Monday, March 18, 2019

Grams to Grams Stoichometry

The gist of this concept is fairly simple to understand. But, there are a lot of details between the concept and answering the questions.

The question will take the form of something like…
You have this much of something. How much of something else do you need for a complete reaction without anything being left over.
Simple enough on face value. But, did you catch the part about there being a lot of steps?

Let's do it.

Step One: Start with a balanced equation.

The balanced chemical equation is like a recipe. It tells the ratio of ingredients in the compound. The principle works for all types of reactions, but a simple synthesis reaction makes the simplest example, so we'll look at that.

And, the mention of recipe evokes the idea of food, so let's start there.

(Credit to my colleague, M. Peterson, for this example.)

Let's make some s'mores. You know…

Now, just to be fair, there are two types of s'mores (just as some compounds form in different ratios): there is the single-chocolate layer s'more and the double-chocolate layer s'more.

So, let's look at the balanced recipe for those:

Cracker = C

Marshmallow = M

Ch = Chocolate squares.

With these basic components two different varieties of S'more can be created!

Single Chocolate S'more

M + 2C + 3Ch --> MC2Ch4


Double Chocolate S'more

M + 2C + 6Ch --> MC2(Ch3)2

The coefficients to the balanced recipe tells how much of each thing is needed in a ratio. For the Single Chocolate S'more, the ratio is:

1:2:3 -->1

1 marshmallow : 2 crackers : 3 squares of chocolate --> 1 s'more

The ratio works for individual items, dozens of items, scores of items, bazillions of items, or…

…or moles of items.

With chemistry, the concept of the mole prevails because of its connection to grams through the atomic mass of the elements. A mole of atoms weighs in grams the atomic mass. This is super, super convenient!

Understanding that the balanced chemical reaction gives us a set, fixed ratio of the atoms that must combine is the first, vital concept needed for the grams to grams stoichometry process.

So, let's make some water? And some hydrogen peroxide? Sure…

2H2 + O2 --> 2H2O   (water)
H2 + O2 --> H2O     (hydrogen peroxide)
The balanced reactions above give us the ratio recipe for the molecules in the reaction. Granted, it is a simple synthesis reaction, but, as stated before, the principle that applies here and the method is the same regardless of the complexity of the reaction.

Step 1 is to find the balanced reactions. Done.

Step 2: Extract the ratio recipe from the coefficients.


For the first reaction, the ratio is 2:1 --> 2

For the second reaction, the ratio is 1:1 --> 1

What does that even mean?

This is actually pretty awesome in that chemistry is cool sort of way. (Just nod and follow along. Trust me, it's awesome!)

Like the s'mores, it is the ratio of "things" needed for a complete reaction without leftovers.

In s'mores, if you have only 2 crackers, you need only 1 marshmallow. Done. Period. Having more than one marshmallow means you'll have leftovers.

In many cases of introductory chemistry, you are looking for complete reactions without leftovers.

So, for the H2O reaction (water), starting with 2 somethings of H2 requires 1 something of O2. The something can be molecules, dozens of molecules, thousands of molecules, or bazillions of molecules. However, counting molecules requires very, very, very small fingers #sarcasm! Counting them a bazillion times is a bazillion times harder!

This is where the mole concept comes to the rescue! The relationship between moles, atomic mass, and grams saves us.

Time to chant (seriously):
Moles to grams you multiply!
Grams to moles divide!
Repeat 10 X
So, if you know the number of moles, multiply by the atomic mass OF THE MOLECULE to find out how many grams you have.

If you know how many grams you have, divide by the atomic mass OF THE MOLECULE to find out how many moles you have.

Now, that we have that figured out, we are ready to go to the next step.

Step 3: Figure out how many moles of the given thing you have.

This is why we chanted.

Say you have 37 grams of O2 and you are doing the water reaction. How many grams of H2 do you need (to react completely without left over).

Okay, start with the balance reaction (step 1)

2H2 + O2 --> 2H2O

If you have 37 grams of Ohow many moles is that?  Okay, back up…

Step 3a: Find the MOLECULAR MASS for the reaction.  Let's put those numbers below the molecules in the reaction.

This is the mass of the molecules Not the mass of the reactants or products. 

2H2 + O2 --> 2H2O
  2 g          32 g              18 g 
A more complete look at the masses…

                    Atom Mass                                 1 g         16 g          2g 16g
                    Balanced Equation                    2H2 + O2 --> 2H2O
                    Molecule Mass                             2 g        32 g            18 g
                    Total Reactant/Product Mass       4 g         32 g            36 g  

Step 3b: Find the moles of what's given.  Chant!

Divide what you have by the mass of one molecule:

37 g / 32 g/mole  =  1.156 moles   (If the units confuse you, just chant again.)

Step 4: Use what you have (Step 3b) and the recipe ratio (Step 2) and find out what you need for the other thing.

That sounds more confusing than it is.

The recipe calls for:

2 : 1 --> 2

We have more than the 1 in the ratio, so we need to figure out what all of the numbers are. Easy.

Multiply the ratio (which is 2 : 1 --> 2)
by what you have (which is 1.156 moles)
divided by what you need (which is 1 mole)


1.156/1 • (2 : 1 --> 2
2.312 : 1.156 --> 2.312
Now what?

That multiplication tells us how much of the other thing we need. In the case above, since we have 1.156 moles of O2, we need 2.312 moles of H2.

Step 5: Convert the moles needed of the other thing to grams.

This is why we chanted.

How much did one mole of H2 weigh? It's up there somewhere!

So, multiply!
2.312 moles • 2 g/mole = 4.624 grams.

Step 6: Bask in satisfaction that you are now done.

That's it. That's the answer. If you have 37 grams of O2 you need 4.624 grams of H2


Let's do the peroxide without all of the discussion.

Starting with .75 grams of H2, how much O2 is needed to completely react without excess?

Step 1: Balance the equation:

H2 + O2 --> H2O

Step 2: Extract the recipe ratio:

1: 1 --> 1

Step 3: How many moles do you have to begin with?

H2 + O2 --> H2O
2 g          32 g             34 g

Grams to moles divide…

.75 g / 2 g/mole = .375 moles of H2

Step 4: Use the ratio to find moles of "other" thing.

.375/1 • ( 1: 1 --> 1 ) 
.375 : .375 --> .375

Step 5: Convert moles needed of the "other" thing to grams.

So, multiply!
.375 moles • 32 g/mole = 12 grams.




The process is long, the concept is sorta complex, but it's really not all that hard. Just more tedious than anything.

Here are the steps:

Step 1: Balance the equation:

Step 2: Extract the recipe ratio:

Step 3: How many moles do you have to begin with?

Step 4: Use the ratio to find moles of "other" thing.

Step 5: Convert moles needed of the "other" thing to grams.

(Chanting is optional.)

Step 6: Bask in satisfaction.

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