1. Chapter 9 The Mole Concept by Christopher Hamaker
© 2011 Pearson Education, Inc.
Chapter 9

2. Mass and Moles of a Substance
Chemistry requires a method for determining the numbers of molecules in a given mass of a substance.
This allows the chemist to carry out “recipes” for compounds based on the relative numbers of atoms involved.
The calculation involving the quantities of reactants and products in a chemical equation is called stoichiometry. 2

3. Avogadro’s Number
Avogadro’s number (symbol N) is the number of atoms in grams of carbon.
Its numerical value is 6.02 x 1023.
Therefore, a g sample of carbon contains 6.02 x 1023 carbon atoms.
Chapter 9

4. 1 mol = Avogadro’s number = 6.02 x 1023 units
The Mole
The mole (mol) is a unit of measure for an amount of a chemical substance.
A mole is Avogadro’s number of particles, which is 6.02 x 1023 particles.
1 mol = Avogadro’s number = 6.02 x 1023 units
We can use the mole relationship to convert between the number of particles and the mass of a substance.
Chapter 9

5. Analogies for Avogadro’s Number
The volume occupied by one mole of softballs would be about the size of Earth.
One mole of Olympic shot put balls has about the same mass as that of Earth.
One mole of hydrogen atoms laid side by side would circle Earth about 1 million times.
Chapter 9

6. One Mole of Several Substances
C12H22O11
H2O
lead
mercury
K2Cr2O7
copper
sulfur
NaCl
Chapter 9

7. One Mole of: S C Hg Cu Fe

8. Example Video The Mole and Avogadro's Number (9:44 min)
© 2011 Pearson Education, Inc.
Chapter 4

9. Mole Calculations We will be using the unit analysis method again.
Recall the following steps:
Write down the unit requested.
Write down the given value.
Apply unit factor(s) to convert the given units to the desired units.
Chapter 9

10. Mole Calculations I How many sodium atoms are in 0.120 mol Na?
We want atoms of Na.
We have mol Na.
1 mole Na = 6.02 x 1023 atoms Na.
= 7.22 x 1022 atoms Na
0.120 mol Na x
1 mol Na
6.02 x 1023 atoms Na
Chapter 9

11. Do You Understand? How many atoms are in 0.551 mol of potassium (K) ?
1 mol K = x 1023 atoms K x
6.022 x 1023 atoms K
1 mol K =
0.014 mol K
8.49 x 1021 atoms K

12. Mole Calculations I, Continued
How many moles of potassium are in 1.25 x 1021 atoms K?
We want moles K.
We have 1.25 x 1021 atoms K.
1 mole K = 6.02 x 1023 atoms K.
= 2.08 x 10-3 mol K
1.25 x 1021 atoms K x
1 mol K
6.02 x 1023 atoms K
Chapter 9

13. Example Video moles to atoms, atoms to moles (4:15 min)
© 2011 Pearson Education, Inc.
Chapter 4

14. Let's Practice 1. Calculate the number of atoms in 1.7 mol B.
Calculate the number of atoms in 5.0 mol Al.
Calculate the moles of Au if there are 5,000,000 atoms of Au.

15. Mass and Moles of a Substance
The molar mass of a substance is the mass of one mole of a substance.
For all substances, molar mass, in grams per mole, is numerically equal to the formula weight in atomic mass units.
That is, one mole of any element weighs its atomic mass in grams. 2

16. Molar Mass
The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance.
The atomic mass of iron is amu.
Therefore, the molar mass of iron is g/mol.
Since oxygen occurs naturally as a diatomic, O2, the molar mass of oxygen gas is two times g or g/mol.
Chapter 9

17. Calculating Molar Mass
The molar mass of a substance is the sum of the molar masses of each element.
What is the molar mass of magnesium nitrate, Mg(NO3)2?
The sum of the atomic masses is as follows:
( ) =
(62.01) = amu
The molar mass for Mg(NO3)2 is g/mol.
Chapter 9

18. Molecular Weight and Formula Weight
The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not.
For example, one formula unit of NaCl contains 1 sodium atom (23.0 amu) and one chlorine atom (35.5 amu), giving a formula weight of 58.5 amu. 2

19. formula mass (amu) = molar mass (grams)
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
NaCl
1Na
22.99 amu
1Cl
amu
NaCl
58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = amu
1 mole NaCl = g NaCl

20. Do You Understand Formula Mass?
What is the formula mass of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
3 Ca
3 x 40.08
2 P
2 x 30.97
8 O
+ 8 x 16.00
amu

