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April 21, 2014

April 21, 2014

Posted by **sara** on Monday, October 15, 2012 at 6:53pm.

from 2 mL of 5:95 ethyl acetate:hexane, using pure hexane and pure ethyl

acetate.

- college orgo -
**DrBob222**, Monday, October 15, 2012 at 8:42pmI'm sure this is not the shortest way to work this problem but I think this will work.

5+95 is 1 part EtAc + 19 parts hexane. Therefore, 20 parts is 2 mL.

2/20 = 1/10 = 0.1; i.e., it is 0.1 part EtAc + 1.9 parts hexane (total of 2 mL).

If we are to increase the concn (5% to 15%) the only way we can do that is to add pure EtAc; therefore I let x = amount EtAc we need to add to obtain 15%.

[(0.1+x)/(2+x)] = 0.15

Solve for x. I get 0.235 part (0.235 mL) EtAc. [Note: To check this out we would have [(0.235+0.1)/(2.00+0.235)] = 0.15 so we're on the right track. Now we want to scale it up how much? By a factor of 10 mL we want/2.235 we have which is a factor of 4.47.

4.47*0.335 = 1.497 mL EtAc and

4.47*2.235 = 9.99 EtAc + hexane.

Check it to make sure it is 15%.

1.497/9.99 = 0.15

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