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# mol calculation confusion

mol calculation

5 replies to this topic

### #1 kedar

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Posted 08 October 2013 - 03:55 AM

Hi friends,

I need to use KN-93 which is a CAMKII inhibitor. I ordered from sigma and is 0.1 mg!  I cannot see anything in bottle and that's ok. It's in powder form. Although i used to feel comfortable with mol calculations, here i am totally confused.

Formula : C26H29ClN2O4S · H3PO4
Molecular Weight : 599,03 g/mol

I need a working concentration of 10 micromol/litre. Can anyone guide me on this?

After some calculation, i see that 1 mg= 1.67*10-6 mol i.e. 1.67*10-3 mM.

I need guidance after that. Need to know how much solvent i need to inject in this bottle to have 10 micromol/litre working conc.

thanks a lot,

K

### #2 phage434

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Posted 08 October 2013 - 05:18 AM

First, figure out how many moles of your compound is in your vial:

(1e-4 grams)/(6e3 grams/mole) = 1.7e-8 moles

Next, figure out the amount of liquid to be added to make a 10 micromolar solution:

(1.7e-8 moles) / (x liters) = 10e-6 moles/liter

Rewriting,

x liter = (1.7e-8 moles) / (10e-6 moles/liter) = 1.7e-3 liter = 1.7 milliliters

This is high school algebra. You need to understand about carrying units and cancelling them along with your calculations.

You also need to understand why this statement is meaningless:

"After some calculation, i see that 1 mg= 1.67*10-6 mol i.e. 1.67*10-3 mM."

The units of moles (mol) and the units of moles/liter (mM) can never be equal.

### #3 kedar

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Posted 08 October 2013 - 06:27 AM

Hi, thanks very much for the explanation . But why 6e3 grams/mole. Shouldn't it be 6e2 grams/mole ?   the mol weight is 600 gm, not 6000. Or did i miss something?

So, i get

1.67e-7 moles in 0.1 mg powder.

Then i again didn't understand the part     - (1.7e-8 moles) / (x liters) = 10e-6 moles/liter.  could u please elaborate?

"After some calculation, i see that 1 mg= 1.67*10-6 mol i.e. 1.67*10-3 mM."

The units of moles (mol) and the units of moles/liter (mM) can never be equal.------------ point noted

### #4 phage434

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Posted 08 October 2013 - 06:57 AM

You are correct, it should have been 6e2 g/mole.

So, i get

1.67e-7 moles in 0.1 mg powder.

This is correct.

What part of this don't you understand:

(1.7e-7 moles) / (x liters) = 10e-6 moles/liter.

I want to figure out how many liters (x) are needed to make a concentration of 10e-6 moles/liter.

I divide the number of moles I have (1.7e-7) by my unknown number of liters (x) and set it equal to 10e-6 moles/liter.

I solve for x by multiplying by x liter on both sides, and dividing by 10e-6 moles/liter.

I'm left with:

(1.7e-7 moles) / (10e-6 moles/liter) = x liters = 17e-3 liters = 17 ml

Note how the moles cancels and the 1/ (1/liter) becomes liter. It's helpful to explicitly carry units in your equations to get them right.

### #5 Missle

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Posted 09 October 2013 - 04:23 AM

As an aside to the calculation conversation, I'm guessing your vial is probably not large enough to hold 17ml of solvent.  I often create at 100x or 1000x stock from which to prepare my working solutions.  If you were to make a 100x stock (1mM), you could add 170ul of solvent to your 0.1mg and then prepare a simple 1:100 dilution into your working solution when needed.

### #6 kedar

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Posted 09 October 2013 - 12:01 PM

thanks both. It was really helpful. thank you very much.