1. Reactions, Equations, and Formulas (Chapter 11) Dr. Walker

2. What will you learn? SOL Ch 1 g, 3 b, c, e Classifying reaction types
Balancing chemical reactions based on the law of conservation of mass
Differentiating between molecular and empirical formulas
Calculating percent composition of elements within compounds

3. Vocabulary Reactants – left side of a reaction; what you start with
Products – right side of a reaction; what you finish with
NaOH + HCl NaCl + H2O
Reactants
Products

4. Common chemical shorthand
(l) – liquid
(s) – solid
(g) – gas
(aq) – dissolved in water
D – add heat
DH – energy change (if necessary)
Temperature – react at a specific temperature

5. Six Reaction Types Synthesis (or Combination) Decomposition Combustion
Single Replacement
Double Replacement
Oxidation/Reduction

6. Synthesis Reactions Multiple Reactants One Product Examples
H2O + CO H2CO3
4 Al + 3 O Al2O3
2 KCl + 3O KClO3

7. Decomposition Reactions
One Reactant
Multiple Products
Examples
2 H2O H2O + O2
Ba(ClO3) BaCl2 + 3 O2

8. Combustion Reactions
Reactants = hydrocarbon (or alcohol) + oxygen (CHO)
Products = carbon dioxide + water
Example
2 C8H O CO H2O
C3H8 + 5 O CO2 + 4 H2O
C6H12O6 + 6 O CO2 + 6 H2O

9. Single Replacement Reactions
Reactants = Element + Compound
Products = Different Element + Different Compound
Examples
2 Na + MgCl NaCl + Mg
Ca + 2 ZnBr CaBr2 + 2 Zn

10. Double Replacement Reactions
Two Products
Two Reactants
Metals “switch” compounds
Examples
K2SO4 + 2 HCl KCl + H2SO4
ZnCl + NaBr ZnBr + NaCl

11. Oxidation-Reduction Reactions (aka Redox Reactions)
Oxidation – Loses electrons
When atoms become cations, they are oxidized
Zn Zn+2 + 1e-
Reduction – Gains electrons
When atoms become anions, they are reduced
Reduces the charge
Cu+2 + 1e Cu+1

12. So How Do We Know What Kind of Reaction Is Taking Place?????
Ask a series of questions where reaction types can be eliminated.

13. First Question How many reactants? One Two
You have a decomposition reaction
Two
Keep Going

14. Second Question We have multiple reactants. How many products? One
This is a synthesis reaction
Two or more
Keep Going

15. Third Question
Are the reactants oxygen gas and a hydrocarbon (or alcohol)?
Yes
This is a combustion reaction No
Keep Going

16. Fourth Question Are the reactants and products two compounds? NO Yes
This is a single replacement reaction.
Yes
This is a double replacement reaction.

17. Fourth Question Are the reactants and products two compounds? Yes
This is a double replacement reaction.

18. Examples
Zn + O ZnO2
Fe2O3 + Mg MgO + Fe

19. Examples Zn + O2 ZnO2 Fe2O3 + Mg MgO + Fe Synthesis Zn + O2 ZnO2
Single Replacement
Fe2O3 + 3 Mg MgO + 2 Fe

20. Examples
Ca(ClO3) CaCl2 + O2
C5H12 + O CO2 + H2O

21. Examples Ca(ClO3)2 CaCl2 + O2 C5H12 + O2 CO2 + H2O Decomposition
Combustion
C5H O CO2 + 6 H2O

22. Examples
KI + MgO K2O + MgI2
Ca + H2O H2 + CaO

23. Examples KI + MgO K2O + MgI2 Ca + H2O H2 + CaO Double Replacement
Single Replacement
Ca + H2O H2 + CaO

24. A little different Redox reactions Gaining or Losing Electrons
Easy to distinguish from other reaction types

25. Redox Examples
Fe Fe3+ + 2e-
Oxidation
Sn+4 + 2e Sn2+
Reduction

26. A little later… One more type… Acid-Base (Neutralization) Reactions
We’ll talk about them in more detail during chapter 15 in a few weeks
Will be on the SOL

