Quantities in Chemical Reactions

Stoichiometry (Greek):

Stoicheion = “element”Metron = “to measure”

Why are we learning this?

Stoichiometry can allow you to:• determine amount of product from known quantity of reactant• determine amount of starting material needed to yield desired quantity of product• identify reaction efficiency (and make amendments in reaction conditions)• determine purity of starting materials • and more

What is required to perform stoichiometric analysis?

Balanced chemical equations!

What is required to perform stoichiometric analysis?

Balanced chemical equations!

Like a recipe, reaction components must interact with one another in fixed proportions!

What information is conveyedIn balanced chemical equation?

• Numbers of particles• Numbers of moles• Mass of reaction components• Volumes (gasses)

Example: Production of gaseous ammonia by Haber process

Balanced Equation

N2(g) + 3 H2(g) → 2 NH3(g)

Particles(Atoms †)(Molecules *)

2 atoms N 1 molecule N2(g)

x NA

8 atoms

6 atoms H3 molecules H2(g) x NA

reactants

2 atoms N6 atoms H2 molecules NH3(g)

x NA

8 atoms products

Moles * 1 mole N2(g) 3 moles H2(g) 2 moles NH3(g)

Mass † 28.02 g N2(g)

34.08 g

3 x 2.02 g H2(g)

reactants2 x 17.04 g NH3(g)

34.08 g products

Volume * 1 volume 3 volumes 2 volumes† Total values are equal before and after the reaction.* Total values may or may not be equal before and after the reaction.

MOLES MOLESMoleRatio

MASS MASS

MOLECULES MOLECULES

÷ M (molar mass) x M (molar mass)

÷ NA (Avogadro’s constant)

x NA (Avogadro’s constant)

mole ratio – obtained from the balanced chemical equation

Overall Strategy for Solving Stoichiometric Problems

Relating particles in balanced equations

Example (Modelled after textbook, page 237, #2)

2 Al(s) + 3 Cl2(g) → 2 AlCl3(S)

Relating particles in balanced equations

Example (Modelled after textbook, page 237, #2)

2 Al(s) + 3 Cl2(g) → 2 AlCl3(S)

A) How many atoms of Al are needed to react with three molecules of Cl2?

Relating particles in balanced equations

Example (Modelled after textbook, page 237, #2)

2 Al(s) + 3 Cl2(g) → 2 AlCl3(S)

A) How many atoms of Al are needed to react with three molecules of Cl2?Answer: 2 Al atoms (directly from balanced equation)

Relating particles in balanced equations

Example (Modelled after textbook, page 237, #2)

2 Al(s) + 3 Cl2(g) → 2 AlCl3(S)

A) How many atoms of Al are needed to react with three molecules of Cl2?Answer: 2 Al atoms (directly from balanced equation)

B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2?

Relating particles in balanced equations

Example (Modelled after textbook, page 237, #2)

2 Al(s) + 3 Cl2(g) → 2 AlCl3(S)

A) How many atoms of Al are needed to react with three molecules of Cl2?Answer: 2 Al atoms (directly from balanced equation)

B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2?

Answer: 250 atoms Al x 2 molecules AlCl3 = 250 molecules AlCl3

2 atoms Al

Relating particles in balanced equations

Example (Modelled after textbook, page 237, #2)

2 Al(s) + 3 Cl2(g) → 2 AlCl3(S)

A) How many atoms of Al are needed to react with three molecules of Cl2?Answer: 2 Al atoms (directly from balanced equation)

B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2?

Answer: 250 atoms Al x 2 molecules AlCl3 = 250 molecules AlCl3

2 atoms Al

C) How many Cl2 molecules are needed to react with 1.834 x 1024 atoms of Al?

Relating particles in balanced equations

Example (Modelled after textbook, page 237, #2)

2 Al(s) + 3 Cl2(g) → 2 AlCl3(S) A) How many atoms of Al are needed to react with three molecules of Cl2?

Answer: 2 Al atoms (directly from balanced equation)B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2?

Answer: 250 atoms Al x 2 molecules AlCl3 = 250 molecules AlCl3

2 atoms AlC) How many Cl2 molecules are needed to react with 1.834 x 1024 atoms of Al?

Answer: 1.834 x 1024 atoms Al x 3 molecules Cl2 = 2.751 x 1024 molecules Cl2

2 atoms Al

Relating particles in balanced equations

Example (Modelled after textbook, page 237, #2)

2 Al(s) + 3 Cl2(g) → 2 AlCl3(S)

A) How many atoms of Al are needed to react with three molecules of Cl2?Answer: 2 Al atoms (directly from balanced equation)

B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2?

