1. Changing Matter Matter can be changed two ways: Physically Chemically
Physical reaction
Physical change
Chemically
Chemical reaction
Chemical change

2. Physical Changes Do NOT CHANGE THE TYPE OF MATTER
Nothing new or different is formed
Could be a change in:
Mass
Volume
Density
Change in state
Color
Shape
Size

3. Examples of Physical Changes
Boiling, vaporization…. Any state change
Dissolving
Breaking
Making a mixture
2 or more types of matter (substances) mixed together
Not in specific amounts
Can be separated physically

4. Chemical Changes
Atoms have electrons arranged in energy levels or energy shells
Electrons in the last (outermost) shell are called valence electrons
Valence electrons let atoms bond with other atoms
Ionic bonding
Gaining or losing electrons
Covalent bonding
Sharing electrons

5. Chemical Changes Atoms that bond form molecules
May be the same type (nonmetals) of atom or,
Different types (metal + nonmetal) of atoms
Different types  compounds

6. Chemical Changes Molecules can bond and “unbond”
Atoms can re-arranged in different combinations
For example:
CaCO3 (1 atom Ca, 1 atom C, 3 atoms O)
Add heat to re-arranged the atoms:
CaO
CO2

7. Chemical Changes Evidence of a chemical reaction Formation of gas
Formation of precipitate
Change in color
Change in energy
Endothermic
Absorbs heat energy (gets cold)
Exothermic
Releases heat energy (gets hot)

8. Chemical Changes Chemical reactions can be represented by equations
CaCO3  CaO + CO2
Reactants Products

9. Chemical Changes Atoms are re-arranged, NOT created or destroyed
Law of Conservation of Matter
Law of Conservation of Mass

10. Chemical Changes Matter is conserved  type of atoms does not change
Nothing is created or destroyed
Mass is conserved  amount of atoms cannot change

11. Chemical Changes To show conservation of mass  Balance equations
Make sure there are the same number of each type of atom in the products and in the reactants

12. The equation for the burning of methane gas in oxygen is:
Balancing Equations
The equation for the burning of methane gas in oxygen is:
CH4 + 2 O2 → CO2 + 2 H2O
Subscript
Shows # of atoms
Coefficient
Shows # of molecules

13. Balancing Equations 1 C  1 C 4 H  4 H 4 O  4 O
No subscript or coefficient is understood to be 1
CH4 + 2 O2 → CO2 + 2 H2O =
C1H4 + 2 O2 → C1O2 + 2 H2O1
1 C  C
4 H  H
4 O  O

14. Periodic Properties Post Lab
Periodic Trends affect properties because of Zeff
Ionization energy
Energy required to remove an electron
Electronegativity
Ability to attract electrons
Atomic Radius
Distance of the valence electrons to the nucleus

15. Other Periodic Trends Groups have similar properties Valence electrons
Members of the same representative family have the same number of valence electrons
Reactivity
Because they have the same number of electrons, they react similarly.
CaCl2, MgCl2, SrCl2, etc.
Density = mass/volume
% error = I actual – theoretical I
theoretical
Sn = 7.265
Pb = 11.34
Si = 2.33
Ge = 5.323

16. Solubility Trends
Common in Double Replacement Rxns

17. Double Replacement Rxns
Two compounds react to form two new compounds
Metal replaces metal
Remember basic formula writing rules
Ex: Ca(NO3) Na2CO3 
CaCO NaNO3 2
ppt

18. Some Types of Chemical Reactions
Type of Reaction
Definition
 Equation
Synthesis
Decomposition
Single Replacement
Double Replacement
Two or more elements or compounds combine to make a more complex substance
A + B → AB
Compounds break down
into simpler substances
AB → A + B
Occurs when one element replaces another one in a compound
AB + C → AC + B
Occurs when different atoms in two different compounds trade places
AB + CD → AC + BD

19. Identifying Chemical Reactions
S = Synthesis D = Decomposition SR = Single Replacement DR = Double Replacement
____ P O2 → P4O10 ____ Mg O2 → MgO
____ HgO → Hg O2 ____ Al2O3 → Al O2
____ Cl NaBr → NaCl Br2 ____ H N2 → NH3

20. MOLAR MASS Molar Mass is shown below the element
symbol on the Periodic Table.
Units: grams
mole
Use molar mass to convert between:
Number of Moles
Mass (grams)
Number of Molecules or Atoms

21. Molar Mass Examples NaHCO3
sodium bicarbonate
sucrose
NaHCO3
(16.00) = g/mol
C12H22O11
12(12.01) + 22(1.01) + 11(16.00) = g/mol

22. MOLES MASS in GRAMS MOLECULESATOMS
Multiply moles by
Molar Mass to get
Divide the mass (in grams) by Molar Mass to get the number of moles.
MASS in
GRAMS
MOLAR
MASS
grams/mol
Multiply the number of moles by Avogadro’s Constant to get the number of atoms or molecules.
MOLES
Multiply the moles
by the Molar Mass to get the mass
(in grams).
AVOGADRO’S CONSTANT
6.022x1023 particles/mol
MOLECULESATOMS
Divide the number of molecules or atoms by Avogadro’s Constant
to get the number of moles.

