# Amount of Blue Dye in Gatorade

Purpose:

The purpose of this lab is to find the amount of blue dye #1 in the blueberry pomegranate gatorade.

Procedure:

We created dilutions using stock solution(10mL stock/0 mL water, 8mL/2mL, 6mL/4mL,4mL/6mL, 3mL/7mL, 2mL/8mL,1mL/9mL, 0mL/0mL). Before we put each dilution in the SPEC 20, we put water in a cuvette and put it in the SPEC 20 so we could zero out the percent transmittance. We collected the percent transmittance of each dilution using the SPEC20 with a wavelength of 630nm. We calculated the molar concentration of each dilution using the formula M1V1=M2V2 We determined the linear relationship between transmittance and molarity by taking the inverse log(-logT) of percent transmittance(in decimal form). We determined the relationship between transmittance and molarity by graphing the data table.

Data:

Solution

Dilution Ratio mL stock/mL water

Molar concentration (M)

Measure percent Transmittance

Measure %T in Decimal Form

1. (stock solution)

10mL/0mL

6.30E-06

0.10%

0.001

2.

8mL/2mL

5.10E-06

0.20%

0.002

3.

6mL/4mL

3.80E-06

0.40%

0.004

4.

4mL/6mL

2.50E-06

6.60%

0.066

5.

3mL/7mL

1.90E-06

19%

0.19

6.

2mL/8mL

1.30E-06

49.30%

0.493

7.

1mL/9ml

6.00E-07

64%

0.64

8.

0mL/10mL

0

100%

1

Graph:

Calculations:

To find molar concentration of each distilled stock solution: M1V1=M2V2

M1=6.33E-6M

For solution #1; (6.33E-6)(10)=M2(10) =6.3E-6M

For solution #2; (6.33E-6)(8)=M2(10) =5.1E-6M

For solution #3; (6.33E-6)(6)=M2(10) =3.8E-6M

For solution #4; (6.33e-6)(4)=M2(10) =2.5E-6M

For solution #5; (6.33E-6)(3)=M2(10) =1.9E-6M

For solution #6; (6.33E-6)(2)=M2(10) =1.3E-6M

For solution #7; (6.33E-6)(1)=M2(10) =.6E-6M

For solution #8; (6.33E-6)(0)=M2(10) =0M

Using the equation of the line and the absorbance of the gatorade find the molar concentration of the gatorade y=5.372E5x-0.132

1.636=5.372E5x-0.132

1.636+.132=5.372E5x

1.767/5.372E5=x

=concentration of...

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