27. Chemical Reaction Vocabulary
4 P + 5 O P2O5
coefficients
subscripts
P 4 x 1 = 4 atoms
O 5 x 2 = 10 atoms
P 2 x 2 = 4 atoms
O 2 x 5 = 10 atoms
Number of atoms for each element = coefficient x subscript
Lack of a coefficient or subscript denotes a value of 1

28. Chemical Reaction Vocabulary
4 P + 5 O P2O5
coefficients
subscripts
P 4 x 1 = 4 atoms
O 5 x 2 = 10 atoms
P 2 x 2 = 4 atoms
O 2 x 5 = 10 atoms
Law of conservation of mass – matter cannot be
or destroyed in a chemical reaction
Mass in = Mass out
Atoms (reactants) = Atoms (products)

29. Balancing Chemical Reactions
Balanced reactions are required to satisfy the law of conservation of mass
You will balance reactions by adding coefficients to reactions
Goal: Have the same number of atoms on each side of the reaction
You may NOT change subscripts or you have a different reaction

30. Steps To Balance Reactions
___ Na + ___Cl2 ___ NaCl Step 1: Count all atoms on each side Remember, atoms = coefficient x subscript

31. Steps To Balance Reactions
___ Na + ___Cl2 ___ NaCl Na = 1 Na = 1 Cl = 2 Cl = 1 Step 1: Count all atoms on each side Remember, atoms = coefficient x subscript

32. Steps To Balance Reactions
___ Na + ___Cl2 _2_ NaCl Na = 1 Na = 2 Cl = 2 Cl = 2 Step 2: Add coefficients to balance atoms that are not equal - 2 is added in front of NaCl to balance Cl

33. Steps To Balance Reactions
2 Na + ___Cl _2_ NaCl
Na = Na = 2
Cl = Cl = 2
Step 3: Adjust other coefficients for unequal elements
2 is added in front of Na to balance Na
Verify that equation is balanced

34. Additional Examples
___ N2 + ___ H2 ___ NH3 Count atoms N N H H

35. Additional Examples
___ N2 + ___ H2 ___ NH3 Count atoms 2 N 1 N 2 H 3 H Common Factor method: 2 and 3 aren’t divisible. Make the total # of H’s a number that can be divided by 2 and 3

36. Additional Examples
N2 + _3_ H2 _2_ NH3 Add coefficients 2 N 2 N 6 H 6 H Common Factor method: Notice that 2 and 3 are both divisible by 6. Balancing H works to balance N as well.

37. Additional Examples
___ H2O ___ H2 + ___ O2 Count atoms ___ H ____ H ___ O ____ O

38. Additional Examples
___ H2O ___ H2 + ___ O2 Count atoms _2_ H 2_ H _1_ O _2_ O

39. Additional Examples
_2_ H2O _2_ H2 + ___ O2 Add coefficients _4_ H 4_ H _2_ O _2_ O Even-odd method Make all atoms even – multiply odd atoms by two

40. Additional Examples ___ CuNO3 + ___K2CO3 → ___ Cu2CO3 + ___ KNO3
Count atoms
___ Cu ___ Cu
___ NO3 ___ NO3
___K ___ K
___CO3 ___ CO3
Keep intact polyatomics together – don’t split them apart!

41. Additional Examples ___ CuNO3 + ___K2CO3 → ___ Cu2CO3 + ___ KNO3
Count atoms
_1_ Cu _2_ Cu
_1_ NO3 _1_ NO3
_2_ K _1_ K
_1_ CO3 _1_ CO3
Keep intact polyatomics together – don’t split them apart!

42. Additional Examples _2_CuNO3 + ___K2CO3 → ___ Cu2CO3 + _2_ KNO3
Add coefficients
_2_ Cu _2_ Cu
_2_ NO3 _2_ NO3
_2_ K _2_ K
_1_ CO3 _1_ CO3
Keep intact polyatomics together – don’t split them apart!

