Stoichiometry is the study of quantitative relationships in chemical formulas and reactions. It allows chemists to calculate amounts of reactants and products using moles, molar mass, and balanced chemical equations. Chemical reactions have fixed ratios of reactants and products that can be used to determine limiting reactants and calculate mass changes. Sample problems demonstrate how stoichiometric calculations are used to find amounts of substances involved in chemical reactions.

1. Stoichiometry

2. What is
Stoichiometry?

3. Stoichiometry
Stoicheion [Gr. “element” or “part”]
Metron [Gr. “measure”]
Study of quantitative aspects of
chemical formulas and reactions

4. Cooking/Baking

5. Recipes
+

6. Recipes
+
Meat + Boiling Water =
Stock

7. Recipes
+
Water + Sugar = Soft
Candy

8. Recipes
+
Flour + Eggs + Milk + Sugar =
Cake

9. Similarly…
• Chemical reactions have
ingredients (reactants) and
finished products.
• Some are simple while
others are complex.
• Both obey the Law of
Conservation of Mass.
(What you put in is what you
get)

10. ChemicalReactions
In chemical reactions, we use moles and
molar mass as our measurement.
Moles to measure amount of substance.
Molar mass to measure the ratio of mass
and amount of substance
Chemical reactions are expressed as
chemical equations

11. ChemicalReactions
Chemical reactions are expressed in
chemical equations (our “recipe”)
2 Mg(s)
+
O2(g
)
--- 2
MgO(s)
Indicates the state of matter;
solid (s), liquid (l), aqueous
solution (aq), gas (g)

12. ChemicalReactions
Chemical reactions are expressed in
chemical equations (our “recipe”)
2 Mg(s)
+
O2(g
)
--- 2
MgO(s)
coefficient
Both sides of the equation must have
equal
amounts of atoms.
Mg: 2
atoms
O: 2 atoms
Mg: 2
atoms
O: 2 atoms

13. Measurement
in Chemical
Reactions

14. Measurement
In baking, we measure ingredients
using measuring cups
Instead, in chemistry, we measure reactants
and products in moles

15. TheMole(mol)
SI unit for amount of substance
Defined as the amount of a substance that
contains the same number of entities as there
are atoms in exactly 12 grams of carbon-12.
That is
6.022 x1023 [called Avogrado’s number]
∴1 mol of carbon-12 contains 6.022
x1023 carbon-12 atoms
1 mol of H2O contains 6.022 x1023 H2O
molecules

16. MolarMass(M)
Ratio of amount of substance (mol) and
mass (g)
Expressed in grams per mol (g/mol)
Molar mass (M)of element: listed in
the Periodic Table
Of compound: sum of the molar masses of
the atoms of the elements in the
compound

17. MolarMass(M)
MM of O2 = 32
g/mol O = 16 g/mol
16x2 = 32
MM of CO2 =
44.01 g/mol

18. Problems
Related
toChemical
Reactions

19. ProblemsRelatedtoChemReactions
Balancing reactions
Converting moles to
grams
Finding the limiting (and excess)
reactant

20. Balancing
N2O5 --- NO2 +
O2

21. Balancing
2
N2O5
--- 4 NO2
+
1
O2

22. Balancing
C8H1
8
+ O2 --- CO2 +
H2O

23. Balancing
2
C8H18
+ 25
O2
---16CO2 +18
H2O

24. Balancing
Cu
(s)
+ S8(
s)
---

Cu2S(
s)

25. Balancing
16 Cu (s) + 1
S8(s)
--- 8
Cu2S(s)

26. Conversio
n
How many moles are there in 7.00 grams of
Cu?
7 g Cu x mol Cu/63.55 g Cu = 0.110 moles

27. Conversio
n
How many grams are there in of 1 mol
O2?
32 grams

28. ReactionsInvolvingaLimiting
Reactant
+

29. ReactionsInvolvingaLimiting
Reactant
+

30. Sample
Problem
A fuel mixture is composed of two liquids,
hydrazine (N2H4) and dinitrogen tetraoxide
(N2O4), which ignite on contact to form
nitrogen gas and water vapour. How many
grams form when 1.00 x102 g of N2H4 and
2.00 x102 g of N2O4 are mixed?
MM of N = 14.01
g/mol MM of H =
1.01 g/mol MM of O
= 16 g/mol

31. Sample
Problem
• Plan:
– Write the balanced
equation
– 2 N2H4 (l) +
N2O4(l)
--- 3 N2(g) + 4
H2O(g)
– To determine the limiting reagent/reactant, we
calculate the mass of N2 from each reactant
assuming an excess of the other. Whichever
yields less N2 is the limiting reactant.

32. Sample
Problem
• 2 N2H4 (l) +
N2O4(l)
--- 3 N2(g) + 4
H2O(g)
• Assuming N2H4 is limiting, 4.68 mol
N2 is produced
• Assuming N2O4 is limiting, 6.51 mol
N2 is produced
• ∴N2H4 is the limiting reactant

33. Sample
Problem
• If N2H4 is the limiting reactant, 131 g
N2 is produced

34. Sample
Problem
Nuclear engineers use chlorine trifluoride in
the processing of uranium fuel for power
plants. This extremely reactive substance is
formed as a gas in special metal containers
by the reaction of elemental chlorine and
fluorine.
With 0.750 mol of Cl2 and 3.00 mol of F2,
what mass of chlorine trifluoride will be
prepared?
MM Cl = 35.45
g/mol MM F =

35. Sample
Problem
Cl2(g) + 3F2(g) --> 2
ClF3(g)
Cl2 is limiting
Mass (g) of ClF3 =
139 g

37. Sources
• Dragon Cave Holiday Cooking 2012. [Photos]
• Silberberg, Martin S. “Stoichiometry of Formulas and
Equations”. Principles of General Chemistry.
McGraw-Hill, 2010.