1. Introductory Chemistry, 3rd Edition Nivaldo Tro
Chapter 6
Chemical
Composition
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2009, Prentice Hall

2. Why Is Knowledge of Composition Important?
All matter is either chemically or physically combined into substances.
Knowing the fraction of material you have can tell you:
the amount of sodium in sodium chloride for diet.
the amount of iron in iron ore for steel production.
the amount of hydrogen in water for hydrogen fuel.
the amount of chlorine in freon to estimate ozone depletion.
Tro's "Introductory Chemistry", Chapter 6

3. How much seed do you plant?
In a garden you count the seeds by hand. How many seeds would you know to plant in a field?

4. Tro's "Introductory Chemistry", Chapter 6
Counting by Weighing
Building a house requires a lot of nails.
If you know that a single nail weighs g, than 100 nails weigh 1.22 g, a 1000 nails weigh 12.2 g and so on.
Analogy:
You want to make 100 lbs of Al2O3, how much aluminum do you use
Tro's "Introductory Chemistry", Chapter 6

5. Counting Nails by the Pound, Continued
A hardware store customer buys 2.60 pounds of nails. A dozen nails has a mass of pounds. How many nails did the customer buy?
1 dozen nails = lbs.
12 nails = 1 dozen nails
Solution map:
Tro's "Introductory Chemistry", Chapter 6

6. Counting Nails by the Pound, Continued
The customer bought 2.60 lbs of nails and received 208 nails. He counted the nails by weighing them!
Tro's "Introductory Chemistry", Chapter 6

7. Counting Nails by the Pound, Continued
What if he bought a different size nail?
Would the mass of a dozen be lbs?
Would there still be 12 nails in a dozen?
Would there be 208 nails in 2.60 lbs?
How would this effect the conversion factors?
Tro's "Introductory Chemistry", Chapter 6

8. Counting Atoms by Moles
If we can find the mass of a particular number of atoms, we can use this information to convert the mass of an element sample to the number of atoms in the sample.
The number of atoms we will use is x 1023 and we call this a mole.
1 mole = x 1023 things.
Like 1 dozen = 12 things.
Avogadro’s number.
Like a kilo = 1000 or a Google = 1×10100
Tro's "Introductory Chemistry", Chapter 6

9. Chemical Packages—Moles
Mole = Number of carbon atoms “in” 12 g of C-12.
1 mole protons or 1 mole of neutrons = 1 amu
C-12 exactly 6 protons and 6 neutrons
since 1 mole × 1 amu = 1 g.
1 mole of C-12 (which is 12 amu) weighs exactly 12 g.
In 12 g of C-12 there are x1023 C-12 atoms.
Tro's "Introductory Chemistry", Chapter 6

10. Tro's "Introductory Chemistry", Chapter 6
Example 6.1:
A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Tro's "Introductory Chemistry", Chapter 6

11. Tro's "Introductory Chemistry", Chapter 6
Example: A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Write down the given quantity and its units.
Given: 1.1 x 1022 Ag atoms
Tro's "Introductory Chemistry", Chapter 6

12. Tro's "Introductory Chemistry", Chapter 6
Example: A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Information:
Given: 1.1 x 1022 Ag atoms
Write down the quantity to find and/or its units.
Find: ? moles
Tro's "Introductory Chemistry", Chapter 6

13. Tro's "Introductory Chemistry", Chapter 6
Example: A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Information:
Given: 1.1 x 1022 Ag atoms
Find: ? moles
Collect needed conversion factors:
1 mole Ag atoms = x 1023 Ag atoms.
Tro's "Introductory Chemistry", Chapter 6

14. Tro's "Introductory Chemistry", Chapter 6
Example: A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Information:
Given: 1.1 x 1022 Ag atoms
Find: ? moles
Conversion Factor:
1 mole = x 1023
Write a solution map for converting the units:
atoms Ag
moles Ag
Tro's "Introductory Chemistry", Chapter 6

15. Tro's "Introductory Chemistry", Chapter 6
Example: A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Information:
Given: 1.1 x 1022 Ag atoms
Find: ? moles
Conversion Factor:
1 mole = x 1023
Solution Map: atoms  mole
Apply the solution map:
= x 10-2 moles Ag
Significant figures and round:
= 1.8 x 10-2 moles Ag
Tro's "Introductory Chemistry", Chapter 6