43. Additional Examples
___ C3H8 + ___ O2 → ___ CO2 + ___ H2O Combustion reactions: Save oxygen for last!!

44. Additional Examples
___ C3H8 + ___ O2 → ___ CO2 + ___ H2O Count atoms 3 C 1 C 8 H 2 H 2 O 3 O Combustion reactions: Save oxygen for last!!

45. Additional Examples
___ C3H8 + ___ O2 → 3 CO2 + 4 H2O Balance, part 1 3 C 3 C 8 H 8 H 2 O 10 O Combustion reactions: Save oxygen for last!!

46. Additional Examples
___ C3H8 + 5 O2 → 3 CO2 + 4 H2O Balance, part 2 (oxygen) 3 C 3 C 8 H 8 H 10 O 10 O Combustion reactions: Save oxygen for last!!

47. Additional Examples
___ C8H18 + ___ O2 → ___ CO2 + ___ H2O

48. Additional Examples
___ C8H18 + ___ O2 → ___ CO2 + ___ H2O Count atoms 8 C 1 C 18 H 2 H 2 O 3 O

49. Additional Examples
___ C8H18 + ___ O2 → _8_ CO2 + 9 H2O Balance C and H 8 C 8 C 18 H 18 H 2 O 27 O Problem: Can’t balance both sides with a whole #

50. Additional Examples
___ C8H18 + ___ O2 → _8_ CO2 + 9 H2O Balance C and H 8 C 8 C 18 H 18 H 27 O 27 O Problem: Can’t balance both sides with a whole #

51. Additional Examples
___ C8H O2 → _8_ CO2 + 9 H2O Balance oxygen 8 C 8 C 18 H 18 H 25 O 25 O Problem: Temporarily use a decimal to balance

52. Additional Examples
2 C8H O2 → _16_ CO H2O Get rid of decimal coefficient 16 C 16 C 36 H 36 H 50 O 50 O Problem: Multiply ALL coefficients by 2 to get rid of decimal

53. Molar Mass
The atomic mass of any element is the weighted average of all isotopes of an element
These are shown in the periodic table
This is also the molar mass of the element
Examples
H = 1.01 g/mole
O = g/mole
C = g/mole
P = g/mole
Cl = g/mole

54. Molar Mass
The molar mass of a compound is the sum of all atomic weights of all atoms in the compound!
Remember
Compounds have more than one element
When there are multiple atoms, you simply add the masses of each atom together to get the mass of a compound

55. Molar Mass Examples For compounds H2O H = 1.01 g/mole x 2
O = g/mole x 1
=18.02 g/mole

56. Molar Mass Examples Another example CH4 C = 12.01 g/mole = 12.01
H = 1.01 g/mole x 4 = 4.04
16.05 g/mole

57. Molar Mass Examples Another Example C6H12O6
C = g/mole x 6 = 72.06
H = 1.01 g/mole x 12 = 12.12
O = g/mole x 6 = 96.00
g/mole

58. Molar Mass Examples Another Example Polyatomics Pb(SO4)2 Pb = 207.20
S = g/mole x 2 = 64.14
O = g/mole x 8 =
g/mole
Notice you must multiply the subscript inside the parentheses by the number outside to get the right number of atoms

59. Practice Find the molar mass of the following compounds: LiOH SO2
Mg(NO3)2
Al2(CO3)3
C9H8O4
C8H10N4O2
C22H30N6O4S

60. Practice Find the molar mass of the following compounds: LiOH
= g/mole
SO2
(16.00) = g/mole
Mg(NO3)2
(14.01) + 6(16.00) = g/mole
Al2(CO3)3
2(26.98) + 3(12.01) + 9(16.00) = g/mole
C9H8O4
9(12.01) + 8(1.01) + 4(16.00) = g/mole
C8H10N4O2
8(12.01) + 10(1.01) + 4(14.01) + 2(16.00) = g/mole
C22H30N6O4S
22(12.01) + 30(1.01) + 6(14.01) + 4(16.01) + 1(32.06) = g/mole

61. Empirical Formulas
Molecular Formula – True number of atoms in a compound
Empirical Formula – SIMPLEST ratio of atoms in a compound
Benzene
Molecular Formula = C6H6
Empirical Formula = CH
Structural Formula
Visual representation of a chemical formula showing bonds

62. Empirical and Molecular Formulas

63. Empirical and Molecular Formulas

64. Molecular Formulas from Empirical Formulas
As previously demonstrated, multiple compounds can have the same empirical formula
How do we figure out molecular formulas?
We need another piece of information…the molar mass of the molecular formula

65. Molecular Formulas from Empirical Formulas
Example
What is the molecular formula of a compound with an empirical formula of CH2 and a molar mass of 70 g/mole?

