12.1 STOICHIOMETRY Deals with amounts of reactants

used & products formed.

What is Stoichiometry?

• The study of the relationship between amounts of reactants used and products formed in a chemical reaction

• Based on the Law of Conservation of Mass and Matter– Matter is neither created nor destroyed– Mass of reactants equals the mass of the

products

MOLE-MASS RELATIONSHIPS

• Balanced equation: 4Fe (s) + 3O2(g) 2Fe2O3(s)

• Interpret: 4 atoms 3 molecules 2 formula units (Particles)

• Mole ratio: 4 moles 3 moles 2 moles (Coefficients)

Note: Particles of ionic compounds are called “formula units”

Calculate the mass of each reactant and product by multiplying the number of moles by the molar mass

SHOW MASS IS CONSERVED4Fe (s) + 3O2(g) 2Fe2O3(s)

• Mass reactants:

• 4 mol Fe (55.8g Fe) = 223.2 g Fe

(1 mole Fe)

• 3 mol O2 (32.00 g O2) = + 96.0 g O2

(1 mol O2) ____________

• 319.2 g Total

MASS OF PRODUCTS:4Fe (s) + 3O2(g) 2Fe2O3(s)

• 2 mol Fe2O3 (159.6 g Fe2O3) = 319.2 g (1 mol Fe2O3)

• Equals the mass of the reactants

(319.2g)

• Law of Conservation of Matter

PROBLEM

• Interpret in terms of particles, moles, and mass. Show that mass is conserved:

• (Hint: look at coefficients for particles & moles)

4 Al + 3O2 2 Al2O3

• Particles:

• Moles:

• Mass:

• Conserved?

SOLUTION: 4Al + 3O2 2 Al2O3

• Particles: 4 molecules 3 molecule 2 molecule

• Moles: 4 mole 3 mole 2 moles

• Mass: 4 (27.0 g) + 3 (26.0 g) = 2 (102.0 g)

• Conserved? 204.0 g = 204.0 g YES!

• Law of Conservation of Matter shown.

MOLE RATIO – is a ratio between the number of moles of any two substances in a balanced

chemical equation

• There are six mole ratios for the following:

• Ex. 4 Al(s) + 3 O2(g) 2 Al2O3(s)

• Note: 4 moles 3 moles 2 moles

• 4 mol Al and 3mol O2

3 mol O2 4 mol Al

MOLE RATIO – is a ratio between the number of moles of any two substances in a balanced

chemical equation

• 4 mol Al and 2 mol Al2O3

2 mol Al2O3 4 mol Al

• 3 mol O2 and 2 mol Al2O3

2 mol Al2O3 3 mol O2

• All stoichiometry calculations begin with a balanced equation and mole ratios!!

PROBLEM - FIND MOLE RATIOS FOR:

• 2 NH3 N2 + 3 H2

• 2 mol 1 mol 3 mol

• Ans. 2 mol NH3 or 1 mol N2

1 mol N2 2 mol NH3

• 2 mol NH3 or 3 mol H2

3 mol H2 2 mol NH3

Ans. 2 NH3 N2 + 3 H2

• 1 mole N2 or 3 mole H2

3 mole H2 1 mole N2

12.2 STOICHIOMETRIC CALCULATIONS

There are 3 Basic Stoichiometry Calculations

1. Mole to Mole Conversions

A piece of magnesium burns in the presence

of oxygen forming magnesium oxide (MgO).

How many moles of oxygen are needed to produce 12 moles of magnesium oxide?

Step 1: Write a balanced equation• 2 Mg (s) + O2 (g) 2 MgO (s)

Write mole ratios

2

2

1 mol O 2 mol Mgand

2 mol Mg 1 mol O

Choose the correct mole ratio needed for this problem

21 mol O

2 mol MgO

Mole to Mole Conversion cont’d

Multiply the known number of moles of MgO by the mole ratio

6 mols of oxygen is needed to produce 12 mols of magnesium oxide

22

12 mol MgO 1 mol O= 6 mol O

2 mol MgO

2. Mole to Mass Conversions

• The following reaction occurs in plants undergoing photosynthesis

• CO2(g) + H2O(l) C6H12O6(s)+ O2(g)

• How many grams of glucose (C6H12O6) are produced when 24.0 mols CO2 reacts in excess water?

Mole to Mass Conversion cont’d Write a balanced equation

• 6 CO2(g) + 6 H2O(l) C6H12O6(s) + 6 O2(g)

Use mole ratios to determine the number of moles of glucose produced by the given amount of carbon dioxide

6 12 626 12 6

2

1 mol C H O24 mol CO4.00 mol C H O

6 mol CO

Mole to Mass Conversion cont’d

Multiply by the molar mass

721 g glucose is produced from 24.0 moles carbon dioxide

6 12 6 6 12 66 12 6

6 12 6

4 mol C H O 180.2g C H O721g C H O

1 mol C H O

3. Mass to Mass ProblemsThe only new step is first step:

Convert grams of given substance to moles!

Massgiven Molesgiv Molesdesired Massdes

• ÷ molar massgiven X mole ratio X molar massdesired

• 3 step problem!

Label: 25.0 g ? g

NH4NO3 N2O + 2H2O Mole Ratio: 1 mol 1 mol 2 mol

• Ammonium nitrate (NH4NO3) produces N2O gas and H2O when it decomposes. Determine the mass of water produced from the decomposition of 25.0 g of NH4NO3.