16. Example: A silver ring contains 1. 1 x 1022 silver atoms
Example: A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Information:
Given: 1.1 x 1022 Ag atoms
Find: ? moles
Conversion Factor:
1 mole = x 1023
Solution Map: atoms  mole
Check the solution:
1.1 x 1022 Ag atoms = 1.8 x 10-2 moles Ag
The units of the answer, moles, are correct.
The magnitude of the answer makes sense
since 1.1 x 1022 is less than 1 mole.
Tro's "Introductory Chemistry", Chapter 6

17. Practice—Calculate the Number of Atoms in 2.45 Mol of Copper.
Tro's "Introductory Chemistry", Chapter 6

18. Practice—Calculate the Number of Atoms in 2
Practice—Calculate the Number of Atoms in 2.45 Mol of Copper, Continued.
Given:
Find:
2.45 mol Cu
atoms Cu
Solution Map:
Relationships:
1 mol = x 1023 atoms
mol Cu
atoms Cu
Solution:
Check:
Since atoms are small, the large number of atoms makes sense.
Tro's "Introductory Chemistry", Chapter 6

19. Relationship Between Moles and Mass
The mass of one mole of atoms is called the molar mass.
The molar mass of an element, in grams, is numerically equal to the element’s atomic mass,
in amu.
The lighter the atom, the less a mole weighs.
The lighter the atom, the more atoms there are
in 1 g.
Tro's "Introductory Chemistry", Chapter 6

20. Mole and Mass Relationships
carbon
12.01 g
1 mole
sulfur
32.06 g
Tro's "Introductory Chemistry", Chapter 6

21. Example 6.2—Calculate the Moles of Sulfur in 57.8 g of Sulfur.
Given:
Find:
57.8 g S
mol S
Solution Map:
Relationships:
1 mol S = g
g S
mol S
Solution:
Check:
Since the given amount is much less than 1 mol S, the number makes sense.
Tro's "Introductory Chemistry", Chapter 6

22. Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead.
Tro's "Introductory Chemistry", Chapter 6

23. Tro's "Introductory Chemistry", Chapter 6
Practice—Calculate the Moles of Carbon in g of Pencil Lead, Continued.
Given:
Find:
g C
mol C
Solution Map:
Relationships:
1 mol C = g
g C
mol C
Solution:
Check:
Since the given amount is much less than 1 mol C, the number makes sense.
Tro's "Introductory Chemistry", Chapter 6

24. Tro's "Introductory Chemistry", Chapter 6
Example 6.3:
How many aluminum atoms are in an aluminum can with a mass of 16.2 g?
Tro's "Introductory Chemistry", Chapter 6

25. Example 6.3—How Many Aluminum Atoms Are in a Can Weighing 16.2 g?
Given:
Find:
16.2 g Al
atoms Al
Solution Map:
Relationships:
1 mol Al = g, 1 mol = x 1023
g Al
mol Al
atoms Al
Solution:
Check:
Since the given amount is much less than 1 mol Cu, the number makes sense.
Tro's "Introductory Chemistry", Chapter 6

26. Molar Mass of Compounds
The relative weights of molecules can be calculated from atomic weights.
Formula mass = 1 molecule of H2O
= 2(1.01 amu H) amu O = amu.
Since 1 mole of H2O contains 2 moles of H and 1 mole of O.
Molar mass = 1 mole H2O
= 2(1.01 g H) g O = g.
Tro's "Introductory Chemistry", Chapter 6

27. Example 6.4—Calculate the Mass of 1.75 Mol of H2O.
Given:
Find:
1.75 mol H2O
g H2O
Solution Map:
Relationships:
1 mol H2O = g
mol H2O
g H2O
Solution:
Check:
Since the given amount is more than 1 mol, the mass being > 18 g makes sense.
Tro's "Introductory Chemistry", Chapter 6

28. Practice—How Many Moles Are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00)
Tro's "Introductory Chemistry", Chapter 6

29. Tro's "Introductory Chemistry", Chapter 6
Practice—How Many Moles Are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00), Continued
Given:
Find:
50.0 g mol PbO2
moles PbO2
Solution Map:
Relationships:
1 mol PbO2 = g
g PbO2
mol PbO2
Solution:
Check:
Since the given amount is less than g, the moles being < 1 makes sense.
Tro's "Introductory Chemistry", Chapter 6

