1. Chapter 7 Chemical Quantities
The Mole
Atomic Mass and Formula Mass
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2. The Mole Copyright © 2008 by Pearson Education, Inc.
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3. Collection Terms A collection term states a specific number of items.
1 dozen donuts = 12 donuts
1 ream of paper = 500 sheets
1 case = 24 cans
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4. A Mole of Atoms A mole (mol) is a collection that contains
The same number of particles as there are carbon atoms in g of carbon.
6.022 x 1023 atoms of an element (Avogadro’s number).
1 mol C = x 1023 C atoms
1 mol Na = x 1023 Na atoms
1 mol Au = x 1023 Au atoms

5. A Mole of A Compound A mole
Of a covalent compound has Avogadro’s number of _________________.
1 mol CO2 = x 1023 CO2 molecules
1 mol H2O = x 1023 H2O molecules
Of an ionic compound contains Avogadro’s number of ___________ units.
1 mol NaCl = x 1023 NaCl formula units
1 mol K2SO4 = x 1023 K2SO4 formula units

6. Samples of One Mole Quantities
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7. Avogadro’s Number Avogadro’s number 6.022 x 1023 can be written as
an equality and two conversion factors.
As an equality:
1 mol particles = x 1023 particles
As conversion Factors:
6.022 x 1023 particles and mol particles
1 mol particles x 1023 particles

8. Using Avogadro’s Number
Converts moles of a substance to the number of particles.
How many Cu atoms are in 0.50 mol
Cu?
0.50 mol Cu x x 1023 Cu atoms
1 mol Cu
= 3.0 x 1023 Cu atoms
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9. Using Avogadro’s Number
Converts particles of a substance to moles.
How many moles of CO2 are
2.50 x 1024 CO2 molecules?
2.50 x 1024 CO2 x 1 mol CO2
6.022 x 1023 CO2
= mol CO2

10. Learning Check The number of atoms in 2.0 mol Al is
2.0 mol Al x x 1023 Al atoms =
1 mol Al
1.2 x 1024 Al atoms

11. Learning Check The number of moles of S in 1.8 x 1024 atoms S is
1.8 x 1024 S atoms x mol S =
6.022 x 1023 S atoms
3.0 mol S atoms

12. Subscripts and Moles The subscripts in a formula state
The relationship of atoms in the formula.
The moles of each element in 1 mol of compound.
Glucose
C6H12O6
In 1 molecule: 6 atoms C 12 atoms H 6 atoms O
In 1 mol: 6 mol C 12 mol H 6 mol O

13. Subscripts State Atoms and Moles
1 mol C9H8O4 = mol C 8 mol H 4 mol O
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14. Factors from Subscripts
Subscripts used for conversion factors
Relate moles of each element in 1 mol compound.
For aspirin C9H8O4 can be written as:
9 mol C mol H mol O
1 mol C9H8O mol C9H8O4 1 mol C9H8O4
and
1 mol C9H8O mol C9H8O mol C9H8O4
9 mol C mol H mol O

15. Learning Check How many moles O are in 0.150 mol aspirin C9H8O4?
0.150 mol C9H8O4 x 4 mol O = mol O
1 mol C9H8O4
subscript factor

16. Learning Check How many O atoms are in 0.150 mol aspirin C9H8O4?
0.150 mol C9H8O4 x 4 mol O x x 1023 O atoms
1 mol C9H8O4 1 mol O
subscript Avogadro’s
factor number
= x 1023 O atoms

17. Atomic Mass Atomic mass is the
Mass of a single atom in atomic mass units (amu).
Mass of an atom compared to a 12C atom.
Number below the symbol of an element. (the average atomic mass)

18. Periodic Table and Atomic Mass
Ag has atomic mass = amu
S has atomic mass = amu
C has atomic mass = amu

19. Atomic Mass Factors The atomic mass Can be written as an equality.
Example: 1 P atom = amu
Can be written as two ___________ ________.
Example: 1 P atom and amu
30.97 amu P atom

20. Uses of Atomic Mass Factors
The atomic mass factors are used to convert
A specific number of atoms to mass (amu).
An amount in amu to ___________ of atoms.

