posted by solution chemistry .
In an experiment, ethanol standards and an n-propanol standard were to be prepared in water. The ethanol standards were weight% and the n-propanol could be either v/v or weight%. One of the questions for the report asks why it was ok to make the n-propanol a v/v%.
I overheard someone mention that it's because the density of ethanol is less than that of water but I'm unsure how that applies to how the solutions are made.
the ethanol standards made were mixed with the n-propanol and ran through a gas chromatograph in order to plot calibration curves using both peak area and peak height data obtained from the chromatograms. the goal of the lab was to determine the %ethanol in an unknown solution which was prepared the same as the standard ethanol mixed with n-propanol
I don't believe density has anything to do with it. The density of n-propanol is less than that of water, also (although it is slightly greater than that of ethanol).
Thanks for the clarification of the experiment in the follow up. If I read this right, you made standards of ethanol in n-propanol; i.e., the concn of each standard for ethanol in propanol varied and that's how the calibration curve was made for ethanol. Again, if I read it right, the concn of n-propanol is the same throughout. In other words, I think you simply made up varying concns of ethanol in the n-propanol in which the n-propanol was the solvent. That being said, it makes little difference about the EXACT concn of the n-propanol since the concn of n-propanol was the same in the standards and unknown. I suspect the prof was trying to save you a little time since v/v % for propanol can be made up so much fster than w/w %. (unless of course you want to make the calculation using density for mL needed for grams and using volumes then to measure out the mass.)
Ah I see, so because the same n-propanol solution was used for all the standards, the concentration remained constant regardless of what it was. And since the ethanol standards were used for the calibration curve they need to be more exact?
You're on the ball. Go to the head of the class. :-)
Thanks for you help.