30. Example 6.5—What Is the Mass of 4.78 x 1024 NO2 Molecules?
Given:
Find:
4.78 x 1024 NO2 molecules
g NO2
Solution Map:
Relationships:
molecules
mol NO2
g NO2
1 mol NO2 = g, 1 mol = x 1023
Solution:
Check:
Since the given amount is more than Avogadro’s number, the mass > 46 g makes sense.
Tro's "Introductory Chemistry", Chapter 6

31. Counting and ratio’s
It takes me .2 gal of gas to get to IVC. It is a very simple ratio: =
What if I only had .1 gal
200 2X4’s, 3 sinks, 2 showers, you can make a house with 3 bathrooms and 3 bedrooms.
What if you had 12 sinks…how many houses could you make. =

32. Chemical Formulas as Conversion Factors
1 spider  8 legs.
1 chair  4 legs.
1 H2O molecule  2 H atoms  1 O atom.
Tro's "Introductory Chemistry", Chapter 6

33. Mole Relationships in Chemical Formulas
Since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound.
Moles of compound
Moles of constituents
1 mol NaCl
1 mol Na, 1 mol Cl
1 mol H2O
2 mol H, 1 mol O
1 mol CaCO3
1 mol Ca, 1 mol C, 3 mol O
1 mol C6H12O6
6 mol C, 12 mol H, 6 mol O
Tro's "Introductory Chemistry", Chapter 6

34. Example 6.6—Calculate the Moles of Oxygen in 1.7 Moles of CaCO3.
Given:
Find:
1.7 mol CaCO3
mol O
Solution Map:
Relationships:
1 mol CaCO3 = 3 mol O
mol CaCO3
mol O
Solution:
Check:
Since the given amount is much less than 1 mol S, the number makes sense.
Tro's "Introductory Chemistry", Chapter 6

35. Tro's "Introductory Chemistry", Chapter 6
Example 6.7:
Carvone (C10H14O) is the main component in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liqueurs, soaps, and perfumes. Find the mass of carbon in 55.4 g of carvone.
Tro's "Introductory Chemistry", Chapter 6

36. Example: Find the mass of carbon in 55.4 g of carvone, (C10H14O).
Write down the given quantity and its units.
Given: 55.4 g C10H14O
Tro's "Introductory Chemistry", Chapter 6

37. Example: Find the mass of carbon in 55.4 g of carvone, (C10H14O).
Information:
Given: 55.4 g C10H14O
Write down the quantity to find and/or its units.
Find: ? g C
Tro's "Introductory Chemistry", Chapter 6

38. Example: Find the mass of carbon in 55.4 g of carvone, (C10H14O).
Information:
Given: 55.4 g C10H14O
Find: g C
Collect needed conversion factors:
Molar mass C10H14O = 10(atomic mass C) + 14(atomic mass H) + 1(atomic mass O)
= 10(12.01) + 14(1.01) + (16.00)
= g/mol
1 mole C10H14O = g C10H14O
1 mole C10H14O  10 mol C
1 mole C = g C
Tro's "Introductory Chemistry", Chapter 6

39. Example: Find the mass of carbon in 55.4 g of carvone, (C10H14O).
Information:
Given: 55.4 g C10H14O
Find: g C
Conversion Factors:
1 mol C10H14O = g
1 mol C10H14O  10 mol C
1 mol C = g
Write a solution map for converting the units: g
C10H14O
mol
C10H14O
mol C g C
Tro's "Introductory Chemistry", Chapter 6

40. Example: Find the mass of carbon in 55.4 g of carvone, (C10H14O).
Information:
Given: 55.4 g C10H14O
Find: g C
Conversion Factors:
1 mol C10H14O = g
1 mol C10H14O  10 mol C
1 mol C = g
Solution Map:
g C10H14O  mol C10H14O 
mol C  g C
Apply the solution map:
= g C
Significant figures and round:
= 44.3 g C
Tro's "Introductory Chemistry", Chapter 6

41. Tro's "Introductory Chemistry", Chapter 6
Percent Composition
Percentage of each element in a compound.
By mass.
Can be determined from:
The formula of the compound.
The experimental mass analysis of the compound.
The percentages may not always total to 100% due to rounding.
Tro's "Introductory Chemistry", Chapter 6

42. Example 6.9—Find the Mass Percent of Cl in C2Cl4F2.
Given:
Find:
C2Cl4F2
% Cl by mass
Solution Map:
Relationships:
Solution:
Check:
Since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense.

