1. The Mole

2. Chapter Objectives
Use the mole and molar mass to make conversions among moles, mass, and number of representative particles.
Determine the percent composition of the components of compounds.
Calculate the empirical and molecular formulas for compounds and determine the formulas for hydrates.

3. Objectives Describe how a mole is used in chemistry.
Relate a mole to common counting units.
Convert moles to number of particles and number of particles to moles.

4. Measuring Matter
We use words everyday to count objects around us: a pair of roses, a dozen eggs, a ream of paper, a gross of pencils, etc.

5. Counting Particles, The Mole
The SI base unit used to measure the amount of particles in a substance: atoms, molecules, formula units, electrons or ions.
1 mol is exactly 12 g of carbon-12 or x1023 atoms.
The number is named after Lorenzo Romano Amedeo Carlo Avogadro, Conte di Quaregna e Ceretto. e-
MgCl2
Na+ e-
H2O

6. Moles to Particles 1 mol = 6.02x1023 particles
Conversion factor = 6.02x1023 particles mol
Number of moles x 6.02x1023 particles mol
= number of particles

7. Moles to Particles
3.50 mol water x 6.02x1023 molecules water mol water
= 2.11x1024 molecules water

8. Practice Problems Determine the number of atoms in 2.50 mol Zn.
Given 3.25 mol AgNO3, determine the
number of formula units.
Calculate the number of molecules in 11.5 mol
H2O.

9. Particles to Moles 1 mol = 6.02x1023 particles
Conversion factor = 6.02x1023 particles mol
Number of particles x mol x1023 particles
= number of moles

10. Particles to Moles 4.50x1024 atoms Mg x 1 mol Mg 6.02x1023 atoms Mg

11. Practice Problems How many moles contain each of the following?
5.75 x 1024 atoms Al
3.75 x 1024 molecules CO2
3.58 x 1023 formula units ZnCl2
2.50 x 1020 atoms Fe

12. Review of Key Ideas Describe how a mole is used in chemistry.
Relate a mole to common counting units.
Convert moles to number of particles and number of particles to moles.

13. Objectives Relate the mass of an atom to the mass of a mole of atoms.
Calculate the number of moles in a given mass of an element and the mass of a given number of moles of an element.
Calculate the number of moles of an element when given the number of atoms of the element.
Calculate the number of atoms of an element when given the number of moles of the element.

14. Mass of a Dozen The mass of a dozen eggs does not
equal the mass of a dozen roses, does it?
So would the mass of a mole of carbon equal the mass of a mole of zinc?

15. Boron 5 B
10.811
Average Atomic Mass
Each atom of carbon-12 has a mass of 12 amu (atomic mass unit).
The atomic masses of all other elements are established relative to carbon-12.
The masses on the periodic table are not whole numbers. This is due to the weighed averages of the masses of all the naturally occurring isotopes of each element.
Iron 26 Fe
55.845
Carbon 6 C
12.011

16. Mass of a Mole = 6.02x1023 atoms of Ag = 107.868 g Ag
The mass of one mole is defined as the mass of atoms in exactly 12 g of carbon-12.
1 mole of carbon-12 atoms is 12 g.
The masses of all other atoms are established relative to the mass of carbon-12.
The mass in grams of one mole of any pure substance is called its molar mass.
The mass of any element is numerically equal to its atomic mass and has the units g/mol.
An atom of Au has an atomic mass of amu and a molar mass of g/mol g of gold is 6.02x1023 atoms of gold which equals 1 mol Au.
1 mole Au
= 6.02x1023 atoms of Au
= g Au

17. m&m® Mass
m&m
Your club is having a candy sale to raise money for a trip, and you buy m&m’s in bulk.
You find that 1 dozen is equal to 30 g.
If someone wants 10 dozen, you can use your conversion factor to weigh out the correct amount instead of counting out 10 dozen m&m’s.
10 dozen x 30 g m&m’s = 300 g m&m’s
1 dozen m&m’s
m&m
m&m

