1. Atoms: The Building Blocks of Matter
Chapter 3
Atoms: The Building Blocks of Matter

2. Objectives Define the terms atomic mass and molar mass.
Define the terms mole and Avogadro’s number.
Write the name for common elements, given the symbol, or the symbol, given the name.
Calculate the molar mass of an element or compound, given its formula.
Calculate the mass of an element or compound given the number of moles, or the number of moles of a given mass of an element or compound.
Calculate the number of atoms or molecules of an element or compound given the number of moles, or the number of moles given the number of atoms or molecules.

3. Chapter 3
Section 1
The Atom

4. The modern definition of an element is a substance that cannot be further broken down by ordinary chemical means – H, C, O
Elements also combine to form compounds that have different physical and chemical properties than those of the elements that form them – H2O.
The transformation of a substance into one or more new substances is a chemical reaction.

5. Relative Atomic Masses
Masses of atoms expressed in grams are very small and not useful. (O: 2.66 x g)
It is more convenient to use a new scale of relative atomic masses. (amu)
The standard used to govern units of atomic mass is the carbon-12 atom.
It has been assigned a mass of exactly atomic mass units, or 12 amu.

6. The atomic mass of any other atom is
determined by comparing it with the mass of the carbon-12 atom.
Examples:
The hydrogen atom has an atomic mass of 1/12 that of the carbon-12 atom or 1 amu.
Oxygen has an atomic mass of 16/12 the mass of a carbon-12 atom or 16 amu.

7. Relating Mass to Numbers of Atoms
Introduction of three very important concepts:
The mole
Avogadro’s number
Molar mass

8. The Mole
Mole – The mole is the SI unit for amount of substance. It is a counting unit.
Based on the number of carbon atoms that are in exactly 12 g of carbon.
One mole of carbon weighs 12 grams.
Mole is related to the counting term dozen.

9. What is a counting unit?
You’re already familiar with one counting unit…a “dozen”
A dozen = 12
“Dozen” 12
A dozen doughnuts
12 doughnuts
A dozen books
12 books
A dozen cars
12 cars
A dozen people
12 people

10. A Mole of Particles Contains 6.02 x 1023 particles
Avogadro’s Number
1 mole C = x 1023 C atoms
1 mole H2O = x 1023 H2O molecules
1 mole NaCl= x 1023 NaCl molecules

11. Avogadro’s Number – 6.02 x 1023
is the number of particles in exactly one mole of a pure substance.
1 mole of gold = 6.02 x 1023 particles
1 mole of uranium = 6.02 x 1023 particles
1 mole of water = 6.02 x 1023 particles

12. How big is a mole?
Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles.
If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

13. Molar Mass
Molar Mass – The mass (in grams) of one mole of a pure substance.
Molar masses are written in units g/mol.
The molar mass of an element is equal to the atomic mass of the element.
Look on the periodic table for the atomic masses.

14. Atomic Weight (elements)
Other terms commonly used for the same meaning of molar mass:
Molecular Weight
Molecular Mass
Atomic Weight (elements)

15. Molar Mass A molar mass of an element contains one mole of atoms.
1mole = 4.00 g helium = 6.02 x 1023 atoms.
1mole = 6.94 g lithium = 6.02 x 1023 atoms.
1mole = g mercury = 6.02 x atoms.

16. Molar Mass Examples: Molar mass of oxygen (O) = 15.99 g/mol
Molar mass of iron (Fe) = g/mol
Molar mass of gold (Au) = 197 g/mol

17. One mole of carbon (12 grams) and one mole of copper (63.5 grams)
Both contain 6.02 x 1023 atoms

18. Molar Mass for Compounds
The molar mass for a compound = the sum of the molar masses of all the elements in the compound.

