The grams to moles formula can be used for two things: finding the number of moles of a different substance by using the grams you have given, or using the grams given to find moles of the same substance. We know this formula will do that because stoichiometry uses the laws of conservation of mass, definite proportions, and multiple proportions. Basically this all means that element A as a reactant/product must be equal to element A as the opposite. It also means that elements in a reaction can not create, destroy or transmute mass with each other.
A few terms so you understand what I mean in some places:
“Given” – what we are given in the problem
“Required” – what we are looking for
“Mole ratio” – always the ratio of required/given
“Molar Mass” – the sum of the atomic masses of all of the elements in the substance/solution
The formula that we will be using here, and that works for all grams to grams equations is this:
Moles Required (X) = Grams Given x (1 mol/molar mass given) x mole ratio.
This is saying that to find moles required, you take the given grams times one mol over the molar mass of the given substance times the mole ratio. The unit is put over molar mass given to simply show that the units will be changing. Here’s what you have with your units:
GRAMS/1 x MOLES/GRAMS x MOLES/MOLES
In the first half, both of the grams cross out.
GRAMS/1 x MOLES/GRAMS x MOLES/MOLES
And then, some of the moles cross out with each other, leaving you with moles.
GRAMS/1 x MOLES/GRAMS x MOLES/MOLES
Let’s dive into some problems:
1. If you have 2.5 grams of Zn, then how many moles will you have of ZnCl2 in the following reaction.
Zn + 2HCl ———-> ZnCl2 + H2
Now, first of all, this equation is already given to you, and it is already balanced. So using the formula, we need to plug in the information that we have.
Moles Required (X) = 2.5g Zn x (1 mol/65.39g Zn) x (1/1)
We already had the grams given, the ratio of the two substances was 1:1, and the molar mass of Zn is 65.39g. All that has to be done now is type it in the calculator and it’s finished! It should come out to be 0.038 moles of ZnCl2. Not so bad, huh? Let’s try one more.
2. Given 200g of AlCl3, find the number of moles produced of Al(NO3)3.
3 AgNO3 + AlCl3 –> 3 AgCl + Al(NO3)3
This equation is already balanced, luckily. So let’s plug in our number and see what happens.
Moles Required (X) = 200g AlCl3 x (1 mol/133.33g AlCl3) x (1/3)
We had 200g given, the ratio was 1:3 (remember, required over given!), and the molar mass given was 133.33g (Al = 26.98 plus Cl3 = 106.35). The answer that you should have gotten is0.50 moles of Al(NO3)3.