43. Practice—Determine the Mass Percent Composition of the Following:
CaCl2 (Ca = 40.08, Cl = 35.45)
Tro's "Introductory Chemistry", Chapter 6

44. Tro's "Introductory Chemistry", Chapter 6
Practice—Determine the Percent Composition of the Following, Continued:
CaCl2
Tro's "Introductory Chemistry", Chapter 6

45. Mass Percent as a Conversion Factor
The mass percent tells you the mass of a constituent element in 100 g of the compound.
The fact that NaCl is 39% Na by mass means that 100 g of NaCl contains 39 g Na.
This can be used as a conversion factor.
100 g NaCl  39 g Na
Tro's "Introductory Chemistry", Chapter 6

46. Tro's "Introductory Chemistry", Chapter 6
Empirical Formulas
The simplest, whole-number ratio of atoms in a molecule is called the empirical formula.
Can be determined from percent composition or combining masses.
The molecular formula is a multiple of the empirical formula.
Tro's "Introductory Chemistry", Chapter 6

47. Empirical (CH2O) -vs- Molecular Formula
Formaldehyde CH2O g/mol
Acetic Acid C2H4O g/mol
Lactic Acid C3H6O g/mol
Erythrose C4H8O g/mol
Ribose C5H10O g/mol
Glucose C6H12O g/mol

48. Empirical Formulas, Continued
Hydrogen Peroxide
Molecular formula = H2O2
Empirical formula = HO
Benzene
Molecular formula = C6H6
Empirical formula = CH
Glucose
Molecular formula = C6H12O6
Empirical formula = CH2O

49. Example 6.11—Finding an Empirical Formula from Experimental Data
Tro's "Introductory Chemistry", Chapter 6

50. Tro's "Introductory Chemistry", Chapter 6
Example:
A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
Tro's "Introductory Chemistry", Chapter 6

51. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of aspirin with the given mass percent composition.
Write down the given quantity and its units.
Given: C = 60.00%
H = 4.48%
O = 35.53%
Therefore, in 100 g of aspirin there are g C, g H, and g O.
Tro's "Introductory Chemistry", Chapter 6

52. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: g C, 4.48 g H, g O
Write down the quantity to find and/or its units.
Find: empirical formula, CxHyOz
Tro's "Introductory Chemistry", Chapter 6

53. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: g C, 4.48 g H, g O
Find: empirical formula, CxHyOz
Collect needed conversion factors:
1 mole C = g C
1 mole H = 1.01 g H
1 mole O = g O
Tro's "Introductory Chemistry", Chapter 6

54. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: g C, 4.48 g H, g O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g; 1 mol H = 1.01 g; 1 mol O = g
Write a solution map:
g C
mol C
whole
number
ratio
mole
ratio
empirical
formula
pseudo-
formula
g H
mol H
g O
mol O
Tro's "Introductory Chemistry", Chapter 6

55. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: g C, 4.48 g H, g O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g;
1 mol H = 1.01 g; 1 mol O = g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
Apply the solution map:
Calculate the moles of each element.
Tro's "Introductory Chemistry", Chapter 6

56. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: mol C, 4.44 mol H,
2.221 mol O
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g;
1 mol H = 1.01 g; 1 mol O = g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
Apply the solution map:
Write a pseudoformula.
C4.996H4.44O2.221
Tro's "Introductory Chemistry", Chapter 6

57. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: C4.996H4.44O2.221
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g;
1 mol H = 1.01 g; 1 mol O = g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
Apply the solution map:
Find the mole ratio by dividing by the smallest number of moles.
Tro's "Introductory Chemistry", Chapter 6

58. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of aspirin with the given mass percent composition.
Information:
Given: C2.25H2O1
Find: empirical formula, CxHyOz
Conversion Factors:
1 mol C = g;
1 mol H = 1.01 g; 1 mol O = g
Solution Map: g C,H,O  mol C,H,O 
mol ratio  empirical formula
Apply the solution map:
Multiply subscripts by factor to give whole number.
{ } x 4
C9H8O4
Tro's "Introductory Chemistry", Chapter 6

59. Example 6.12—Finding an Empirical Formula from Experimental Data
Tro's "Introductory Chemistry", Chapter 6

60. Tro's "Introductory Chemistry", Chapter 6
Example:
A 3.24-g sample of titanium reacts with oxygen to form 5.40 g of the metal oxide. What is the formula of the oxide?
Tro's "Introductory Chemistry", Chapter 6

61. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of oxide of titanium with the given elemental analysis.
Write down the given quantity and its units.
Given: Ti = 3.24 g
compound = 5.40 g
Tro's "Introductory Chemistry", Chapter 6

62. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of oxide of titanium with the given elemental analysis.
Information:
Given: g Ti, 5.40 g compound
Write down the quantity to find and/or its units.
Find: empirical formula, TixOy
Tro's "Introductory Chemistry", Chapter 6

63. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of oxide of titanium with the given elemental analysis.
Information:
Given: g Ti, 5.40 g compound
Find: empirical formula, TixOy
Collect needed conversion factors:
1 mole Ti = g Ti
1 mole O = g O
Tro's "Introductory Chemistry", Chapter 6

64. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of oxide of titanium with the given elemental analysis.
Information:
Given: g Ti, 5.40 g compound
Find: empirical formula, TixOy
Conversion Factors:
1 mol Ti = 47.88g;1 mol O = 16.00g
Write a solution map:
g Ti
mol Ti
whole
number
ratio
mole
ratio
empirical
formula
pseudo-
formula
g O
mol O
Tro's "Introductory Chemistry", Chapter 6

65. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of oxide of titanium with the given elemental analysis.
Information:
Given: g Ti, 5.40 g compound
Find: empirical formula, TixOy
Conversion Factors:
1 mol Ti= 47.88g;1 mol O= 16.00g
Solution Map: g Ti,O  mol Ti,O 
mol ratio  empirical formula
Apply the solution map:
Calculate the mass of each element.
5.40 g compound − 3.24 g Ti = 2.16 g O
Tro's "Introductory Chemistry", Chapter 6

66. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of oxide of titanium with the given elemental analysis.
Information:
Given: g Ti, 2.16 g O
Find: empirical formula, TixOy
Conversion Factors:
1 mol Ti= 47.88g;1 mol O= 16.00g
Solution Map: g Ti,O  mol Ti,O 
mol ratio  empirical formula
Apply the solution map:
Calculate the moles of each element.
Tro's "Introductory Chemistry", Chapter 6

67. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of oxide of titanium with the given elemental analysis.
Information:
Given: mol Ti, mol O
Find: empirical formula, TixOy
Conversion Factors:
1 mol Ti= 47.88g;1 mol O= 16.00g
Solution Map: g Ti,O  mol Ti,O 
mol ratio  empirical formula
Apply the solution map:
Write a pseudoformula.
Ti0.0677O0.135
Tro's "Introductory Chemistry", Chapter 6

68. Tro's "Introductory Chemistry", Chapter 6
Example: Find the empirical formula of oxide of titanium with the given elemental analysis.
Information:
Given: mol Ti, mol O
Find: empirical formula, TixOy
Conversion Factors:
1 mol Ti= 47.88g;1 mol O= 16.00g
Solution Map: g Ti,O  mol Ti,O 
mol ratio  empirical formula
Apply the solution map:
Find the mole ratio by dividing by the smallest number of moles.
Tro's "Introductory Chemistry", Chapter 6

69. Tro's "Introductory Chemistry", Chapter 6
Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00).
Tro's "Introductory Chemistry", Chapter 6

70. Tro's "Introductory Chemistry", Chapter 6
Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00), Continued.
Given: 75.7% Sn, (100 – 75.3) = 24.3% F 
in 100 g Tin (II) fluoride there are 75.7 g Sn and 24.3 g F.
Find: SnxFy
Conversion Factors: 1 mol Sn = g; 1 mol F = g
Solution Map:
whole
number
ratio
g Sn
mol Sn
mole
ratio
pseudo-
formula
empirical
formula
g F
mol F
Tro's "Introductory Chemistry", Chapter 6

71. Tro's "Introductory Chemistry", Chapter 6
Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00), Continued.
Apply solution map:
Sn0.638F1.28
SnF2
Tro's "Introductory Chemistry", Chapter 6

72. Tro's "Introductory Chemistry", Chapter 6
Practice—Determine the Empirical Formula of Hematite, which Contains 72.4% Fe (55.85) and the Rest Oxygen (16.00).
Tro's "Introductory Chemistry", Chapter 6

