help with mol/liter

Gold Refining Forum

Help Support Gold Refining Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

Wyndham

Well-known member
Joined
Nov 29, 2007
Messages
111
Location
Seagrove NC
I have found a formula that tells the mols/liter and I have found sulfuric acid at 98.1 gram/mol. I guess thatyou multiple the # of mols x the grams per mol per liter. Is there any ref site that list most of the materials used with the mol weight?
If I've misunderstood mol/l to grams, please correct me
Thanks Wyndham
 
Molarity is the concentration, usually given in moles per liter.

I hope you noticed that there is a difference between moles and molarity. I'll cover them both.

A mole is 6.022 x 10^23 molecules (or atoms or ions, just a number). It is selected so that one mole of carbon 12 is exactly 12 grams. That makes it easier to do chemical mathematics, one mole of of gold for example is 196.966569g as the gold atom have atomic mass 196.966569 u. (It is 196.966569/12 times heavier than carbon 12.)
One mole of iron is just as many atoms as one mole of gold, 6.022 x 10^23 molecules.

To find the weight of a mole of a certain substance, only add the atomic weights of the different atoms in the molecule. For example one mole of sulphuric acid, H2SO4, gives
2*H+S+4*O = 2*1.00794 + 32.065 + 4*15.9994 = 98.07848 g

Molarity (M=moles/liter, pronounced molar) is just another way to measure concentrations.
One mole per liter of sulphuric acid is 98.07848 grams of H2SO4 mixed with water to one liter volume. Two moles per liter is twice the amount. One mole of gold ions (196.96g) dissolved in one liter of water would give you 1 mole per liter, a 1M Au solution
So a liter of 1M sulphuric acid contains 98.07848 g of H2SO4 per liter of solution.

When mixing chemicals it is easier to calculate the ratios to mix if you have known solutions. If you for example wants to mix one gold atom per two sulphuric acid atoms then you would just mix two parts 1M H2SO4 with one part 1M gold solution resulting in a mixture with 2/(2+1)M H2SO4 and 1/(2+1)M gold = 1.33M H2SO4 + 0.33 M Au in solution.

Another example the other way around. We want a solution with 2M H2SO4, 0.5M H2O2 and 0.1M HCl. At our disposal we have 5M solutions of all chemicals we need. Then we could just mix two parts of 5M H2SO4, 0.5 part of 5M H2O2 and a tenth of a part of 5M HCl. Add water until the volume is 5 parts total.

I hope this helps you on your way.

Standard disclaimer, I'm no chemist, I'm a physicist and reserve the right to make mistakes in chemistry. :mrgreen:
 

Latest posts

Back
Top