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2. Standardized Test Prep
Resources
Chapter Presentation
Bellringer
Transparencies
Sample Problems
Visual Concepts
Standardized Test Prep

3. Chapter 7 Table of Contents
The Mole and Chemical Composition
Chapter 7
Table of Contents
Section 1 Avogadro’s Number and Molar Conversions
Section 2 Relative Atomic Mass and Chemical Formulas
Section 3 Formulas and Percentage Composition

4. Chapter 7 Chapter 7 Bellringer
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Chapter 7
Bellringer
List as many common counting units as you can.
Determine how many groups of each unit in your list are present in each of the following amounts:
500 goldfish
150 unicycles
50 jet planes

5. Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Objectives
Identify the mole as the unit used to count particles, whether atoms, ions, or molecules.
Use Avogadro’s number to convert between amount in moles and number of particles.
Solve problems converting between mass, amount in moles, and number of particles using Avogadro’s number and molar mass.

6. Avogadro’s Number and the Mole
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Avogadro’s Number and the Mole
The SI unit for amount is called the mole (mol). A mole is the number of atoms in exactly 12 grams of carbon-12.
Scientists use the mole to make counting large numbers of particles easier.
The number of particles in a mole is called Avogadro’s Number. Avogadro’s number is  1023 units/mole.

7. Avogadro’s Number and the Mole, continued
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Avogadro’s Number and the Mole, continued
The Mole Is a Counting Unit
The mole is used to count out a given number of particles, whether they are atoms, molecules, formula units, ions, or electrons.
The mole is just one kind of counting unit:
1 dozen = 12 objects
1 hour = 3600 seconds
1 mole =  1023 particles

8. Chapter 7
Visual Concepts
The Mole

9. Avogadro’s Number and the Mole, continued
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Avogadro’s Number and the Mole, continued
Amount in Moles Can Be Converted to Number of Particles
Counting units are used to make conversion factors.
The definition of one mole is
6.022  1023 particles = 1 mol
The conversion factor is

10. Avogadro’s Number and the Mole, continued
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Avogadro’s Number and the Mole, continued
Choose the Conversion Factor That Cancels the Given Units
All conversion factors are equal to 1, so you can use them to convert among different units.
You can tell which conversion factor to use, because the needed conversion factor should cancel the units of the given quantity to give you the units of the answer or the unknown quantity.

11. Converting Between Amount in Moles and Number of Particles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Between Amount in Moles and Number of Particles

12. Converting Amount in Moles to Number of Particles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Amount in Moles to Number of Particles
Sample Problem A
Find the number of molecules in 2.5 mol of sulfur dioxide.

13. Converting Amount in Moles to Number of Particles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Amount in Moles to Number of Particles
Sample Problem A Solution
2.5 mol SO2  ? = ? molecules SO2
You are converting from the unit mol to the unit molecules. The conversion factor must have the units of molecules/mol.
You use  1023 molecules/1 mol.

14. Converting Amount in Moles to Number of Particles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Amount in Moles to Number of Particles
Sample Problem A Solution, continued
2.5 mol SO2  ? = ? molecules SO2
2.5 mol SO2 = 1.5  1024 molecules SO2

15. Avogadro’s Number and the Mole, continued
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Avogadro’s Number and the Mole, continued
Number of Particles Can Be Converted to Amount in Moles
The reverse calculation is similar to that in Sample Problem A but the conversion factor is inverted to get the correct units in the answer.
example: How many moles are 2.54  1022 iron(III) ions?

16. Converting Number of Particles to Amount in Moles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Number of Particles to Amount in Moles
Sample Problem B
A sample contains 3.01  1023 molecules of sulfur dioxide, SO2. Determine the amount in moles.

17. Converting Number of Particles to Amount in Moles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Number of Particles to Amount in Moles
Sample Problem B Solution
3.01 × 1023 molecules SO2  ? = ? mol SO2
You are converting from the unit molecules to the unit mol. The conversion factor must have the units of mol/molecules.
You use 1 mol/6.022  1023 molecules.

