Suppose you performed a distillation of a solution containing 1-pentanol (C5H12O), 2-pentanol (C5H12O), 3-pentanol (C5H12O), and methyl alcohol (CH4O). All three pentanols here have the same chemical formula, but they do have a different arrangement of their atoms (i.e. they are shaped differently), so this makes them have slightly different properties from one another.
During the distillation, you collect the FIRST product that comes out, and set it aside. Then, you collect the SECOND product that comes out, and set it aside Then, you collect the THIRD product, analyze it, and determine that it has a formula mass of approximately 88 amu.
What is the identity of this second product that you collected? How do you know (discuss ALL relevant data)? (HINT: You will find the table at
http://www2.stetson.edu/~wgrubbs/datadriven/petermurphy/flame/table1.html very helpful when answering this question!)
based on the boiling points my guess would be -3 pentanol. But for some reason, i think that would be too easy of an answer. is there a way to find the answer based on the amu?
Please help. ! am lost on this.
Ok, so this is the answer I came up with. Is this correct? Is there a better or more effective way to explain it? Thanks!
Answer: The second product that was collected is the 3-pentanol (C5H12O0).
I concluded this answer, by comparing the boiling points of each sample.
The boiling points (in increasing order) are as follows:
(1st to boil) 4- menthyl alcohol = 147 * F
(2 nd to boil) 3- pentanol = 239 * F
(3rd to boil) 2- pentanol = 246 * F
(4th to boil) 1- pentanol = 280 * F
Based on this data, the substances would reach their boiling points in this increasing order. Therefore, using the distillation process, the samples would also be collected in the order above.
To confirm this answer I used information from the question stating that the third sample to be collected would be the 2- pentanol with a formula mass of approximately 88 amu.
I confirmed this by calculating the formula mass of 2- pentanol (C5H12O):
(5 x 12.01 amu) + (12 x 1.01 amu) + ( 1 x 16.00 amu) = 88.17 amu or 88 amu
Using the on the law of constant composition, it is known that all samples of a given compound have the same proportions of their constituent elements. I used this info to confirm that the second product that was collected is the 3-pentanol (C5H12O0) because it also has the same formula mass.
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