19. Molar Mass
A molar mass of a compound is the sum of the molar masses of the elements.
Example: Water, H2O:
2 H = 2 x 1g/mole = 2g/mole
1 O = 1 x 16g/mole = 16g/mole
molar mass of H20 =18g/mol

20. Molar Mass
A molar mass of a compound is the sum of the molar masses of the elements.
Example: methane, CH4:
4 H = 4 x 1 g = 4 g/mole
1 C = 1 x 12 g = 12 g/mole
molar mass of CH4 =16 g/mol

21. Example: Molar Mass & Parenthesis
Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis.
Example:
Find the molar mass for Sr(NO3)2

22. Find the molar mass for Al(OH)3
Example
Example:
Find the molar mass for Al(OH)3

23. Homework
Worksheet C.5 – molar masses of compounds
Due:

24. Gram/Mole Conversions
How many roses are in 3 ½ dozen roses?
Relationship: 1 dozen roses = 12 roses
3.5 dozen x 12 roses = roses
1 dozen

25. Gram/Mole Conversions
Molar masses can be used as a conversion factor in chemical calculations.
Example: The molar mass of helium is 4.00 g/mol. How many grams of helium are in 2 moles of helium:
amount of He in moles amount of He in grams
2.00 mol He x = 8.00 g He
4.00 g He mol He

26. Amount in moles x molar mass (g/mol) = mass in grams
Gram/Mole Conversions
Molar masses can be used as a conversion factor in chemical calculations.
Amount in moles x molar mass (g/mol) = mass in grams
Example: What is the mass in grams of 2.50 mol of oxygen gas?
2.50 mol O2 x = 80.0 g O2
32.00 g O mol O2

27. Gram/Mole Conversions
A chemist produced 11.9 g of aluminum, Al. How many moles of aluminum were produced?
mass of Al in grams amount of Al in moles
1 mol Al
27 g Al
11.9 g Al x = 0.44 mol Al

28. Gram/Mole Conversions
How many moles are present in 352 g of iron(III) oxide, Fe2O3?
mass of Fe2O3 in grams amount of Fe2O3 in moles
Calculate the molar mass of Fe2O3 = 160 g/mol
1 mol Fe2O3
160 g
352 g Fe2O3 x = 2.2 mol Fe2O3

29. Gram/Mole Conversions
Review sample problem – page 82
Practice problems – Top of page 83, 1-4
(worksheet)

30. Gram/Mole Conversions
Practice problems – page 83 (bottom), 1-2
(worksheet)

31. Classwork Worksheet: C-7 Grams to mole conversions
Moles to gram conversions

32. Atoms / Gram Conversions

33. Atoms/Molecules and Grams
Since 6.02 X 1023 molecules = 1 mole AND 1 mole = molar mass (grams)
You can convert atoms/molecules to moles and then moles to grams! (Two step process)
You can’t go directly from atoms to grams!!!! You MUST go thru MOLES.

34. Everything must go through
Calculations
molar mass Avogadro’s number Grams Moles atoms
Everything must go through
Moles!!!

35. Atoms/Molecules and Grams
How many atoms of Cu are present in 35.4 g of Cu?
35.4 g Cu mol Cu X 1023 atoms Cu g Cu mol Cu
= 3.4 X 1023 atoms Cu

36. Problem
How many atoms of K are present in 78.4 g of K?

37. Atoms/Molecules and Grams
How many atoms of K are present in 78.4 g of K?
78.4 g K mol K X 1023 atoms K g K mol K
= 12.1 X 1023 atoms K

38. Problem
What is the mass (in grams) of 1.20 X 108 atoms of copper (Cu)?

39. Problem
What is the mass (in grams) of 1.20 X 108 atoms of copper (Cu)?
1.20 x 108 atoms mole Cu g Cu
1 mole Cu
6.02 x 1023 atoms
= 1.27 x grams copper

40. Problem
What is the mass (in grams) of 1.20 X 1024 molecules of glucose (C6H12O6)?

41. Problem
What is the mass (in grams) of 1.20 X 1024 molecules of glucose (C6H12O6)?
1.20 x 1024 mol mole glucose g glucose
1 mole glucose
6.02 x 1023 mol.
= 359 grams glucose

42. Classwork Problems – page 88
Questions: 21, 24 (a, c, e and f), (a-e)

43. Classwork
Test Review
Practice problems – page 226
Questions 1 - 3

44. Chemical Formulas and Chemical Compounds
Chapter 7
Chemical Formulas and Chemical Compounds

45. Chapter 7
Section 3
Chemical Formulas

46. A chemical formula indicates which elements are in a compound and how many of each element.
Example:
Water – H2O
Subscripts indicate there are two atoms of hydrogen and one atom of oxygen in water.

