Basic molarity/dilution problems etc

[jaykay]

Hi all,

Unfortunately I never was really taught some math skills I need to know..my high school teacher was fired in the middle of the year and never was replaced with a good teacher..I'm not happy about that because I get so confused with basic math skills. For example, now I do research in a lab for a summer job and can do the experiments just fine but get stuck for a long time on stupid math problems!

Anyone have an easy way of explaining these things? Or a book that teaches it?

I get stuff like what a mole is (concentration of mass/volume) but have trouble when it comes to making stock solutions from the basic material, serial dilutions, all kinds of stuff.

It's frustrating, please help.

[ecoli]

C1V1 = C2V2

 

Useful little equation. It means the concentration of the initial solution times the initial volume your using is equal to the concentration of the second solution you need times the volume of that solution. Generally, in a lab, you are going to need to solve for V1.

 

This equation is useful esp. for diluting stock solutions. You have the concentration of the stock solution, and you know the concentration you need for the working solution, so this will tell you how much volume of stock solution you should add (and therefore, how much water you need to dilute it.)

 

Sorry to hear about your unfortunate situation. My younger brother actually went through something similar this year (chemistry, too actually). Glad to see that didn't crush your science appreciation. Where are you working, if you don't mind me asking?

[ultma]

n = C x V

V = n/C

C = n/V

 

amonunt(mol) = concentration(mol/L) x volume(L)

 

ie. mol/L x L = mol

 

n = m/M

M = m/n

m = nxM

 

amount(mol) = mass(g) / molar mass(g/mol)

 

ie. g / (g x 1/mol) = 1/(1/mol) = mol

 

You can use these two in combination to make up a stock solution.

 

Say I want to make a 2M NaOH solution in a 2L volumetric flask

 

n = 2 (mol/L) x 2(L) --------*( (molxL)/L = mol )

n = 4 (mol)

 

4 = m/M

m = 4 x M ----------*(M = molar mass of Na+O+H = 40(g/mol))

m = 4(mol) x 40(g/mol) ---------*( (molxg)/mol = g )

m = 160g

 

So 160g in 2L of water will make up the 2M NaOH solution

 

Intersesting fact on a side note if I add 500mL of Ethanol to 500mL of water what will my final volume be?

[jaykay]

So I have this instruction sheet:

stock is 2mg/ml

it says for a concentration of 4 ug/ml, use 2 ul stock plus 48 ul water.

how did they do this calculation?