Lab Repeort 12 - Lab Report Experiment 12.Microfluidic Paper Chromatography PDF

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Lab Report Experiment 12.Microfluidic Paper Chromatography...

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Experiment 12.Microfluidic Paper Chromatography

Purpose: The purpose of this experiment was to determine which solvent (H2O, 0.10% NaCl, 2.0% 2propanol) will be the best to separate our five dyes by using a microfluidic paper chromatography to identify the compounds in our three unknowns. Theory/Principles: By using a microfluidic test we can determine which solvent is best fitted to separate our five dyes (Blue 1, Blue 2, Red 3, Yellow 5, Green 3) and to identify the compound in our three unknowns (E,N,C). We are told in the lab experiment worksheet that there are other microfluidic devices used before in a glass and plastic form. However, for our case we used a chromatographic paper that is printed with the designed diagrams such as in Figure 1. in our lab experiment worksheet. The microfluidic paper chromatography has been printed with the diagrams using a wax printer which is then heated for the wax to diffuse through the thickness of the paper that forms wax barriers. In our lab experiment worksheet it also states that wax is hydrophobic and will restrict aqueous solvents within the channels. We also know from our lab instructor that since the paper is polar and depending on how the dye reacts to both paper and solvent then it disperses better. However, depending on the dyes interaction with paper it can also stay with it instead of dispersing further or any better compared to any other dyes. During the experiment our dyes will either go through a stationary or mobile phase on the paper depending on their interactions with the paper and solvents. The stationary phase is when the dyes don't disperse and stay put which can happen because of their interaction with the paper. While in the mobile phase the dyes are dispersing within the channels either by the help of our solvents H2O, 0.10% NaCl, or 20% 2- propanol. Once we figure out which solvent was the best to disperse the five dyes we would then use that solvent to determine the food dyes in our three unknowns.

From here we can then calculate the retention factors (Rf) values from our knowns and unknowns given in the equation of R= d(a)/d(solvent). We determine d(a) by obtaining the distance moved by the solute and we obtain d(solvent) by the distance moved by the solvent. Experimental Procedures: Didn’t use lab manuals procedure for this experiment instead we followed the lab experiment handout given to us by our professor. Experiment 12. Microfluidic Paper Chromatography (2-3) Data Tables/Summary: -

Table 1: Pictures of the experiment right after procedure so we can see/measure the distance travelled by the solvents (H2O, 0.10% NaCl, 20% 2-propanol) and dyes (each channel labeled with designated color). To figure out unknown we used the solvent 0.10% NaCl as it was the best to disperse the dyes.

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Table 2: Same experiment from above but a clearer image but this is after the solvents and dyes dried down and was signed by our instructor.

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Table 3: Measurements in cm and Rf values for each solvents chromatograms

20 % 2-Proponal (solvent distance= 4.5 cm) Dyes

Distance (cm)

Rf Values

G3

4.4 cm

0.97

Y5

3.7 cm

0.82

R3

3.0 cm

0.66

B1

4.0 cm

0.88

B2

3.9 cm

0.86

R3 + G3

R3= 3.3 cm, G3=3.5 cm

R3= 0.73, G3= 0.77

R3 + Y5

R3= 3.8 cm, Y5= 3.9 cm

R3= 0.84, Y5= 0.86

R3 + B1

R3= 3.9 cm, B1= 4.1 cm

R3= 0.86, B1= 0.91

H2O (solvent distance= 3.5 cm) Dyes

Distance (cm)

Rf Values

G3

3.4 cm

0.97

Y5

3.5 cm

1

R3

3.3 cm

0.94

B1

4.1 cm

1.17

B2

3.7 cm

1.05

R3 + G3

R3= 3.2 cm, G3= 3.4 cm

R3= 0.91, G3= 0.97

R3 + Y5

R3= 3.1 cm, Y5= 3.3 cm

R3= 0.88, Y5= 0.94

R3 + B1

R3= 3.6 cm, B1= 3.8 cm

R3= 1.02, B1= 1.08

0.10 % NaCl (solvents distance= 4.0 cm)

Dyes

Distance (cm)

Rf Values

G3

3.9 cm

0.97

Y5

3.5 cm

0.87

R3

1.1 cm

0.27

B1

4.0 cm

1

B2

3.4 cm

0.85

R3 + G3

R3= 1.8 cm, G3= 3.6 cm

R3= 0.45, G3= 0.90

R3 + Y5

R3= 1.7 cm, Y5= 3.5 cm

R3= 0.42, Y5= 0.87

R3 + B1

R3= 1.5 cm. B1= 5.1 cm

R3= 0.37, B1= 1.27

Unknown (solvents distance= 4.7 cm) Dyes

Distance (cm)

Rf Values

G3

4.1 cm

0.87

Y5

3.7 cm

0.78

R3

1.1 cm

0.23

B1

4.5 cm

0.95

B2

3.4 cm

0.72

C

B2= 3.2 cm, G3= 4.4 cm

B2= 0.68, G3= 0.93

N

R3= 1.8 cm, B2= 2.9 cm

R3= 0.38, B2= 0.61

E

R3= 1.9 cm, G3= 4.3 cm

R3= 0.40, G3= 0.91

Results and Discussion: Sample Calculations:

- Rf (a) = d(a)/d(solvent) 20 % 2-Proponal (solvent distance= 4.5 cm) Rf(G3)= 4.4 cm/ 4.5 cm= 0.97 Rf(Y5)= 3.7 cm/ 4.5 cm= 0.82 Rf(R3)= 3.0cm/4.5 cm= 0.66 Rf(B1)= 4.0cm/4.5 cm=0.88 Rf(B2)= 3.9cm/4.5 cm= 0.86 R3+ G3: Rf(R3)= 3.3cm/4.5 cm= 0.73 Rf(G3)=3.5 cm /4.5 cm= 0.77 R3+Y5: Rf(R3)= 3.8cm/4.5 cm= 0.84 Rf(Y5)= 3.9cm/4.5 cm= 0.86 R3 + B1: Rf(R3)= 3.9cm/4.5 cm= 0.86 Rf(B1)= 4.1cm/4.5 cm= 0.91 -

For the rest of the devices (Unknown, H2O, 0.10% NaCl) I used the same equation and same process shown above to solve for the rest of the Rf values

Logical Explanation:

By using a microfluidic paper chromatography test we were able to determine which solvent is best fitted to separate our five dyes and figure out the dyes involved in our unknowns (E, N, C). By looking at each of our devices we were able to notice that 0.10 % NaCl dispersed/separated each of the dyes better through the channel and it even dispersed R3+G3 to the end point. When looking at our devices for H2O and 20 % 2- propanol we can see that the solvents are not separating the mix dyes but instead moving the mix dyes at a close distance which will make it difficult if we used either one of the solvents to help us identify our unknowns. Also NaCl has a better interaction with the dyes because they have more ionic groups and has more of an ion interaction. Given that 0.10% NaCl was the better solvent for disperse/separation we ended up using it to help us identify our unknowns. Looking at the unknown device we used 0.10% NaCl for, we can clearly see that our unknown E is a mixture of R3 + G3, for unknown N we see a mixture of R3 + B2, and for unknown C we see a mixture of B2 + G3. We were able to identify the color dyes for our unknowns by comparing their separation and colors to the other devices we used as well. Another way I identify the unknowns was comparing their relative retention factors were I was able to see some close similar values but some differences as well because each solvent either didn’t dispersed/separated some of the dyes well enough. However, the best way for me to identify the unknowns was by direct comparison of each device and the dyes interactions with each solvent. Discussion Ques. 1. a) Yes, the time required for running different spot sizes increases with the size a spot. When spot is bigger than it will take more time for it to disperse compared to the smaller spots. Through the equation of speed= d/t = √3RT/M we can also see that if the molar mass is bigger then the speed is smaller and time will be bigger.

b) Yes, the size of the end spot does depend upon the mobile phase because as polarity goes up the size of the spot becomes smaller as it moves faster while with low polarity the spots become larger. It all depends on the interaction between the dye with the solvent and paper. c) The end spot size correlates with solvent polarity as mentioned in answer part b. For lower polarity the spot size is larger and for higher polarity the spot size is smaller. d) As the end spot size increases then the more time it would require for the run (similar to a). 2. a) The best solvent system that would be better for washing B1 out from a cotton shirt would be 2-propanol because it is an organic solvent along with polar hydroxyl group. b) 2- propanol would be the best for G3. 3. Known uncertainties of Rf values for device Unknown: G3= 0.05 cm/ 4.1 cm x 100%= 1.21 % Y5= 0.05 cm/ 3.7 cm x 100%= 1.35 % R3= 0.05 cm/ 1.1 cm x 100%= 4.54 % B1= 0.05 cm/ 4.5 cm x 100% = 1.11 % B2= 0.05 cm/ 3.4 cm x 100%= 1.47 % For unknown: C: B2= 0.05 cm/ 3.2 cm x 100%=1.56 %

G3=0.05 cm/4.4 cm x 100%=1.13 %

N: R3= 0.05 cm/1.8 cm x 100%= 2.77 %

B2=0.05 cm/2.9cm x 100%=1.72 %

E: R3= 0.05 cm/1.9 cm x 100%= 2.63 %

G3= 0.05 cm/ 4.3 cm x 100%= 1.16 %

4. a) H2O: ion-dipole , dipole- dipole, hydrogen bonding 0.10 % NaCl: ion-ion, dipole- dipole, ion- dipole, hydrogen bonding 20 % 2-proponal: dipole-dipole, ion-dipole, hydrogen bonding b) i) Large Rf value for B1 in 0.10% NaCl device is in mobile phase ii) Small Rf value for R3 in Unknown device (0.10% NaCl solvent used) is in stationary phase The relative importance of the types of forces that result from the observed Rf values show us that the smaller the Rf value is then it will most likely be in a stationary phase and the dyes interaction with the solvent was weaker while the dye interaction was stronger with the paper which is why it didn’t disperse. We can also notice that the bigger the Rf value than the more likely the dye is in a mobile phase. Conclusion: By understanding and knowing how to use a microfluidic paper chromatography test we were able to determine that 0.10 % NaCl is the best solvent to disperse/separate each of the food dyes through the channels and was able to help us identify our unknowns to be the mix dyes R3 + G3 (E), R3 + B2 (N), and B2 + G3 (C). References: Experiment 12. Microfluidic Paper Chromatography (2-3) Goldwhite, H.; Tikkanen, W. Experiment 12 .Paper Chromatographic Separation of Food Dyes, Experiments in General Chemistry, 4th ed.;The McGraw Hill Companies. (81-85)...