18. number of moles x atomic mass of element, g = mass (g)
Sodium 11 Na
22.99
Molar Mass
Your are now working in the lab, and you need 5.00 moles of sodium (Na) for a reaction. How can you measure that quantity?
The number of moles of sodium can be converted to an equivalent mass to be measured by using the molar mass.
Remember: The mass in grams of one mole of any pure substance is called its molar mass, and the mass of any element is numerically equal to its atomic mass and has the units g/mol.
number of moles x atomic mass of element, g = mass (g)
1 mol element
5.00 moles Na x g Na = 115 g Na
1 mol Na

19. Practice Problems
Determine the mass in grams of each of the following.
3.57 mol Al
42.6 mol Si
3.45 mol Co
2.45 mol Zn

20. Molar Mass Ruthenium 44 Ru 101.07
Your are now working in the lab, and you use g of Ruthenium (Ru) for a reaction. You want to determine how many moles of Ru you are using.
The mass of ruthenium used can be converted to equivalent moles by using the molar mass.
mass x (1 mol of element/atomic mass of element, g) = number of moles of element
100.0 g Ru x 1 mol Ru = mol Ru
g Ru

21. Practice Problems Determine the number of moles in each of
the following of each of the following.
25.5 g Ag
300.0 g S
125 g Zn
1.00 kg Fe

22. Mass to Atoms
m&m
At the end of your candy sale, you find that 750 g of m&m®’s have not sold.
Without counting, you can determine how many m&m®’s are left.
1 dozen = 30 g, and 1 dozen contains 12 m&m’s
750 g m&m’s x 1 dozen m&m’s = 25 dozen m&m’s
30 g m&m’s
25 dozen m&m’s x 12 m&m’s = 300 m&m’s
1 dozen m&m’s
m&m
m&m

23. You need a conversion factor:
Mass to Atoms
Just as you cannot directly convert the mass of m&m®’s to the number of m&m®’s, you cannot directly convert the mass of a substance to the number of particles contained in the substance.
You need a conversion factor:
Avogadro’s number, 6.02x1023

24. Mass to Atoms
Sulfur 16 S
32.065
You are now working in the lab, and you use g of Sulfur (S) for a reaction. You want to determine how many atoms of S you are using.
The mass of sulfur used can be converted to equivalent atoms by using Avogadro’s number.
mass x (1 mol/atomic mass, g) = number of moles
100.0 g S x 1 mol S = mol S g
number of moles x 6.02x1023 particles = particles
1 mol
3.119 mol S x 6.02x1023 atoms = 18.8x1023 atoms S
1 mol S

25. Practice Problems How many atoms are in each of the following samples?
55.2 g Li
0.230 g Pb
11.5 g Hg
45.6 g Si
0.120 kg Ti

26. Atoms to Mass
If you have 32.1x1023 atoms of Sulfur (S), what is the mass of sulfur you have?
Atoms of sulfur used can be converted to equivalent mass by using Avogadro’s number.
particles x (1 mol/6.02x1023 particles) = number of moles
32.1x1023 atoms S x (1 mol S/6.02x1023 atoms) = mol S
number of moles x atomic mass, g = mass
1 mol
5.33 mol S x g S = 171 g S
1 mol S
Sulfur 16 S
32.065

27. Practice Problems What is the mass in grams of each of the following?
6.02 x 1024 atoms Bi
1.00 x 1024 atoms Mn
3.40 x 1022 atoms He
1.50 x 1015 atoms N
1.50 x 1015 atoms U

28. Conversion Summary Particles Mass Moles 1 mol 6.02x1023 particles
atomic mass, g
1 mol
atomic mass, g
1 mol
1 mol
6.02x1023 particles

29. Review of Key Ideas
Relate the mass of an atom to the mass of a mole of atoms.
Calculate the number of moles in a given mass of an element and the mass of a given number of moles of an element.
Calculate the number of moles of an element when given the number of atoms of the element.
Calculate the number of atoms of an element when given the number of moles of the element.

