Extraction of caffeine:

You had extracted 2 g of crude caffeine and you were in a hurry. Instead of doing the 6 successive extractions with 2ml portions of methylene chloride, you only did 2 successive extractions with 6 ml portions of solvent. Now, calculate the total mass of pure caffeine recovered in these 2 operations.

You didn't provide any data.

K = (organic phase)/(aqueous phase), solve to see how much stays in the aqueous phase, then do the extraction again. I think you will find that 2 extractions of 6 mL each will not be as efficient as 6, 2-mL extractions.

To calculate the total mass of pure caffeine recovered from the 2 operations, you need to know the percentage yield of caffeine in each extraction.

First, we need to determine the dilution factor caused by using larger extraction volumes. The original procedure calls for 2 ml portions of methylene chloride, but you used 6 ml portions instead. Thus, the dilution factor is 6 ml divided by 2 ml, which equals 3.

Now, let's assume that the percentage yield of caffeine in the original procedure (using 2 ml portions) is y%. This means that for every 100 g of crude caffeine extracted, y g of pure caffeine is obtained.

In the first extraction with a 6 ml portion of solvent, the total amount of caffeine extracted from the original 2 g of crude caffeine is 2 g multiplied by (6 ml / 2 ml) = 6 g. Since the percentage yield is y%, the amount of pure caffeine recovered in this extraction is y% of 6 g, which is (y/100) * 6 g.

In the second extraction with another 6 ml portion of solvent, the total amount of caffeine extracted is again 6 g because we didn't receive any new crude caffeine. The amount of pure caffeine recovered in this extraction is also (y/100) * 6 g.

Finally, to calculate the total mass of pure caffeine recovered in these 2 operations, we need to sum up the pure caffeine obtained in both extractions:

Total mass of pure caffeine recovered = (y/100) * 6 g + (y/100) * 6 g = (y/100) * 12 g.

So, the total mass of pure caffeine recovered in these 2 operations is (y/100) multiplied by 12 g.