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 –

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 = 40.08 g/mol– MM of Chlorine = 35.45g g/mol x 2 = 70.9 g/mol– Molar Mass of CaCl2=

40.08 g/mol Ca + 70.9 g/mol Cl = 110.98 g/mol CaCl2

20

Ca  40.08

17Cl

35.45

FlowchartAtoms or Molecules

Moles

Mass (grams)

Divide by 6.02 X 1023

Multiply by 6.02 X 1023

Multiply by atomic/molar mass from periodic table

Divide by atomic/molar mass from periodic table

Practice• Calculate the Molar Mass of calcium phosphide

– Formula = – Masses elements:Ca = 40.08 g/mol * 3P = 30.97 g/mol * 2

– Molar Mass = 182.18 g/mol

Ca3P2

Calculations molar mass Avogadro’s number

Grams Moles particles

22.4 L

Volume

Everything must go through Moles!!!

Chocolate Chip Cookies!!1 cup butter 1 cup packed brown sugar 2 eggs 2 1/2 cups all-purpose flour 2 cups semisweet chocolate

chipsMakes 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

Cookies and Chemistry…Huh!?!?• 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

• The Law of Conservation of Mass applies• We can check mass by using moles.

Be + 2F BeF2

1 mole Be 9.01 g Be

1 mole Be= 9.01 g Be

2 mole F 19.00 g F

1 mole F= 38.00 g F

36.04 g H2 + O247.01 g Be + 2F

+

reactants

In terms of Mass (for products)

Be + 2F BeF2

1 moles BeF247.01 g BeF2

1 mole BeF2

= 47.01 g BeF2

47.01 g Be + 2F = 47.01 g BeF2

Mass of reactants and products. must be equal

reactant = product

#4. In terms of Volume

• At STP, 1 mol of any gas = 22.4 L 2H2 + O2 ® 2H2O (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 = 4 Al and 6 O

Mass: Reactant: 312 g/mol = Products: 108 g/mol + 204 g/mol

Practice1). Balance the equation and interpret it in terms of

atoms, moles and mass. Show that the law of conservation is observed.

- N2 + 3 H2--> 2 NH3

Atoms: 2 N and 6 H = 2 N and 6 H

Moles: 1 N2 and 3 H2 ≠ 2 NH3

Mass: Reactants: 28.02 g/mol + 6.06 g/mol = Products: 34.08 g/mol

Section 12.2Chemical 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

or2 moles Al2O3

3 mole O2

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 + 3O23.34 mol Al2O3

2 mol Al2O3

3 mol O2 = 5.01 mol O2

If you know the amount of ANY chemical in the reaction,

you can find the amount of ALL the other chemicals!

Conversion factor from balanced equation

Practice: 2C2H2 + 5O2 ® 4CO2 + 2H2O• If 3.84 moles of C2H2 are burned, how many moles of O2 are

needed?

9.6 mol O2

•How many moles of C2H2 are needed to produce 8.95 mole of H2O?

8.95 mol C2H2

3.84 moles C2H2

2 moles C2H2

5 moles O2

8.95 moles H2O

2 moles H2O

2 moles C2H2

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 Cl2 70.90g 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

0.50 mol Al

1 mol Al

3 mol I

1 mol I

126.90 g I= 190.35 g I

Mass-Mole• We can also start with mass and convert to moles• 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

= 0.185 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 O21 mol Al2O3

101.96 g Al2O3

0.29 g O2

Mass-Mass Problem:

6.50 grams of aluminum reacts with oxygen. How many grams of aluminum oxide are formed?

4Al + 3O2 2Al2O3

=6.50 g Al

? g Al2O3

1 mol Al

26.98 g Al 4 mol Al

2 mol Al2O3

1 mol Al2O3

101.96 g Al2O3

(6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) = 12.3 g Al2O3

are formed

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

Answer = 17.2 g Cu

10.1 g Fe 1 mol Fe

55.85 g Fe 2 mol Fe

3 mol Cu

1 mol Cu

63.55 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

17.5 L O2 22.4 L O2 1 mol O2

2 mol O2

1 mol CH4

1 mol CH4 22.4 L CH4

= 8.75 L CH4

22.4 L O2 1 mol O2

1 mol CH4 22.4 L CH4

Stoichiometry Song - Mark Rosengarten

Section 12.3Limiting 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

• 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

10.6 g Cu 63.55g Cu 1 mol Cu

2 mol Cu 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 13.3 g Cu2S

3.83 g S 32.06g S 1 mol S

1 mol S 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 19.0 g Cu2S

Cu is the Limiting

Reagent, since it

produced less product.

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 I2 1 mol I2 2 mol K 39.1 g K

254 g I2 1 mol I2 1 mol K= 4.62 g K USED!

15.0 g K – 4.62 g K = 10.38 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:• 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 Al26.98 g Al1 mol Al

2 mol Al

3 mol Cu

1 mol Cu

63.55 g Cu= 36.39 g Cu

51.7 g CuSO4

159.62 g CuSO4

1 mol CuSO4

3 mol CuSO4

3 mol Cu1 mol Cu

63.55 g Cu

= 20.58 g Cu

How much excess reactant do we have from the last problem?

• CuSO4 limiting reactant

• Excess Al = 10.3 g – 5.83 g =

51.7 g CuSO4

159.62 g CuSO4

1 mol CuSO4

3 mol CuSO4

2 mol Al

1 mol Al

26.98 g Al

= 5.83 g Alactually used

4.47 g Al excess

Limiting Reactant: Recap1. You can recognize a limiting reactant problem

because there is MORE THAN ONE GIVEN AMOUNT.2. Convert ALL of the reactants to the SAME product

(pick any product you choose.)3. The lowest answer is the limiting reactant4. The other reactant(s) are in EXCESS.5. To find the amount of excess, subtract the amount

used from the given amount.6. 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 made2. Actual yield- what you actually get in the lab

when the chemicals are mixed3. Percent yield = Actual

Theoreticalx 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

= 13.8 g Cu

= 49.1 %

3.92 g Al

26.98 g Al1 mol Al

2 mol Al

3 mol Cu

1 mol Cu

63.55 mol Cu

6.78 g Cu

13.8 g CuX 100

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