21. Using Atomic Mass Factors
1. What is the mass in amu of 75 P atoms?
75 P atoms x amu = 2323 amu (2.323 x103 amu)
1 P atom
2. How many Cu atoms have a mass of x amu?
4.500 x 105 amu x 1 Cu atom = Cu atoms
63.55 amu

22. Learning Check What is the mass in amu of 75 silver atoms?
75 Ag atoms x amu = 8093 amu
1 Ag atom

23. Learning Check How many gold atoms have a mass of 1.85 x 105 amu?
1.85 x 105 amu x 1 Au atom = 939 Au atoms
197.0 amu

24. Formula Mass The formula mass is The mass in amu of a compound.
The _____ of the atomic masses of the elements in a formula.
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25. Calculating Formula Mass
To calculate the formula mass of Na2SO4
Multiply the atomic mass of each element by its subscript, then total the masses of the atoms.
2 Na x amu = amu
1 Na
1 S x amu = amu amu
1 S
4 O x amu = amu
1 O

26. Learning Check
Using the periodic table, calculate the formula mass of aluminum sulfide Al2S3.
2 Al x amu = amu
1 Al
3 S x amu = amu
1 S amu

27. Molar Mass Copyright © 2008 by Pearson Education, Inc.
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28. Molar Mass The molar mass
Is the mass of one mol of an element or compound.
Is the atomic mass expressed in grams.
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29. Molar Mass from the Periodic Table
Is the atomic mass expressed in grams.
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30. Learning Check Give the molar mass for: A. 1 mol K atoms =
B. 1 mol Sn atoms =

31. Molar Mass of a Compound
The molar mass of a compound is the sum of the molar masses of the elements in the formula.
Example: Calculate the molar mass of CaCl2.
Element
Number of Moles
Atomic Mass
Total Mass Ca 1
40.08 g/mol
40.08 g Cl 2
35.45 g/mol
70.90 g
CaCl2 g

32. Molar Mass of K3PO4 Calculate the molar mass of K3PO4. Element
Number of Moles
Atomic Mass
Total Mass in K3PO4 K 3
39.10 g/mol g P 1
30.97 g/mol
30.97 g O 4
16.00 g/mol
64.00 g
K3PO4 g

33. Some One-Mol Quantities
32.07 g g g g g
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34. Learning Check A. K2O 94.20 g/mol
2 mol K (39.10 g/mol) + 1 mol O (16.00 g/mol)
78.20 g g
= g
B. Al(OH) g/mol
1 mol Al (26.98 g/mol) + 3 mol O (16.00 g/mol)
+ 3 mol H (1.008 g/mol)
26.98 g g g
= g

35. Calculations Using Molar Mass
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36. Molar Mass Factors Molar mass conversion factors
Are written from molar mass.
Relate grams and moles of an element or compound.
Example: Write molar mass factors for methane CH4 used in gas cook tops and gas heaters.
Molar mass:
1 mol CH4 = g
Conversion factors:
16.04 g CH and mol CH4
1 mol CH g CH4

37. Learning Check Acetic acid C2H4O2 gives the sour taste to vinegar.
Write two molar mass factors for acetic acid.
Calculate molar mass:
= g/mol
1 mol of acetic acid = g acetic acid
Molar mass factors
1 mol acetic acid and g acetic acid
60.05 g acetic acid mol acetic acid

38. Calculations Using Molar Mass
Molar mass factors are used to convert between the grams of a substance and the number of moles.
Molar mass factor
Grams
Moles

39. Moles to Grams Aluminum is used to build lightweight bicycle
frames. How many grams of Al are 3.00 mol Al?
Molar mass equality: 1 mol Al = g Al
Setup with molar mass as a factor:
3.00 mol Al x g Al = g Al
1 mol Al
molar mass factor
for Al