30. Objectives
Recognize the mole relationships shown by a chemical formula.
Calculate the molar mass of a compound.
Calculate the number of moles of a compound from a given mass of the compound, and the mass of a compound from a given number of moles of the compound.
Determine the number of atoms or ions in a mass of a compound.

31. Chemical Formulas and the Mole
A chemical formula indicates the types of atoms and their quantity.
CHFCl2 has 1 carbon atom, 1 hydrogen atom, 1 fluorine atom, and 2 chlorine atoms.
Ratio of 1:1:1:2.
1 mole of CHFCl2 would contain Avogadro’s number of molecules of CHFCl2.
1 mole of carbon atoms, 1 mole of hydrogen atoms, 1 mole of fluorine atoms and 2 moles of chlorine atoms.

32. Moles of Compound to Moles of Atoms
You may need to convert between the moles of a compound to the moles of the individual atoms in the compound.
For CHFCl2, you may use the following conversion factors:
1 mol C atoms 2 mol Cl atoms
1 mol CHFCl2 1 mol CHFCl2

33. Moles of Compound to Moles of Atoms
1 mol C atoms 2 mol Cl atoms
1 mol CHFCl2 1 mol CHFCl2
6.00 moles of CHFCl2 has how many moles of Cl atoms?
moles of compound x # moles of atom = moles of atom
1 mol of compound
6.00 mol CHFCl2 x 2 mol Cl atoms =
1 mol CHFCl2
12.0 mol Cl atoms

34. Practice Problems
Determine the number of moles of chloride ions in 2.50 mol ZnCl2.
Calculate the number of moles of each element in 1.25 mol glucose, C6H12O6.
Determine the number of moles of sulfate ions present in 3.00 moles iron(III) sulfate.
How many moles of oxygen atoms are present in 5.00 mol diphosphorous pentoxide?
Calculate the number of moles of hydrogen atoms in 11.5 mol water.

35. Molar Mass of a Compound
The mass of a mole of a compound equals the sum of the masses of every atom that makes up the compound.
You know the molar mass of an element is equal to its average atomic mass (g/mol).
You know that a chemical formula indicates the number of moles of each element in the compound.
Now you can determine the molar mass of a compound!

36. Determining Molar Mass of a Compound
You want to determine the molar mass of NaHSO4.
1 mol of Na, 1 mol of H, 1 mol of S, 4 moles of O
Number of moles of each element x molar mass of each element = number of grams of each element in the compound
1.00 mol Na x g Na/1 mol Na = 23.0 g Na
1.00 mol H x 1.01 g H/1 mol H = 1.01 g H
1.00 mol S x g S/1 mol S = 32.1 g S
4.00 mol O x g O/1 mol O = 64.0 g O
Molar mass = sum of masses of individual elements
Molar mass of NaHSO4 =
23.0g g g g = 120.1g

37. Practice Problems
Determine the molar mass of each of the following ionic compounds: NaOH, CaCl2, KC2H3O2, Sr(NO3)2
Determine the molar mass of each of the following molecular compounds: C2H5OH, C12H22O11, HCN, CCl4, H2O

38. Moles of a Compound to Mass of a Compound
Determine the molar mass of the compound.
Convert moles of compound to mass of compound needed.
moles of compound x molar mass of compound
1 mole compound
= mass of compound

39. Determining Mass of a Compound
You are given 5.00 mol of C2H6 and must convert the moles to mass using the molar mass as a conversion factor.
2.00 mol C x g C/1 mol C = 24.0 g C
6.00 mol H x 1.01 g H/1 mol H = 6.06 g H
Molar mass of C2H6= 24.0g g = 30.1 g
5.00 mol C2H6 x 30.1 g C2H6 = 151 g C2H6
1 mol C2H6

40. Practice Problems What is the mass of 3.25 moles of sulfuric acid?
What is the mass of 4.35 x 10-2 moles of zinc chloride?
How many grams of potassium permanganate are in 2.55 moles?