47. Example:
Aluminum sulfate – Al2(SO4)3
Parenthesis are used to surround the polyatomic group to identify it as a unit. The subscript 3 refers to everything inside the parenthesis.
Al – 2 atoms
S – 3 atoms
O – 12 atoms

48. Percentage Composition
It is often useful to know the percentage by mass of a particular element in a chemical compound.
Percentage Composition – the percentage by mass of each element in a compound.
Mass of element in a compound x = % element
molar mass of compound in cmpd

49. Example:
Find the percentage composition of each element in copper (I) sulfide, Cu2S.
The molar mass of Cu2S = g/mol
2 mole Cu x g Cu
mole
1 mole S x g S = g S
= g Cu

50. Example:
Find the percentage composition of each element in copper (I) sulfide, Cu2S.
The molar mass of Cu2S = g/mol
% Cu: g Cu
159.2 g Cu2S
% S: g S X 100 = 20.15% S
X 100 = 79.85% Cu

51. Find the percentage composition of each element in glucose, C6H12O6.
Molar mass of glucose = 180 g/mole

52. Classwork
Practice problem – page 228
Questions 1 A and B

53. Determining Chemical Formulas
Chapter 7
Section 4
Determining Chemical Formulas

54. Empirical Formula
When a new substance is discovered, it is analyzed to reveal its percent composition of elements.
This allows a chemist to determine the ratio of elements in the compound.
From this data, the empirical formula can be determined.

55. Empirical Formula – consists of the symbols for the elements in a compound, with subscripts showing the smallest whole-number mole ratio of the different elements in the compound.
Example: H2O not H4O2

56. Empirical formula does not necessarily indicate the actual numbers of each element in a molecule.
Example:
Diborane gas
Empirical formula is BH3
Molecular formula is B2H6

57. Calculating Empirical Formula
To determine a compound’s empirical formula from its percentage composition, begin by converting percentage composition to a mass composition.
% comp grams moles mole ratio

58. The percentage composition of a new compound containing boron and hydrogen was found to be:
Convert the percentages to grams:
78 g of B and 22 g of H.

59. Next the mass composition of each element is converted to moles:
78 g B x = mol B
1 mol B
10.81 g B
1 mol H
1.01 g H
22 g H x = mol H

60. These values give a mole ratio of 7.22 mol B to 21.7 mol H.
Convert these numbers to the smallest whole numbers:
7.22 mol B
7.22
= 1
21.7 mol H
7.22
= 3.01

61. Thus, diborane contains a mole ratio of
1 B : 3 H to give an empirical formula of:
BH3
Review Sample Problem 7-12 and 7-13,
Page 230 and 231

62. Calculating Molecular Formula
The molecular formula is the actual formula
for the compound.
To know the molecular formula you must know the compounds molar mass.

63. Example:
Experimentation shows the molecular weight for diborane is g/mol.
The molecular weight for the empirical formula BH3 is g/mol.
Dividing the formula mass by the empirical mass yields:
= 2
13.84

64. Take this number and times it by the subscripts in the empirical formula to get the molecular formula:
2 X BH3 = B2H6

65. Classwork Do Practice Problems 1-3 , page 231 Page 233: Section Review
Questions 1- 3

66. Find the mass percentage of water in in ZnSO4 ● 7H2O
Analysis of a compound contains 4.0 g calcium and 16 g of bromide. What is the empirical formula of the compound?

67. Homework Page 237: Chapter 7 Review Problems
Questions 34, 35, 36, 37, 38

68. Volume-Mass Relationships of Gases
Chapter 11
Section 1
Volume-Mass Relationships of Gases

69. Objectives
Calculate the volume of a gas at STP (standard temperature and pressure), or the number of moles of a given volume of an element or compound.

70. Gases
Monoatomic Gases – gases which are composed of only one atom of the element. Ex: He, Ar, Xe – Noble gases
Diatomic Gases – gases which are composed of only two atom of the element. Ex: H2, N2, O2, F2, Cl2

71. Molar Volume of Gases
Recall that one mole of any substance contains a number of molecules equal to Avogadro's number – 6.02 x 1023
One mole of oxygen gas (O2), contains x 1023 diatomic oxygen molecules and has a mass of g/mol.

72. One mole of hydrogen gas (H2), contains 6
One mole of hydrogen gas (H2), contains x 1023 diatomic hydrogen molecules and has a mass of 2.0 g/mol.
One mole of helium gas (He), contains x 1023 monatomic helium atoms and has a mass of 4.0 g/mol.

