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Simple distillation of compounds with similar boiling points

  1. Apr 20, 2017 #1
    I just had a lab, and the results really aren't adding up to what i expected. And this is specifically about simple distillation, not fractional distillation.

    Lets say i have a solution of two liquids, one's boiling point is 90°C and the other is 110°C, a 20°C difference which is pretty close. So i heat up the plate, and that flask heats up to hot enough to boil, then the vapor reaches the thermometer and lets say it reads between 85°C - 95°C the entire process.

    Now, heres what im not sure about. I'm thinking even though the temperature on the thermometer on top of the flask is between 85°C - 95°C, i'm pretty sure if i measured the temperature of the boiling liquid in the flask directly, it would be more than 95°C. Is this true? I'm pretty sure this is true because during the lab, the thermometer read well below the lowest boiling point and i was still collecting liquid in the collection vial.

    2nd question, will i have any of the 110°C compound go into the collection vial? For compounds that have very different boiling points, the answer is no (or at least i'm pretty sure). But for close boiling points, if my 1st idea is correct, then that might mean theres a chance of the 110°C compound going into the collection vial in small amounts (i think?) If this is possible, can someone explain to me how this is happening?
    Last edited: Apr 20, 2017
  2. jcsd
  3. Apr 20, 2017 #2


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    Always good. Arouses your curiosity.

    Somehow material is transported between a and b and the driving force is a temperature difference here. You want to investigate the relationship between temperature and composition of the liquid as givens, and the composition of the vapour as a result. In short: read up on distillation a little bit, preferably in a textbook or your notes.
  4. Apr 20, 2017 #3


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    Distillation is not a simple process .

    The agitation of the mixed liquid caused by the boiling action can cause droplets of both liquids to be carried over into the receiver .

    Usually micro sized droplets (like the water in wet steam) but it is possible for larger droplets and even quite large slugs of liquid to be carried over .
  5. Apr 25, 2017 #4
    I suggest you start with a search on the thermodynamics of a phase change. Once the pot reaches the boiling point of the first component, all additional heat goes into the enthalpy of vaporization, The pot will not exceed the first BP. After the the first component distills off, the pot will increase in T to the BP of the second component.

    Also look for azeotropes and colligative properties.
  6. Apr 27, 2017 #5
    Have you tried to apply Raoult's law to the mixture to determine the boiling point and the composition of the vapor initially coming off? Have you tried to develop a model of the system to estimate the boiling temperature and the vapor composition as a function of the mass that went overhead?
  7. Apr 27, 2017 #6
    None of this is correct. If the equilibrium vapor pressure of species A as a function of temperature is ##P_A(T)## and the equilibirum vapor pressure of species B as a function of temperature is ##P_B(T)##, then, according to Raoult's law (assuming an ideal solution), the total pressure of the solution at temperature T will be $$P(T)=x_AP_A(T)+x_BP_B(T)$$ where the x's are the mole fractions. The boiling point of the solution is reached when P(T) is equal to the imposed total pressure. So the initial boiling point will lie somewhere between the boiling point of the high boiler and the boiling point of the low boiler. The mole fraction of the low boiler in the vapor will always be higher than the mole fraction of the high boiler in the liquid. So, as the liquid boils away, the liquid will become enriched with the high boiler. The mole fractions of the two species in the vapor at any time will be equal to $$y_A=\frac{x_AP_A(T)}{x_AP_A(T)+x_BP_B(T)}$$and$$y_B=\frac{x_BP_B(T)}{x_AP_A(T)+x_BP_B(T)}$$The boiling point of the liquid will gradually rise as the liquid boils away and the mole fraction of the high boiler in the liquid increases.
    Last edited: Apr 30, 2017
  8. May 1, 2017 #7
    My apologies to the OP for any confusion. Thanks for setting me straight Chet.
  9. May 18, 2017 #8
    When distilling a mixture of liquids, you'll always* distill over a mixture, but the fraction of each in the vapor will change with temperature. That's why some things are 'triple distilled' - by distilling multiple times you'll get a fraction of a fraction of a fraction of what you're trying to get rid of - less and less.

    That's where a fractionating column helps: a giant surface on which the mix can condense and re-vaporize, giving, in essence, a very large amount of individual distillations.

    You'll probably get some ( a bit) of the higher boiling liquid over, I'd think yeah.

    *I actually think this is 'always', and not 'usually', which is pretty rare in chemistry
  10. May 18, 2017 #9
    Yes. It's pretty straightforward to design a distillation column (countercurrent plate column) for a system like this using the McCabe Thiele method. This is a standard chemical engineering calculation.
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