• 1) Find moles NH4NO3 (molar mass): 80.04 g/mol

• Use the inverse of the molar mass to convert grams of NH4NO3 to moles of NH4NO3

4 3 4 3

4 3

25.0g NH NO 1 mol NH NO 0.312 mol

80.04g NH NO

Label: 25.0 g ? g

NH4NO3 N2O + 2H2OMole ratio: 1 mol 1 mol 2 mol

• Determine the mole ratio of mol H2O to mol NH4NO3 from the chemical equation. The desired substance is the numerator.

• Multiply mol NH4NO3 by the mole ratio.

• Calculate the mass of H2O using the molar mass.

2

4 3

2 mol H O

1 mol NH NO

4 3 22

4 3

0.312 mol NH NO 2 mol H O = 0.624 mol H O

1 mol NH NO

2 22

2

0.624 mol H O 18.02g H O = 11.2 g H O

1 mol H O

Label: 5.6g ? g

PROBLEM: N2 + 3H2 2 NH3

• If 5.6 g nitrogen reacts completely with hydrogen, what mass of ammonia is formed?

• 1) Find moles of nitrogen (given) – molar mass:

• 5.6 g N2|_____ mass N2 = 2 x 14.0 = 28.0 g/mol

• 1 | g N2

• 5.6 mol N2|_1mol N2

• 1 | 28.0 g N2

Label: 5.6 g ? g N2 + 3H2 2 NH3

Mole Ratio: 1mol 3mol 2mol

• 2) Find moles ammonia (desired) - mole ratio:

• 5.6 g N2 |1mol N2 |_2 mol NH3___• 1 | 28 g N2 | 1 mol N2

• 3) Find grams ammonia (desired) – • molar mass: NH3 = 14 + 3 = 17 g/mole

• 5.6 g N2 |1mol N2|_2 mol NH3 | 17.0 g NH3

• 1 | 28g N2 | 1 mol N2 | 1 mol NH3

• given ÷ MM given mole ratio x MM desired

• = 6.8 g NH3

• Mass H2 = 6.8 g – 5.6 g = 1.2 g

LIMITING REACTANTS

Limiting and Excess Reactants• When a chemical reaction occurs, the reactants

are not always present in the exact ratio indicated by the balanced equation.

• What usually happens is that a chemical reaction will run until the reactant that is in short supply is used up.

Which reactant will be used up first?

What is a Limiting Reactant?

• A limiting reactant is the reactant that limits (stops) a reaction and determines the amount of product

• An excess reactant is any reactant that is left over after the reaction stops (all reactants except the limiting reactant)

IN ORDER FOR CHEMICAL REACTION TO OCCUR, YOU MUST HAVE A COMPLETE

SET OF REACTANTS (REAGENTS):

• You don’t always have the exact amounts.• Ex. Let’s make some McBurgers!!!

• Ingredients:– 2 buns; 1 beef patty; 1cheese

slice; 1 tomato slice; 1 lettuce leaf; 3 pickles

YOU HAVE AVAILABLE:

• 6 buns• 3 burger patties• 5 cheese slices• 6 tomato slices• 5 lettuce leafs• 6 pickles

RECIPE CALLS

FOR: 2 buns

1 burger patty

1 cheese slice

1 tomato slice

1 lettuce leaf

3 pickles

How many McBurgers can you make?

Based on the individual ingredients you have available:

You have enough:• Buns for (3)

• Burger Patties for (3)• Cheese Slices for (5)• Tomato Slices for (6)• Lettuce Leafs for (5)• Pickles for (2)

The Limiting Reactant (LR) for the McBurger making process in the _______!

• This is the ingredient we ran out of first and were unable to continue with the McBurger making process

• We were only able to make 2 McBurger from the 6 pickles we had available to use–(Remember each McBurger requires

3 pickles)

pickles!

Excess reactants (XR) are the ingredients not used in the McBurger making

process:• 2 buns

• 1 burger patty

• 3 cheese slices

• 4 tomato slices

• 3 lettuce leafs

• Note: You made 2 McBurgers

N2 + 3H2 2 NH3

Using the above equation you are given 3.0 mols N2 and 5.0 mols H2

• Determine the limiting reactant• Determine the excess reactant

• Determine the amount of NH3 produced

PRACTICE PROBLEM:

• Draw 2 dimensional analysis problems. Smaller product is your answer. (Always use L.R. to find answer!)

3.0 mol N2 x 2 mol NH3 = 6.0 mol NH3

1 mol N2

5.0 mol H2 x 2 mol NH3 = 3.3 mol NH3

3 mol H2 .

Label: 3.0mol 5.0 mol ?mol N2 + 3H2 2 NH3

M.R. 1 mol 3 mol 2 mol

ANSWER: 3.3 mol ammonia formed. Hydrogen (H2) is L.R. Nitrogen (N2) is the XS.

IF A PAPER BURNS IN A ROOM:

• What is the limiting reactant?

• What is in excess?

• What would happen if the paper burned in a closed jar?

• LR?• XS?

+ +

+ + +

4.0mol 7.0mol

PROBLEM: 2 Al + 3 Cl2 2 AlCl3

• Given 4.0 mol Al and 7.0 mol Cl2, what is the maximum amount of aluminum chloride formed?

• Step 1: 2 set ups:

• 4.0 mol Al _____________ mol Al

• 7.0 mol Cl2 ________

mol Cl2

Label: 4.0mol 7.0 mol ? Mol

2 Al + 3 Cl2 2 AlCl3

• 4.0 mol Al 2 mol AlCl3 = 4.0 mol AlCl3

2 mol Al

• 7.0 mol Cl2 2 mol AlCl3 = 4.7 mol AlCl3 3 mol Cl2

• Which is the limiting reactant?• *Al is L.R. Ans. 4.0 mol AlCl3

• *Cl2 is XS

THE END