1. Chapter 3 Molecules, Compounds, and Chemical Equations

2. Elements and Compounds
Elements combine together to make an almost unlimited number of compounds
The properties of a compound are totally different from its constituent elements

3. Formation of Water from Its Elements

4. Chemical Bonds
Compounds are made of atoms held together by chemical bonds
attractive forces between each atoms’ protons and the bonding electrons

5. Two general Bond Types Ionic and Covalent
Ionic bonds result when electrons have been transferred between a metal and a nonmetal atom, resulting in oppositely charged ions that attract each other
Covalent bonds result when two atoms share one or more of their electrons
generally found when nonmetal atoms bond together

6. Formulas Describe Compounds
Each element is represented by its letter symbol
The number of atoms of each element is written to the right of the element’s symbol as a subscript
A compound is a distinct substance
composed of atoms from two or more elements
the compound is defined by the number and type of each atom
compounds can be a neutral molecule or an ion

7. An element is made up of Multiple types of atoms Compounds
A single type of atom

8. Representing Compounds with Chemical Formulas
Compounds are generally represented with a chemical formula
All chemical formulas tell what elements are in the compound through use of the letter symbol of the element
All formulas and models convey a limited amount of information – none are perfect representations

9. Types of Formulas: Empirical Formula
A compound’s empirical formula gives the relative number of atoms of each element reduced to the lowest possible denominator
it doesn’t describe the # atoms, the order of attachment, or the shape
formulas of ionic compounds are empirical
Fluorspar (CaCl2) is an ionic compound. There is 1 Ca2+ ion for every 2 Cl− ions in the compound.
molecular compounds can also be written as an empirical formula
e.g. the molecular formula for oxalic acid is C2H2O4 but the empirical formula is CHO2

10. Types of Formula: Molecular Formula
A molecular formula gives the actual number of atoms of each element in a molecule
it does not describe the order of attachment, or the shape
The molecular formula C2H2O4.
tells you that there are 2 carbon atoms, 2 hydrogen atoms, and 4 oxygen atoms
Doesn’t tell you how the carbon, oxygen and hydrogen atoms are attached together

11. Types of Formula: Structural Formula
A structural formula tells what atoms are present and how the atoms in a molecule are attached together.
It uses lines to represent covalent bonds
it does not describe the 3-dimensional shape, but an experienced chemist can make a good guess at it
each line describes the number of electrons shared by the bonded atoms
single line = 2 shared electrons, a single covalent bond
double line = 4 shared electrons, a double covalent bond
triple line = 6 shared electrons, a triple covalent bond

12. Representing Compounds Molecular Models
Molecular models show the 3-d structure of the molecule plus all the information given by the structural formula
Ball-and-stick models use balls to represent the atoms and sticks to represent the bonds between them
Space-filling models use interconnected spheres to show the electron clouds of atoms connecting together

13. Structural Formula of Oxalic Acid
Space filling
Model
Ball and Stick Model

14. The following formulas represent:
/ C2O4H2
/ CO2H
Empirical / Molecular / Structural
Molecular / Empirical / Structural
Structural / Molecular / Empirical
Empirical / Structural / Molecular
Molecular / Structural / Empirical
Structural / Empirical / Molecular

16. Write the empirical formula for each of the following
The ionic compound that has two aluminum ions for every three oxide ions
arabinose, C5H10O5
pyrimidine
ethylene glycol
Al2O3
CH2O
C2H2N
CH3O

17. Classification of Elements and Compounds

18. Classifying Elements more than 2 atoms (e.g. P4, S8, Se8)
Atomic elements
Single atoms
Most elements are atomic elements
Diatomic elements
2 atoms (e.g. H2)
Polyatomic elements
more than 2 atoms (e.g. P4, S8, Se8)

19. Molecular Elements N7 O8 F9 H1 Cl17 Br35 I53
7 elements occur as diatomic molecules (2 atoms) H1
Cl17
Br35
I53
5A A A
N O F9 1A
shaped like a 7 on the periodic table…
start with element 7, N – group V, go across the table to group VIIA, F, then go down group VIIA to I
Hydrogen stands alone
Other elements occur as polyatomic molecules
P4, S8, Se8

20. Molecular Elements

21. Classifying Compounds
Ionic compounds a 3-D array of anions surrounded by cations and vice-versa
Made from combining metals (which form cations) with non-metals (which form anions)
No individual molecular units, only formula units
Can contain polyatomic ions
2+ atoms attached together by covalent bonds with an overall charge on the entire ion
Table salt – contains
an array of Na+ ions
and Cl- ions

