1. Chapter 12
Stoichiometry

2. 12.1 The Arithmetic of Equations
Objectives
Explain how balanced equations apply to both chemistry and everyday life
Interpret balanced chemical equations in terms of moles, representative particles, mass and gas volume at STP
Identify the quantities that are always conserved in chemical reactions

3. COOKIES! Who makes the best chocolate chip cookies? WHY?
When baking cookies, a recipe is usually used, telling the exact amount of each ingredient
If you need more, you can double or triple the amount of each ingredient
Thus, a recipe is much like a balanced equation!

4. Using Balanced Chemical Equations
We use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction
Stoichiometry
Greek for “measuring elements”
Defined as the calculation of quantities in chemical reactions based on a balanced equation

5. Using Balanced Chemical Equations
Stoichiometry allows chemists to tally the amounts of reactants and products using ratios of moles or representative particles
There are four ways to interpret a balanced chemical equation.
In terms of:
Particles
Moles
Mass
Volume

6. In Terms of Particles
At the atomic level, a balanced equation indicates that the number and type of each atom that makes up each reactant also makes up each product
Reminders
Element = atoms
Molecular compound = molecules
Ionic compound – formula units

7. In Terms of Particles 2H2 + O2  2H2O
Two molecules of hydrogen and one molecule of oxygen form two molecules of water
2Al2O3  4Al + 3O2
2 formula units of Al2O3 form 4 atoms of Al and 3 molecules of O2
Try this
2Na + 2H2O  2NaOH + H2

8. ** Most important information **
In Terms of Moles
A balanced chemical equation tells us the number of moles of reactants and products
The coefficients tell us how many moles of each substance
2Al2O3  4Al + 3O2
2Na + 2H2O  2NaOH + H2
A balanced equation is a Molar Ratio
** Most important information **

9. In Terms of Mass 2.02 g H2 2 moles H2 = 4.04 g H2 1 mole H2 32.00 g O2
A balanced chemical equation obeys the law of conservation of mass
Using mole relationship, we can relate mass to the number of atoms in a chemical equation
2.02 g H2
2 moles H2 =
4.04 g H2
1 mole H2 +
32.00 g O2
32.00 g O2
1 mole O2 =
1 mole O2
36.04 g H2 + O2
36.04 g H2 + O2

10. In Terms of Mass 2H2 + O2 ® 2H2O 2H2 + O2 ® 2H2O 18.02 g H2O =
2 moles H2O
1 mole H2O
2H2 + O2 ® 2H2O
36.04 g H2 + O2 =
36.04 g H2O
Although the number of moles of reactants does not equal the number of moles of product, the total number of grams of reactant does equal total number of grams of product

11. In Terms of Volume At STP, one mole of any gas = 22.4 L
2H O  H2O
(2 x 22.4 L H2) + (1 x 22.4 L O2)  (2 x 22.4 L H2O)
NOTE – Mass and atoms are always conserved – however, molecules, formula units, moles and volume will not necessarily be conserved

12. 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

13. Section 12.2 Chemical Calculations
Objectives:
Construct mole ratios from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations
Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP

14. Writing and Using Mole Ratios
Conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles
In chemical calculations, mole ratios are used to convert-
between moles of reactant and moles of product
between moles of reactants
between moles of products

15. Writing and Using Mole Ratios
N2(g) + 3H2(g)  2NH3(g)
3 mole ratios:
1 mole N2 2 mole NH3 3 mole H2
3 mole H2 1 mole N2 2 mole NH3

16. Mole to Mole Conversions
2 Al2O3  4Al + 3O2
each time we use 2 moles of Al2O3 we will also make 3 moles of O2
2 moles Al2O3
3 mole O2 or
3 mole O2
2 moles Al2O3
These are the two possible conversion factors

17. Mole to Mole Conversions
How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?
2 Al2O3  4Al + 3O2
3 mol O2
3.34 mol Al2O3 =
5.01 mol O2
2 mol Al2O3

18. Practice:
2C2H2 + 5 O2  4CO2 + 2 H2O
If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?
(9.6 mol)
How many moles of C2H2 are needed to produce 8.95 mole of H2O?
(8.95 mol)
If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?
(4.94 mol)

19. How do you get good at this?

20. Calculating Stoichiometric Problems
No laboratory balance can measure substances directly in moles
Instead the amount of a substance is usually determined by measuring its mass in grams
Using the molar mass of the substances we can convert grams to moles and then back to grams

21. Calculating Stoichiometric Problems
Balance the equation
Convert mass in grams to moles
Set up mole ratios
Use mole ratios to calculate moles of desired chemicals
Convert moles back into grams, if necessary

22. Mass-Mass Problem: 4 Al + 3 O2  2Al2O3 12.3 g Al2O3
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?
4 Al O2  2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3
g Al2O3
? g Al2O3 =
1 mol Al2O3
26.98 g Al
4 mol Al
12.3 g Al2O3
(6.50 x 2 x ) ÷ (26.98 x 4) =

23. Another example: Answer = 17.2 g Cu 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu
If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?
2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu
Answer = 17.2 g Cu

24. 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
22.4 L O2
1 mol O2
1 mol CH4
1 mol CH4
22.4 L CH4
1 mol O2
22.4 L CH4
17.5 L O2
22.4 L O2
2 mol O2
1 mol CH4
= 8.75 L CH4
Notice anything concerning these two steps?

25. Avogadro told us…
Equal volumes of gas, at the same temperature and pressure contain the same number of particles
Moles are numbers of particles
You can treat reactions as if they happen liters as a time, as long as you keep the temperature and pressure the same
1 mole = 22.4 STP

26. Shortcut for Volume-Volume:
How many liters of CH4 at STP are required to completely react with 17.5 L of O2?
CH4 + 2O2  CO2 + 2H2O
1 L CH4
17.5 L O2
= 8.75 L CH4
2 L O2
Note: This only works for Volume-Volume problems.

27. Other Stoichiometric Calculations
In a typical stoichiometric problem, the given quantity is -
first converted to moles
Then the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance
Finally, the moles are converted to any other unit of measurement(grams, number of particles, liters) related to the unit mole, as the problem requires
Remember your conversion factors from chap. 10
1 mole = molar mass (masses of elements from Periodic Table)
1 mole = 6.02 x 1023 representative particles
1 mole = 6.02 x 1023 molecules
1 mole = 6.02 x 1023 atoms
1 mole = 22.4 L