1. Stoichiometry: Calculations with Chemical Formulas and Equations

2. Moles

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

4. Moles
Very small macroscopic samples contain VERY many atoms, molecules, etc... (ex. 1 tsp. H2O contains 2 x 1023 molecules). [Need convenient counting unit]
The number of carbon atoms in 12 g of 12C is called Avogadro’s number.
One Mole (latin “mole” = a mass) is the amount of material that contains Avogadro’s number
Note: a mole refers to a fixed number of any type of particles! (atoms, molecules, f.u., e-)
Avogadro’s number = x 1023
(All of the following masses have the same number of atoms: 1 g of H g of O g of C)

5. Molar Mass
The molar mass of an element is the mass number for the element that we find on the periodic table
The formula mass (in amu’s) will be the same number as the molar mass (in g/mol)

6. Avagadro’s Number and the Mole
1 mole of 12C atoms = 6.02 x 1023 atoms
1 mole of 11B atoms = 6.02 x atoms
1 mol of PCl3 molecules = 6.02 x 1023molecules
1 mol of Na+ ions = 6.02 x Na ions
1 mol of toasters = 6.02 x toasters
1 mole of students = x 1023 students

7. Problems
Find the mass of 5.2 moles of calcium

8. Problems Find the mass of 5.2 moles of calcium
5.2 mol x 40g/1 mol = 208g

9. Problems
Find the mass of 5.2 moles of CaCl2

10. Problems Find the mass of 5.2 moles of CaCl2
5.2 mol x 111g/1 mol = 577.2g

11. Problems
Find the number of atoms in 5.2 moles of Ca

12. Problems Find the number of atoms in 5.2 moles of Ca
5.2mol x 6.02 x 1023/1mol = 31.3 x 1023atoms

13. Problems
How many O2 molecules are in 0.25 moles
of O2

14. Problems How many O2 molecules are in 0.25 moles of O2
.25mol x x 1023 molecules
1 mol
= 1.5 x 1023 molecules

15. Problems
How many grams are in .25 moles of O2

16. Problems 1 mol = 8 grams How many grams are in .25 moles of O2
.25 mol x 32g
1 mol = 8 grams

17. Problems
How many molecules are in 8 grams of O2

18. Problems How many molecules are in 8 grams of O2
8g x 1 mol x 6.02 x 1023molecules =
32g mol
1.5 x 1023molecules

19. How many atoms are in 8 grams of O2

20. Problems How many atoms are in 8 grams of O2
8g x 1 mol x 6.02 x 1023molec. x 2 atoms =
32g mol molec.
3.0 x 1023molecules

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

22. 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

23. Standard Temperature and Pressure (STP)
0ºC ( 32ºF) and 1 atm pressure
abbreviated “STP”
At STP 1 mole of any gas occupies 22.4 L
Called the molar volume
We now have another conversion factor: 1 mole = 22.4 L of any gas at STP

24. Percentage Composition
Percentage Composition - percentage by mass contributed by each element in the substance. May be used to verify the purity or identity of a particular compound.
Part/whole x 100
total mass of element/formula mass x 100 = % composition
Percentage Composition of C6H12O6
(formula mass = 180)
% C = (6)(12)/180 x 100 = 40.0% carbon
% O = (6)(16)/180 x 100 = 53.3% oxygen
% H = (12)(1)/180 x 100 = 6.7% hydrogen
2 example calculations follow

25. Percentage Composition
Sample exercise: Calculate the percentage of nitrogen, by mass, in Ca(NO3)2.

26. Percentage Composition
Sample exercise: Calculate the percentage of nitrogen, by mass, in Ca(NO3)2.
Formula Mass:
Ca 1(40.1) = 40.1
N 2(14.0) = 28.0
O 6(16.0) = 96.0
164.1 amu

27. Percentage Composition
Sample exercise: Calculate the percentage of nitrogen, by mass, in Ca(NO3)2.
Formula Mass: part x 100
Ca 1(40.1) = whole
N 2(14.0) = 28.0
O 6(16.0) = u x 100 =
164.1 amu u

