[Violagirl]

In O Chem lab, I know that the solubilities of your compounds are important when performing an extraction and that it's also important to consider the densities to determine your aqueous and organic substances. For my lab the other day, we had to do an extraction with camphor and ethyl acetate. To prepare for the lab, I had looked up the densities of both and determined that camphor would be the aqueous solution as it had a higher density and also because its density is similar to that of water. When looking up the true definition of aqueous, I found that it means that a solution is similiar to water and has polar-like tendencies. I tried explaining it to my TA when he questioned my preperation for that portion of lab but he did not seem to understand my stance on it. I wanted to ask on here, is density, along with polarity, an important factor to consider when determining if a solution is aqueous when it comes to performing extractions?

[DrCMS]

"What makes a solution aqueous?" Water.

An aqueous solution is a water solution that is it.

To prepare for the lab, I had looked up the densities of both and determined that camphor would be the aqueous solution as it had a higher density and also because its density is similar to that of water.

No idea why you think density of camphor has anything to do with unless you thought you were going to use molten camphor?

I tried explaining it to my TA when he questioned my preperation for that portion of lab but he did not seem to understand my stance on it.

I can see why you have been complaining here about TA's now.  You are wrong, they are right. There is no stance on this question there is the simple fact that an aqueous solution is a water solution.  Just because you think you've got a nice argument why you are correct it does not alter the fact that you are wrong.  If you'd prepared better by looking up the definition of aqueous solution and checked the solubility of camphor you might have been properly prepared.  As it was you wasted your time and the TA's.

I wanted to ask on here, is density, along with polarity, an important factor to consider when determining if a solution is aqueous when it comes to performing extractions?

No it is not, density will determine if the aqueous layer ends up on the top or bottom but which ever layer has the water is the aqueous layer.

[Violagirl]

I have been prepared. I should have explained this in my first post but up to this point with extractions, we've done extractions with liquids that were completely insoluble with one another. Hence, why I primarily focused on the densities of both ethyl acetate and camphor in that camphor, with a density of .990 would end up on the bottom and why ethyl acetate, with a density of .897 would be the top layer. Before, when doing extractions, the less dense liquid would be the organic liquid and the more dense liquid would end up on the bottom as the liquid to be drained from the bottom. In this case, it was different which I determined when comparing the densities of both when I did my prelab and having two organic solutions together whereas before, we had an organic solution and an aqueous solvent and was not sure how to connect the information that I had together. After this extraction, I know now that it's also important to consider the solubilities of compounds that will be combined together to gain an idea of what's present, what potentially gets mixed together, and what needs to be obtained. Maybe my TA did not emphasize on that fact as much and I should have asked for further clarification on misunderstandings that I had. Thank you so much for your help, I have a better understanding now it.  

[Dan]

Hence, why I primarily focused on the densities of both ethyl acetate and camphor in that camphor, with a density of .990 would end up on the bottom and why ethyl acetate, with a density of .897 would be the top layer. Before, when doing extractions, the less dense liquid would be the organic liquid and the more dense liquid would end up on the bottom as the liquid to be drained from the bottom.

As has been said, water solubility has nothing to do with density. If in a previous experiment the more dense compound happened to be water soluble it was coincidence, nothing more. The bottom layer is not necessarily aqueous - dichloromethane and chloroform are both organic solvents immiscible with water and with a greater density than water. Water floats on these solvents. Camphor is very soluble in ethyl acetate and almost insoluble in water - this is easily explained by polarity.

[Violagirl]

I see what you mean now. The density is only relevant when determining which layer will be on top or on the bottom but is different when it comes to determining aqueous solutions. Thank you! 

[fledarmus]

I still don't think you are getting the concept of solutes and solvents.

Aqueous solutions are solutions in which water is the solvent. Many things can be dissolved in the water, but it is still an aqueous solution. Dissolving things in the water may change the density of the solution, but it is still an aqueous solution.

In organic chemistry, we frequently find it useful to use immiscible solvents to separate components of a reaction mixture. Water is usually one of the two solvents, and the other can be chloroform, dichloromethane, toluene, hexane, ethyl acetate, or almost anything else that isn't soluble in water. When you mix the two solvents together, they separate (that is what "immiscible" means) and form two phases, an aqueous phase (the water) and an organic phase (the other solvent). The denser of the two will be on the bottom - usually that is the aqueous phase, but chloroform and dichloromethane are denser than water and if you are using those for your organic phase, they will usually be on the bottom.

When you run a reaction, you typically have a lot of different molecules mixed together at the end of the reaction - your reaction product, remaining starting materials, any salts, acids, or bases that were required to run the reaction, the reaction solvent, any side products that were formed, and so on. You can separate many of those components by pouring the reaction mixture into your two immiscible solvents. The water soluble components of your reaction mixture will dissolve in the aqueous phase, and the organic soluble components will dissolve in the organic phase. Any insoluble components can be filtered out. Now  when you separate the aqueous phase from the organic phase, you can keep the layer that has the product you want, and dispose of the other layer, purifying your product by that much. If you are really clever and the reaction is nice to you, you can get all of your product in one phase, and all of the other crap in the other phase. (Can we say "crap" on this forum?)

So when you are doing an extraction, you really aren't concerned about the densities of any of the components of your reaction, only of your extraction solvents. In your case, the extraction solvents were water and ethyl acetate. Camphor is not soluble in water but it is soluble in ethyl acetate, so it will be in the ethyl acetate layer. Since ethyl acetate is less dense than water, it will be the top layer.