1. Chapter 7 Section 2 Relative Atomic Mass and Chemical Formulas
The Mole and Chemical Composition
Chapter 7
Section 2
Relative Atomic Mass and Chemical Formulas

2. Chapter 7 Chapter 7 Objectives
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Chapter 7
Objectives
Use 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.

3. Average Atomic Mass and the Periodic Table
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Average Atomic Mass and the Periodic Table
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.

4. Calculating Average Atomic Mass
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Calculating Average Atomic Mass
Sample Problem 1:
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%

5. Calculating Average Atomic Mass
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Calculating Average Atomic Mass
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

6. Calculating Average Atomic Mass
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Calculating Average Atomic Mass
Sample Problem 2:
The mass of a Cl - 35 atom is amu, and makes up 75.8% of Cl atoms. The rest of naturally occurring chlorine is Cl- 37, with atom of amu. What is the average atomic mass of chlorine?

7. Calculating Average Atomic Mass
Section 2 Relative Atomic Mass and Chemical Formulas
Chapter 7
Calculating Average Atomic Mass
U-234 makes up % of uranium atoms amd has a mass of amu. U-235 makes up 0.720% and has a mass of amu. U-238 has a mass of amu and makes up %. What is the average atomic mass of uranium?
Carbon -12 makes up 98.90% of existing carbon. Carbon-13, with a mass of , makes up 1.10%. What is the average atomic mass of carbon?
Page 236, Practice problem # 1, 2

8. 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.

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

10. 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.

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

12. 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.

13. 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

14. 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

15. Chapter 7 Section 3 Formulas and Percentage Composition
The Mole and Chemical Composition
Chapter 7
Section 3
Formulas and Percentage Composition

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

17. 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.

18. 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 also can be used to compare the ratio of masses contributed by the elements in two different substances.

19. Chapter 7
Percentage Composition of Iron Oxides

20. 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.

21. Chapter 7
Empirical and Actual Formulas

22. 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.

23. 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.

24. 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

25. 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.

26. 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.

27. Using Analytical Data, continued
Section 3 Formulas and Percentage Composition
Chapter 7
Using Analytical Data, continued
Molecular Formulas Are Multiples of Empirical Formulas
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.

28. 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.

29. 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

30. 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

31. 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.

32. 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.

33. 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.

34. 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

35. 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

36. 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

37. 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

38. 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

39. 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

40. 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

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

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

43. 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?

44. 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

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

46. 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.

47. 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?

48. 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

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

50. 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

51. 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

52. 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.

53. 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.

54. 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

55. 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

56. 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

57. 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