Solubility Calculations

Member
edited August 2016
Hi:

I'm looking to find out how much ethyl lactate is needed to dissolve 1 gram of ferulic acid.

I don't mind working it out experimentally but since I don't have the materials on hand just yet and I came across the table below, I thought I might as well ask if someone is willing to show me how to interpret it.

Any help is appreciated. Thanks.

• Member, PCF student
edited August 2016
T/K is temperature in Kelvin. I wish people would use a more straightforward system, I have no idea what they mean by "1.40 mass% water" it is nonsense. Do they mean 1.4 grams water + 1 gram anhydrous??? There are much plainer ways of putting this.
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• Member, Professional Chemist
edited August 2016
in the second column of both tables, x stands for the mole fraction of ferulic acid; in other words, what proportion of the molecules in solution is ferulic acid

e.g. the first line states that at 298.2 Kelvin (25 °C), 8.03% of the molecules in the solution are ferulic acid, and 91.97% are ethyl lactate

to convert this to % w/w, you have to multiply the mole fraction of each component by its molecular weight, then determine what proportion of the total weight is ferulic acid

0.0803 mol × 194 g/mol = 15.5782 g ferulic acid
0.9197 mol × 118 g/mol = 108.5246 g ethyl lactate

total weight = (15.5782 + 108.5246) g = 124.1028 g

%w/w ferulic acid = 15.5782 g ÷ 124.1028 g = 0.12553... = 12.6%

@Belassi that's academics for you - I took it to mean a mixture of 98.6% w/w ethyl lactate and 1.4% w/w water (as a residual impurity from the manufacturing process)
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• Member
Thanks @Belassi ... that gets me closer. Since the product will be anhydrous, I can disregard the left side of the chart. Now to find out the meaning of solute mole fraction...
• Member

I was busy playing around with these free tutorials to see if I could teach myself...
http://www.ausetute.com.au/massmole.html
http://www.ausetute.com.au/molpcent.html

Now I know the right answer so I can go back and fix my calculating errors. Sweet.