Question

 

What Is the Concentration of the dye in a sports drink?

■■Context for This Investigation
Measuring how much of which wavelengths of light are absorbed by a substance, and getting useful information about that substance from the results, is the scientific discipline of spectroscopy. The visible spectrum is the only part of the electromagnetic spectrum that we can access with the equipment found in a typical high school chemistry laboratory. The basic principles of spectral analysis that you will learn in this lab can also be applied to the more sophisticated instrumentation required to access the ultraviolet, infrared, and X-ray regions. What you learn by performing the lab will help you to understand more sophisticated instruments that you may encounter at a college or university.

In a visible spectrophotometer, you shine a beam of light into a solution containing the sample, and detect how much of it comes out of the other side of the solution. By comparing the amount of light transmitted by the pure solvent to the amount transmitted when the sample is dissolved in it, we can calculate a quantity called the absorbance. Spectrophotometers can report measurements as percent transmittance (%T) or directly as absorbance. In this investigation, you will be guided to discover the relationship between transmittance and concentration and ultimately the relationship between transmittance, absorbance, and concentration of a solution.

■■Skills
Prior Skills
Students should be able to: ■■ Measure with a metric ruler; ■■ Construct a line graph; and

■■ Perform dilutions of solutions prior to engaging in this lab.

Developing Skills of Science Practices, Instrumentation, and Procedure

The student will be able to read (SPEC 200) a UV-VIS spectrum or create a UV-VIS spectrum (SPEC 20).
The student will be able to apply the information of the graph to determine the molar concentration of a sports drink and then determine the mass of the dye in the container.

The student will be able to apply mathematical routines to quantities that describe natural phenomena. The student will be able to analyze data to identify patterns or relationships.
The student will be able to discover which plot of transmittance vs. molar concentration will produce a linear relationship.

The student will be able to design a procedure/data-collection strategy to determine the concentration of dye in a sports drink.
The student will be able to design a plan for collecting data to answer a particular scientific question. The student will be able to justify the selection of the kind of data needed to answer a particular scientific question.

The student will be able to make claims and predictions about natural phenomena based on scientific theories and models.

Materials

Spectrophotometer SPEC 20 or SPEC 200 Stock solution of the blue liquid food dye with known concentration.
Distilled water
Test tube racks

Safety and Disposal

Cuvettes Disposable pipettes
Blue-colored sports drink (with ONLY blue #1 dye)

10-Test tubes (20 mL)
Graduated cylinders (10 mL) or pipets

All of the food dyes can be flushed down the sink with plenty of water.

Data Table 1. %Tranmittance of diluted solutions of known concentration

Solution	Dilution Ratio mL stock/mL Water	Molar Concentration (μM)	Measured %Transmittance	Measured Transmittance as a Decimal
1. (stock solution)	10 mL/0 mL	..................................	52.2	0.522
2.	9 mL/1 mL	..................................	54.1	0.541
3.	8 mL/2 mL	..................................​	75.4	0.754
4.	7 mL/3 mL	..................................	79.7	0.797
5.	6 mL/4 mL	..................................	81.1	0.811

6.

5 mL/5 mL

..................................

85.0 0.

850

7. 4 ml/ 6ml .................................. 91.1 0.911

8. 3ml/ 7ml .................................. 93.9 0.939

9. 2ml/ 8ml     .................................. 93.8 0.938

10. 1ml/ 9ml     .................................. 98.8 0.988

*I don't know how to calculate molar concentration..

Before doing your graphs – add a data point that is 0.00 M concentration, 100 % Transmittance Plot your data %T vs. [dye]

It would be really useful if your group data was a straight line that goes through zero. Scientists try to find linear relationships because such relationships make it easier to identify unknowns and predict outcomes of investigations. It is often helpful to have the slope of the relationship between the dependent and independent variable be positive. “What mathematical routine could we choose to plot to get a linear relationship between transmittance and molarity that has a positive slope, and which line goes through zero?”

Here are some options to try graphing:
1/T vs. [dye]; 1 x 10T vs. [dye]; logT vs. [dye]; –logT vs. [dye]

The relationship that gives a linear relationship that goes through the zero is called absorbance vs. conc.

You can use this calibration line to obtain the concentration of the same absorbing species in a solution of unknown concentration. If you hold the instrument the same way, and the wavelength and cuvette the same, they should be able to measure transmittance, manipulate data as needed, and, along with the slope of the line, have the information needed to determine the concentration of the absorbing species in solution.

■■Investigation
Many common sports drinks contain blue #1 dye. Students will use the relationship between transmittance, absorbance, and concentration (as well as their calibration line from the prelab) to determine the concentration of this dye in the sports drink.

Procedure

Give the students a sample of the blue-colored sports drink. Students will use the skills and information gained in the prelab to design a data-collection and data analysis procedure to determine the molarity or concentration of blue #1 dye in the sports drink.