18. Converting Number of Particles to Amount in Moles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Number of Particles to Amount in Moles
Sample Problem B Solution, continued
3.01  1023 molecules SO2  ? = ? mol SO2
3.01  1023 molecules SO2 = mol SO2

19. Molar Mass Relates Moles to Grams
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Molar Mass Relates Moles to Grams
Amount in Moles Can Be Converted to Mass
The molar mass is the mass in grams of one mole of an element or compound.
Molar mass is numerically equal to the atomic mass of monatomic elements and the formula mass of compounds and diatomic elements.
The units for molar mass are g/mol.
Molar mass can be used as a conversion factor in problems converting between mass and amount.

20. Chapter 7
Visual Concepts
Molar Mass

21. Molar Mass as a Conversion Factor
Chapter 7
Visual Concepts
Molar Mass as a Conversion Factor

22. Molar Mass Relates Moles to Grams, continued
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Molar Mass Relates Moles to Grams, continued
The Mole Plays a Central Part in Chemical Conversions
To convert from number of particles to mass, you must use a two-part process:
First, convert number of particles to amount in moles.
Second, convert amount in moles to mass in grams.
One step common to many problems in chemistry is converting to amount in moles.

23. Converting Between Mass, Amount, and Number of Particles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Between Mass, Amount, and Number of Particles

24. Converting Number of Particles to Mass
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Number of Particles to Mass
Sample Problem C
Find the mass in grams of 2.44  1024 atoms of carbon, whose molar mass is g/mol.

25. Converting Number of Particles to Mass
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Number of Particles to Mass
Sample Problem C Solution
First part:
2.44  1024 atoms  ? = ? mol
Select the conversion factor that will take you from number of atoms to amount in moles.
You use 1 mol/6.022  1023 atoms.

26. Converting Number of Particles to Mass
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Number of Particles to Mass
Sample Problem C Solution, continued
Second part:
? mol  ? = ? g
Select the conversion factor that will take you from amount in moles to mass in grams.
You use the molar mass of carbon, g C/1 mol.

27. Converting Number of Particles to Mass
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Number of Particles to Mass
Sample Problem C Solution, continued
= 48.7 g C

28. Molar Mass Relates Moles to Grams, continued
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Molar Mass Relates Moles to Grams, continued
Mass Can Be Converted to Amount in Moles
Converting from mass to number of particles is the reverse of the operation in the previous problem.
To convert from mass to number of particles, you must use a two-part process:
First, convert mass in grams to amount in moles.
Second, convert amount in moles to number of particles.

29. Converting Mass to Number of Particles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Mass to Number of Particles
Sample Problem D
Find the number of molecules present in
47.5 g of glycerol, C3H8O3. The molar mass of glycerol is g/mol.

30. Converting Mass to Number of Particles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Mass to Number of Particles
Sample Problem D Solution
First part:
47.5 g  ? = ? mol
Select the conversion factor that will take you from mass in grams to amount in moles.
You use the inverse of the molar mass of glycerol:

31. Converting Mass to Number of Particles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Mass to Number of Particles
Sample Problem D Solution, continued
Second part:
? mol  ? = ? molecules
Select the conversion factor that will take you from amount in moles to number of particles.
You use

32. Converting Mass to Number of Particles
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Converting Mass to Number of Particles
Sample Problem D Solution, continued
= 3.11  1023 molecules

33. Molar Mass Relates Moles to Grams, continued
Section 1 Avogadro’s Number and Molar Conversions
Chapter 7
Molar Mass Relates Moles to Grams, continued
Remember to Round Consistently
Remember that an answer must never be given to more significant figures than is appropriate.
Round molar masses from the periodic table to two significant figures to the right of the decimal point.

34. Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Bellringer
Compare the masses of a roll of pennies and a roll of dimes. Both contain 50 coins.
Why are the masses of the rolls different when both rolls contain the same number of coins?
If given a roll of mixed coins, what information would you need to determine the mass of the roll?

35. Chapter 7 Chapter 7 Objectives
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Chapter 7
Objectives
Use a periodic table or isotopic composition data to determine the average atomic masses of elements.
Infer information about a compound from its chemical formula.
Determine the molar mass of a compound from its formula.

