# Chapter 12 Stoichiometry

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Chapter 12 Stoichiometry
Mr. Mole

Molar Mass of Compounds
Molar mass (MM) of a compound - determined by adding up the atomic masses of each element Ex. Molar mass of CaCl2 MM of Calcium = g/mol MM of Chlorine = 35.45g g/mol x 2 = 70.9 g/mol Molar Mass of CaCl2= g/mol Ca g/mol Cl = g/mol CaCl2 20 Ca  17 Cl 35.45

Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 1023
Multiply by 6.02 X 1023 Moles Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table Mass (grams)

Practice Calculate the Molar Mass of calcium phosphide Formula =
Masses elements: Ca = g/mol * 3 P = g/mol * 2 Molar Mass = g/mol Ca3P2

Calculations molar mass Avogadro’s number Grams Moles particles 22.4 L Volume Everything must go through Moles!!!

1 cup butter 1 cup packed brown sugar 2 eggs 2 1/2 cups all-purpose flour 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How many eggs to make 9 dozen cookies? How much brown sugar would I need if I had 4 eggs? 2 6 2

Just like chocolate chip cookies have recipes, chemists have recipes called equations Instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use compounds

Chemistry Recipes Balanced reactions tell us how much reactant will react to get a product like the cookie recipe Be sure you have a balanced equation before you start! Ex: 2 Na + Cl2  2 NaCl Reaction tells us by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H2 + O2  2 H2O How many moles of reactants are needed? What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much hydrogen would we need to react, and how much water would we get? What if we had 50 moles of hydrogen, how much oxygen would we need, and how much water produced? 2 H2, 1 O2 4 H2, 2 O2 6 H2, 6 H2O 25 O2, 50 H2O

Stoichiometry is… Greek for “measuring elements”
Pronounced “stoy kee ah muh tree” Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. There are 4 ways to interpret a balanced chemical equation

#1. In terms of Particles An Element is made of atoms
A Molecular compound (made of only nonmetals) is made up of molecules Ionic Compounds (made of a metal and nonmetal parts) are made of formula units

Example: 2H2 + O2 → 2H2O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. Another example: 2Al2O3 ® 4Al + 3O2 2 formula units Al2O3 form 4 atoms Al and 3 molecules O2 Now read this: 2Na + 2H2O ® 2NaOH + H2

#2. In terms of Moles Coefficients tell us how many moles of each substance 2Al2O3 ® 4Al + 3O2 2 mol Al2O3, 4 mol Al, 3 mol O2 2Na + 2H2O ® 2NaOH + H2 Remember: A balanced equation is a Molar Ratio

#3. In terms of Mass Be + 2F  BeF2
The Law of Conservation of Mass applies We can check mass by using moles. Be + 2F  BeF2 1 mole Be 9.01 g Be = 9.01 g Be 1 mole Be + 19.00 g F 2 mole F 38.00 g F = 1 mole F 47.01 g Be + 2F 36.04 g H2 + O2 reactants

In terms of Mass (for products)
Be + 2F  BeF2 1 moles BeF2 47.01 g BeF2 = 47.01 g BeF2 1 mole BeF2 47.01 g Be + 2F = 47.01 g BeF2 reactant = product Mass of reactants and products. must be equal

67.2 Liters of reactant ≠ 44.8 Liters of product!
#4. In terms of Volume At STP, 1 mol of any gas = 22.4 L 2H O ® H2O (2 x 22.4 L H2) + (1 x 22.4 L O2) ® (2 x 22.4 L H2O) **Mass and atoms are ALWAYS conserved however, molecules, formula units, moles, and volumes will not necessarily be conserved! 67.2 Liters of reactant ≠ 44.8 Liters of product!

Practice: Show that the following equation follows the Law of Conservation of Mass (show the atoms balance, and the mass on both sides is equal) 2Al2O3 ® 4Al + 3O2 Atoms: 4 Al and 6 O = Al and 6 O Mass: Reactant: g/mol = Products: 108 g/mol g/mol

Practice Atoms: 2 N and 6 H = 2 N and 6 H Moles: 1 N2 and 3 H2 ≠ 2 NH3
1). Balance the equation and interpret it in terms of atoms, moles and mass. Show that the law of conservation is observed. N H2--> NH3 Atoms: 2 N and 6 H = N and 6 H Moles: 1 N2 and 3 H ≠ NH3 Mass: Reactants: g/mol g/mol = Products: g/mol

Section 12.2 Chemical Calculations

Mole to Mole conversions
2Al2O3 ® 4Al + 3O2 each time we use 2 moles of Al2O3 we will also make 3 moles of O2 2 moles Al2O3 3 mole O2 or 3 mole O2 2 moles Al2O3 This is how we can convert from mols to mols. This is why we need balanced equations.

Mole to Mole conversions
How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2Al2O3 ® 4Al + 3O2 3.34 mol Al2O3 3 mol O2 = 5.01 mol O2 2 mol Al2O3 Conversion factor from balanced equation If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals!

Practice: 2C2H2 + 5O2 ® 4CO2 + 2H2O 8.95 mol C2H2 9.6 mol O2
If 3.84 moles of C2H2 are burned, how many moles of O2 are needed? 3.84 moles C2H2 5 moles O2 9.6 mol O2 2 moles C2H2 How many moles of C2H2 are needed to produce 8.95 mole of H2O? 2 moles C2H2 8.95 moles H2O 8.95 mol C2H2 2 moles H2O

Steps to Calculate Stoichiometric Problems
1. Correctly balance the equation. 2. Convert the given amount into moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into final unit.