Answer: 250 atoms Al x 2 molecules AlCl3 = 250 molecules AlCl3

2 atoms AlC) How many Cl2 molecules are needed to react with 1.834 x 1024 atoms of Al?

Answer: 1.834 x 1024 atoms Al x 3 molecules Cl2 = 2.751 x 1024 molecules Cl2

2 atoms Al D) How many molecules of AlCl3 are formed when 2.751 x 1024 molecules of Cl2 react with sufficient Al?

Relating particles in balanced equations

Example (Modelled after textbook, page 237, #2)

2 Al(s) + 3 Cl2(g) → 2 AlCl3(S)

A) How many atoms of Al are needed to react with three molecules of Cl2?Answer: 2 Al atoms (directly from balanced equation)

B) How many molecules of AlCl3 are formed when 250 atoms of Al react with sufficient Cl2?

Answer: 250 atoms Al x 2 molecules AlCl3 = 250 molecules AlCl3

2 atoms AlC) How many Cl2 molecules are needed to react with 1.834 x 1024 atoms of Al?

Answer: 1.834 x 1024 atoms Al x 3 molecules Cl2 = 2.751 x 1024 molecules Cl2

2 atoms Al D) How many molecules of AlCl3 are formed when 2.751 x 1024 molecules of Cl2 react with sufficient Al?

Answer: 2.751 x 1024 molecules Cl2 x 2 molecules AlCl3= 1.834 x 1024 molecules AlCl3

3 molecules Cl2

In previous example, we saw that coefficients for reactants and products in balanced equation represented the number of particles (atoms or molecules) participating directly in chemicalreaction.

Equation coefficients representParticles and moles

But…

Equation coefficients representParticles and moles

Multiply ratio of coefficients by NA :

e.g. (2 atoms Al(s) : 3 molecules Cl2(g) : 2 molecules AlCl3(s)) x NA

2 NA Al(s) : 3 NA Cl2(g) : 2 NA AlCl3(s)

= 2 mol Al(s) : 3 mol Cl2(g) : 2 mol AlCl3(s)

Equation coefficients representParticles and moles

2 NA Al(s) : 3 NA Cl2(g) : 2 NA AlCl3(s)

= 2 mol Al(s) : 3 mol Cl2(g) : 2 mol AlCl3(s)

Reactants Total 5 mol

Products Total2 mol

Note: Because atoms are rearranged in different proportions, there is no law of conservation of moles.

Equation coefficients representParticles and moles

By analogy, imagine leaving 3 moles of puddles of water on the floor, and then using one mole of paper towel sheets to clean it up.

1 mol dry paper towel sheets

3 mol puddlesof water

1 mol wetpaper towel sheets

Think about product as 1 paper towel . 3 H2O puddles. Where have we seen such a formulation recently?

Reactants total 4 moles Products total 1 mole

MASS TO MASS STOICHIOMETRICCALCULATIONS

Most likely, situations will involve being given the mass (or moles, or particles) of one participant in a reaction, and being tasked with finding the mass of another reaction species.

Use knowledge of the molar mass of species present in the reaction to interconvert between mass and moles, using the molar ratio of species from the balanced chemical equation to identify the number of molesof the desired chemical entity.

Example #1: (Mass of one reactant to another)

2 NiO(OH)(s) + 2 H2O(l) + Cd(s) → 2 Ni(OH)2(s) + Cd(OH)2(s)

Find the mass of solid Cadmium required to fully react with 0.45 grams Nickel (III) oxide hydroxide in a rechargeable nickel-cadmium battery. The overall reaction is:

Example #1: (Mass of one reactant to another)

2 NiO(OH)(s) + 2 H2O(l) + Cd(s) → 2 Ni(OH)2(s) + Cd(OH)2(s)

Find the mass of solid Cadmium required to fully react with 0.45 grams Nickel (III) oxide hydroxide in a rechargeable nickel-cadmium battery. The overall reaction is:

0.45 g NiO(OH) x 1 mol NiO(OH) = 0.0049073 mol NiO(OH) 91.70 g NiO(OH)

Mass → Moles

Example #1: (Mass of one reactant to another)

2 NiO(OH)(s) + 2 H2O(l) + Cd(s) → 2 Ni(OH)2(s) + Cd(OH)2(s)

Find the mass of solid Cadmium required to fully react with 0.45 grams Nickel (III) oxide hydroxide in a rechargeable nickel-cadmium battery. The overall reaction is:

0.45 g NiO(OH) x 1 mol NiO(OH) = 0.0049073 mol NiO(OH) 91.70 g NiO(OH)

0.0049073 mol NiO(OH) x 1 mol Cd = 0.0024537 mol Cd 2 mol NiO(OH)

Mass → Moles

Moles → Moles

Example #1: (Mass of one reactant to another)