36. Average Atomic Mass and the Periodic Table
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Average Atomic Mass and the Periodic Table
Most Elements are a Mixture of Isotopes
Isotopes are atoms that have different numbers of neutrons than other atoms of the same element do.
Average atomic mass is a weighted average of the atomic mass of an element’s isotopes.
If you know the abundance of each isotope, you can calculate the average atomic mass of an element.

37. Chapter 7
Visual Concepts
Average Atomic Mass

38. Calculating Average Atomic Mass
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Calculating Average Atomic Mass
Sample Problem E
The mass of a Cu-63 atom is amu, and that of a Cu-65 atom is amu. Using the data below, find the average atomic mass of copper.
abundance of Cu-63 = 69.17%
abundance of Cu-65 = 30.83%

39. Calculating Average Atomic Mass
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Calculating Average Atomic Mass
Sample Problem E Solution
The contribution of each isotope is equal to its atomic mass multiplied by the fraction of that isotope.
contribution of Cu-63: amu ×
contribution of Cu-65: amu ×
Average atomic mass is the sum of the individual contributions:
(62.94 amu × ) + (64.93 amu × )
= amu

40. Chemical Formulas and Moles
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Chemical Formulas and Moles
Formulas Express Composition
A compound’s chemical formula tells you which elements, as well as how much of each, are present in a compound.
Formulas for covalent compounds show the elements and the number of atoms of each element in a molecule.
Formulas for ionic compounds show the simplest ratio of cations and anions in any pure sample.

41. Chemical Formulas and Moles, continued
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Chemical Formulas and Moles, continued
Formulas Express Composition, continued
Any sample of compound has many atoms and ions, and the formula gives a ratio of those atoms or ions.

42. Chemical Formulas and Moles, continued
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Chemical Formulas and Moles, continued
Formulas Give Ratios of Polyatomic Ions
Formulas for polyatomic ions show the simplest ratio of cations and anions.
They also show the elements and the number of atoms of each element in each ion.
For example, the formula KNO3 indicates a ratio of one K+ cation to one anion.

43. Understanding Formulas for Polyatomic Ionic Compounds
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Understanding Formulas for Polyatomic Ionic Compounds

44. Chemical Formulas and Moles, continued
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Chemical Formulas and Moles, continued
Formulas Are Used to Calculate Molar Masses
The molar mass of a molecular compound is the sum of the masses of all the atoms in the formula expressed in g/mol.
The molar mass of an ionic compound is the sum of the masses of all the atoms in the formula expressed in g/mol.

45. Chapter 7
Visual Concepts
Formula Mass

46. Calculating Molar Mass for Ionic Compounds
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Calculating Molar Mass for Ionic Compounds

47. Calculating Molar Mass of Compounds
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Calculating Molar Mass of Compounds
Sample Problem F
Find the molar mass of barium nitrate, Ba(NO3)2.

48. Calculating Molar Mass of Compounds
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Calculating Molar Mass of Compounds
Sample Problem F Solution
Find the number of moles of each element in 1 mol Ba(NO3)2:
Each mole has 1 mol Ba, 2 mol N, and 6 mol O.
Use the periodic table to find the molar mass of each element in the formula:
molar mass of Ba: g/mol
molar mass of N: g/mol
molar mass of O: g/mol

49. Calculating Molar Mass of Compounds
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Calculating Molar Mass of Compounds
Sample Problem F Solution, continued
Multiply the molar mass of each element by the number of moles of each element. Add these masses to get the total molar mass of Ba(NO3)2.
mass of 1 mol Ba = 1  g/mol = g/mol
mass of 2 mol N = 2  g/mol = g/mol
+ mass of 6 mol O = 6  g/mol = g/mol
molar mass of Ba(NO3)2 = g/mol

50. Section 3 Formulas and Percentage Composition
Chapter 7
Bellringer
Brainstorm a list of what you know about percentages.

51. Section 3 Formulas and Percentage Composition
Chapter 7
Objectives
Determine a compound’s empirical formula from its percentage composition.
Determine the molecular formula or formula unit of a compound from its empirical formula and its formula mass.
Calculate percentage composition of a compound from its molecular formula or formula unit.

52. Chapter 7 Using Analytical Data
Section 3 Formulas and Percentage Composition
Chapter 7
Using Analytical Data
The percentage composition is the percentage by mass of each element in a compound.
Percentage composition helps verify a substance’s identity.
Percentage composition also can be used to compare the ratio of masses contributed by the elements in two different substances.