28. Percentage Composition
Sample exercise: Calculate the percentage of nitrogen, by mass, in Ca(NO3)2.
Formula Mass: part x 100
Ca 1(40.1) = whole
N 2(14.0) = 28.0
O 6(16.0) = u x 100 = 17.1%
164.1 amu u nitrogen

29. Calculate the percentage composition
of sodium perchlorate

30. Calculate the percentage composition
of sodium perchlorate NaClO4
Na /122 x 100 = 18.8%
Cl /122 x 100 = 28.7%
O 4(16)/122 x 100 = 52.4%

31. Percentage Composition
Problem: In 1987, the first substance to act as a superconductor at a temperature above that of liquid nitrogen (77 K) was discovered. The approximate formula of the substance is YBa2Cu3O7. Calculate the percent composition by mass of this material.
M W of YBa2Cu3O7 = (88.9) + 2(137.3) + 3(63.6) + 7(16)
= amu
atomic masses: Y = 88.9; Ba =137.3; Cu = 63.6; O = 16.0

32. Percentage Composition
M W of YBa2Cu3O7 = amu
Y = 1(88.9) = (88.9) = 13.3 %
666.0
Ba = 2(137.3) = (274.6) = 41.3 %
Cu = 3(63.5) = (190.5) = 28.6 %
O = 7(16.0) = (112) = 16.8 %
AW: Y = 88.9; Ba =137.3; Cu = 63.6; O = 16.0

33. Empirical Formulas
Empirical Formula - Relative number of each element in a compound.
Using moles and percent weight (elemental analysis by chemical means), we can calculate an empirical formula
Steps:
assume a 100 g (convenient since working with % because the elements % can be thought of as g)
calculate moles of element present in 100g sample
find ratios of moles (approx) to lead to integral formula subscripts.

34. Empirical Formulas Mass Percent Empirical of Elements Formula Moles of
assume
100 g sample
calculate mole
ratio
Grams of
each Element
Moles of
each Elements
Use
atomic
masses

35. Empirical Formulas, Examples
Determine the empirical formula for a compound which
contains 87.5 % N and 12.5% H by mass.
% g in 100g moles ratio
87.5 % N = g N 1 mole = moles N = 1
14 g
12.5% H = g H 1 mole H = moles H = 2
1 g
Empirical Formula = NH2

36. Empirical Formulas, Examples
Sample exercise: A g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance?

37. Empirical Formulas, Examples
Sample exercise: A g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance?
3.758 g x 100 = 70.58% C -> g C 1 mol C = 5.88 mol C
5.325 g g C

38. Empirical Formulas, Examples
Sample exercise: A g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance?
0.316 g x 100 = 5.93% H -> 5.93 g H 1 mol H = 5.93 mol H
5.325 g g H

39. Empirical Formulas, Examples
Sample exercise: A g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance?
1.251 g x 100 = 23.49% O -> g O 1 mol O = 1.47 mol O
5.325 g g O

40. Empirical Formulas, Examples
Sample exercise: A g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance?
5.88 mol C ; mol H ; mol O

41. Empirical Formulas, Examples
Sample exercise: A g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance?
5.88 mol C ; mol H ; mol O
1.47 mol 1.47 mol 1.47 mol

42. Empirical Formulas, Examples
Sample exercise: A g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance?
5.88 mol C ; mol H ; mol O C4H4O
1.47 mol 1.47 mol 1.47 mol

43. Coefficient  subscript = # atoms
Chemical Equations (Shorthand communication for a great deal of information)
1. Write the unbalanced equation making sure each chemical formula is correctly written.
2. Count atoms on each side.
3. Add coefficients to make #s equal.
Coefficient  subscript = # atoms
4. Reduce coefficients to lowest possible ratio, if necessary.
5. Double check atom balance!!!

44. Law of Conservation of Mass
In a chemical reaction, mass (or matter) is not created or destroyed.
Atoms can only rearrange to form new substances
“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

45. Chemical Equations
Are concise representations of chemical reactions

46. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

47. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Reactants appear on the left side of the equation.

48. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Products appear on the right side of the equation.

49. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
The states of the reactants and products are written in parentheses to the right of each compound.

50. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.

51. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule

52. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule
Coefficients tell the number of molecules

53. Chemical Equations 2 H2 + O2 2 H2O N2O5(g) + H2O 2 HNO3
Must have equal numbers of atoms of each element on each side of the equation = BALANCED EQUATION
NOTE
The coefficients in front of the formula for a compound refers to the number of molecules (intact) involved while a subscript refers to the ratio of atoms within the molecule
2 H2 + O2 2 H2O
4 hydrogen 4 hydrogen
2 oxygen 2 oxygen
N2O5(g) + H2O 2 HNO3
2 nitrogen 2 nitrogen
6 oxygen 6 oxygen
2 hydrogen 2 hydrogen

54. Chemical Equations Chemical Equations
balancing equations often requires some trial and error of coefficients
NOTE
NOTE
Never change subscripts in formulas when balancing chemical reactions!
subscripts change compounds; coefficients change amounts
PCl3(l) + 3 H2O(l) H3PO3(aq) + 3 HCl
6 hydrogen 6 hydrogen
3 oxygen 3 oxygen
1 phosphorus 1 phosphorus
3 chloride 3 chlorine
C6H12(l) + 9 O2(g) CO2(g) + 6 H2O(l)
6 carbon 6 carbon
18 oxygen 18 oxygen
12 hydrogen 12 hydrogen

55. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
C2H4 + O2  CO2 + H2O

56. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
C2H4 + O2  CO2 + H2O
C C
H H
O O

57. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
C2H4 + O CO2 + H2O
C 2 C 1
H H 2
O 2 O 3

58. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
C2H4 + O CO2 + H2O
C 2 C (1)2= 2
H H 2
O 2 O 3 5

59. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
C2H4 + O CO H2O
C 2 C (1)2= 2
H H (2)2 = 4
O 2 O

60. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
C2H4 + 3O CO H2O
C 2 C (1)2= 2
H H (2)2 = 4
O (2)3 = 6 O

61. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
Al + HCl AlCl3 + H2

62. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
Al + HCl AlCl3 + H2
Al Al
H H
Cl Cl

63. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
Al + HCl AlCl3 + H2
Al 1 Al 1
H 1 H 2
Cl 1 Cl 3

64. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
Al + 3HCl AlCl3 + H2
Al 1 Al 1
H (1)3 = 3 H 2
Cl (1)3 = 3 Cl 3

65. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
Al + 6HCl AlCl H2
Al 1 Al 1
H (1)6 = 6 H (2)3 = 6
Cl (1)6 = 6 Cl 3

66. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
Al + 6HCl AlCl H2
Al 1 Al (1)2 = 2
H (1)6 = 6 H (2)3 = 6
Cl (1)6 = 6 Cl (3)2 = 6

67. Chemical Equations
Sample exercise: Balance the following equations by providing the missing coefficients:
2Al + 6HCl AlCl H2
Al (1)2 = 2 Al (1)2 = 2
H (1)6 = 6 H (2)3 = 6
Cl (1)6 = 6 Cl (3)2 = 6

68. Chemical Reactivity
Chemical Reactions
The course of a chemical reaction can often be predicted by recognizing general patterns of reactivity through similar reactions previously observed. Elements in same family (column of table) have similar reactions.
The periodic table is helpful in predicting products of reactions. Atoms like to assume electron configurations of the Noble Gases.

