[LabRat01]

I'm a little stuck on this chemistry question!


What's the minimum no# of extractions needed to remove at least 97% of compound X from 68cm3 of an aq.solution containing 0.35g of compound X, if each extraction is carried out with 9 cm3 of dichloromethane, and the distribution coefficient is 8.66?

I got...over 600...that can't be right surely???

[sjb]

I'm a little stuck on this chemistry question!


What's the minimum no# of extractions needed to remove at least 97% of compound X from 68cm3 of an aq.solution containing 0.35g of compound X, if each extraction is carried out with 9 cm3 of dichloromethane, and the distribution coefficient is 8.66?

I got...over 600...that can't be right surely???

Does sound a little high. What's your working?

[rucik5]

Hi. It is possible that I will make some wrong assumptions here, but this is how I'd solve this: First, you've got the 9:68 ratio of your organic to aqueous. Next, you've got the coefficient, which is 8.66, which means you will have 8.16 times more of your compound in your organic vs your aqueous. Overall, that gives 8.66 parts of compound in 9 parts solution plus 1 part compound in 68 parts of solution. Hence, that gives 8.66*9 vs 1*68 ratio of the X between Org and Aq. If you calculate what's the % distribution in this case and simply use that number to work out how much you're taking out with each wash, how much is left and what is the goal (97% of 0.35).

So if T is total amount, and A, B, C ... are the amounts taken in washes 1,2,3... then you get this:
T ----wash---> A (A grams washed out)
T - A (grams left) ------wash---> B
T-A-B ----wash---> C
Your total mass washed out is A after first, A+B after second, A+B+C after third. How many times do you need then?
PS: If my working is crap then the whole post goes to hell;)