Hi, sorry for another question, just want to clarify what you have said. Steam distillation is recommended under vacuum, so that the temperature is even lower and no oxidation occurs, right? Also in the second post you advice to go 1 drop per second, so about 3ml/minute, otherwise some oil will also come through with the steam. Is this not the main goal of steam distillation, I mean we are trying to get the pue prduct (oil) to come over to the reveiving flask. What am i getttig wrong? thank you
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Hi, sorry for another question, just want to clarify what you have said. Steam distillation is recommended under vacuum, so that the temperature is even lower and no oxidation occurs, right? Also in the second post you advice to go 1 drop per second, so about 3ml/minute, otherwise some oil will also come through with the steam. Is this not the main goal of steam distillation, I mean we are trying to get the pue prduct (oil) to come over to the reveiving flask. What am i getttig wrong? thank you
Which oil do you mean? Amphetamine/meth? Steam distillation is assumed that you distill them as azeotropic mixture.oil will also come through with the steam
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diogenesHi, thank you for interesting theory link. An answer to your question is in a linked topic:
P.S. @KokosDreams and @ImOutAlso You will be interested to read this topic (link above).
G.PattonThanks brother! Saved this to my library
G.PattonThe link recommends vigorous stirring to keep the oil/water mix `swirling` continuously, which will prevent the oil from `locking` the water underneath.
In this way the vapour pressure of the water, which is (roughly) 1 bar = 760Hgmm at 100C. Now the overall vapour pressure of your`swirling` mixture will be the sum of the vapour pressure of the oil and water. The oil`s vapour pressure at 100C is nowhere near 760Hgmm so will contribute very little to the sum, and only reduce the boiling point a tiny bit. So in case there is an unwanted, but volatile component in your oil, with similar boiling point as your product, it will be difficult to tell when your desired product and when this undesired product is coming over to the collection flask.
E.g. the vapour pressure of benzaldehyde is 0.074bar at 97C water`s vapour pressure is 0.91 so the sum is just short of 1 bar, and benzaldehyde will distill with the water at 98C. I haven`t found a calculator for amphetamine base vapour pressure, but it is lower than benzadehyde because it`s boiling point is higher and its weight is higher as well. So it will boil between 98 and 100C. There are a lot of factors, such as altitude etc. and these are just estimated values, but the point is that most molecules we are dealing with will boil with water/steam in a narrow temperature range and it not possible to tell what is coming over based on temperature. The steam temperature difference will be minuscule even if the compounds` boiling points are very different e.g. benzaldehyde and methamphetamine. With steam distillation it is only possible to separate the volatile components from the non-volatile `pollutions` in your oil. (Which is what we need as most of the volatile components is our product.)
G.Patton
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In theory boiling point drops to below 100C, right? I have done this without vacuum and the boiling point kept at 98-99. The amph base-water azeotrope has lower boiling point than water alone. On the other hand the small change in boiling point makes it impossible to distinguish between different oils.
e.g. https://chem.libretexts.org/Bookshe...ria/Immiscible_Liquids_and_Steam_Distillation
What I don`t understand is why most sources generate steam in a separate flask. Is it not easier to just put the water and the oil together in one flask, stir and distill? This can be supplemented by an addition funnel to replace water if it runs out while there is still some oil left.
What do you want to say, I don't understand your idea? Side products will have same bp shift with steam. Hence you can separate them by this distillation from amine.Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapor pressure you measure will simply be the vapor pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapor. The top one is sealing it in.
On the other hand the small change in boiling point makes it impossible to distinguish between different oils.
P.S. @KokosDreams and @ImOutAlso You will be interested to read this topic (link above).
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Hi, thank you for interesting theory link. An answer to your question is in a linked topic:
What do you want to say, I don't understand your idea? Side products will have same bp shift with steam. Hence you can separate them by this distillation from amine.
P.S. @KokosDreams and @ImOutAlso You will be interested to read this topic (link above).
Hi, thank you for interesting theory link. An answer to your question is in a linked topic:
What do you want to say, I don't understand your idea? Side products will have same bp shift with steam. Hence you can separate them by this distillation from amine.
P.S. @KokosDreams and @ImOutAlso You will be interested to read this topic (link above).
In this way the vapour pressure of the water, which is (roughly) 1 bar = 760Hgmm at 100C. Now the overall vapour pressure of your`swirling` mixture will be the sum of the vapour pressure of the oil and water. The oil`s vapour pressure at 100C is nowhere near 760Hgmm so will contribute very little to the sum, and only reduce the boiling point a tiny bit. So in case there is an unwanted, but volatile component in your oil, with similar boiling point as your product, it will be difficult to tell when your desired product and when this undesired product is coming over to the collection flask.
E.g. the vapour pressure of benzaldehyde is 0.074bar at 97C water`s vapour pressure is 0.91 so the sum is just short of 1 bar, and benzaldehyde will distill with the water at 98C. I haven`t found a calculator for amphetamine base vapour pressure, but it is lower than benzadehyde because it`s boiling point is higher and its weight is higher as well. So it will boil between 98 and 100C. There are a lot of factors, such as altitude etc. and these are just estimated values, but the point is that most molecules we are dealing with will boil with water/steam in a narrow temperature range and it not possible to tell what is coming over based on temperature. The steam temperature difference will be minuscule even if the compounds` boiling points are very different e.g. benzaldehyde and methamphetamine. With steam distillation it is only possible to separate the volatile components from the non-volatile `pollutions` in your oil. (Which is what we need as most of the volatile components is our product.)