23. Molar Conversion Examples
How many moles of carbon are in 26 g of carbon?
26 g C
1 mol C
12.01 g C
= 2.2 mol C

24. Molar Conversion Examples
How many molecules are in 2.50 moles of C12H22O11?
6.02  1023
molecules
1 mol
2.50 mol
= 1.51  1024
molecules
C12H22O11

25. Molar Conversion Examples
Find the mass of 2.1  1024 molecules of NaHCO3.
2.1  1024
molecules
1 mol
6.02  1023
molecules
84.01 g
1 mol
= 290 g NaHCO3

26. Sample Problems – Moles and Atoms
Determine the number of atoms present in 2.50 moles of strontium.
Convert 5.01 x 1024 atoms of strontium to moles of strontium.

27. Sample Problems – Moles and Mass
Determine the mass in grams of 2.50 moles of strontium.
Determine the number of moles represented by grams of strontium.

28. Sample Problems – Moles, Atoms, Mass
Determine the mass in grams of (exactly) 5 atoms of strontium.
Determine the number of atoms represented by 43.5 grams of strontium.

29. Percentage Composition
the percentage by mass of each element in a compound

30. Percentage Composition
Find the % composition of Cu2S.
g Cu
g Cu2S
79.852% Cu
%Cu =
 100 =
32.07 g S
g Cu2S
20.15% S
%S =
 100 =

31. Percentage Composition
Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.
28 g
36 g
%Fe =
 100 =
78% Fe
8.0 g
36 g
%O =
 100 =
22% O

32. Percentage Composition
How many grams of copper are in a 38.0-gram sample of Cu2S?
Cu2S is % Cu
(38.0 g Cu2S)( ) = 30.3 g Cu

33. Percentage Composition
Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O?
36.04 g g
%H2O =
 100 =
24.51%
H2O

34. C2H6 CH3 Empirical Formula
Smallest whole number ratio of atoms in a compound
C2H6
reduce subscripts
CH3

35. Empirical Formula 1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to find subscripts.
4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

36. Empirical Formula
Find the empirical formula for a sample of 25.9% N and 74.1% O.
25.9 g
1 mol
14.01 g
= 1.85 mol N
= 1 N
1.85 mol
74.1 g
1 mol
16.00 g
= 4.63 mol O
= 2.5 O

37. N2O5 N1O2.5 Empirical Formula
Need to make the subscripts whole numbers
 multiply by 2
N2O5

38. CH3 C2H6 Molecular Formula
“True Formula” - the actual number of atoms in a compound
CH3
empirical
formula ?
C2H6
molecular
formula

39. Molecular Formula 1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molecular mass by the empirical mass.
4. Multiply each subscript by the answer from step 3.

40. (CH2)2  C2H4 Molecular Formula
The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is g/mol?
empirical mass = g/mol
28.1 g/mol
14.03 g/mol
= 2.00
(CH2)2  C2H4

41. Using the Basics… The empirical formula of a
compound is found to be P2O5.
The molar mass of the compound
is 284 grams/mole.
What is the molecular formula for the compound?

42. Proportional Relationships
Stoichiometry
mass relationships between substances in a chemical reaction
based on the mole ratio
Mole Ratio
indicated by coefficients in a balanced equation
2 Mg + O2  2 MgO

43. N2 + 3H2  2NH3
1 mol
2 mol
3 mol

44. N2 + 3H2  2NH3
1 mol
2 mol
3 mol

45. Stoichiometry Steps 1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
Mole ratio - moles  moles
Molar mass - moles  grams
Molarity - moles  liters soln
Molar volume - moles  liters gas
Mole ratio - moles  moles
Core step in all stoichiometry problems!!
4. Check answer.

46. Standard Temperature & Pressure
Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature & Pressure
0°C and 1 atm

47. Molar Volume at STP (g/mol) particles/mol LITERS OF GAS AT STP
Molar Volume (22.4 L/mol)
MASS IN
GRAMS
MOLES
NUMBER OF
PARTICLES
Molar Mass
(g/mol)
6.02  1023
particles/mol
Molarity (mol/L)
LITERS OF
SOLUTION

48. Mole-Mole Calculations

49. Phosphoric Acid
Phosphoric acid (H3PO4) is one of the most widely produced industrial chemicals in the world.
Most of the world’s phosphoric acid is produced by the wet process which involves the reaction of phosphate rock, Ca5(PO4)3F, with sulfuric acid (H2SO4).
Ca5(PO4)3F(s) + 5H2SO4  3H3PO4 + HF + 5CaSO4

50. Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4).
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 Moles starting substance: mol H2SO4
Step 2 The conversion needed is moles H2SO4  moles H3PO4
Mole Ratio

51. Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of sulfuric acid (H2SO4) that react when 10 moles of Ca5(PO4)3 react.
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 The starting substance is 10.0 mol Ca5(PO4)3F
Step 2 The conversion needed is moles Ca5(PO4)3F  moles H2SO4
Mole Ratio

52. Stoichiometry Problems
How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
2KClO3  2KCl + 3O2
? mol
9 mol
9 mol O2
2 mol KClO3
3 mol O2
= 6 mol
KClO3

53. Mole-Mass Calculations

54. Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4.
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Method 1 Step by Step
Step 1 The starting substance is 784 grams of H3PO4.
Step 2 Convert grams of H3PO4 to moles of H3PO4.
Step 3 Convert moles of H3PO4 to moles of H2SO4 by the mole-ratio method.
Mole Ratio

55. Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Method 2 Continuous
grams H3PO4  moles H3PO4  moles H2SO4
The conversion needed is
Mole Ratio

56. Stoichiometry Problems
How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP?
2KClO3  2KCl + 3O2
? g
9.00 L
9.00 L O2
1 mol O2
22.4 L
2 mol
KClO3
3 mol O2
122.55
g KClO3
1 mol
KClO3
= 32.8 g
KClO3

57. Mass-Mass Calculations

58. Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 1 Step by Step
Step 1 The starting substance is 112 grams of H2. Convert 112 g of H2 to moles.
grams  moles
Step 2 Calculate the moles of NH3 by the mole ratio method.

59. Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 1 Step by Step
Step 3 Convert moles NH3 to grams NH3.
moles  grams

60. Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 2 Continuous
grams H2  moles H2  moles NH3  grams NH3

61. Stoichiometry Problems
How many grams of silver will be formed from 12.0 g copper?
Cu + 2AgNO3  2Ag + Cu(NO3)2
12.0 g
? g
12.0
g Cu
1 mol Cu
63.55
g Cu
2 mol Ag
1 mol Cu
107.87
g Ag
1 mol Ag
= 40.7 g Ag

62. Stoichiometry Problems
How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?
Cu + 2AgNO3  2Ag + Cu(NO3)2
? g
1.5L
0.10M
1.5 L
.10 mol
AgNO3
1 L
1 mol Cu
2 mol
AgNO3
63.55
g Cu
1 mol Cu
= 4.8 g Cu

63. Limiting Reactant

64. The limiting reactant limits the amount of product that can be formed.
The limiting reactant is one of the reactants in a chemical reaction.
It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present.
The limiting reactant limits the amount of product that can be formed.

65. How many bicycles can be assembled from the parts shown?
From eight wheels four bikes can be constructed.
From three pedal assemblies three bikes can be constructed.
From four frames four bikes can be constructed.
The limiting part is the number of pedal assemblies.
9.2

66. H2 + Cl2  2HCl  + Cl2 is the limiting reactant
4 molecules Cl2 can form 8 molecules HCl
Cl2 is the limiting reactant
3 molecules of H2 remain H2 is in excess
7 molecules H2 can form 14 molecules HCl
9.3

67. Steps Used to Determine the Limiting Reactant

68. Calculate the amount of product (moles or grams, as needed) formed from each reactant.
Determine which reactant is limiting. (The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess.
Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. This gives the amount of the substance that remains unreacted.

69. Examples

70. How many moles of HCl can be produced by reacting 4. 0 mol H2 and 3
How many moles of HCl can be produced by reacting 4.0 mol H2 and 3.5 mol Cl2? Which compound is the limiting reactant?
H2 + Cl2 → 2HCl
Step 1 Calculate the moles of HCl that can form from each reactant.
Step 2 Determine the limiting reactant.
The limiting reactant is Cl2 because it produces less HCl than H2.

71. Step 1 Calculate the grams of AgBr that can form from each reactant.
How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Calculate the grams of AgBr that can form from each reactant.
The conversion needed is
g reactant → mol reactant → mol AgBr → g AgBr

72. Step 2 Determine the limiting reactant.
How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Determine the limiting reactant.
The limiting reactant is MgBr2 because it forms less Ag Br.

73. How many grams of the excess reactant (AgNO3) remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 3 Calculate the grams of unreacted AgNO3. First calculate the number of grams of AgNO3 that will react with 50 g of MgBr2.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3
The amount of MgBr2 that remains is
100.0 g AgNO3 -
92.3 g AgNO3 =
7.7 g AgNO3

74. Reaction Yield

75. The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.

76. Many reactions fail to give a 100% yield of product.
This occurs because of side reactions and the fact that many reactions are reversible.

77. The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant.
The actual yield is the amount of product finally obtained from a given amount of reactant.

78. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.

79. Silver bromide was prepared by reacting 200
Silver bromide was prepared by reacting g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if g of silver bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Determine the theoretical yield by calculating the grams of AgBr that can be formed.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgBr → g AgBr

80. MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Silver bromide was prepared by reacting g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if g of silver bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Calculate the percent yield.
must have same units