Mole-Mass Conversions
Most of the time in chemistry, the amounts are given in grams instead of moles We still use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl2  2 NaCl 5.00 moles Na 1 mol Cl g Cl2 2 mol Na 1 mol Cl2 = 177g Cl2

Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. Al + 3I  AlI3 g I 0.50 mol Al 3 mol I = g I 1 mol I 1 mol Al

We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water 2 C2H6 + 7 O2  4 CO2 + 6 H20 10.0 g H2O 1 mol H2O 2 mol C2H6 18.0 g H2O 6 mol H20 = mol C2H6

Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide 2Al + 3O  Al2O3 10.0 g Al2O3 1 mol Al2O3 3 mol O2 g Al2O3 1 mol Al2O3 0.29 g O2

Mass-Mass Problem: 4Al + 3O2  2Al2O3 12.3 g Al2O3 are formed
6.50 grams of aluminum reacts with oxygen. How many grams of aluminum oxide are formed? 4Al O2  2Al2O3 6.50 g Al 1 mol Al 2 mol Al2O3 g Al2O3 = ? g Al2O3 26.98 g Al 4 mol Al 1 mol Al2O3 12.3 g Al2O3 are formed (6.50 x 1 x 2 x ) ÷ (26.98 x 4 x 1) =

Another example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu 10.1 g Fe 1 mol Fe 3 mol Cu 63.55 g Cu 55.85 g Fe 2 mol Fe 1 mol Cu Answer = 17.2 g Cu

Volume-Volume Calculations:
How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ? CH4 + 2O2 ® CO2 + 2H2O 22.4 L O2 1 mol O2 1 mol CH4 1 mol CH4 22.4 L CH4 1 mol O2 22.4 L CH4 17.5 L O2 22.4 L O2 2 mol O2 1 mol CH4 = 8.75 L CH4

Stoichiometry Song - Mark Rosengarten

Section 12.3 Limiting Reagent & Percent Yield
OBJECTIVES: Identify the limiting reagent in a reaction.

“Limiting” Reagent If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make? The limiting reagent is the reactant you run out of first. The excess reagent is the one you have left over. The limiting reagent determines how much product you can make

How do you find out which is limited?
The chemical that makes the least amount of product is the “limiting reagent”. Limiting reagent problems will give you 2 amounts of chemicals You must do two stoichiometry problems; one for each reagent given. If 2 products are given, pick one and use it for both calculations

Cu is the Limiting Reagent, since it produced less product.
If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) will be formed? 2Cu + S ® Cu2S Cu is the Limiting Reagent, since it produced less product. 1 mol Cu 1 mol Cu2S g Cu2S 10.6 g Cu 63.55g Cu 2 mol Cu 1 mol Cu2S = 13.3 g Cu2S 1 mol S 1 mol Cu2S g Cu2S 3.83 g S 32.06g S 1 mol S 1 mol Cu2S = 19.0 g Cu2S

Finding the Amount of Excess
By calculating the amount of the reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the next problem?

Finding Excess Practice
15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2  2 KI We found that Iodine is the limiting reactant. 15.0 g I mol I mol K g K 254 g I mol I mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

Another example: 2Al + 3CuSO4 → 3Cu + Al2(SO4)3
If 10.3 g of aluminum are reacted with 51.7 g of CuSO4, how much copper (grams) will be produced? 2Al + 3CuSO4 → 3Cu + Al2(SO4)3 CuSO4 limiting reactant 10.3 g Al 1 mol Al 3 mol Cu 63.55 g Cu = g Cu 26.98 g Al 2 mol Al 1 mol Cu 51.7 g CuSO4 1 mol CuSO4 3 mol Cu 63.55 g Cu g CuSO4 3 mol CuSO4 1 mol Cu = g Cu

How much excess reactant do we have from the last problem?
CuSO4 limiting reactant Excess Al = g – 5.83 g = 51.7 g CuSO4 1 mol CuSO4 26.98 g Al 2 mol Al g CuSO4 3 mol CuSO4 1 mol Al = 5.83 g Al actually used 4.47 g Al excess

Limiting Reactant: Recap
You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. Convert ALL of the reactants to the SAME product (pick any product you choose.) The lowest answer is the limiting reactant The other reactant(s) are in EXCESS. To find the amount of excess, subtract the amount used from the given amount. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

The Concept of: A little different type of yield than you had in Driver’s Education class.

What is Yield? Yield is the amount of product made in a chemical reaction. There are three types: 1. Theoretical yield- what the balanced equation tells should be made 2. Actual yield- what you actually get in the lab when the chemicals are mixed 3. Percent yield = Actual Theoretical x 100%

Example: 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu What is the actual yield? What is the theoretical yield? What is the percent yield? = 6.78 g Cu 3.92 g Al 1 mol Al 3 mol Cu 63.55 mol Cu = 13.8 g Cu 26.98 g Al 2 mol Al 1 mol Cu 6.78 g Cu X 100 = 49.1 % 13.8 g Cu

Details on Yield Percent yield tells us how “efficient” a reaction is.
Percent yield can not be bigger than 100 %. Theoretical yield will always be larger than actual yield! Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring