Calculations: Calculated Bandwidths Of Runs
Calculations: a. 1. W = Vend − Vstart W = 1.87 mL – 0.92 mL
W = .95 mL Wave = W1+W22 Wave = .95 mL+ .88 mL2 Wave = .915 mL 2. Vstartave = Vstart1+Vstart22 Vstartave = .92 mL+ .90mL2 Vstartave = .910 mL VRave = Vstartave + Wave2 VRave = .910 mL + .915 mL2 VRave = 1.37 mL 3. k’ = Vrave-VmVm k’ = 1.37 mL-.49 mL.49 mL k’ = 1.79 4. α = k'bluek'red α = 4.141.79 α = 2.31 5. R = (VRaveBlue-VRaveRed).5 (Wave(Blue)+Wave(Red)) R = (2.52 mL-1.37 mL).5 (2.04 mL+.915 mL) R = .778
d. Table #1: Calculated Bandwidths of Runs
| Red Dye | Blue Dye | | Run #1 | Run #2 | Run #1 | Run #2 | Bandwidth (mL) | .95 | .88 | 2.15 | 1.92 | …show more content…
For good separation of the dyes, the resolution should be greater than one. What was the value you calculated? Did the two dyes overlap as they emerged from the column, or was the separation a good one?
The resolution calculated was .778 (see calculations), which made it difficult to perceive where exactly one dye began and the other ended. Thus, when the dyes were emerging from the column, they mixed and formed a blend of the two colors to form a light purple color, which made it difficult to ascertain where the red ended and to discern where the blue section began. Thus, it was not a perfect separation as the dyes mixed when emerging.
5. In the step gradient, four separate fractions were collected. How were these related to the polarities of the column and the eluting solution?
The four eluting solvents decreased in polarity from one run to the next, starting water and then isopropyl alcohol starting from 5%, to 28%, to 70%. The water removed the most polar compounds from the column, and as the eluting solvent decreased in polarity (increasing in their percentage of concentrations), the less polar compounds emerged since “like dissolves
W = .95 mL Wave = W1+W22 Wave = .95 mL+ .88 mL2 Wave = .915 mL 2. Vstartave = Vstart1+Vstart22 Vstartave = .92 mL+ .90mL2 Vstartave = .910 mL VRave = Vstartave + Wave2 VRave = .910 mL + .915 mL2 VRave = 1.37 mL 3. k’ = Vrave-VmVm k’ = 1.37 mL-.49 mL.49 mL k’ = 1.79 4. α = k'bluek'red α = 4.141.79 α = 2.31 5. R = (VRaveBlue-VRaveRed).5 (Wave(Blue)+Wave(Red)) R = (2.52 mL-1.37 mL).5 (2.04 mL+.915 mL) R = .778
d. Table #1: Calculated Bandwidths of Runs
| Red Dye | Blue Dye | | Run #1 | Run #2 | Run #1 | Run #2 | Bandwidth (mL) | .95 | .88 | 2.15 | 1.92 | …show more content…
For good separation of the dyes, the resolution should be greater than one. What was the value you calculated? Did the two dyes overlap as they emerged from the column, or was the separation a good one?
The resolution calculated was .778 (see calculations), which made it difficult to perceive where exactly one dye began and the other ended. Thus, when the dyes were emerging from the column, they mixed and formed a blend of the two colors to form a light purple color, which made it difficult to ascertain where the red ended and to discern where the blue section began. Thus, it was not a perfect separation as the dyes mixed when emerging.
5. In the step gradient, four separate fractions were collected. How were these related to the polarities of the column and the eluting solution?
The four eluting solvents decreased in polarity from one run to the next, starting water and then isopropyl alcohol starting from 5%, to 28%, to 70%. The water removed the most polar compounds from the column, and as the eluting solvent decreased in polarity (increasing in their percentage of concentrations), the less polar compounds emerged since “like dissolves