53. Chapter 7
Percentage Composition of Iron Oxides

54. Using Analytical Data, continued
Section 3 Formulas and Percentage Composition
Chapter 7
Using Analytical Data, continued
Determining Empirical Formulas
An empirical formula is a chemical formula that shows the simplest ratio for the relative numbers and kinds of atoms in a compound.
An actual formula shows the actual ratio of elements or ions in a single unit of a compound.
For example, the empirical formula for ammonium nitrate is NH2O, while the actual formula is NH4NO2.

55. Chapter 7
Empirical and Actual Formulas

56. Using Analytical Data, continued
Section 3 Formulas and Percentage Composition
Chapter 7
Using Analytical Data, continued
Determining Empirical Formulas, continued
You can use the percentage composition for a compound to determine its empirical formula.
Convert the percentage of each element to g.
Convert from g to mol using the molar mass of each element as a conversion factor.
Compare these amounts in mol to find the simplest whole-number ratio among the elements.

57. Percentage Composition
Chapter 7
Visual Concepts
Percentage Composition

58. Determining an Empirical Formula from Percentage Composition
Section 3 Formulas and Percentage Composition
Chapter 7
Determining an Empirical Formula from Percentage Composition
Sample Problem G
Chemical analysis of a liquid shows that it is 60.0% C, 13.4% H, and 26.6% O by mass. Calculate the empirical formula of this substance.

59. Determining an Empirical Formula from Percentage Composition
Section 3 Formulas and Percentage Composition
Chapter 7
Determining an Empirical Formula from Percentage Composition
Sample Problem G Solution
Assume that you have a g sample, and convert the percentages to grams.
for C: 60.0%  g = 60.0 g C
for H: 13.4%  g = 13.4 g H
for O: 26.6%  g = 26.6 g O

60. Determining an Empirical Formula from Percentage Composition
Section 3 Formulas and Percentage Composition
Chapter 7
Determining an Empirical Formula from Percentage Composition
Sample Problem G Solution, continued
Convert the mass of each element into the amount in moles, using the reciprocal of the molar mass.

61. The empirical formula is C3H8O.
Section 3 Formulas and Percentage Composition
Chapter 7
Determining an Empirical Formula from Percentage Composition
Sample Problem G Solution, continued
The formula can be written as C5H13.3O1.66, but you divide by the smallest subscript to get whole numbers.
The empirical formula is C3H8O.

62. Using Analytical Data, continued
Section 3 Formulas and Percentage Composition
Chapter 7
Using Analytical Data, continued
Molecular Formulas Are Multiples of Empirical Formulas
The formula for an ionic compound shows the simplest whole-number ratio of the large numbers of ions in a crystal of the compound.
A molecular formula is a whole-number multiple of the empirical formula.
The molar mass of any compound is equal to the molar mass of the empirical formula times a whole number, n.

63. Comparing Molecular and Empirical Formulas
Chapter 7
Visual Concepts
Comparing Molecular and Empirical Formulas

64. Determining a Molecular Formula from an Empirical Formula
Section 3 Formulas and Percentage Composition
Chapter 7
Determining a Molecular Formula from an Empirical Formula
Sample Problem H
The empirical formula for a compound is P2O5. Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound.

65. Determining a Molecular Formula from an Empirical Formula
Section 3 Formulas and Percentage Composition
Chapter 7
Determining a Molecular Formula from an Empirical Formula
Sample Problem H Solution
Find the molar mass of the empirical formula P2O5.
2  molar mass of P = g/mol
+ 5  molar mass of O = g/mol
molar mass of P2O5 = g/mol

66. n (empirical formula) = 2 (P2O5) = P4O10
Section 3 Formulas and Percentage Composition
Chapter 7
Determining a Molecular Formula from an Empirical Formula
Sample Problem H Solution, continued
n (empirical formula) = 2 (P2O5) = P4O10

67. Using Analytical Data, continued
Section 3 Formulas and Percentage Composition
Chapter 7
Using Analytical Data, continued
Chemical Formulas Can Give Percentage Composition
If you know the chemical formula of any compound, then you can calculate the percentage composition.
From the subscripts, determine the mass contributed by each element and add these to get molar mass.
Divide the mass of each element by the molar mass.
Multiply by 100 to find the percentage composition of that element.