40. Learning Check Grams Moles Molar mass factor
Allyl sulfide C6H10S is a compound that has the odor of garlic. How many moles of C6H10S are in 225 g C6H10S?
Action Plan: Calculate the molar mass, and convert 225 g to moles.
Molar mass factor
Grams
Moles
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41. Solution Calculate the molar mass of C6H10S.
(6 x 12.01) + (10 x 1.008) + (1 x 32.07)
= g/mol
Set up the calculation using a mole factor.
225 g C6H10S x mol C6H10S
g C6H10S
molar mass factor(inverted)
= 1.97 mol C6H10S

42. Grams, Moles, and Particles
A molar mass factor and Avogadro’s number
convert
Grams to particles
molar mass Avogadro’s
number
(g mol particles)
Particles to grams
Avogadro’s molar mass
(particles mol g)

43. Learning Check How many H2O molecules are in 24.0 g H2O?
A) x 1023
B) x 1025
C) x 1023

44. Learning Check How many H2O molecules are in 24.0 g H2O?
24.0 g H2O x 1 mol H2O x x 1023 H2O molecules
18.02 g H2O mol H2O
= 8.02 x 1023 H2O molecules

45. Learning Check If the odor of C6H10S can be detected from
2 x g in one liter of air, how many molecules
of C6H10S are present?
2 x g x mol x x 1023 molecules
g mol
= 1 x 109 molecules C6H10S

46. Percent Composition and
Empirical Formulas
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47. Percent Composition Percent composition
Is the percent by mass of each element in a formula.
Example: Calculate the percent composition of CO2.
CO2 = 1 C(12.01g) + 2 O(16.00 g) = g/mol)
12.01 g C x = % C
44.01 g CO2
32.00 g O x = % O

48. Learning Check
What is the percent composition of lactic acid, C3H6O3, a compound that appears in the blood after vigorous activity? →

49. Solution STEP 1 3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol
36.03 g C g H g O = g/mol
STEP 2
%C = g C x 100 = 40.00% C
90.08 g cpd
%H = g H x 100 = 6.714% H
%O = g O x 100 = % O
C3H6O3

50. Learning Check
The chemical isoamyl acetate C7H14O2 contributes to the odor of pears. What is the percent carbon in isoamyl acetate?
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51. Solution Molar mass C7H14O2 = 7C(12.01) + 14H(1.008)
+ 2O(16.00) = g/mol
Total C = 7C(12.01) = g
% C = total g C x 100
total g cpd
% C = g C x 100 = % C
g cpd

52. Empirical Formulas The empirical formula
Is the simplest whole number ratio of the atoms.
Is calculated by dividing the subscripts in the actual (molecular) formula by a whole number to give the lowest ratio.
C5H10O5  5 = C1H2O1 = CH2O
actual (molecular) empirical formula
formula

53. Some Molecular and Empirical Formulas
The molecular formula is the same or a multiple of the empirical.
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54. Learning Check P-1 1. What is the empirical formula for _________?
A) C2H4 B) CH2 C) CH
2. What is the empirical formula for _________?
A) C4H7 B) C6H12 C) C8H14
3. Which is a possible molecular formula for ______?
A) C4H4O4 B) C2H4O C) C3H6O3

55. Learning Check
A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula.
Lets ask, what does the empirical formula tell us?
The ratio of atoms of each element,
ratio of moles of each element.
Action Plan: Convert mass (g) to moles.

56. Solution Convert 7.31 g Ni and 20.0 g Br to moles.
7.31 g Ni x 1 mol Ni = mol Ni
58.69 g Ni
20.0 g Br x 1 mol Br = mol Br
79.90 g Br
Divide by smallest:
0.125 mol Ni = 1 Ni mol Br = 2 Br
Write ratio as subscripts: NiBr2

57. Converting Decimals to Whole Numbers
When the number of moles for an element is a
decimal, all the moles are multiplied by a small
integer to obtain whole number.
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58. Learning Check
Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula.
We have percentages, not grams.
However, the percentages are really an expression of grams of an element in 100. (exact) grams of compound.