69. Example, if you know that
Chemical Reactivity
Chemical Reactions
Example, if you know that
2Li + 2H LiOH + H2
then you should be able to predict the products from the reaction of Na, K and the other members of group 1 (alkali metals) with water. Thus a general reaction would be;
2 M H2O 2 MOH + H2

70. Types of Chemical Reactions: Direct Combination/Synthesis Reactions
Decomposition/Analysis Reactions
Single Replacement Reactions
Double Replacement Reactions
AND COMBUSTION REACTIONS

71. Direct Combination/Synthesis
the combination of 2 or more substances to form a compound
only one product forms
A + B  AB
ex: 2P + 3Br2  2PBr3

72. Decomposition/Analysis
a compound breaks down into 2 or more simpler substances
only one reactant
AB  A + B
ex: 2H2O2  2H2O + O2

73. A + BC  AC + B ex: Zn + 2HCl  ZnCl2 + H2 Single Replacement
one element replaces another in a compound
metal replaces metal (+) OR
nonmetal replaces nonmetal (-)
A + BC  AC + B
ex: Zn + 2HCl  ZnCl2 + H2

74. ex: 2KOH + CuSO4  K2SO4 + Cu(OH)2
Double Replacement
ions in two compounds “change partners”
cation(+) of one compound combines with anion(-) of the other
AB + CD  AD + CB
ex: 2KOH + CuSO4  K2SO4 + Cu(OH)2

75. CxHy+O2  CO2+H2O ex: CH4 + 2O2  CO2 + 2H2O Combustion
the burning of a substance in O2 to produce heat
hydrocarbons (CxHy) always form CO2 and H2O
CxHy+O2  CO2+H2O
ex: CH4 + 2O2  CO2 + 2H2O

76. Direct Combination/Synthesis Reactions
- when two or more elements or compound(s) react to form a product (opposite of decomposition reactions)
A + B C
(often heat released)
2 Na(s) + 3 N2(g) NaN3(s)
(ex. Air Bag Inflator)

77. Decomposition/Analysis Reactions
when one compound reacts to form two or more elements or compounds (opposite of synthesis reactions)
C A + B
(often heat required)
2 NaN3(s) Na(s) N2(g)
B(OH)3 (heat) HBO2 + H2O
(ex. Air Bag Inflator)

78. Chemical Reactivity
Sample Exercise: Write balanced chemical equations for the following reactions:
Solid mercury (II) sulfide decomposes into its component elements when heated.

79. Chemical Reactivity
Sample Exercise: Write balanced chemical equations for the following reactions:
Solid mercury (II) sulfide decomposes into its component elements when heated.
Hg+2 S-2
HgS  Hg + S

80. Chemical Reactivity
Sample Exercise: Write balanced chemical equations for the following reactions:
The surface of aluminum metal undergoes a combination reaction with oxygen in air.

81. Chemical Reactivity
Sample Exercise: Write balanced chemical equations for the following reactions:
The surface of aluminum metal undergoes a combination reaction with oxygen in air.
Al+3 O-2
Al + O2  Al2O3

82. Chemical Reactivity
Sample Exercise: Write balanced chemical equations for the following reactions:
The surface of aluminum metal undergoes a combination reaction with oxygen in air.
Al+3 O-2
4Al + 3O2  2Al2O3

83. Chemical Reactivity
Sample Exercise: Write balanced chemical equations for the following reactions:
Si2H6 burns when exposed to air.

84. Chemical Reactivity
Sample Exercise: Write balanced chemical equations for the following reactions:
Si2H6 burns when exposed to air. Hint, Si is in the same group as C, and therefore reacts similarly.
Si2H6 + O2  SiO2 + H2O

85. Chemical Reactivity
Sample Exercise: Write balanced chemical equations for the following reactions:
Si2H6 burns when exposed to air. Hint, Si is in the same group as C, and therefore reacts similarly.
2Si2H6 + 7O2  4SiO2 + 6H2O

86. Single Replacement Reactions
- a free element becomes an ion and an ion in solution becomes a neutral atom
A + BC  B + AC
Cl2(g) + 2NaBr(aq)  Br2(l) + 2NaCl(aq)

87. Double Replacement Reactions
when ionic “partners” switch
AB + CD AD + BC
(often in aqueous solutions)
Ag(NO3) + KCl AgCl(s) + KNO3
BaCl2 + Na2SO4 BaSO4(s) + 2 NaCl