1. Stoichiometry Chapter 9

2. Balanced Equations  Coefficients tell you how many times that particular molecule is needed in a reaction  Subscripts tell you the number of times that atom appears in a particular compound  So when reading the following equation: C 3 H 8 + 5 O 2 -> 3 CO 2 + 4 H 2 O You would say 1 molecule of C 3 H 8 reacts with 5 molecules of oxygen gas to produce 3 molecules of carbon dioxide and 4 molecules of water

3. Equations are like receipes  The reactants are your ingredients, the products are what you make  The coefficients tell you how much of each ingredient you need  Ex. 1 cup chocolate chips + 2 eggs + 3 cups sugar + 4 cups flour -> 1 cake + 3 cookies  You can compare weights of ingredients because an egg weighs a different amount than a chocolate chip, you can only compare the relative amounts

4. The mole revisited  1 mole = 6.023 x 1023 atoms (or molecules)  1 mole = 1 atomic mass (or molar mass)  You can only compare 2 parts of an equations by looking at the number of moles because any 2 compounds don’t weigh the same amount (ex. 3 Mrs. Jordans aren’t the same weight as 3 chemistry books) –Just because they are present in the same number doesn’t make the amount of grams equal

5. Mole to Mole Conversions  The coefficients in a balanced equation let you compare the amount of 2 different compounds.  Ex. How many cups of chocolate chips would be needed to make 2 cakes?  Ex. How many cups of chocolate chips would be needed to use up 6 eggs?  Ex. How many moles of CO 2 are made from 10 moles O 2 ?

6. Mole to Mole Conversions, Cont’d  2 H 2 O -> 2 H 2 + O 2  How much O 2 will be produced from 4.5 moles of water? 4.5 mol H 2 O 1 mol O 2 = 2.25 mol O 2 4.5 mol H 2 O 1 mol O 2 = 2.25 mol O 2 2 mole H 2 O 2 mole H 2 O

7. Mass to Mass Conversions  You must first convert grams to moles, using the molar mass of the starting compound  Then switch compounds, using the coefficients in the balanced equation  Then convert moles to grams for the new compound, using the molar mass of the new compound

8. Mass to Mass, An Example  ___C 3 H 8 + ___ O 2 -> ___ CO 2 + ___ H 2 O How many grams of O2 are needed to react fully with 10 grams of C 3 H 8 –C 3 H 8 + 5 O 2 -> 3 CO 2 + 4 H 2 O –10 g C 3 H 8 1 mol C 3 H 8 5 mol O 2 32 g O 2 44 g C 3 H 8 1 mol C 3 H 8 1 mol O 2 = 36.26 g O 2

9. Limiting Reactants  If you don’t have enough of all the ingredients to keep going, you’ll have to stop when you run out of the very first ingredient  Ex. If you have 4 dozen eggs, but only 4 cups of flour, you can only make 1 cake, even though there are plenty of eggs left  We call this thing the limiting reactant  In order to solve these problems, you need to do 2 stoichiometry problems and the smaller answer is correct (because at that point you run out of one of your ingredients)

10. Limiting Reactants, An Example  N 2 + 3 H 2 -> 2 NH 3 How much NH 3 can be made from 2 grams of N 2 and 2 grams of H 2 ? 2 g N 2 1 mol N 2 2 mol NH 3 = 0.143 2 g N 2 1 mol N 2 2 mol NH 3 = 0.143 28 g N 2 1 mol N 2 mol NH 3 28 g N 2 1 mol N 2 mol NH 3 2 g H 2 1 mol H 2 2 mol NH 3 = 0.660 2 g H 2 3 mol H 2 mol NH 3 2 g H 2 3 mol H 2 mol NH 3 So N 2 is limiting and only 0.143 mol NH 3 can be made with some H 2 being left over.

11. Percent Yield  How much you actually get versus how much you would have expected –% yield = actual x 100 expected expected  A measure of how well your experiment did  You have to be told how much you actually got  You do a stoichiometry problem to figure out how much you would expect

12. Percent Yield, An Example  If you recover 7.57 g of CH 3 OH from the reaction of 8.6 g of H 2, what was your percent yield? 2 H 2 + CO -> CH 3 OH Since 7.57 g is what you actually got, we use 8.6 in the stoichiometry problem 8.6 g H 2 1 mol H 2 1 mol CH 3 OH 32 g CH 3 OH 2 g H 2 2 mol H 2 1 mol CH 3 OH = 68.8 g CH 3 OH So % yield = 7.57 x 100 = 11.002 % 68.8 68.8