Asked by Joseph
If you have 5.6 ml of 95:5 ethyl acetate, how many ml of 85:15 ethyl acetate can you make? Assume unlimited supply of pure ethyl acetate.
Answers
Answered by
DrBob222
I don't know how this fits into TLC chromatography but here is what I would do.
5.6 mL x 0.95% = mL x 85%
mL = 6.2588 mL is what you can make.
So you take 5.6 mL 95:5 ethyl acetate and add pure ethyl acetate to make 6.2588 mL total. Check it. You have 5.6 mL x 0.95 = 5.32 mL pure ethyl acetate in the 5.6 mL sample. What percent is that of 6.2588?
(5.32/6.2588)*100 = 85. I know I've used too many significant figures but on such small numbers it wouldn't come out to quite 85%.
This is a strange problem and I may have interpreted it incorrectly. You can check the % ethyl acetate, too.
mL pure ethyl acetate in the 5.6 mL initially = 5.6-5.32 = 0.28.
We have added pure ethyl acetate of 6.2588-5.6 = 0.6588 added.
Total = 0.6588 + 0.28 = 0.9388.
Then (0.9388/6.2588)* 100 = 14.99968.
Again the s.f. are not right but I wanted the numbers to come out to 85 and 15.
5.6 mL x 0.95% = mL x 85%
mL = 6.2588 mL is what you can make.
So you take 5.6 mL 95:5 ethyl acetate and add pure ethyl acetate to make 6.2588 mL total. Check it. You have 5.6 mL x 0.95 = 5.32 mL pure ethyl acetate in the 5.6 mL sample. What percent is that of 6.2588?
(5.32/6.2588)*100 = 85. I know I've used too many significant figures but on such small numbers it wouldn't come out to quite 85%.
This is a strange problem and I may have interpreted it incorrectly. You can check the % ethyl acetate, too.
mL pure ethyl acetate in the 5.6 mL initially = 5.6-5.32 = 0.28.
We have added pure ethyl acetate of 6.2588-5.6 = 0.6588 added.
Total = 0.6588 + 0.28 = 0.9388.
Then (0.9388/6.2588)* 100 = 14.99968.
Again the s.f. are not right but I wanted the numbers to come out to 85 and 15.
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