22. Ions with predictable charges
= Alkali metals
= Alkali earth metals
= Halogens
Ions with predictable charges -3 1A 2A 7A 8A +3 3A 4A 5A 6A +1 +2 Cs Rb K Na Li N O I Br Cl F -1 Ba Sr Ca Mg Al S Se
add pictures of elements from text Te -2

24. Compounds that Contain Ions
A neutral compound must have no total charge, therefore the number of cations and anions multiplied by their charges, must balance to get 0 charge
If Na+ is combined with S2−, you will need 2 Na+ ions for every S2− ion to balance the charges, therefore the formula must be Na2S

25. atomic element ionic compound molecular element molecular compound
Classify the Following as Either an Atomic Element, Molecular Element, Molecular Compound, or Ionic Compound
atomic element
ionic compound
molecular element
molecular compound
Aluminum, Al
Aluminum chloride, AlCl3
Chlorine, Cl2
Acetone, C3H6O
Carbon monoxide, CO
Cobalt, Co

26. Ex 3.4: Write a formula for ionic compound that forms between calcium and oxygen
Write the symbol for the metal cation and the nonmetal anion. Obtain the ion’s charges from the element’s group number on the periodic table and write it as a superscript on the symbol.
Adjust the number of cations and anions using subscripts to balance the overall charge.
Check that the sum of the charges of the cations equals the sum of the charges of the anions.
Ca O2−
CaO
cations: +2
anions: −2
The charges cancel

27. Example 3.3: Write the formula of a compound made from aluminum ions and oxide ions
Write the symbol for the metal cation and its charge
Write the symbol for the nonmetal anion and its charge
Charge (without sign) becomes subscript for other ion
Reduce subscripts to smallest whole number ratio
Check that the total charge of the cations cancels the total charge of the anions
Al3+ column 3A
O2− column 6A
Al+3 O2−
Al2O3
no change required
Al = (2)∙(+3) = +6
O = (3)∙(−2) = −6

28. Write the formulas for compounds made from the following ions
Potassium ion with a nitride ion
K+ with N3−
Calcium ion with a bromide ion
Ca2+ with Br1−
Aluminum ion with a sulfide ion
Al3+ with S2−
K3N
CaBr2
Al2S3

29. Rules for Naming Ionic Compounds
Some compounds have common names that can only be learned by memorization
NaCl = table salt
NaHCO3 = baking soda
Others have a systematic name based on naming the component ions

30. Naming Binary Ionic Compounds for Metals with Invariant Charge
First, use the metal’s elemental name for the cation
Next, add the nonmetal anion’s name
assign the proper negative charge based its group number on the Periodic Table
the nonmetal name’s suffix changes to -ide

31. CsF: A Binary Ionic Compound with a Metal having Invariant Charge
1. Identify cation and anion
Cs = Cs+ because it is Group 1A
F = F− because it is Group 7A
2. Name the cation Cs+ = cesium
3. Name the anion F− = fluoride
4. Write the cation name first, then the anion name
cesium fluoride

32. Naming Binary Ionic Compounds for Metals with Variable Charge
Again, use the elemental metal name for the metal cation
Determine the anion’s charge  cation’s charge
A (Roman Numeral) follows the metal ion to indicate its charge
Next add the nonmetal anion’s name
the nonmetal name’s suffix changes to -ide

33. CuF2: A Binary Ionic Compound with a Metal having Variable Charge
Identify cation and anion
F = F− because it is Group 7A
Cu = Cu2+ to balance the two (−) charges from 2 F−
2. Name the cation
Cu2+ = copper(II)
3. Name the anion
F− = fluoride
4. Write the cation name first, then the anion name
copper(II) fluoride

34. Name the compound composed of chlorine and potassium: KCl
potassium (I) chloride
chlorine potassiumide
potassium chloride
potassium chlorine

35. Name the compound composed of magnesium and bromine: MgBr2
magnesium bromide
magnesium (II) dibromide
bromine magneside
bromine magnesiumide
magnesium dibromine

36. Name the compound composed of aluminum and sulfur: Al2S3
trisulfur dialuminide
aluminum (III) trisulfide
sulfur magnesiumide
aluminum sulfuride
aluminum sulfide

37. Determining the charge on a metal cation having variable charge in a compound composed of gold and sulfur: Au2S3
Determine the invariant charge on the anion
Au2S3 : the anion is S = Group 6A, its charge is 2−
Determine the total negative charge
3 S in the formula, the total negative charge is 3 x 2 − = 6−
Set the total positive charge = total negative charge
total negative charge = 6−, total positive charge = 6+
Divide by the number of cations in the formula
with 2 Au in the formula and a total positive charge of 6+, each Au must have a 3+ charge
The compound name is Gold (III) Sulfide

