1. Chapter 3: Stiochiometry

2. Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2 ® 2CO H2O
reactants
products
When the equation is balanced it has quantitative significance:
1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

3. Balancing Chemical Equations
Law of Conservation of Matter
Must have an equal number of each type of atom on both sides of the equation, and thus an equal mass
Use coefficients to balance

4. Balancing Examples Na (s) + H2O (l)  NaOH (aq) + H2 (g)
2. C2H O2  CO H2O

5. Types of Reactions Combination (Synthesis) Decomposition Combustion
Rapid reaction that produces a flame
Use O2 as a reactant
When organic material combusts, it forms CO2(g) and H2O(l)
When inorganic material combusts, forms an oxide of whatever combusted

6. Combustion Examples
Write out the reaction for the combustion of lead (will form a lead(II) compound).
Write out the reaction for the combustion of ethane.

7. Calculating Formula Mass
Calculate the formula mass of magnesium carbonate, MgCO3.
24.31 g g + 3(16.00 g) =
84.32 g

8. Calculating Percentage Composition
Calculate the percentage composition of magnesium carbonate, MgCO3.
First find the formula mass (from previous slide):
24.31 g g + 3(16.00 g) = g
Then find the percentage composition
100.00

9. The Mole 1 dozen = 12 1 gross = 144 1 ream = 500 1 mole = 6.022 x 1023
There are exactly 12 grams of carbon-12 in one mole of carbon-12.

10. Avogadro’s Number
6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro ( ).
Amadeo Avogadro

11. Why are moles so special?
Moles are THE comparative unit in chemistry!
We do NOT compare mass, we do NOT compare volume, because those can change depending on substance!
BUT one MOLE of something is always the same… 6.02 x 1023 particles!

12. Calculations with Moles: Converting moles to grams
How many grams of lithium are in 3.50 moles of lithium?
3.50 mol Li
6.94 g Li
= g Li
45.1
1 mol Li

13. Calculations with Moles: Converting grams to moles
How many moles of lithium are in 18.2 grams of lithium?
18.2 g Li
1 mol Li
= mol Li
2.62
6.94 g Li

14. Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of lithium?
3.50 mol
6.02 x 1023 atoms
= atoms
2.07 x 1024
1 mol

15. Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li
1 mol Li
6.022 x 1023 atoms Li
6.94 g Li
1 mol Li
(18.2)(6.022 x 1023)/6.94
= atoms Li
1.58 x 1024

16. Formulas
Empirical formula: the lowest whole number ratio of atoms in a compound.
Molecular formula: the true number of atoms of each element in the formula of a compound.

17. Formulas (continued)
Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3

18. Formulas (continued)
Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11

19. Empirical Formula Determination
Base calculation on 100 grams of compound.
Convert grams into moles
Divide each value of moles by the smallest of the values.
Multiply each number by an integer to obtain all whole numbers. (if needed)

20. Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

21. Empirical Formula Determination (part 2)
Divide each value of moles by the smallest of the values.
Carbon:
Hydrogen:
Oxygen:

22. Empirical Formula Determination (part 3)
Multiply each number by an integer to obtain all whole numbers.
Carbon: 1.50
Hydrogen: 2.50
Oxygen: 1.00
x 2
x 2
x 2 3 5 2
Empirical formula:
C3H5O2

23. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = g
2. Divide the molecular mass by the mass given by the emipirical formula.
(C3H5O2) x 2 =
C6H10O4

24. Combustion Analysis
Combust a compound (usually hydrocarbon), problem gives you grams H2O and CO2, asks for empirical formula of the hydrocarbon
Remember: combustion always yields CO2 and H2O
Given grams H2O and CO2, how do you find EF?
g CO2  mol CO2  mol C
g H2O  mol H2O  mol H (2 mol H in 1 mol H2O!)
NOW we have all moles, we can follow our empirical formula steps

25. Combustion Analysis Example
An unknown substance is known to contain only C, H, and O. Combustion of g of the substance produces g CO2 and g H2O. Calculate the empirical formula of the substance.

26. Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.
1. Identify reactants and products and write the balanced equation. 4 Al + 3 O2 2
Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?

27. Working a Stoichiometry Problem
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 x 2 x ÷ ÷ 4 =
12.3 g Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3
g Al2O3 =
? g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3

28. Stiochiometry Example
Solid lithium hydroxide is used in space vehicles to remove exhaled CO2. The lithium hydroxide reacts with the gaseous CO2 to form solid lithium carbonate and liquid water. How many grams of CO2 can be absorbed by each 1.00 g of lithium hydroxide?

29. Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the amounts of products formed.

30. Limiting Reactant
Why would you purposely add an excess of some reactants to a reaction?
How to ID LR problems: will give you amounts for 2 or more reactants, ask you to find out how much product is made.
WARNING: MIGHT NOT CONTAIN THE WORDS “LIMITING REACTANT!”

31. Limiting Reactant Example
Part of the SO2 that is introduced into the atmosphere ends up being converted to sulfuric acid. The net reaction is :
2 SO2 (g) + O2 (g) + 2H2O (l)  2 H2SO4 (aq)
How much sulfuric acid can be formed from 5.0 mol of SO2, 1.0 mol of O2, and an unlimited quantity of H2O?

32. Yields
Theoretical Yield = The quantity of product produced when all of the limiting reagent is consumed.
Actual Yield = The amount of the product actually produced.
Percent Yield = actual yield x 100
theoretical yield

33. Yield Example
Adipic acid, H2C6H8O4, is a raw material used for the production of nylon. It is made commercially by a controlled reaction between cyclohexane, C6H12, and O2:
2 C6H O2  2 H2C6H8O H2O
a. Assume that you carry out this reaction starting with 25.0 g of cyclohexane, and the cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid?
b. If you obtain 33.5 g of adipic acid for your reaction, what is the percent yield of adipic acid?

34. These are all very similar problems!
All stoichiometry problems (and nearly all problems in chemistry) hinge on a few main ideas:
The MOLE is the unit of comparison (NOT the gram!)
Balanced chemical equations tell us how many moles (or particles) of each atom are involved through the subscripts and coefficients
Molecular weight allows you to easily convert between grams and moles, and vice versa.

35. Chapter 3 Homework Read: p. 77-103
Do # 4, 5, 6, 7, 10, 11, 14, 15, 18, 19, 21, 24, 25, 30, 31, 33, 36, 38, 41, 43, 45, 48, 49, 52, 54, 55, 58, 61, 66, 68, 71, 72, 74, 75, 78, 79, 84, 110