Concentration from density and purity

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Homework Statement


This isn't homework, rather it's regular work. I have a >97% solution of 1-thioglycerol, the density of which is 1.25 g/mL. I am trying to get a 1.5 x 10-4 M solution at the end, and I suspect the instructions I have been given are incorrect.
Molar mass = 108.16

Homework Equations


Concentration = Density / molar mass.

The Attempt at a Solution


The molarity of my stock solution is (1.25 * 1000) = 1250 g/L. 1250 / 108.16 = 11.5 M.
I am instructed to make a 10-fold dilution, giving me a molarity of 1.15 M.
I should then take 6.5 microlitres and add this to a final volume of 100 millilitres.
But 1.15 x (6.5 x 10^-6) = 7.5 x 10^-6. Divided by (100 x 10^-3) = 7.5 x 10^-5 M, which is half the concentration I need?

Thanks, Adam.
 
on Phys.org
No, I'm looking for 1.5 x 10-4 M monothioglycerol.
 
I calculated this again from the end concentration, and would just like anyone to proof-check it.
If we start with the concentration I want at the end (1.5 x 10^-4 moles per litre), and if I want this in a volume of 100 millitlitres:
(1.5 x 10^-4) x (0.1) = 1.5 x 10^-5 moles.
Multiply the moles needed by the molecular mass to obtain the grams needed: (1.5 x 10^-5) x 108.16 = 0.00162 grams.
The density is 1.25 grams per millilitre, but we need the grams per litre amount, so 1.25 x 1000 = 1250 grams per litre.
Divide the grams by the density in grams per litre to obtain the volume needed: 0.00162 / 1250 = 1.3 x 10^-6 L.
This is 1.3 microlitres. Therefore, if I makea 10-fold dilution of my stock solution, I should need 13 microlitres per 100 millilitres of media to make.
This is the same answer as I got in my original post, but I'm uncertain if I've used the density correctly...