38. Find the charge on the cation
CrO3
Fe3N2
each O = 2−
Name the compound
3 O = 6−
chromium (VI) oxide
 Cr = 6+
each N = 3−
Name the compound
2 N = 6−
iron (II) nitride
3 Fe = 6+
 Fe = 2+

39. Find the charge on the cation in TiCl4-1 -4 +1 +4

40. Find the charge on the cation
TiCl4
each Cl = 1-
4 Cl = 4−
 Ti = 4+
Name the compound
titanium(IV) chloride

41. Name the compound PbBr2 Fe2S3 each Br = -1 2 x -1= -2  Pb 2+
lead(II) bromide
iron(III) sulfide
each S = -2
3 x -2= -6
-6/2 Fe = -3
 Fe 3+

42. Mn4+ S2- Mn4+ S2− Mn2S4 MnS2 Mn = (1)∙(4+) = +4 S = (2)∙(2−) = −4
To Write the formula for a binary ionic compound containing metal with variable charge manganese(IV) sulfide
1. Write the symbol for the cation and its charge 2. Write the symbol for the anion and its charge 3. The charge (without the sign) becomes subscript for other ion 4. Reduce subscripts to smallest whole number ratio 5. Check that the total charge of the cations cancels the total charge of the anions
Mn4+
S2-
Mn4+ S2−
Mn2S4
MnS2
Mn = (1)∙(4+) = +4
S = (2)∙(2−) = −4

43. What are the formulas for compounds made from copper (II) nitride, Cu2+ with N3−
Cu2N3
Cu3N2
N3Cu2
N2Cu3

44. What are the formulas for compounds made from iron(III) with bromide: Fe3+ with Br-
FeBr3
Fe3Br
FeBr
Br3Fe
BrFe3

45. Compounds Containing Polyatomic Anions
Name the cation first, then the polyatomic anion
The name and the charge of the polyatomic ion do not change
If more than one polyatomic ion is present, the polyatomic ion is surrounded by parentheses
a subscript tells how many polyatomic ions there are.
NO (NO3-) (NO3-)3

46. Periodic Pattern of Polyatomic Ions Elements in the same column form similar polyatomic ions - ate groups

47. Patterns for Polyatomic Ions
-ate ion
chlorate = ClO3−
remove 2 oxygen
with same charge
hypochlorite = ClO−
add 1 oxygen
with same charge
perchlorate = ClO4−
remove 1 oxygen
with same charge
chlorite = ClO2−

48. Some Common Polyatomic Ions
If the polyatomic ion has a H in the front, add hydrogen- as a prefix and adjust the overall charge by +1

49. Example: Naming ionic compounds containing a polyatomic ion - Na2SO4
1. Identify the ions
2Na = 2Na+ because in Group 1A
SO4 = SO42− a polyatomic ion
2. Name the cation
Na+ = sodium, metal with invariant charge
3. Name the anion
SO42− = sulfate
4. Write the name of the cation followed by the name of the anion
sodium sulfate

50. Example: Naming ionic compounds containing a polyatomic ion Fe(NO3)3
1. Identify the ions
NO3 = NO3− a polyatomic ion
Fe = Fe3+ to balance the charge of the 3 NO3−
2. Name the cation
Fe3+ = iron(III), a metal with variable charge
3. Name the anion
NO3− = nitrate
4. Write the name of the cation followed by the name of the anion
iron(III) nitrate

51. Name the Following Compounds
1. Ca(C2H3O2)2
2. Cu(NO3)2
calcium acetate
copper(II) nitrate
Polyatomic Ammonium Cations – NH4+
1. NH4Cl
ammonium chloride

52. Example – Writing formula for ionic compounds containing polyatomic ion Iron(III) phosphate
1. Write the symbol for the cation and its charge 2. Write the symbol for the anion and its charge 3. Charge (without sign) becomes subscript for other ion 4. Reduce subscripts to smallest whole number ratio 5. Check that the total charge of the cations cancels the total charge of the anions
Fe3+
PO43−
Fe3+ PO43−
Fe3(PO4)3
FePO4
Fe = (1)∙(3+) = +3
PO4 = (1)∙(3−) = −3

53. Practice — What are the formulas for compounds made from the following ions?
aluminum ion with a sulfate ion
chromium(II) with hydrogen carbonate
Al3+ with SO42−
Al2(SO4)3
Cr2+ with HCO3−
Cr(HCO3)2

54. Hydrates
Hydrates are ionic compounds containing a specific number of waters in each formula unit
Water of hydration can be driven off by heating
In formula, attached waters written as
CoCl2 ∙ 6H2O
Named with hydrate suffix. The # of attached waters is indicated by the prefixes to the right
CoCl2 ∙ 6H2O = cobalt(II) chloride hexahydrate
CaSO4 ∙ ½H2O = calcium sulfate hemihydrate

