# Fractional distillation

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#### arc5005

##### Full Member
7+ Year Member

If the vapor pressure above a 50% mixture of methanol with butanol at 22° C were collected and placed into a flask where it condenses, what is the mole percent of methanol in vapor above the new (second) flask? [Pure methanol has a vapor pressure of 87 torr and pure butanol has a vapor pressure of 29 torr at 22° C]

A. Less than 25%
B. Between 25% and 50%
C. Between 50% and 75%
D. Greater than 75%

D) Greater than 75%

From a 50% by moles mixture of the two liquids, the vapor due to methanol above the first flask is 87/(87 + 29). When the vapor is condensed, collected, and placed into the second flask, the condensed solution is 75% methanol by moles. The vapor pressure of methanol above the second flask is greater than 75%, because methanol evaporates more readily than butanol evaporates. This question presents the principle behind fractional distillation. By continuously condensing and re-distilling, the distillate becomes richer with regard to the component with the lower boiling poing.

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I guess I don't understand the math here or why we use this math equation for this type of problem, (forgive me, my math skills are poor because my last math course was over 8 years ago). When I see this problem, why am I supposed to intuitively think that I need to do 87 / (87 + 29)? Maybe this is because I have not reviewed the fractional distillation section of my review books yet, and this problem came up in a phase change section.

87 torr / (87 torr methanol + 29 ethanol) = 87 / 116

(116/116) = 100% > (87 / 116) > 58/108 = 50%

87/116 = 0.75 = 75%

#### aldol16

##### Full Member
5+ Year Member
Here's how fractional distillation works:

You have a solution of A and B together. Say A boils at a lower temperature (more readily) than B. The important thing to note here is that one component is more volatile than the other. If the two were equally volatile, fractional distillation wouldn't work. You put the solution of A and B into a closed container at some set temperature. Molecules of A and B will enter the gas phase and a rapid equilibrium will be reached. Since A is more volatile than B, you would expect there to be more A gas particles than B gas particles. In other words, say you start out with 50 molecules of A and 50 molecules of B in liquid form. You seal them in the container and allow equilibrium to be reached. In the gas phase above the liquid, would you expect more A or B? A, of course, since it's more volatile. If you take that gas phase out and condense it, you're going to have more than a 50:50 ratio of A to B. Then when you allow it to evaporate as well, you keep enriching the gas phase in A.

Mathematically, just think about it this way. In a 50/50 mixture of A and B, the percentage in the gas phase that A will contribute is just its percentage of the total vapor pressure. So if A has a vapor pressure in the pure form of 87 torr and B has a vapor pressure in the pure form of 29 torr, then if you put A and B together in equal amounts, A will contribute 87/(87+29) of the total vapor pressure and B will contribute the rest of it.

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#### arc5005

##### Full Member
7+ Year Member
Here's how fractional distillation works:

You have a solution of A and B together. Say A boils at a lower temperature (more readily) than B. The important thing to note here is that one component is more volatile than the other. If the two were equally volatile, fractional distillation wouldn't work. You put the solution of A and B into a closed container at some set temperature. Molecules of A and B will enter the gas phase and a rapid equilibrium will be reached. Since A is more volatile than B, you would expect there to be more A gas particles than B gas particles. In other words, say you start out with 50 molecules of A and 50 molecules of B in liquid form. You seal them in the container and allow equilibrium to be reached. In the gas phase above the liquid, would you expect more A or B? A, of course, since it's more volatile. If you take that gas phase out and condense it, you're going to have more than a 50:50 ratio of A to B. Then when you allow it to evaporate as well, you keep enriching the gas phase in A.

Mathematically, just think about it this way. In a 50/50 mixture of A and B, the percentage in the gas phase that A will contribute is just its percentage of the total vapor pressure. So if A has a vapor pressure in the pure form of 87 torr and B has a vapor pressure in the pure form of 29 torr, then if you put A and B together in equal amounts, A will contribute 87/(87+29) of the total vapor pressure and B will contribute the rest of it.

thank you 