73. Tro's "Introductory Chemistry", Chapter 6
Practice—Determine the Empirical Formula of Hematite, which Contains 72.4% Fe (55.85) and the Rest Oxygen (16.00), Continued.
Given: 72.4% Fe, (100 – 72.4) = 27.6% O 
in 100 g hematite there are 72.4 g Fe and 27.6 g O.
Find: FexOy
Conversion Factors: 1 mol Fe = g; 1 mol O = g
Solution Map:
whole
number
ratio
g Fe
mol Fe
mole
ratio
pseudo-
formula
empirical
formula
g O
mol O
Tro's "Introductory Chemistry", Chapter 6

74. Tro's "Introductory Chemistry", Chapter 6
Practice—Determine the Empirical Formula of Hematite, which Contains 72.4% Fe (55.85) and the Rest Oxygen (16.00), Continued.
Apply solution map:
Fe1.30O1.73
Tro's "Introductory Chemistry", Chapter 6

75. Tro's "Introductory Chemistry", Chapter 6
All These Molecules Have the Same Empirical Formula. How Are the Molecules Different?
Name
Molecular
Formula
Empirical
Glyceraldehyde
C3H6O3
CH2O
Erythrose
C4H8O4
Arabinose
C5H10O5
Glucose
C6H12O6
Tro's "Introductory Chemistry", Chapter 6

76. Tro's "Introductory Chemistry", Chapter 6
All These Molecules Have the Same Empirical Formula. How Are the Molecules Different?, Continued
Name
Molecular
Formula
Empirical
Molar
Mass, g
Glyceraldehyde
C3H6O3
CH2O (30) 90
Erythrose
C4H8O4
CH2O
120
Arabinose
C5H10O5
150
Glucose
C6H12O6
180
Tro's "Introductory Chemistry", Chapter 6

77. Tro's "Introductory Chemistry", Chapter 6
Molecular Formulas
The molecular formula is a multiple of the empirical formula.
To determine the molecular formula, you need to know the empirical formula and the molar mass of the compound.
Molar massreal formula
Molar massempirical formula
= Factor used to multiply subscripts
Tro's "Introductory Chemistry", Chapter 6

78. Tro's "Introductory Chemistry", Chapter 6
Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8.
Determine the empirical formula.
May need to calculate it as previous.
C5H8
Determine the molar mass of the empirical formula.
5 C = 60.05, 8 H = 8.064
C5H8 = g/mol
Tro's "Introductory Chemistry", Chapter 6

79. Tro's "Introductory Chemistry", Chapter 6
Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued.
Divide the given molar mass of the compound by the molar mass of the empirical formula.
Round to the nearest whole number.
Tro's "Introductory Chemistry", Chapter 6

80. Tro's "Introductory Chemistry", Chapter 6
Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued.
Multiply the empirical formula by the factor above to give the molecular formula.
(C5H8)3 = C15H24
Tro's "Introductory Chemistry", Chapter 6

81. Tro's "Introductory Chemistry", Chapter 6
Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C5H3. What is its Molecular Formula? (C = 12.01, H=1.01)
Tro's "Introductory Chemistry", Chapter 6

82. Tro's "Introductory Chemistry", Chapter 6
Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C5H3. What is its Molecular Formula? (C = 12.01, H=1.01), Continued
C5 = 5(12.01 g) = g
H3 = 3(1.01 g) = g
C5H3 = g
Molecular formula = {C5H3} x 4 = C20H12
Tro's "Introductory Chemistry", Chapter 6

83. Tro's "Introductory Chemistry", Chapter 6
Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N. (C=12.01, H=1.01, N=14.01)
Tro's "Introductory Chemistry", Chapter 6

84. Tro's "Introductory Chemistry", Chapter 6
Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N, Continued
Given: 74.0% C, 8.7% H, {100 – ( )} = 17.3% N 
in 100 g nicotine there are 74.0 g C, 8.7 g H, and 17.3 g N.
Find: CxHyNz
Conversion Factors:
1 mol C = g; 1 mol H = 1.01 g; 1 mol N = g
Solution Map:
g C
mol C
whole
number
ratio
mole
ratio
pseudo-
formula
empirical
formula
g H
mol H
g N
mol N
Tro's "Introductory Chemistry", Chapter 6

85. Tro's "Introductory Chemistry", Chapter 6
Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N, Continued.
Apply solution map:
C5 = 5(12.01 g) = g
N1 = 1(14.01 g) = g
H7 = 7(1.01 g) = g
C5H7N = g
C6.16H8.6N1.23
{C5H7N} x 2 = C10H14N2
Tro's "Introductory Chemistry", Chapter 6