66. Molecular Formulas from Empirical Formulas
Example
What is the molecular formula of a compound with an empirical formula of CH2 and a molar mass of 70 g/mole?
Step 1 – Figure out molar mass of empirical formula. Molar mass of CH2 = 12 + (2 x 1) = 14 g/mole
Mass from periodic table for carbon
(12) and H (1)

67. Molecular Formulas from Empirical Formulas
Example
What is the molecular formula of a compound with an empirical formula of CH2 and a molar mass of 70 g/mole?
Step 2 – Divide given mass by the molar mass of the empirical formula you found in the previous step
(70 g/mole)/(14 g/mole) = 5

68. Molecular Formulas from Empirical Formulas
Example
What is the molecular formula of a compound with an empirical formula of CH2 and a molar mass of 70 g/mole?
Step 3 – Multiply subscripts of empirical formula by number found in step 2
CH2 x 5 = C5H10
You can double check this formula – if it doesn’t have the molar mass given in the problem, it’s wrong

69. Another Example
A compound is found to have an empirical formula of CH2O and a molar mass of g/mole. What is the molecular formula?

70. Another Example
A compound is found to have an empirical formula of CH2O and a molar mass of g/mole. What is the molecular formula?
Step 1 – CH2O – molar mass = g/mole
Step 2 – ( g/mole)/(30.03 g/mole) = 4
Step 3 – 4 x (CH2O) = C4H8O4

71. Percentage Composition
The percentage by mass of each element in a compound
“Mass of whole” = the molar mass of the
whole compound
“Mass of part” = the molar mass of a single element

72. How do we find percentage composition?
Example
Give the percentage composition of each element of H2O.

73. How do we find percentage composition?
Example
Give the percentage composition of each element of H2O.
Step 1 = Find molar mass of compound
H = 1.01 g/mole x 2 = 2.02 g/mole
O = g/mole x 1 = g/mole
18.02 g/mole

74. How do we find percentage composition?
Example
Give the percentage composition of each element of H2O.
Step 2: Divide the mass of each element by the total mass and multiply by 100.
H = [ (2.02 g/mole)/(18.02 g/mole)] x 100 = 11.2%
O = [ (16.00 g/mole)/(18.02 g/mole)] x 100 = 88.8%

75. Percentage Compositon
Example:
Find the percentage composition of (NH4)2SO4

76. Percentage Compositon
Example – (NH4)2SO4
Molar mass
N = g/mole x 2 = g/mole
H = 1.01 g/mole x 8 = 8.08 g/mole
S = g/mole x 1 = g/mole
O = g/mole x 4 = g/mole
Total molar mass = g/mole

77. Percentage Compositon
Example – (NH4)2SO4
Percentage Composition
N = [(28.02 g/mole)/( g/mole) x 100 = 21.2%
H = [(8.08 g/mole)/( g/mole) x 100 = 6.1 %
S = [(32.06 g/mole)/( g/mole) x 100 = 24.3 %
O = [(64.00 g/mole)/( g/mole) x 100 = 48.4 %
Total molar mass = g/mole
Double check – what is our total percentage? It should be 100%

78. Terms To Know Reactants Products Synthesis (or Combination)
Decomposition
Combustion
Single Displacement (Replacement)
Double Displacement (Replacement)
Oxidation/Reduction
Law of conservation of mass
Empirical formula
Structural formula
Percentage composition

79. Skill To Master Determine reaction type from given chemical equation
Balance chemical reactions by changing coefficients
Determine empirical and molecular formulas from the other
Determine percentage composition of each element from a chemical formula