Guide to Stoichiometry

What is Stoichiometry and why do we use it?

Stoichiometry is a the part of chemistry that deals with the ratio of moles of different elements in a chemical equation. I know that sounds confusing but think of it like a sandwich.

2 slices of bread + 1 slice of cheese + 1 slice of ham= 1 sandwich

Now, say you have 6 slices of bread, 3 slices of cheese, and 3 slices of ham; you are able to make 3 sandwiches. However, if you have 6 slices of bread, 3 slices of cheese, and only 2 slices of ham you only have enough ingredients to make 2 sandwiches because there is not enough ham.

Stoichiometry uses these ratios in order to figure out how much of each reactant you're going to need in order to make a chemical reaction, or vice versa, how much product you will end up making in the chemical reaction.

Starting a stoichiometric equation

Before you start doing actual stoichiometry, you're gonna have to find out the basics of your chemical equation.

Say we are given sodium sulfate + Barium hydroxide. This is double replacement because there are two compounds that react together.

Here is the balanced equation: Na2SO4 + Ba(OH)2 > 2NaOH + BaSO4.

So we won't have to stop and do this during the actual stoichiometry part, we need to find the molar masses of all the reactants and products.

Na2SO4 = 142.042 g/mole NaOH = 39.997 g/mole

Ba(OH)2 = 171.342 g/mole BaSO4 = 233.39 g/mole

Mole to Mole Conversions

9.24 g Na2SO4| 2 moles NaOH

| 1 mole Na2SO4

Then you multiply the top and divide by the bottom and you end up with 18.48 moles of NaOH.

Mass to Mass Conversions, Limiting/Excess Reactants, and Theoretical/Percent Yield

Now we need to find the limiting and excess reactants. To do this, you need to do mass to mass conversions on both of your reactants. You are given 12.1 grams of Na2SO4.

12.1 g Na2SO4 | 1 mole Na2SO4 | 1 mole BaSO4 | 233.39 g BaSO4

| 142.042 g Na2SO4 | 1 mole Na2SO4 | 1 mole BaSO4

You multiply by the top and divide by the bottom and you end up with 19.86 g of BaSO4.

You are given 12.3 grams of Ba(OH)2

12.3 g Ba(OH)2 | 1 mole Ba(OH)2 | 1 mole BaSO4 | 233.39 g BaSO4

| 171.342 g Ba(OH)2 | 1 mole Ba(OH)2 | 1 mole BaSO4

You multiply by the top and divide by the bottom and you end up with 16.75 g of BaSO4.

The limiting reactant is Ba(OH)2 because it yields less than Na2SO4. The excess reactant is Na2SO4 because it yields more than Ba(OH)2. The theoretical yield is 16.75 g BaSO4 because it is less than 19.86 g BaSO4.

To find the percent yield, all you have to do is divide the actual yield by the theoretical yield and multiply by 100.

Say the actual yield was 12.4 g BaSO4.

12.4/16.75 * 100= 74.03% The percent yield is 74.03%.