73. Avogadro’s Law – equal volumes of gases at the same temperature and pressure contain equal number of molecules.
According to Avogadro’s law, one mole of any gas will occupy the same volume as one mole of any other gas at the same temperature and pressure, despite mass differences.
One mole of H2, O2 and He occupy the same volume at same temp. and pressure.

74. Standard Molar Volume of a Gas – the volume occupied by one mole of a gas at STP (standard temperature and pressure).
The volume = 22.4 L
Standard temperature = 273K = 0oC
Standard pressure is 1 atm.

75. mass = 32.0 g/mol mass = 2.0 g/mol
Knowing the volume of a gas, you can use 1 mol/22.4 L as a conversion factor to find the number of moles and mass of a given volume of gas at STP.
1 mol of O vs mol of H2
volume = 22.4 L volume = 22.4 L
mass = 32.0 g/mol mass = 2.0 g/mol

76. Problem:
A chemical reaction produces mol of oxygen gas. What volume in liters is occupied by this gas at STP?
moles of O volume of O2 in L
mol x = volume of O2 in L
22.4 L
1 mol

77. moles of O2 volume of O2 in L
mol x = volume of O2 in L
0.068 mol x = L of O2
22.4 L
1 mol
22.4 L
1 mol

78. Problem:
A chemical reaction produced 98.0 mL of sulfur dioxide gas, SO2, at STP. What is the mass in grams of the gas produced?
liters of SO moles of SO grams of SO2
L x x = g SO2
1 mol SO2
22.4 L
g SO2__
1 mol SO2

79. L x x = g SO2 1 mol SO2 22.4 L g SO2__ 1 mol SO2 64.07g SO2__
liters of SO moles of SO grams of SO2
L x x = g SO2
1 mol SO2
22.4 L
g SO2__
1 mol SO2
1 mol SO2
22.4 L
64.07g SO2__
1 mol SO2
0.098 L x x = g SO2

80. We will use the concept of molar volume of gases when we talk about the gas laws later in the year!

81. Classwork Page 336-337: Practice Problems Questions 1-3

82. Chapter 13
Section 3
Concentrations of
Solutions

83. Objectives Define a solution’s concentration in terms of molarity.
Calculate the concentration of a solution, given the mass of solute and volume of solution, or the mass of solute needed to prepare a solution of a given concentration.

84. What is solubility? Imagine dissolving a spoonful of salt in water
What is the solute?
What is the solvent? 84

85. Concentration of a solution – a measure of the amount of solute in a given amount of solvent.
Common solution terms:
Dilute – small amount of solute in a solvent.
Concentrated – a large amount of solute in a solvent

86. Molarity
Molarity (M) – the number of moles of solute in one liter of solution.
To determine the molarity of a solution you must know the molar mass (molecular weight) of the solute.

87. The relationship between molarity, moles and volume is the following:
Amount of solute (mol)
Volume of solution (L)
Molarity (M) =

88. For example, a one molar solution of sodium hydroxide, (NaOH), contains one mole of NaOH in one liter of solution.
1 M NaOH =
Since 1 mole of NaOH = 40 g
1M NaOH =
1 mol NaOH
1 L
40.0g NaOH
1 L

89. Example problem:
What is the molarity of 20 g of NaOH in 1 L of solution? (NaOH = 40 g/mol)
Amount of solute (mol)
Volume of solution (L)
Molarity (M) =
0.50 mol NaOH
1.0 L
Molarity (M) =
Molarity (M) = M

90. Example problem:
What is the molarity of a 3.50 L solution that contains 90.0 g of sodium chloride, NaCl?
Amount of solute (mol)
Volume of solution (L)
Molarity (M) =
You first need to convert 90.0 g of NaCl to moles of NaCl.
90.0 g___
58.5 g/mol
Moles of NaCl = = mol

91. Amount of solute (mol)
Volume of solution (L)
Molarity (M) =
1.54 mol NaCl
3.50 L
Molarity (M) =
Molarity (M) = M NaCl

92. You have 0. 8 L of a 0. 5 M HCl solution
You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? (HCl = 36.5 g/mol)
Amount of solute (mol)
Volume of solution (L)
Molarity (M) =
moles of HCl
0.8 L
0.5 M =
moles of HCl = (0.50 M)(0.8 L) = 0.4 mol
HCl

94. Classwork
Page 415:
Questions 1-3

95. Homework Page 421: Chapter 13 Review Problems
Questions 15 (b), 16 (a-b), 17, 18 (a,c), 19