41. Mass of a Compound to Moles of a Compound
Determine the molar mass of the compound.
Convert mass of compound to moles of compound needed.
mass of compound x 1 mole of compound
molar mass of compound
= moles of compound

42. Determining Moles of a Compound
You produced 10.0 g of C2H6 and must convert the mass to moles using the molar mass as a conversion factor.
2.00 mol C x g C/1 mol C = 24.0 g C
6.00 mol H x 1.01 g H/1 mol H = 6.06 g H
Molar mass of C2H6= 24.0g g = 30.1 g
10.0 g C2H6 x 1 mol C2H6 = mol C2H6
30.1 g C2H6

43. Practice Problems
Determine the number of moles present in each of the following.
22.6 g AgNO3
6.50 g ZnSO4
35.0 g HCl

44. Mass of a Compound to Number of Particles of a Compound
Convert mass to moles.
Convert moles to number of molecules.
Convert molecules to number of atoms.
Mass of BF3 = 25.5 g & Molar mass = g/mol
25.5 g x (1 mol BF3/ g) = moles BF3
0.376 mol x (6.02x1023 molecules/1 mol BF3) = 2.26x1023 molecules BF3
2.26x1023 molecules BF3 x (1 B atom/1 BF3 molecule) =
2.26x1023 B atoms
2.26x1023 molecules BF3x (3 F atoms/1 BF3 molecules) =
6.78x1023 F atoms
Calculate the mass in grams of one molecule of BF3
67.807g/mol BF3 x 1 mol/6.02x1023 molecules
= 1.13x10-22 g/molecule

45. Practice Problems A sample of silver chromate has a mass of 25.8 g.
How many Ag+ ions are present?
How many CrO42- ions are present?
What is the mass is grams of one
formula unit of silver chromate?

46. Practice Problems A sample of C2H5OH has a mass of 45.6 g.
How many C atoms are present?
How many H atoms are present?
How many O atoms are present?

47. Practice Problems A sample of sodium sulfite has a mass of 2.25 g.
How many Na+ ions are present?
How many SO32- ions are present?
What is the mass is grams of one
formula unit of sodium sulfite?

48. Practice Problems A sample of carbon dioxide has a mass of 52.0 g.
How many C atoms are present?
How many O atoms are present?
What is the mass is grams of one molecule
of CO2?

49. Conversion Summary Moles of atoms or ions Particles Mass Moles
1 mol of compound
1 mol of compound
Particles
Mass
1 mol
6.02x1023 particles
1 mol
Number of g
Moles

50. Review of Key Ideas
Recognize the mole relationships shown by a chemical formula.
Calculate the molar mass of a compound.
Calculate the number of moles of a compound from a given mass of the compound, and the mass of a compound from a given number of moles of the compound.
Determine the number of atoms or ions in a mass of a compound.

51. Objectives
Explain what is meant by the percent composition of a compound.
Determine the empirical and molecular formulas for a compound from mass percent and actual mass data.

52. Percent Composition
The synthetic chemist makes the compounds and the analytical chemists analyzes their content.
A g sample of a new compound contains 25.0 g of X, 50.0 g of Y, and
25.0 g of Z.
The percent by mass of any element can be found by dividing the mass of the element by the mass of the compound and multiplying by 100.

53. Percent Composition Equation
Percent means parts per 100, so the percents by mass of all of the elements must always add up to 100.
mass of element x 100 = percent by mass
mass of compound

54. Determining Percent Composition
The chemical formula for carbon dioxide is CO2. Its molar mass is g/mol.
For 1 mol of CO2, there is 1 mol of C and 2 mol of O.
The mass of C in CO2 is (1 mol)(12.01 g/mol) = g
The mass of O in CO2 is (2 mol)(16.00 g/mol) = g
The percent composition of C in CO2 is
(12.01 g C/44.01 g CO2) x 100 = 27.29% C
The percent composition of O in CO2 is
(32.00g O/44.01 g CO2) x 100 = 72.71% O

55. Practice Problems Determine the percent by mass of each element in
calcium chloride.
Calculate the % composition of sodium sulfate.
Which has the larger percent by mass of sulfur, H2SO3 or H2S2O8?
What is the % composition of phosphoric acid?