55. Cobalt(II) chloride hexahydrate

56. nickel(II) chloride hexahydrate
What is the formula of magnesium sulfate heptahydrate?
What is the name of NiCl2•6H2O?
Mg2+ + SO42−
MgSO4
MgSO47H2O
2Cl−  Ni2+
nickel(II) chloride
nickel(II) chloride hexahydrate

57. Classifying Compounds
Molecular compounds
compounds made of only nonmetals
covalent compounds – composed of individual molecule units where atoms of different elements are chemically attached by covalent bonds
Propane – contains
individual C3H8
molecules

58. Naming Binary Molecular Compounds of 2 nonmetals
1. Write the name of the first element in the formula
a) Which is the element furthest left and down on the Periodic Table
2. Write the second element’s name in the formula with an -ide suffix
as if it were an anion, but these compounds do not contain ions!
3. Use a prefix in front of each name to indicate the number of atoms (slide 59)
a) never use the prefix mono- for the first element

59. Drop the final “a” if the element’s name begins with a vowel
Subscript – Prefixes
1 = mono-
2 = di-
3 = tri-
4 = tetra-
5 = penta-
6 = hexa-
7 = hepta-
8 = octa-
9 = nona-
10 = deca-
Drop the final “a” if the element’s name begins with a vowel

60. Naming binary molecular BF3
1. Name the first element: boron
2. Name the second element with an –ide
fluorine  fluoride
Add a prefix to the 2nd nonmetal’s name to indicate the subscript
a) Do not use the prefix mono on the first element
monoboron trifluoride
4. Combine the first element’s name without the mono- prefix with the second element’s name with a prefix
boron trifluoride

61. Name the Following NO2 PCl5 I2F7 nitrogen dioxide
phosphorus pentachloride
diiodine heptafluoride

62. Binary Molecular Compound dinitrogen pentoxide
Identify the symbols of the elements
nitrogen = N
oxide = oxygen = O
Write the formula using the prefix numbers as subscripts
di = 2, penta = 5
N2O5

63. Write Formulas for the Following
dinitrogen tetroxide
sulfur hexafluoride
diarsenic trisulfide
N2O4
SF6
As2S3

64. 2x(1.01 amu/H) + 1x(16.00 amu/O) = 18.02 amu/H2O
Formula Mass = the sum of the mass of all atoms in a single molecule or formula unit
Formula mass, aka molecular mass, aka molecular weight (MW), aka molar mass
the formula mass of 1 H2O molecule, which is made up of 2 hydrogen and 1 oxygen =
2x(1.01 amu/H) + 1x(16.00 amu/O) = amu/H2O
Molar mass can be used as a conversion factor
to convert grams to moles
and moles to grams

65. How many moles are in 50.0 g of PbO2? (Pb = 207.2 amu, O = 16.00 amu)
Given:
Find:
50.0 g mol PbO2
moles PbO2
Conceptual Plan:
Relationships:
1 mol PbO2 = g
g PbO2
mol PbO2
Calc molar mass
of PbO2
207.2 amu
Solution:
Remember, adding decimals use the least
number of decimal places for sig. figs.
Check:
because the given amount is less than g, the moles being < 1 makes sense

66. Find the number of CO2 molecules in 10.8 g of dry ice
Given:
Find:
10.8 g CO2
molecules CO2
g CO2
mol CO2
molecules CO2
Conceptual Plan:
Relationships:
1 mol CO2 = g,
1 mol = x 1023
Solution:
Check:
because the given amount is much less than 1 mol CO2, the number makes sense

67. How many formula units are in 50.0 g of PbO2? (PbO2 = 239.2)
Given:
Find:
50.0 g PbO2
formula units PbO2
Conceptual Plan:
Relationships:
1 mol PbO2 = g,1 mol = x 1023
g PbO2
mol PbO2
units PbO2
Solution:
Check:
because the given amount is less than 1 mol PbO2, the number makes sense

68. What is the mass of 4.78 x 1024 NO2 molecules?
Given:
Find:
4.78 x 1024 NO2 molecules
g NO2
Conceptual Plan:
Relationships:
1 mol NO2 = g, 1 mol = x 1023
molecules
mol NO2
g NO2
Solution:
Check:
because the given amount is more than Avogadro’s number, the mass > 46 g makes sense

69. Percent Composition Mass percent of each element in a compound
Can be determined from the formula of the compound, or the experimental mass analysis of the compound
The percentages may not always total to 100% due to rounding

70. Example 3.13: Find the mass percent of Cl in C2Cl4F2
Given:
Find:
C2Cl4F2
% Cl by mass
Conceptual Plan:
Relationships:
Solution:
Check:
because the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense

71. Determine % