar mass of the Yellow 5 dye is equal to 534.36 g/mol, how many grams of Yellow 5 food dye were in the 0.235 L solution of drink mix? Remember that you diluted your original solution tenfold before measuring its absorbance. Choose the closest answer.

A.0.018 g

B.0.0050 g

C.0.0075 g

D.0.0018 g

Given that the molar mass of the Red 40 dye is equal to 496.42 g/mol, how many grams of Red 40 food dye were in the 0.235 L solution of drink mix? Remember that you diluted your original solution tenfold before measuring its absorbance. Choose the closest answer.

A.0.085 g

B.0.0045 g

C.0.0011 g

D.0.011 g

What is the mass percent of the Yellow 5 food dye in the orange powder? % mass = (mass of dye / mass of powder) x 100

A.0.36%

B.0.036%

C.12%

D.2.2%

What is the mass percent of the Red 40 food dye in the orange powder? % mass = (mass of dye / mass of powder) x 100  

A.1.8%

B.0.061%

C.0.022%

D.0.22%

Answer 1

Ans. The critical data related to standard calibration curve is NOT mentioned in question.

You would require to prepare standard calibration curve and generate its linear regression expression separately for both the dyes.

The solution is explained using a hypothetical calibration curve data for red 40 dye.

#A. Calculating mass of red 40 dye in 0.235 L solution.

Step 1: Calculating mass of red 40 dye in 10.0 mL aliquot

Let’s assume, the standard calibration curve yielded the linear regression equation as         “y = 59.8520x + 0.0997”

Equation of the graph is “y = 59.8520x + 0.0 in form of “y = mx + b”

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation y = 59.8520x + 0.0997 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 59.8520 units on X-axis (concentration) plus 0.0997.

Putting y = 0.250 for absorbance of aliquot in trendline equation-

            0.250 = 59.852x + 0.0997

            Or, 59.8520x = 0.250 – 0.0997 = 0.1503

            Or, x = 0.1503 / 59.8520 = 0.0025112

Therefore, [Red 40 dye] in the 10.0 mL aliquot = 0.0025112 M

Now,

Moles of red 40 dye in 10.0 mL (= 0.010 L) aliquot = Molarity x Volume of solution in liters

                                                            = 0.0025112M x 0.010 L

                                                            = (0.0025112 mol/ L) x 0.010 L       ; [1 M = 1 mol/L]

                                                            = 0.0000251 mol

Mass of red 40 dye in 10.0 mL aliquot = moles x Molar mass

                                                            = 0.0000251 mol x (534.36 g/mol)

                                                            = 0.01342 g

Therefore, 10.0 mL of final aliquot (whose OD was recorded) consists of 0.01342 g red 40 dye.

Step 2: Calculating mass of red 40 dye in 0.235 L original solution

10.0 mL of the aliquot (whose OD was recorded) consists 0.01342 g red 40 dye.

10.0 mL of the above aliquot was prepared from 1.0 mL original solution (0.235 L).

So, 1.0 mL of the original solution also consists of 0.01342 g red 40 dye.

Thus, [red 40 dye] in original solution = 0.01342 g/ 1.0 mL

Now,

Mass of red 40 dye in original solution = [red 40 dye] in original solution x Volume in mL

(0.01342 g / 1.0 mL) x 235.0 mL

                                                                        = 0.335 g

Therefore, total red 40 dye content in 0.235 L (= 235.0 mL) original solution = 0.335 g

Note: The calculation is done using a hypothetical calibration curve. Please use the experimental calibration curve data to get exact value.