density of air and acetone vapour

andieje used Ask the Experts™

I recently did an experiment where acetone was heated in a water bath to vaporise it. The acetone was in a Dumas flask and excess vapour escapes through the capillary. I assumed any air in the flask would be expelled before the acetone vapour because air is less dense than acetone.

Acetone liquid has density 0.79g per cm^3 and air is 1.2kg per m^3. how can i find out the density of acetone vapour

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They go through the procedure in some detail here:
Note that the Dumas experiment is for determining the molar weight of a vaporizable

You have asked about the density, which will vary with pressure and temperature.

If all you really need is the density, you just have to divide the mass by the volume.
The link above discusses both measurements.


You will find a good description of how to determine the density of acetone at

When acetone vaporizes, acetone molecules mixes with air in the bottle. Thereby increasing the pressure and forcing the air-acetone gas mixture out of the bottle.  The acetone gas molecules moves freely in the bottle as if the were alone there and mixes totally with the air molecules. The gravitational effect is neglectable. The volume of the vaporized acetone is much much larger than the volume of the bottle. Forcing air and acetone molecules to escape through the capillary. So after a while there are no air molecules (approximately) left in the bottle, only acetone.

Acetone gas density  = Mass of acetone gas in the bottle / Volume of the bottle.

Mass of acetone in bottle = Weight of bottle with air and acetone gas at end of experiment - Weight of bottle with only air in it.  

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The density of any gas can be found from PV=nRT and the molar mass, which for acetone is58 g/mole


I forgot to add on to the end of my question...

How do i find the density of acetone at a particular temp and pressure. I can do that using pv = nrt as ozone says.

How do I find the density of air at the same temp/pressure? I actually asked teh question in the first place because of teh Dumas experiment. I was wondering how you can ensure there is no air left in the flask when you heat the volatile liquid (acetone) and i was presuming it was because air is less dense so I need to find teh density of air under my particular conditions of temp and pressure.

I know that the density of air at 100C is < 1gL^-1


I get the density of acetone at 362.15k and 100238.15Pa to be 1.93g L-1

Does that sound feasible? I worked this out from having 4.231 * 10^-3 moles of acetone in a volume of 127.062cm3

It sounds in the right ball park to me. It's certainly more dense than air
>>I was wondering how you can ensure there is no air left in the flask

The relative densities of the gasses are not as important as the time.  
You are actually mixing the gasses and flushing the air out of the flask.
The time constant for this is on the order of the volume divided by the flow rate.

5 time constants gives you (1-1/e^5) ==> 99.3% accuracy.


What do you mean by 5 time constants?
If you start with 100 ml of air, and gradually add 100 ml of acetone vapor while drawing off the mixed gasses at the same rate, you will wind up with
acetone and air in the ratio (1-1/e)/((1/e) = (63/37).

If you flush the volume with 5x the volume of acetone, you wind up with
(1-1/e^5) = 99.3% acetone.



What is the name of the process you are describing so I can read about it as I'm not reall sure what you mean.

For example, how do you know you are drawing odd the mixed gasses at the same rate in the Dumas experiment. All i know is that they will escape though the capillary tube in the top of teh flask. I'm also not sure how I am 'flushing the volume' in anyway in the experiment.

YOu say: The time constant for this is on the order of the volume divided by the flow rate.
I don't know what the flow rate is or how to find it

many thanks
Sorry for the delay.  Some of the confusion is my fault.  I didn't quite understand the setup of the experiment.

You start with a flask full of air with a small sample of acetone.  You do have to flush out or dilute the air.

The capillary is open to the air, so you start and finish at 1 atm pressure.  So whatever volume of vapor you generate
will eventually be expelled.  But the rates may not match at every instant since pressure may build up in the flask.
One error on my part.

You could measure the gas flow rate by putting the end of capillary under water and capturing the bubbles in a
graduated cylinder.  This is a standard lab practice, but not really necessary here.  Another red herring.

But you do have to make sure the air is flushed out of the flask.  The way to do that in this case is to make sure
you start with enough acetone.  
So after you're done, you should calculate how the total vapor volume of your sample.  It is probably 50x or 100x the volume of the flask.  If it were less than 5x the volume, you would start to see the effects of air contamination.

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