# Amount of Blue Dye in Gatorade

Topics: Concentration, Dilution Pages: 4 (796 words) Published: October 26, 2014

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...

Please join StudyMode to read the full document