68. Using Analytical Data, continued
Section 3 Formulas and Percentage Composition
Chapter 7
Using Analytical Data, continued
Chemical Formulas Can Give Percentage Composition
CO and CO2 are both made up of C and O, but they have different percentage compositions.

69. Using a Chemical Formula to Determine Percentage Composition
Section 3 Formulas and Percentage Composition
Chapter 7
Using a Chemical Formula to Determine Percentage Composition
Sample Problem I
Calculate the percentage composition of copper(I) sulfide, Cu2S, a copper ore called chalcocite.

70. Using a Chemical Formula to Determine Percentage Composition
Section 3 Formulas and Percentage Composition
Chapter 7
Using a Chemical Formula to Determine Percentage Composition
Sample Problem I Solution
Find the molar mass of Cu2S.
2 mol  g Cu/mol = g Cu
mol  g S/mol = g S
molar mass of Cu2S = g/mol

71. Using a Chemical Formula to Determine Percentage Composition
Section 3 Formulas and Percentage Composition
Chapter 7
Using a Chemical Formula to Determine Percentage Composition
Sample Problem I Solution, continued
Calculate the fraction that each element contributes to the total mass by substituting the masses into the equations below and rounding correctly.
79.852% Cu

72. Using a Chemical Formula to Determine Percentage Composition
Section 3 Formulas and Percentage Composition
Chapter 7
Using a Chemical Formula to Determine Percentage Composition
Sample Problem I Solution, continued
20.15% S

73. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
1. Element A has two isotopes. One has an atomic mass of 120 and constitutes 60%; the other has an atomic mass of 122 and constitutes 40%. Which range below includes the average atomic mass of Element A?
A. less than 120
B. between 120 and 121
C. between 121 and 122
D. greater than 122

74. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
1. Element A has two isotopes. One has an atomic mass of 120 and constitutes 60%; the other has an atomic mass of 122 and constitutes 40%. Which range below includes the average atomic mass of Element A?
A. less than 120
B. between 120 and 121
C. between 121 and 122
D. greater than 122

75. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
2. Which of the following can be determined from the empirical formula of a compound alone?
F. the true formula of the compound
G. the molecular mass of the compound
H. the percentage composition of the compound
I. the arrangement of atoms within a molecule of the compound

76. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
2. Which of the following can be determined from the empirical formula of a compound alone?
F. the true formula of the compound
G. the molecular mass of the compound
H. the percentage composition of the compound
I. the arrangement of atoms within a molecule of the compound

77. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
3. How many ions are in 0.5 moles of NaCl?
A  1023
B  1023
C  1023
D  1023

78. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
3. How many ions are in 0.5 moles of NaCl?
A  1023
B  1023
C  1023
D  1023

79. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
4. How many moles of calcium (mass = 40.1) are in a serving of milk containing 290 mg of calcium?

80. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
4. How many moles of calcium (mass = 40.1) are in a serving of milk containing 290 mg of calcium?
Answer: 7.23  10–3 mol

81. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
5. How is Avogadro's number related to moles?

82. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
5. How is Avogadro's number related to moles?
Answer: Avogadro's number is the number of particles in a mole.

83. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
6. Antimony has two isotopes. One, amounting to 57.3% of the atoms, has a mass of amu. The other, 42.7% of the atoms, has a mass of amu. What is the average atomic mass of antimony?

84. Understanding Concepts
Chapter 7
Standardized Test Preparation
Understanding Concepts
6. Antimony has two isotopes. One, amounting to 57.3% of the atoms, has a mass of amu. The other, 42.7% of the atoms, has a mass of amu. What is the average atomic mass of antimony?
Answer: amu

85. Chapter 7 Reading Skills Standardized Test Preparation
Read the passage below. Then answer the questions.
In 1800 two English chemists, Nicholson and Carlisle, discovered that when an electric current is passed through water, hydrogen and oxygen were produced in a 2:1 volume ratio and a 1:8 mass ratio. This evidence helped to support John Dalton's theory that matter consisted of atoms, demonstrating that water consists of the two elements in a constant proportion. If the same number of moles of each gas occupy the same volume, then each molecule of water must consist of twice as much hydrogen as oxygen, even though the mass of hydrogen is only one-eighth that of oxygen.

