1. Stoichiometry
Chapter 12

2. 12.1: Arithmetic of Equations
A balanced equation is the recipe for a reaction.
Chemists use balanced equations as a basis to calculate how much reactant is needed or product is formed
The calculation of these quantities in chemical reactions is called STOICHIOMETRY.

3. Interpreting Chemical Equations
A balanced chemical equation can be interpreted in terms of many different quantities
Number of atoms
Molecules
Moles
Mass
Volume

4. Atoms/ Molecules # of Atoms # of Molecules
on the atomic level, a balanced equation indicates the number of and type of atoms that make up both the reactants and products
The number of atoms is not changed in the reaction.
# of Molecules
A balanced equation also establishes a ratio of how molecules interact
Look at the formation of ammonia
The ratio established is 1:3:2
1 molecule of atmospheric nitrogen reacts with 3 molecules of atmospheric hydrogen to form 2 molecules of ammonia

5. Moles
A balanced equations gives you the number of moles in both the reactants and products
We use the coefficients to establish this
This is the most important information that an equation can give us.
Using this we can calculate exactly how much of each reactant is needed to give a desired product
Due to recombination of elements, however, the # of moles in reactants does not always equal the # of moles in the products.

6. Mass/ Volume Mass Volume
Balanced equations always obey conservation of mass laws
This means the mass of reactants is always equal to mass of the products no matter is the # of moles changes
Volume
If standard pressure is assumed, then a balanced equation can also tell you the volume of products produced
Remember 1 moles = 22.4 Liters

7. Mass Conservation
Mass and Atoms are always conserved inside of a reaction
Volume, Molecules, and # of moles can change
This depends on the reaction

8. Homework
Pg. 358
Questions 5-10
Due tomorrow

9. 12.2 Chemical Calculations
Writing and Using Mole Ratios
Mole ratio – a conversion factor derived from the coefficients of a balanced chemical equation, interpreted in terms of moles.
Mole ratios can be used to calculate conversions between moles of products and reactants, just reactants, or just products

10. Mole to Mole Calculations
How many moles of ammonia are produced from moles of nitrogen??
0.60 mol N x 2 mol NH3/1 mole N = ??

11. Mass to Mass Calculations
No balance/scale can measure directly in moles; instead we work in the world of grams so being able to convert from moles to grams or grams to moles is very important
The process for converting from grams of reactant to grams of product follows this sequence
Grams to moles; moles to moles (using mole ratio from balanced equation); then finally moles to grams of product

12. Mass to Mass cont. Steps in Solving Mass to Mass problems
1. change mass of given to moles by using molar mass
2. change moles of given to moles of wanted using mole ratio
3. change moles of wanted to grams of wanted using molar mass of wanted.

13. Mass to Mass cont.
How many grams of NH3 are produced when 5.40 grams of hydrogen reacts with an excess of nitrogen
Remember our balanced equation: N2 + 3H2  2NH3
Take a shot at solving this follow the steps from the previous slide.

14. Practice Problems Page 361 in textbook #13 and #14
Try to solve these quickly.
13/14 equation: CaC2 + 2H2O  C2H2 + Ca(OH)2

15. Other Stoichiometric Calculations
We can also interconvert moles and grams into representative particles (molecules/atoms) and into volumes (assuming standard pressure)
Think back…….
What number did we use to convert to molecules/particles?
What number did we use to convert to STP?

16. Other Calculations cont.
Representative Particles
The conversion factor we use is AVOGADRO’S NUMBER
6.02 X 1023 = 1 MOLE
Practice #15
Volume
The conversion factor used for volume always assumes standard temperature and pressure.
1 mole = 22.4 Liters
Practice #17 and #19

17. Assignment 12. 2 Section Review Pg. 366 Questions 22-24
Will grade tomorrow

18. 12.3 Limiting Reagent
In any equation an insufficient quantity of any reactant will limit the amount of product produced
Limiting reagent – the reactant that is completely used up in a chemical reaction
Excess reagent – reactant that is not completely used up in a chemical reaction
There is a little bit of this left over.

19. Calculations of Limiting Reagent
2Cu + S  Cu2S
Given quantities:
80.0 grams Copper
25.0 grams Sulfur
Which is limiting?
How do we calculate this?

20. Solution 2Cu + S  Cu2S 80.0 g. Cu x 1mol/63.5gCu = 1.26 mol Cu
25.0 g. S x 1mol/32.1g.S = mol S
1.26 mol Cu x 1mol S/ 2mol Cu = mol S
Copper (Cu) is the limiting reagent
Practice Problems #25 and #26 on page 370

21. Limiting Reagent to find Qty of Product produced
Using same equation and quantities from previous problem: 2Cu + S  Cu2S
L.R. = 1.26 mol Cu
Solution:
1.26 mol Cu x 1mol Cu2S/ 2mol Cu x g. Cu2S/ 1mol Cu2S = 100grams Cu2S
Try Practice Problems #27 and #28 on pg. 371

22. Percent Yield
Percent yield measures the efficiency of a reaction, when carried out under laboratory conditions.
Theoretical vs Actual Yield
Theoretical yield – maximum amount of product that can be produced from given amounts of reactants
Actual yield – The amount of product produced when the reaction is performed

23. Percent Yield Percent Yield = Actual yield/ Theoretical yield x 100
Never is larger than 100%
Side reactions, impure reactants, and inaccurate measurement cause few, if any, reactions to run to 100% completion.

24. Example: Pg. 374 Sample 12.9 Practice Problems pg. 374 (29 and 30)
Given: mass of CaCO3 = 24.8 g
1 mol of CaCO3 = g
1 mol CaO = 56.1 g
Unknown: theoretical yield of CaO in grams
Equation: CaCO3  CaO + CO2
Practice Problems pg. 374 (29 and 30)

25. Example 2
Sample on pg. 375
Givens: actual yield of CaO = 13.1 grams
Theoretical yield of CaO = 13.9 grams
Unknown: Percent Yield of CaO
Practice Problems: pg. 375 (31 and 32)

26. Assignment
12.3 Review
Pg. 375
Questions 33-35