Stoichiometry CALCULATIONS!

Stoichiometry calculations involve particles-moles-mass. That's right, we're back to mole conversions. Woohoo!

A little "road" map to help you do the calculations!

Step 1: Write a BALANCED equation. This is important to get the MOLE RATIO.

Step 2: Make a "road map". This will help to figure out where to start and where to go to get to the answer you need.

Step 3: Do the calculations!

Remeber SIGFIGS!!!

Let's try an example! 

This equation is given: 1 C3H8 + 5 O2 --> 3 CO2 + 4 H2O

What mass of C3H8 is needed to produce 100.0g of H2O?

Step 1: It is already balanced so move on to the next step!

Step 2: Road map: Mass of H2O --> Moles H2O --> Use the ratio of 1mol C3H8=4mol H2O --> Convert moles of C3H8 to mass C3H8.

Step 3: 

100.0gH2O x 1 mol H2O x 1 mol C3H8 x 44.0g C3H8 = 61.1g C3H8

                        18.0g H2O   4 mol H2O      1 mol C3H8

As you can see, you can't just convert grams to grams. In order to get there, you have to go through moles. In stoichiometry, many calculations follow this pattern:

grams (x) <--> moles (x) <--> moles (y) <--> grams (y)

You can start anywhere along this path.

Let's have some more practice!

Using the same equation as above, what is the mass of CO2 produced by reacting 2.00 mol of O2?

2.00 mol O2 x 3 mol CO2 x 44.0g CO2 = 52.8g CO2

                         5 mol O2       1 mol CO2

1.35 x 10^-6 g of C3H8 is extracted from a gas-bearing rock. How many molecules of CO2 are produced if the gas sample is burned in the presence of excess O2?

1.35 x 10^-6gC3H8 x 1mol C3H8 x 3molCO2 x 6.022 x 10^23 moleculesCO2= 5.54 x 10^16 molecules

                               44.0g C3H8   1mol C3H8                 1 mol CO2

Here's a quiz for some extra practice! :D

A quest to perfecting your stoichiometry calculating skills!