56. Empirical Formula
Determine the smallest whole number ratio of the moles of the elements in the compound.
The ratio provides the subscripts in the empirical formula for the compound.
The empirical formula may be equal to the chemical formula as in CO. The ratio of carbon to oxygen is 1:1.
The chemical formula may be a simple multiple of the empirical formula as in H2O2. The ratio of hydrogen to oxygen is 1:1.

57. Determining Empirical Formula
A g sample of NH3 contains 82.22% N and
17.78% H or g N and g H.
82.22g N x (1 mol N/14.01g N) = mol N
17.78g H x (1 mol H/1.01g H) = 17.6 mol H
Can convert the number of moles into a mole ratio.
Take the largest number of moles and divide it by the smallest number of moles.
17.6 mol H/5.869 = 3 mol H
5.869 mol N/5.869 = 1 mol N
Simplest whole number ratio is 3 atoms H: 1 atom N
Therefore the empirical formula is NH3.

58. Empirical Formula Caveat
Often the calculated mole values are
still not whole numbers. In which case,
all of the mole values must be
multiplied by the smallest factor that
will make them whole numbers.

59. Practice Problems Calculating Empirical Formula from
Percent Composition
A blue solid is found to contain 36.84%
N and 63.16% O. What is the empirical
formula for this solid?

60. Practice Problems Calculating Empirical Formula from
Percent Composition
Determine the empirical formula for a
compound that contains 35.98%
aluminum and 64.02% S.

61. Practice Problems Calculating Empirical Formula from
Percent Composition
The chemical analysis of aspirin
indicates that the molecule is 60.00%
C, 4.44% H, and 35.56% O. Determine
the empirical formula.

62. Practice Problems Calculating Empirical Formula from
Percent Composition
What is the empirical formula for a
compound that contains 10.89% Mg,
31.77% Cl, and 57.34% O?

63. Practice Problems Calculating Empirical Formula from Mass Data
When an oxide of potassium is
decomposed, g of K and 4.00 g of
O are obtained. What is the empirical
formula for the compound?

64. Practice Problems Calculating Empirical Formula from Mass Data
Analysis of a compound composed of Fe
and O yields g Fe and g O.
What is the empirical formula for this
compound?

65. Practice Problems Calculating Empirical Formula from Mass Data
The pain reliever morphine contains
g C, g H, g O, and
1.228 g N. Determine the empirical
formula.

66. Practice Problems Calculating Empirical Formula from Mass Data
An oxide of Al contains g Al and
0.485 g O. Find the empirical formula
for the oxide.

67. Molecular Formula
It is possible that two or more substances can have the same empirical formula, since the mole ratio only indicates the smallest whole-number ratio of moles of elements in a compound.
Therefore, we need to go one step further and determine the molecular formula which specifies the actual number of atoms of each element in one molecule or formula unit of the substance.

68. Molecular Formula Equation
molecular formula = (empirical formula)n
n is the factor by which the subscripts in the empirical formula must be multiplied by in order to obtain the molecular formula.

69. Determining Molecular Formula
Methanal and ethanoic acid have the same empirical formula CH2O; however, they have different molecular formulas.
Molar mass of methanal is g/mol
Molar mass of ethanoic acid is g/mol
Dividing the actual molar mass by the mass of the empirical formula indicates the mass of methanal is the same as the empirical formula and that of ethanoic acid is twice the mass of the empirical formula.
(30.03 g/mol methanal)/(30.03 g/mol CH2O) = 1.000
(60.06 g/mol ethanoic acid)/(30.03 g/mol CH2O) = 2.000
Thus the molecular formula for ethanoic acid
contains twice the number of carbon, hydrogen, and oxygen atoms represented by the empirical formula.

