Thin Layer Chromatography

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Isolation of Component Via Extraction

Objective:
Separation of organic acid from a neutral compound using liquid/liquid extraction.

Experimental:
Similar to the “Thin Layer Chromatography” experiment, a TLC plate was ran with 2 ml of EtOAc and 8 mL of hexane in the jar. The Rf value was calculated using the formula located in the data for all the TLC plate. The mixed compound that was dotted on the TLC plate had 0.300 g of the impure benzoic acid and 10 mL of dichloromethane. A 50 mL flask was weighed and recorded. A separation funnel was then filled with 10mL of 3M NaOH and 10 mL of the dichloromethane. The funnel was capped and flipped upside-down while the lid was held. The solvent was mixed inside the separation funnel for a couple
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The Rf value was found in Figure 1 using the equation shown in the data section. It showed that dot “A” had an Rf value of 0.646 and dot “B” had an Rf value of 0.146. The mixture used in the first TLC run contained benzoic acid and biphenyl in the dichloromethane. When the benzoic acid (organic acid) and biphenyl (neutral) was mixed with the 3M NaOH (aqueous solvent), the pH strip indicated that the solvent was base. The 3M NaOH reacted with only the benzoic acid, causing it to produce sodium benzoate while the biphenyl remained the same. The manipulation of the aqueous pH level caused the organic acid and neutral compound to separate, but did not cause the layers to form. The sodium benzoate was soluble within the water (aqueous layer), while the biphenyl was soluble in the dichloromethane (organic layer). Due to Na2OH4 being highly soluble in water, it was used to absorb any water found in the organic layer. Using the equations and the numbers on Table 1 found in the data section, the Rf value for the isolated biphenyl was 0.66, the weight was 0.107 g, and recovery percentage was 35.7%. The pH level of the aqueous solvent was then manipulated from a base (sodium benzoate) back into an acid (benzoic acid), by adding HCl. The Rf value for benzoic acid was 0.115, the weight was 0.092 g and the recovery percentage was 30.7%. Using the equation shown in the data section, the total recovery weight came out to be 0.199 and the total recovery percentage was calculated to be

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