2 NiO(OH)(s) + 2 H2O(l) + Cd(s) → 2 Ni(OH)2(s) + Cd(OH)2(s)

Find the mass of solid Cadmium required to fully react with 0.450 grams Nickel (III) oxide hydroxide in a rechargeable nickel-cadmium battery. The overall reaction is:

0.450 g NiO(OH) x 1 mol NiO(OH) = 0.0049073 mol NiO(OH) 91.70 g NiO(OH)

0.0049073 mol NiO(OH) x 1 mol Cd = 0.0024537 mol Cd 2 mol NiO(OH)

0.0024537 mol Cd x 112.41 g Cd = 0.2758 g Cd(s) = 0.276 g Cd(s)

1 mol Cd

Mass → Moles

Moles → Moles

Moles → Mass

Example #2: (Mass of reactant to product)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4 H2O(g)

Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams.

Example #2: (Mass of reactant to product)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4 H2O(g)

Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams.

1.75 g (NH4)2Cr2O7 x 1 mol (NH4)2Cr2O7 = 0.0069417 mol (NH4)2Cr2O7

252.10 g (NH4)2Cr2O7 Mass → Moles

Example #2: (Mass of reactant to product)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4 H2O(g)

Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams.

1.75 g (NH4)2Cr2O7 x 1 mol (NH4)2Cr2O7 = 0.0069417 mol (NH4)2Cr2O7

252.10 g (NH4)2Cr2O7

a) 0.0069417 mol (NH4)2Cr2O7 x 1 mol Cr2O3 = 0.0069417 mol Cr2O3

1 mol (NH4)2Cr2O7

Mass → Moles

Moles → Moles

Example #2: (Mass of reactant to product)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4 H2O(g)

Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams.

1.75 g (NH4)2Cr2O7 x 1 mol (NH4)2Cr2O7 = 0.0069417 mol (NH4)2Cr2O7

252.10 g (NH4)2Cr2O7

a) 0.0069417 mol (NH4)2Cr2O7 x 1 mol Cr2O3 = 0.0069417 mol Cr2O3

1 mol (NH4)2Cr2O7

0.0069417 mol Cr2O3 x 152 g Cr2O3 = 1.055 g Cr2O3(s) = 1.06 g Cr2O3(s)

1 mol Cr2O3

Mass → Moles

Moles → Moles

Moles → Mass

Example #2: (Mass of reactant to product)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4 H2O(g)

Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams.

1.75 g (NH4)2Cr2O7 x 1 mol (NH4)2Cr2O7 = 0.0069417 mol (NH4)2Cr2O7

252.10 g (NH4)2Cr2O7

b) 0.0069417 mol (NH4)2Cr2O7 x 1 mol N2 = 0.0069417 mol N2

1 mol (NH4)2Cr2O7

Mass → Moles

Moles → Moles

Example #2: (Mass of reactant to product)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4 H2O(g)

Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams.

1.75 g (NH4)2Cr2O7 x 1 mol (NH4)2Cr2O7 = 0.0069417 mol (NH4)2Cr2O7

252.10 g (NH4)2Cr2O7

b) 0.0069417 mol (NH4)2Cr2O7 x 1 mol N2 = 0.0069417 mol N2

1 mol (NH4)2Cr2O7

0.0069417 mol N2 x 28.02 g N2 = 0.1945 g N2(g) = 0.195 g N2(g)

1 mol N2

Mass → Moles

Moles → Moles

Moles → Mass

Example #2: (Mass of reactant to product)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4 H2O(g)

Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams.

1.75 g (NH4)2Cr2O7 x 1 mol (NH4)2Cr2O7 = 0.0069417 mol (NH4)2Cr2O7

252.10 g (NH4)2Cr2O7

c) 0.0069417 mol (NH4)2Cr2O7 x 4 mol H2O = 0.0277668 mol H2O

1 mol (NH4)2Cr2O7

Mass → Moles

Moles → Moles

Example #2: (Mass of reactant to product)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4 H2O(g)

Calculate the masses of all three products formed from the decomposition of ammonium dichromate. The initial mass of (NH4)2Cr2O7(s) is 1.75 grams.

1.75 g (NH4)2Cr2O7 x 1 mol (NH4)2Cr2O7 = 0.0069417 mol (NH4)2Cr2O7

252.10 g (NH4)2Cr2O7

c) 0.0069417 mol (NH4)2Cr2O7 x 4 mol H2O = 0.0277668 mol H2O

1 mol (NH4)2Cr2O7

Mass → Moles

Moles → Moles

0.0277668 mol H2O x 18.02 g H2O = 0.5004 g H2O(g) = 0.500 g H2O(g)

1 mol H2O

Moles → Mass