70. Practice Problems Analysis of a chemical used in
photographic developing fluid indicates
a chemical composition of 65.45% C,
5.45% H, and 29.09% O. The molar
mass is found to be g/mol.
Determine the molecular formula.

71. Practice Problems A compound was found to contain 49.98
g C and g H. The molar mass of
the compound is g/mol.
Determine the molecular formula.

72. Practice Problems A colorless liquid composed of 46.68%
N and 53.32% O has a molar mass of
60.01 g/mol. What is the molecular
formula?

73. Determining Empirical and Molecular Formulas
Percent Composition
Mass of component
elements
Mass of each element
Molar mass
Ratio of moles of elements
If all are
whole numbers
If not all whole numbers,
multiply by the smallest factor
that will produce whole numbers
Empirical Formula
Experimental molar mass = n
Mass of empirical formula
(Empirical formula)n
Molecular formula

74. Review of Key Ideas
Explain what is meant by the percent composition of a compound.
Determine the empirical and molecular formulas for a compound from mass percent and actual mass data.

75. Objectives
Explain what a hydrate is and how its name reflects its composition.
Determine the formulas for a hydrate from laboratory data.

76. What are Hydrates?
Solids in which water molecules are trapped are called hydrates.
A hydrate is a compound that has a specific number of water molecules bound to its atoms.
Opal is made up of silicon dioxide, SiO2, and the color that results is from water trapped in the mineral.

77. Naming Hydrates
The number of water molecules associated with each formula unit of the compound is written following a dot.
Na2CO3·10H20 is sodium carbonate decahydrate.
It has 10 water molecules associated with one formula unit of the compound.
The mass of water must be included in the molar mass calculation.

78. Practice Problem
Pick two of the hydrates in Table 11-1, Formulas for Hydrates, and determine the molar mass and the percent composition in terms of the percent anhydrous, without water, compound and the percent water.

79. Analyzing a Hydrate
In order to analyze a hydrate, you must drive off the water of hydration by heating the hydrate. The remaining substance is anhydrous.
CaSO4·2H20 is a hydrate; after heating, the water is removed, and CaSO4 remains.

80. Coefficient of Water for a Hydrate
You must find the number of moles of water associated with one mole of the hydrate.
You have 5.00 g of barium chloride hydrate.
You know the formula is BaCl2·xH20
How do you determine the coefficient of water, x?

81. Determining the Mass of the Water of Hydration
In order to find x, you must heat the hydrate to drive off the water of hydration.
After heating, the dried substance BaCl2 has a mass of 4.26 g.
BaCl2 is considered to be anhydrous, without water
The mass of the water of hydration is 5.00 g BaCl2 – 4.26 g of BaCl2 = 0.74 g H20

82. Mass to Moles Convert masses to moles
4.26 g BaCl2 x (1 mol BaCl2/ g BaCl2) = mol BaCl2
0.74 g H2O x (1 mol H2O/18.02 g H20) = mol H2O

83. Calculate the Ratio of Moles which is x
x = moles H2O/moles BaCl2
x = (0.041 mol H2O/ mol BaCl2) =
2.0 mol H2O/1.00 mol BaCl2
The ratio of water to BaCl2 is 2:1.
Therefore, the value of x is 2, and the formula for the hydrate is BaCl2·2H20.

84. Practice Problems A hydrate is found to have the following
percent composition: 48.8% MgSO4 and
51.2% H2O. What is the formula and name
for this hydrate?
If g of cobalt(II) chloride hydrate is
heated, 9.25 g of anhydrous cobalt chloride remains.
What is the formula and name for this hydrate?

85. Review of Key Ideas
Explain what a hydrate is and how its name reflects its composition.
Determine the formulas for a hydrate from laboratory data.

86. Review of Key Ideas for Chapter
Use the mole and molar mass to make conversions among moles, mass, and number of representative particles.
Determine the percent composition of the components of compounds.
Calculate the empirical and molecular formulas for compounds and determine the formulas for hydrates.