21. molecular mass (amu) = molar mass (grams)
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
SO2 1S
32.07 amu 2O
+ 2 x amu
SO2
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = amu
1 mole SO2 = g SO2

22. Example Video How to Calculate Molar Mass and Formula Mass (3:00 min)
© 2011 Pearson Education, Inc.
Chapter 4

23. 6.02 x 1023 particles = 1 mol = molar mass
Mole Calculations II
Now we will use the molar mass of a compound to convert between grams of a substance and moles or particles of a substance.
6.02 x 1023 particles = 1 mol = molar mass
If we want to convert particles to mass, we must first convert particles to moles, and then we can convert moles to mass.
Chapter 9

26. Mole–Mole Calculations
What is the mass of 1.33 moles of titanium, Ti?
We want grams.
We have 1.33 moles of titanium.
Use the molar mass of Ti: 1 mol Ti = g Ti.
= 63.7 g Ti
1.33 mole Ti x
47.88 g Ti
1 mole Ti
Chapter 9

27. Mass and Moles of a Substance
Mole calculations
Suppose we have 5.75 moles of magnesium (atomic wt. = 24.3 g/mol). What is its mass? 2

28. Mass and Moles of a Substance
Mole calculations
Suppose we have grams of iron (Fe). The atomic weight of iron is 55.8 g/mol. How many moles of iron does this represent? 2

29. Mass and Moles of a Substance
Mole calculations
This same method applies to compounds. Suppose we have grams of H2O (molecular weight = 18.0 g/mol). How many moles does this represent? 2

30. Mass and Moles of a Substance
Mole calculations
Conversely, suppose we have 3.25 moles of glucose, C6H12O6 (molecular wt. = g/mol). What is its mass? 2

31. Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = x 1023 atoms H
1 mol C3H8O
60 g C3H8O x
8 mol H atoms
1 mol C3H8O x
6.022 x 1023 H atoms
1 mol H atoms x =
72.5 g C3H8O
5.82 x 1024 atoms H

32. Example Video moles to grams conversion (1:07 min)
Gram to Mole Conversion 1 (1:38 min)
© 2011 Pearson Education, Inc.
Chapter 4

33. Mole Calculations III What is the mass of 2.55 x 1023 atoms of lead?
We want grams.
We have atoms of lead.
Use Avogadro’s number and the molar mass of Pb.
2.55 × 1023 atoms Pb x
1 mol Pb
6.02×1023 atoms Pb
207.2 g Pb
1 mole Pb x
= 87.9 g Pb
Chapter 9

34. Mole Calculations III, Continued
How many O2 molecules are present in g of oxygen gas?
We want molecules O2.
We have grams O2.
Use Avogadro’s number and the molar mass of O2.
0.470 g O2 x
1 mol O2
32.00 g O2
6.02x1023 molecules O2
1 mole O2 x
8.84 x 1021 molecules O2
Chapter 9

35. Do You Understand Molar Mass?
How many atoms are in g of potassium (K) ?
1 mol K = g K
1 mol K = x 1023 atoms K
1 mol K
39.10 g K x x
6.022 x 1023 atoms K
1 mol K =
0.551 g K
8.49 x 1021 atoms K

36. Example Video Grams to Molecules (5:01 min)
Molecules (or atoms) into Grams Conversion (4:29 min)
© 2011 Pearson Education, Inc.
Chapter 4

37. Let's Practice Using these Tools
1. What is the mass in grams of 2.5 mol Na.
2. Calculate the mass of 5,000,000 atoms of Ag.

38. Mass of a Single Molecule
What is the mass of a single molecule of sulfur dioxide? The molar mass of SO2 is g/mol.
We want mass/molecule SO2, we have the molar mass of sulfur dioxide.
Use Avogadro’s number and the molar mass of SO2 as follows :
64.07 g SO2
1 mol SO2
6.02 x 1023 molecules SO2
1 mole SO2 x
1.06 x g/molecule
Chapter 9

39. Molar Volume
At standard temperature and pressure, 1 mole of any gas occupies 22.4 L.
The volume occupied by 1 mole of gas (22.4 L) is called the molar volume.
Standard temperature and pressure are 0 C and 1 atm.
Chapter 9

40. Molar Volume of Gases We now have a new unit factor equation:
1 mole gas = 6.02 x 1023 molecules gas = 22.4 L gas
Chapter 9