59. Solution
STEP 1. Calculate the moles of each element in 100. g of the compound.
100. g aspirin contains 60.0% C or 60.0 g C, 4.5% H or 4.5 g H, and 35.5% O or 35.5 g O.
60.0 g C x 1 mol C = mol C
12.01 g C
4.5 g H x 1 mol H = mol H
1.008 g H
35.5 g O x 1mol O = mol O
16.00 g O

60. Solution (continued) STEP 2. Divide by the smallest number of mol.
5.00 mol C = mol C (decimal)
2.22
4.5 mol H = 2.0 mol H
2.22 mol O = mole O

61. Solution (continued)
3. Use the lowest whole number ratio as subscripts
When the moles are not whole numbers, multiply by a factor to give whole numbers, in this case x 4.
C: mol C x 4 = 9 mol C
H: 2.0 mol H x 4 = 8 mol H
O: mol O x 4 = 4 mol O
Using these whole numbers as subscripts the simplest formula is
C9H8O4

62. Molecular Formulas Copyright © 2008 by Pearson Education, Inc.
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63. Relating Molecular and Empirical Formulas
A molecular formula
Is a multiple (or equal) of its empirical formula.
Has a molar mass that is the empirical formula mass multiplied by a whole number.
molar mass = a whole number
empirical mass
Is obtained by multiplying the empirical formula by a whole number.

64. Diagram of Molecular and Empirical Formulas
A small integer links
A molecular formula and its empirical formula.
A molar mass and its empirical formula mass.
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65. Finding the Molecular Formula
Determine the molecular formula of compound that
has a molar mass of g and an empirical
formula of CH.
STEP 1. Empirical formula mass of CH = g
STEP 2. Divide the molar mass by the empirical
mass.
78.11 g = ~ 6
13.02 g
STEP 3. Multiply each subscript in C1H1 by 6.
molecular formula = C1x 6 H1 x 6 = C6H6

66. Some Compounds with Empirical Formula CH2O
formaldehyde
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67. Learning Check
A compound has a molar mass of 176.1g and an empirical formula of C3H4O3. What is the molecular formula?
What do we need to do?
From the empirical formula, determine the
empirical mass.
Then determine the whole number integer.

68. Solution
A compound has a formula mass of and an empirical formula of C3H4O3. What is the molecular formula?
C3H4O3 = g/EF
176.1 g (molar mass) = 2.00
88.06 g (empirical mass)
Molecular formula = 2 x empirical formula
C3 x 2H4 x 2O3 x 2 = C6H8O6

69. Molecular Formula
A compound contains C 24.27%, H 4.07%, and Cl 71.65%. The molar mass is about 99 g. What are the empirical and molecular formulas?
STEP 1. Calculate the empirical formula.
Write the mass percents as the grams in a
g sample of the compound.
C g H g Cl g

70. Finding the Molecular Formula (Continued)
Calculate the number of moles of each element.
24.27 g C x 1 mol C = mol C
12.01 g C
4.07 g H x 1 mol H = mol H
1.008 g H
71.65 g Cl x 1 mol Cl = mol Cl
35.45 g Cl

71. Finding the Molecular Formula (Continued)
Divide by the smallest number of moles:
2.021 mol C = 1 mol C
2.021
4.04 mol H = 2 mol H
2.02 mol Cl = 1 mol Cl
Empirical formula = C1H2Cl1 = CH2Cl
Calculate empirical mass (EM) CH2Cl = g

72. Finding the Molecular Formula (Continued)
STEP 2. Divide molar mass by empirical mass.
Molar mass = 99 g = 2
Empirical mass g
STEP 3. Multiply the empirical formula by the small
integer to determine the molecular formula.
2 x (CH2Cl)
C1 x2 H2 x 2 Cl1x 2 = C2H4Cl2

73. Learning Check
A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula?

74. Solution In 100. g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl.
27.4 g S x 1 mol S = mol S
32.07 g S
12.0 g N x 1 mol N = mol N
14.01 g N
60.6 g Cl x 1mol Cl = mol Cl
35.45 g Cl

75. Solution (continued) Divide by the smallest number of moles
0.854 mol S / = mol S
0.857 mol N/ = mol N
1.71 mol Cl/ = mol Cl
empirical formula = SNCl2 = g
Molar Mass/ Empirical mass
351 g = 3 g
molecular formula = (SNCl2)3 = S3N3Cl6