86. Chapter 7 Reading Skills
Standardized Test Preparation
Reading Skills
7. Based on this experiment, what is the empirical formula of water?
F. HO
G. H2O
H. H2O8
I. O8H

87. Chapter 7 Reading Skills
Standardized Test Preparation
Reading Skills
7. Based on this experiment, what is the empirical formula of water?
F. HO
G. H2O
H. H2O8
I. O8H

88. Chapter 7 Reading Skills
Standardized Test Preparation
Reading Skills
8. How would the experimental result have been different if hydrogen gas existed as individual atoms while oxygen formed molecules with two atoms bound by a covalent bond?
A. The result would be the same.
B. The ratio of hydrogen to oxygen would be 1:1.
C. The ratio of hydrogen to oxygen would be 1:4.
D. The ratio of hydrogen to oxygen would be 4:1.

89. Chapter 7 Reading Skills
Standardized Test Preparation
Reading Skills
8. How would the experimental result have been different if hydrogen gas existed as individual atoms while oxygen formed molecules with two atoms bound by a covalent bond?
A. The result would be the same.
B. The ratio of hydrogen to oxygen would be 1:1.
C. The ratio of hydrogen to oxygen would be 1:4.
D. The ratio of hydrogen to oxygen would be 4:1.

90. Chapter 7 Reading Skills
Standardized Test Preparation
Reading Skills
9. How does the empirical formula for water compare to its molecular formula?

91. Chapter 7 Reading Skills
Standardized Test Preparation
Reading Skills
9. How does the empirical formula for water compare to its molecular formula?
Answer: Both formulas are the same, H2O.

92. Interpreting Graphics
Chapter 7
Standardized Test Preparation
Interpreting Graphics
Use the diagram below to answer questions 10–13.

93. Interpreting Graphics
Chapter 7
Standardized Test Preparation
Interpreting Graphics
10. How many moles of oxygen atoms are there in moles of carbon dioxide?
F. 66.7
G. 72.7 H I

94. Interpreting Graphics
Chapter 7
Standardized Test Preparation
Interpreting Graphics
10. How many moles of oxygen atoms are there in moles of carbon dioxide?
F. 66.7
G. 72.7 H I

95. Interpreting Graphics
Chapter 7
Standardized Test Preparation
Interpreting Graphics
11. Explain why the percentage of oxygen in carbon dioxide is not twice the percentage of oxygen in carbon monoxide, if there are twice as many oxygen atoms.

96. Interpreting Graphics
Chapter 7
Standardized Test Preparation
Interpreting Graphics
11. Explain why the percentage of oxygen in carbon dioxide is not twice the percentage of oxygen in carbon monoxide, if there are twice as many oxygen atoms.
Answer: Although there are twice as many oxygen atoms in carbon dioxide, the percentage composition is based on the mass, not the number of atoms.

97. Interpreting Graphics
Chapter 7
Standardized Test Preparation
Interpreting Graphics
12. If you did not know the true formulas for carbon monoxide and carbon dioxide, what information would you need beyond what is provided in the illustration in order to calculate them?
A. the percentage compositions
B. the atomic masses of carbon and oxygen
C. the melting and boiling points of each compound
D. the number of atoms of each element in the compound

98. Interpreting Graphics
Chapter 7
Standardized Test Preparation
Interpreting Graphics
12. If you did not know the true formulas for carbon monoxide and carbon dioxide, what information would you need beyond what is provided in the illustration in order to calculate them?
A. the percentage compositions
B. the atomic masses of carbon and oxygen
C. the melting and boiling points of each compound
D. the number of atoms of each element in the compound

99. Interpreting Graphics
Chapter 7
Standardized Test Preparation
Interpreting Graphics
13. How many grams of carbon are contained in grams of carbon dioxide? F G H I

100. Interpreting Graphics
Chapter 7
Standardized Test Preparation
Interpreting Graphics
13. How many grams of carbon are contained in grams of carbon dioxide? F G H I