41. One Mole of a Gas at STP
The box below has a volume of 22.4 L, which is the volume occupied by 1 mole of a gas at STP.
Chapter 9

42. Gas Density The density of gases is much less than that of liquids.
We can easily calculate the density of any gas at STP.
The formula for gas density at STP is as follows:
= density, g/L
molar mass in grams
molar volume in liters
Chapter 9

43. Calculating Gas Density
What is the density of ammonia gas, NH3, at STP?
First we need the molar mass for ammonia.
(1.01) = g/mol
The molar volume NH3 at STP is 22.4 L/mol.
Density is mass/volume.
= g/L
17.04 g/mol
22.4 L/mol
Chapter 9

44. Molar Mass of a Gas
We can also use molar volume to calculate the molar mass of an unknown gas.
1.96 g of an unknown gas occupies 1.00 L at STP. What is the molar mass?
We want g/mol; we have g/L.
1.96 g
1.00 L
22.4 L
1 mole x
= g/mol
Chapter 9

45. Mole Unit Factors We now have three interpretations for the mole:
1 mol = 6.02 x 1023 particles.
1 mol = molar mass.
1 mol = 22.4 L at STP for a gas.
This gives us three unit factors to use to convert among moles, particles, mass, and volume.
Chapter 9

46. Calculating Molar Volume
A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present?
We want moles; we have volume.
Use molar volume of a gas: 1 mol = 22.4 L.
4.50 L CH4 x
= mol CH4
1 mol CH4
22.4 L CH4
Chapter 9

47. Calculating Mass Volume
What is the mass of 3.36 L of ozone gas, O3, at STP?
We want mass O3; we have 3.36 L O3.
Convert volume to moles, then moles to mass.
3.36 L O3 x x
22.4 L O3
1 mol O3
48.00 g O3
= 7.20 g O3
Chapter 9

48. Calculating Molecule Volume
How many molecules of hydrogen gas, H2, occupy L at STP?
We want molecules H2; we have L H2.
Convert volume to moles, and then moles to molecules.
0.500 L H2 x
1 mol H2
22.4 L H2
6.02x1023 molecules H2
1 mole H2 x
= 1.34 x 1022 molecules H2
Chapter 9

49. Percent Composition
The percent composition of a compound lists the mass percent of each element.
For example, the percent composition of water, H2O is 11% hydrogen and 89% oxygen.
All water contains % hydrogen and % oxygen by mass.
Chapter 9

50. Determining Chemical Formulas
The percent composition of a compound is the mass percentage of each element in the compound.
We define the mass percentage of “A” as the parts of “A” per hundred parts of the total, by mass. That is, 2

51. Calculating Percent Composition
There are a few steps to calculating the percent composition of a compound. Let’s practice using H2O.
Assume you have 1 mole of the compound.
One mole of H2O contains 2 mol of hydrogen and 1 mol of oxygen. Therefore,
2(1.01 g H) + 1(16.00 g O) = molar mass H2O
2.02 g H g O = g H2O
Chapter 9

52. Calculating Percent Composition, Continued
Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water.
2.02 g H
18.02 g H2O
x 100% = 11.2% H
16.00 g O
18.02 g H2O
x 100% = 88.79% O
Chapter 9

53. Percent Composition Problem
TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Calculate the percent composition of TNT, C7H5(NO2)3.
7(12.01 g C) + 5(1.01 g H) + 3 (14.01 g N g O) = g C7H5(NO2)3
84.07 g C g H g N g O = g C7H5(NO2)3
Chapter 9

54. Percent Composition of TNT
84.07 g C
g TNT
x 100% = 37.01% C
1.01 g H
g TNT
x 100% = 2.22% H
42.03 g N
g TNT
x 100% = 18.50% N
96.00 g O
g TNT
x 100% = 42.26% O
Chapter 9

55. Mass Percentages from Formulas
Let’s calculate the percent composition of butane, C4H10.
First, we need the molecular mass of C4H10.
Now, we can calculate the percents. 2

56. Percent composition of an element in a compound =
n x molar mass of element
molar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the compound
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
C2H6O
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% % % = 100.0%

57. Example Video
Mass Percent Composition of an Element in a Compound (5:28 min)
© 2011 Pearson Education, Inc.
Chapter 4

58. Determining Chemical Formulas
Determining the formula of a compound from the percent composition.
The percent composition of a compound leads directly to its empirical formula.
An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts. 2

59. Empirical Formulas
The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule.
The molecular formula of benzene is C6H6.
The empirical formula of benzene is CH.
The molecular formula of octane is C8H18.
The empirical formula of octane is C4H9.
Chapter 9

