1. 12.1 – What is Stoichiometry?
Objectives:
Describe stoichiometry calculations.
Write mole ratios from balanced equations.
Complete mole-mole stoichiometry problems.

2. 12.1 – Mole-Mass Relationships
Stoichiometry – study of quantitative relationships that exist in chemical formulas and/or chemical reactions.
Law of Conservation of Mass – Matter can neither be created nor destroyed in a chemical reaction
The mass of the reactants must equal the mass of the products

3. 12.1 – Using Mole Ratios
Mole Ratio - The relationship between moles of different substances in a balanced chemical equation.
2H2O  2H2 + 1O2
In this chemical reaction, the mole ratio of H2O to H2 is 2:2 (or simplifying 1:1). For each mole of H2O reacted, 1 mole H2 gas will be produced.
The mole ratio can be expressed as a conversion factor
The mole ratio of H2O to O2 is 2:1.
For each mole of H2O reacted, ½ mole of O2 will be produced.
2 mol H2O or
2 mol H2
2 mol H2O or
1 mol O2

4. 12.1 – Mole Ratios Examples: Given the unbalanced equation,
Sb(s) + Cl2(g)  SbCl3(s)
what is the mole ratio of Cl2 to SbCl3?
2Sb(s) + 3Cl2(g)  2SbCl3(s)
3:2
Pb(C2H3O2)2 (aq) + KI (aq)  PbI2(s) + KC2H3O2 (aq)
what is the mole ratio of Pb(C2H3O2)2 to KI?
1Pb(C2H3O2)2 (aq) + 2KI (aq)  1PbI2(s) + 2KC2H3O2 (aq)
1:2

5. 12.2 – What is Stoichiometry?
Objectives:
Use mole ratios to complete stoichiometric calculations.

6. 12.2 – Solving Stoichiometry Problems
The 4 Steps
Balance the equation.
Convert the amount of known substance into moles if necessary.
Convert the number of moles of known substance into moles of unknown substance using the mole ratio.
Convert the number of moles of the unknown substance into the desired units if necessary.
Amount given x
1 mol
moles of unknown
molar mass
6.02 x 1023 p.
22.4 L =
Amount of unknown
moles of known

7. 12.2 – Mole to Mole Problems The 4 Steps Balance the equation.
Convert the amount of known substance into moles if necessary.
Convert the number of moles of known substance into moles of unknown substance using the mole ratio.
Convert the number of moles of the unknown substance into the desired units if necessary.
moles known x
moles of unknown =
moles of known

8. 12.2 – Mole to Mole Problems Stoichiometry Conversions Mass (g) Mass
molar mass
mass x 1 mol
Mole Ratio Step
1 mol
moles x molar mass
molar 22.4 L
volume x 1 mol
1 mol
moles x L
Volume
(L)
Volume
(L)
moles x moles unknown
moles known
Moles of Known Substance
Moles of Unknown Substance
6.02 x part.
part. x 1 mol
moles x x 1023 part.
This ratio is derived from the coefficients of the balanced equation.
1 mol
Particles
(molecules, ions, atoms, formula units, etc.)
Particles
(molecules, ions, atoms, formula units, etc.)

9. 12.2 – Mole to Mass / Mass to Mole
The 4 Steps
Balance the equation.
Convert the amount of known substance into moles if necessary.
Convert the number of moles of known substance into moles of unknown substance using the mole ratio.
Convert the number of moles of the unknown substance into the desired units if necessary.
Mole to Mass
Mass to Mole
moles given x
moles of unknown
molar mass =
Amount of unknown
moles of known
1 mol
mass given x
1 mol
moles of unknown =
Amount of unknown
molar mass
moles of known

10. 12.2 – Mass to Particle / Particle to Mass
The 4 Steps
Balance the equation.
Convert the amount of known substance into moles if necessary.
Convert the number of moles of known substance into moles of unknown substance using the mole ratio.
Convert the number of moles of the unknown substance into the desired units if necessary.
Mass to Particles
Particles to Mass
mass given x
1 mol
moles of unknown
6.02 x 1023 p. =
particles of unknown
molar mass
moles of known
particles given x
1 mol
moles of unknown
molar mass =
mass of unknown
6.02 x 1023 p.
moles of known

11. 12.2 – Mass to Mass The 4 Steps Balance the equation.
Convert the amount of known substance into moles if necessary.
Convert the number of moles of known substance into moles of unknown substance using the mole ratio.
Convert the number of moles of the unknown substance into the desired units if necessary.
Mass to mass
mass given x
1 mol
moles of unknown
molar mass =
mass of unknown
moles of known

12. 12.3 – Limiting Reactants Objectives:
Describe how a limiting reactant affects a chemical reaction.
Determine the limiting reactant in a chemical reactions.
Calculate the mass of product for a reaction using limiting reactants.
Calculate the excess reactant remaining after a chemical reaction.

13. 12.3 – Limiting Reactants Why Do Reactions Stop?
Reactions usually stop because one reactant has run out.
Stoichiometric Proportion – The reactants are in exactly the right proportion so that one does not limit the other(s).
Limiting Reactant – The reactant that will be used up first in a chemical reaction.
The quantities of products in a chemical formed in a chemical reaction are determined by the amount of the limiting reactant.
All other reactants are considered to be “in excess” meaning some will remain when the reaction has stopped

14. 12.3 – Limiting Reactants Determining the Limiting Reactant
To find the limiting reactant, carry out the stoichiometric calculations with each reactant.
The calculation that provides the least product is based on the limiting reactant
Example: H2 reacts with O2 to form H2O. If 50. grams of H2 reacts with 50. grams of O2, what mass of water will be produced?
2H2 + O2 → 2H2O
50. g
50. g
? g
50. g H2 x
1 mol H2
2 mol H2O
g H2O =
450 g H2O
2.016 g H2
2 mol H2
1 mol H2O
50. g O2 x
1 mol O2
2 mol H2O
g H2O =
56 g H2O
g O2
1 mol H2O

15. 12.4 – Percent Yield Objectives:
Differentiate between theoretical yield and actual yield.
Calculate the percent yield for a reaction given the theoretical yield.

16. 12.4 – Percent Yield
Yields
Theoretical yield – the maximum amount of a product that can be produced based on calculation (also called calculated yield or mathematical yield)
Actual yield – the amount of a product that is actually produced through experiment (also called laboratory yield, experimental yield, or real yield).
Percent yield – the % of the expected yield that was actually produced.
percent yield =
actual yield
x 100
theoretical yield