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Making up various concentraions of ethanol

  1. Feb 20, 2008 #1
    I am setting up a prac for a class next week, and need a little help on making up solutions of ethanol. (I am a microbiologist and havent done any of this sort of stuff for years)

    The prac calls for 1M, 2M, 3M and 4M solutions of ethanol, and I have a 99% solution of ethanol.

    The way I thought of doing this was converting the molarity to % and then worked out what dilutions to make to achieve that percentage.

    Is this correct?

    Thanks

    ian
     
  2. jcsd
  3. Feb 21, 2008 #2
    yeh probably should work out fine... but also count the fact that u have 99%... u should do fine
     
  4. Feb 21, 2008 #3
    Mathematically speaking to make a 1Mol solution you would have to know the molar mass of ethanol (46.06g Mol) get 1Mol of that (46.06g) and make up a 1 Litre solution of it, or if you dont want to work with such large amounts scale it down by a factor 5 or 10 if you want.

    Remember however that ethanols density is not == to water, so putting 46.06cm-3 is not equal to putting 46.06g in the solution. To figure out what sort of volume you must add, you must assume a 100% purity (for an accurate density) and divide through by it, so for ethanol the density is 0.789g/cm3 = 789g/dm3 Therefore you'd want
    [tex]\frac{46.06}{789}*1000= 58.38cm^{3}[/tex]
    I multiplied it by 1000 to convert the answer from dm3 to cm3

    As for eliminating your 99% purity of the ethanol problem, you will need to find out the effective mass of ethanol you must put in, this is simply done by dividing by the percentage of mass that ethanol constitutes. For your case its
    [tex]\frac{46.06}{0.99} = 46.525g[/tex]

    Combining the effect of needing 99% but whilst also assuming the density does not deviate massively for it being 99% rather then 100%, we find:

    [tex]\frac{46.525}{789}*1000 = 58.96cm^{3}[/tex]

    58.96cm3 (Approx to 59cm3 since im guessing you wont have the gear for super accurate concentrations anyway) of ethanol must be added then made up to 1000cm3 (litre), alternatively you could just wait in line and weigh it out ¬¬.
     
    Last edited: Feb 21, 2008
  5. Feb 21, 2008 #4

    chemisttree

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    Just weigh out 46 grams of ethanol into a 1 L vessel and dilute to the mark. Its not that hard and you don't have to know the temperature dependence of molar volume, etc....

    You do have a balance don't you?
     
  6. Feb 21, 2008 #5
    Yeah I have all the equipment, volumetric flasks and pipettes balances etc, it is a fully functioning laboratory. It has just been so long since I have had to work with moles and molarity that I have forgotten it.

    I just wanted to check i was on the right track.

    thanks for all of your replies.

    Ian
     
  7. Feb 22, 2008 #6
    Yeah, I was going off the suggestion that all labs are like the ones I use, I.E. at the beggining of a session you have about 15 people trying to use the 4 scales ¬_¬ damn underfunded education!
     
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