60. Calculating Empirical Formulas
We can calculate the empirical formula of a compound from its composition data.
We can determine the mole ratio of each element from the mass to determine the empirical formula of radium oxide, Ra?O?.
– A g sample of radium metal was heated to produce g of radium oxide. What is the empirical formula?
– We have g Ra and = g O.
Chapter 9

61. Calculating Empirical Formulas, Continued
– The molar mass of radium is g/mol, and the molar mass of oxygen is g/mol.
1 mol Ra
g Ra
1.640 g Ra x
= mol Ra
1 mol O
16.00 g O
0.115 g O x
= mol O
– We get Ra O Simplify the mole ratio by dividing by the smallest number.
– We get Ra1.01O1.00 = RaO is the empirical formula.
Chapter 9

62. Empirical Formulas from Percent Composition
We can also use percent composition data to calculate empirical formulas.
Assume that you have 100 grams of sample.
Acetylene is 92.2% carbon and 7.83% hydrogen. What is the empirical formula?
– If we assume 100 grams of sample, we have g carbon and 7.83 g hydrogen.
Chapter 9

63. Empirical Formula for Acetylene
Calculate the moles of each element.
1 mol C
12.01 g C
92.2 g C x
= 7.68 mol C
1 mol H
1.01 g H
7.83 g H x
= 7.75 mol H
The ratio of elements in acetylene is C7.68H7.75. Divide by the smallest number to get the following formula:
7.68 C
= C1.00H1.01 = CH
7.75 H
Chapter 9

64. Percent Composition and Empirical Formulas
Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O percent.
nK = g K x
= mol K
1 mol K
39.10 g K
nMn = g Mn x
= mol Mn
1 mol Mn
54.94 g Mn
nO = g O x
= mol O
1 mol O
16.00 g O

65. Percent Composition and Empirical Formulas
nK = , nMn = , nO = 2.532
K : ~
1.0
0.6330
0.6329
Mn :
0.6329
= 1.0
O : ~
4.0
2.532
0.6329
KMnO4

66. Determining Chemical Formulas
Determining the empirical formula from the percent composition.
Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula?
In other words, give the smallest whole-number ratio of the subscripts in the formula
Cx HyOz 2

67. Determining Chemical Formulas
Determining the empirical formula from the percent composition.
Our grams of benzoic acid would contain:
This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio might emerge. 2

68. Determining Chemical Formulas
Determining the empirical formula from the percent composition.
Our grams of benzoic acid would contain:
now it’s not too difficult to see that the smallest whole number ratio is 7:6:2.
The empirical formula is C7H6O2 . 2

69. Molecular Formulas
The empirical formula for acetylene is CH. This represents the ratio of C to H atoms on acetylene.
The actual molecular formula is some multiple of the empirical formula, (CH)n.
Acetylene has a molar mass of 26 g/mol. Find n to find the molecular formula: = CH
(CH)n
26 g/mol
13 g/mol
n = 2 and the molecular formula is C2H2.
Chapter 9

70. Example Video Molecular and Empirical Formulas (15:54 min)
Chapter 4

71. Chapter Summary
Avogadro’s number is 6.02 x 1023, and is 1 mole of any substance.
The molar mass of a substance is the sum of the atomic masses of each element in the formula.
At STP, 1 mole of any gas occupies 22.4 L.
Chapter 9

72. Determining Chemical Formulas
Determining the “true” molecular formula from the empirical formula.
An empirical formula gives only the smallest whole-number ratio of atoms in a formula.
The “true” molecular formula could be a multiple of the empirical formula (since both would have the same percent composition).
To determine the “true” molecular formula, we must know the “true” molecular weight of the compound. 2

73. Determining Chemical Formulas
Determining the “true” molecular formula from the empirical formula.
For example, suppose the empirical formula of a compound is CH2O and its “true” molecular weight is 60.0 g/mol.
The molar weight of the empirical formula (the “empirical weight”) is only 30.0 g/mol.
This would imply that the “true” molecular formula is actually the empirical formula doubled, or
C2H4O2 2

74. Chapter Summary, Continued
We can use the following flow chart for mole calculations:
Chapter 9

75. Chapter Summary, Continued
The percent composition of a substance is the mass percent of each element in that substance.
The empirical formula of a substance is the simplest whole number ratio of the elements in the formula.
The molecular formula is a multiple of the empirical formula.
Chapter 9