# Stoichiometry

## Example Reaction:

Our example reaction will be the equation for cellular respiration, which is written as:

## What Kind of Reaction Is It?

Our example is a combustion reaction. Combustion reactions occur when a hydrocarbon (hydrogen and oxygen) is burned, and the products are always carbon dioxide and water.

## What's The IUPAC Name For Our Equation?

The International Union of Pure and Applied Chemistry (or IUPAC) has created guidelines for naming chemical reactions. Were we to write our chemical reaction out in words, it would read "When heated, solid glucose reacts with gaseous oxygen to yield gaseous carbon dioxide and liquid water."

## Molar Masses

To calculate the molar mass of each reactant and product, we take each element separately and multiply their atomic masses by the subscript the possess, adding them all together in the end to get the total molar mass of each reactant and product.

Note: Coefficients are not used when calculating molar mass.

Example

C₆H₁₂O₆ can be rewritten as C₆ + H₁₂ + O₆

Next, we multiply each element's atomic mass by their subscript.
C = 6(12.011) ≈ 72 g/mol

H = 12(1.008) ≈ 12 g/mol

O = 6(15.999) ≈ 96 g/mol

Adding up all of the answers gives us approximately 180 g/mol, the total molar mass of glucose!

Using this method, we are able to calculate each product and reactant's molar masses as:

C₆H₁₂O₆ ≈ 180 g/mol
6O₂ ≈ 32 g/mol

6CO₂ ≈ 44 g/mol

6H₂O ≈ 18 g/mol

## Mole to Mole Conversion

Utilizing the train track method, we are able to calculate how many moles of product can be made with any amount of reactant, and vice-versa.

Example

Say we are given 11.30 moles of glucose. If we wanted to find out how many moles of water it could produce, we simply employ the train track method.

1. After writing 11.30 moles of C₆H₁₂O₆ on top, we move down and to the right one cell and write 1 mole of C₆H₁₂O₆. Why? Because there is only one mole of C₆H₁₂O₆ in the complete balanced equation.

Tip: To find out how many moles of any reactant or product there are in an equation, look at the coefficient!

2. Write the amount of moles of the compound or element you're converting to there is in the equation in the next cell, which is above the bottom right cell. In this case, there are 6 moles of H₂O in our chemical equation, which can be seen from it's coefficient of 6.

3. The last step is to multiply all the numbers across the top and then divide your answer by the product of all the numbers across the bottom of the train track.

Doing this gives us (11.30(6))/1 = 67.8/1 = 67.80 mol H₂O
Tip: Don't forget your sig figs!

Practice

If you have 12.1 moles of CO₂, how many moles of O₂ are needed to make 12.1 moles of CO₂?

Solution:

## Mass to Mass Conversion

Just like the mole to mole conversion, we can determine how much mass is needed to produce/how much mass will be produced when given the mass of a reactant/product! To do this, we must convert grams to moles, and back to grams again.

Example

If we have 6.22 grams of oxygen, we can use the trusty train track method once again to determine how much water can be produced.

1. Write your given (6.22 g O₂) in the first cell. Your given will always go here.

2. Write the molar mass of your given reactant/product in the next cell, and 1 mole of your reactant/product in the cell above it. We do this because the molar mass of your reactant/product is the exact amount of grams that can be found in 1 mole of that reactant/product.

Tip: Because the molar mass of a compound/element is the equivalent to 1 mole, the cell above or below your molar mass will always be 1 mole of the compound/element!

3. For the next two cells, we write the actual amount of moles present in the balanced chemical equation for both elements/compounds. In this example, because both the top and bottom contain 6 moles each, they are able to cancel out, meaning we don't need to include them in our calculation, which saves us a bit of extra time.

4. In the next cell we write 1 mole of whatever element/compound you are looking for. This cell will also always be 1 mole because in the cell just above it we write the molar mass of the element/compound we're calculating for.

5. Finally, we multiply all values across the top and divide them by the product of all the values on the bottom.

(6.22(18.01528))/31.9988 = 112/31.9988 = 3.50 g H₂O

## Limiting and Excess Reactants

Unfortunately, nobody has an infinite amount of supplies to use within chemical reactions. Because of this, we have what are known as limiting and excess reactants.

A limiting reactant is the reactant in a chemical reactant that runs out first. It limits the amount of product you can make because all of it is used up first, even though there may be other reactants left over.

An excess reactant is a reactant that is left over after the limiting reactant has been used up.

To find out what the limiting and excess reactants are, we can use a simple mass to mass conversion starting with each reactant and find out which produces more and which produces less.

Example

In this example, we will start off with 5.25 grams of C₆H₁₂O₆ and 12.18 grams of O₂ and will be converting both of these to CO₂.

Traveling through the steps of a mass to mass conversion as listed above, we find the largest amount of carbon dioxide each reactant could make. This is called the theoretical yield.

If glucose is predicted to produce 7.69 grams of carbon dioxide, and oxygen is predicted to produce 16.75 grams of carbon dioxide, our glucose would be our limiting reactant because it produces less CO₂, meaning it would run out first. This leaves our oxygen to be our excess reactant because there is still some oxygen left over from the reaction that cannot be used up until it finds more glucose.

## Theoretical Yield

Theoretical yield can be easily defined as the maximum amount of product that can be made with a given amount of reactants.

This means that the theoretical yield of a chemical reaction is the same number of grams that can be produced from the limiting reactant.

For example, the theoretical yield of our chemical reaction would be 7.69 grams of CO₂ because that is the largest amount of product we can make without going over the limiting reactant.

Note: The theoretical yield will be the same as the amount of product the limiting reactant can make.

## Percent Yield

Percent yield is the percentage of the maximum possible yield (theoretical yield) that was actually produced.

To find the percent yield, we will use the equation shown above.

Example

To calculate percent yield, we need to know what the actual yield of our reaction is, which is simply the amount of product that was actually made.

Dividing the actual yield by the theoretical yield and multiplying by 100% gives us the percent yield.

For this example we will use 6.1 grams of CO₂ as our actual yield.

Burning Sugar