1. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

2. Reaction Types

3. Combination Reactions
Two or more substances react to form one product
Examples:
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
2 Mg (s) + O2 (g)  2 MgO (s)

4. 2 Mg (s) + O2 (g)  2 MgO (s)
Link to Video

5. Decomposition Reactions
One substance breaks down into two or more substances
Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
Link to Video

6. Combustion Reactions Rapid reactions that produce a flame
Most often involve hydrocarbons reacting with oxygen in the air
Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)

7. Balancing Reactions

8. Law of Conservation of Mass
“We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.”
--Antoine Lavoisier, 1789

9. Dalton’s Postulates
Atoms of an element are not changed into atoms of a different element by chemical reactions; atoms are neither created nor destroyed in chemical reactions. They just rearrange!

10. A process in which one or more substances is changed into one or more new substances is a chemical reaction.
A chemical equation uses chemical symbols to show what happens during a chemical reaction.
reactants
products
3.7

11. How to “Read” Chemical Equations
2 Mg + O MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7

12. Balancing Chemical Equations
Write the correct formula(s) for the reactants and the product(s).
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
2C2H6
NOT
C4H12
3.7

13. Balancing Chemical Equations
Start by balancing those elements that appear in only one reactant and one product.
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
2 carbon
on left
multiply CO2 by 2
C2H6 + O2
2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2
2CO2 + 3H2O
3.7

14. Balancing Chemical Equations
Balance those elements that appear in two or more reactants or products.
multiply O2 by 7 2
C2H6 + O2
2CO2 + 3H2O
2 oxygen
on left
4 oxygen
(2x2)
+ 3 oxygen
(3x1)
= 7 oxygen
on right
C2H O2
2CO2 + 3H2O 7 2
remove fraction
multiply both sides by 2
2C2H6 + 7O2
4CO2 + 6H2O
3.7

15. Balancing Chemical Equations
Check to make sure that you have the same number of each type of atom on both sides of the equation.
2C2H6 + 7O2
4CO2 + 6H2O
14 O (7 x 2)
14 O (4 x 2 + 6)
12 H (2 x 6)
12 H (6 x 2)
4 C (2 x 2)
4 C
Reactants
Products
4 C
12 H
14 O
3.7

16. Remember IS NOT 2 Mg + O2 2 MgO 2 grams Mg + 1 gram O2 makes 2 g MgO
So How do we relate masses of reactants and products? That’s what we can observe!
3.7

17. Formula Weights

18. Formula Weight (FW)
Sum of the average atomic weights for the atoms in a chemical formula
So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
These are generally reported for ionic compounds

19. Molecular Weight (MW)
Sum of the average atomic weights of the atoms in a molecule
For the molecule ethane, C2H6, the molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu

20. Percent Composition
One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:
% element =
(number of atoms)(atomic weight)
(FW of the compound)
x 100

21. Percent Composition So the percentage of carbon in ethane is…
(2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu =
x 100
= 80.0%

22. Moles

23. Using Moles
Moles provide a bridge from the molecular scale to the real-world scale

24. Avogadro’s Number
6.02 x 1023
1 mole of 12C has a mass of 12 g

25. Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly grams of 12C
1 mol = NA = x 1023
Avogadro’s number (NA)
3.2

26. atomic mass (amu) = molar mass (grams/mol)
eggs
shoes
Molar mass is the mass of 1 mole of in grams
marbles
atoms
1 mole 12C atoms = x 1023 atoms = g
1 12C atom = amu
1 mole 12C atoms = g 12C
1 mole lithium atoms = g of Li
For any element
atomic mass (amu) = molar mass (grams/mol)
3.2

27. Percent composition revisited
n x molar mass of element
molar mass of compound
x 100%
n is the subscript number … Assume one mole of substance and the math goes like this.
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
C2H6O
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% % % = 100.0%
3.5

28. Question What is the mass on one atom of carbon 12?
What is the mass in grams of one amu?

29. Mole Relationships
One mole of atoms, ions, or molecules contains Avogadro’s number of those particles
One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound

30. Do You Understand Molar Mass?
How many atoms are in g of potassium (K) ?
3.2

31. Do You Understand Molar Mass?
How many atoms are in g of potassium (K) ?
1 mol K = g K
1 mol K
39.10 g K x
0.551 g K
= moles K
3.2

32. molecular mass (amu) = molar mass (grams)
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
SO2 1S
32.07 amu 2O
+ 2 x amu
SO2
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = amu
1 mole SO2 = g SO2
3.3

33. Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
3.3

34. Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12.01) + (8 x 1.008) =
60.09 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = x 1023 atoms H
1 mol C3H8O
60.09 g C3H8O x
8 mol H atoms
1 mol C3H8O x
72.5 g C3H8O
= 9.65 moles H
or 5.82 x 1024 atoms H
3.3

35. Mass Changes in Chemical Reactions
Write balanced chemical equation
Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the number of moles of the sought quantity
Convert moles of sought quantity into desired units
3.8

36. Methanol burns in air according to the equation
2CH3OH + 3O CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
1 mol CH3OH
32.0 g CH3OH x
4 mol H2O
2 mol CH3OH x
18.0 g H2O
1 mol H2O x =
209 g CH3OH
235 g H2O
3.8

37. Finding Empirical Formulas

38. Calculating Empirical Formulas
One can calculate the empirical formula from the percent composition

39. Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

40. Calculating Empirical Formulas
Assuming g of para-aminobenzoic acid,
C: g x = mol C
H: g x = 5.09 mol H
N: g x = mol N
O: g x = mol O
1 mol
12.01 g
14.01 g
1.01 g
16.00 g

41. Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of moles:
C: =  7
H: =  7
N: = 1.000
O: =  2
5.105 mol
mol
5.09 mol
1.458 mol

42. Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2

43. Combustion Analysis
Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this
C is determined from the mass of CO2 produced
H is determined from the mass of H2O produced
O is determined by difference after the C and H have been determined

44. Elemental Analyses
Compounds containing other elements are analyzed using methods analogous to those used for C, H and O

45. Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products

46. Stoichiometric Calculations
From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

47. Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams

48. Real Life Analysis
What is the percent composition of an organic substance called ethanol?
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
3.6

49. g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
g C
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
g H
1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
3.6

50. Limiting Reactants

51. How Many Cookies Can I Make?
You can make cookies until you run out of one of the ingredients
Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

52. How Many Cookies Can I Make?
In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

53. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount

54. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount
In other words, it’s the reactant you’ll run out of first (in this case, the H2)

55. Limiting Reactants
In the example below, the O2 would be the excess reagent

56. Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed OR
g Fe2O3
mol Fe2O3
mol Al needed
g Al needed
1 mol Al
27.0 g Al x
1 mol Fe2O3
2 mol Al x
160. g Fe2O3
1 mol Fe2O3 x =
124 g Al
367 g Fe2O3
Start with 124 g Al
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
3.9

57. Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
g Al2O3
2Al + Fe2O Al2O3 + 2Fe
1 mol Al
27.0 g Al x
1 mol Al2O3
2 mol Al x
102. g Al2O3
1 mol Al2O3 x =
124 g Al
234 g Al2O3
3.9

58. Theoretical Yield
The theoretical yield is the amount of product that can be made
In other words it’s the amount of product possible as calculated through the stoichiometry problem
This is different from the actual yield, the amount one actually produces and measures

59. Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100
A comparison of the amount actually obtained to the amount it was possible to make
Actual Yield
Theoretical Yield
Percent Yield = x 100