Making up various concentraions of ethanol

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SUMMARY

This discussion focuses on preparing 1M, 2M, 3M, and 4M ethanol solutions from a 99% ethanol stock solution. To create a 1M solution, one must weigh 46.06g of ethanol and dissolve it in 1 liter of solution, adjusting for the density of ethanol (0.789g/cm³). The effective mass of ethanol required from the 99% solution is calculated as 46.525g, which translates to approximately 58.96cm³ of the 99% ethanol needed to achieve the desired concentration. The discussion emphasizes the importance of accurate measurements and proper laboratory techniques.

PREREQUISITES
  • Understanding of molarity and solution preparation
  • Knowledge of the molar mass of ethanol (46.06g/mol)
  • Familiarity with density calculations (density of ethanol: 0.789g/cm³)
  • Basic laboratory skills, including using balances and volumetric flasks
NEXT STEPS
  • Research the preparation of different molar solutions using various stock concentrations
  • Learn about density adjustments for solutions with varying purity levels
  • Explore the principles of dilution and concentration calculations in chemistry
  • Study the impact of temperature on the density of liquids, particularly ethanol
USEFUL FOR

This discussion is beneficial for microbiologists, chemistry students, and laboratory technicians involved in solution preparation and concentration calculations.

iross75
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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 haven't 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
 
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yeh probably should work out fine... but also count the fact that u have 99%... u should do fine
 
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 don't 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
\frac{46.06}{789}*1000= 58.38cm^{3}
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
\frac{46.06}{0.99} = 46.525g

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:

\frac{46.525}{789}*1000 = 58.96cm^{3}

58.96cm3 (Approx to 59cm3 since I am guessing you won't 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 ¬¬.
 
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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?
 
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
 
Yeah, I was going off the suggestion that all labs are like the ones I use, I.E. at the beginning of a session you have about 15 people trying to use the 4 scales ¬_¬ damn underfunded education!
 

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