Data Collection and Computation

1. Determine the molar concentration of blue #1 dye in the sports drink. Show all work.

2. Determine the mass of blue #1 dye found in 500 mL of the drink. Molecular Formula for Blue Dye #1 = C38N2S3O9Na3. Show all work.

■■Postlab Assessment

Suppose a solution was too concentrated for an accurate reading with the spectrophotometer. The

concentrated solution was diluted by placing 1.00 mL of the concentrated solution in 4.00 mL of water. The solution was then placed in the spectrophotometer and an absorbance was obtained and

after a few calculations the molar concentration was calculated to be 3.5 x 10–6 M. What was the concentration of the original solution before creating the stock solution by dilution?

The spectrophotometer really measures the percent of light that is transmitted through the solution. The instrument then converts %T (transmittance) into absorbance by using the equation you determined in the prelab section. If the absorbance of a sample is 0.85, what is the percent of light transmitted through the colored sample at this collected wavelength?

DATA 2..

UNKNOWN @ 630nm %T

Powered 38.8%

laid tropical punch 81.37%

roolaid grape 39.1%

Answer 1

And. Step 1: Calculate [Dye] in standard aliquots

The information provide is insufficient, for example the concentration of stock dye solution is not clearly mentioned. However, it shall not affect the principles or methods of calculation.

Given, the stock [Dye] solution be 1.45 x 10^-5 M = 14.5 uM.

Now, using C1V1 = C2V2   - equation 1

C1= Concentration, and V1= volume of initial solution 1         ; Stock dye

C2= Concentration, and V2 = Volume of final solution 2         ; Diluted dye

For example, we calculate for tube number 2 in table 1. Putting the values in equation 1-

            14.5 uM x 9.0 mL = C2 x 10.0 mL                                       ; V2 = 10.0 mL

            Or, C2 = (14.5 uM x 9.0 mL) / 10.0 mL = 13.05 uM

The calculation is made for rest of the tubes in excel.

Step 2: Plot Abs vs [Dye]

Absorbance = 2 – log (%T)

For example, for tube 2 in table 1-

            Abs = 2 – log 54.1 = 2 – 1.7332 = 0.267

Rest of Abs have been calculated in excel.

Plot Abs vs corresponding [Dye] in excel. Generate the linear trendline equation. It is y = 0.0205x- 0.0612.

Note: Note the large difference between % T of Tube 1, 2 and rest of the others. You should have got a gradual increase in % T while decrease in concentration. It seems that there has been mistake while recording % T of sample. You should get a gradual increase in % T as smooth as gradual decrease in concentration. If so, you would get a better graph, more (nearly) linear with R² closer to 1.00. Also note that closer os R² value to 1.00, more perfect is your data.

Step 3: Calculating [Dye] from unknown’s Abs.

I would do one sample calculation for illustration purpose only. You can do all the calculations again once you begin with original stock [Dye] given to you and ‘correcting’ % T values if any.

Equation of the standard graph is “y = 0.0205x – 0.0612 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 = 0.0205x – 0.0612 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 0.0205 units on X-axis (concentration) minus 0.0612.

Unknown sample 1 : Powdered. Abs = 2 – log 38.8 = 0.411

Putting y = 0.411 in above trendline equation-

            0.411 = 0.0205x – 0.0612

            Or, 0.411 – 0.0612 = 0.0205x

            Or, x = 0.3498 / 0.0205 = 17.06

            Hence, x = 17.06

Note that unit of concentration on X-axis is “uM”. Therefore, [dye] in powdered sample = 17.06 uM.

Unknown sample 2 : Laid Tropical pouch

            Abs = 2 – log T = 2 – log 81.3 = 0.090

Putting y = 0.090 in above trendline equation-

            Or, x = (0.090 – 0.0612) / 0.0205 = 1.40

            Hence, x = 1.40

Hence, [Dye] = 1.40 uM

Unknown sample 3 : Roolaid grape

            Abs = 2 – log T = 2 – log 39.1 = 0.408

Putting y = 0.0408 in above trendline equation-

            Or, x = (0.408 – 0.0612) / 0.0205 = 16.92

            Hence, x = 16.92

Hence, [Dye] = 16.92 uM

Part 2: Accounting dilution [Post lab Assessment]

From the information, we have

            C1 = ?             ; V1 = 1.00 mL

            C2 = 3.5 x 10^-6 M = 3.5 uM             ; V2 = 5.0 mL

Now, concertation of undiluted sample, C1 = (C2 V2) / V1

                        Or, C2 = (3.5 uM x 5.0 mL) / 1.0 mL = 17.5 uM

Hence, concertation of undiluted sample = 17.5 uM = 1.75 x 10^-5 M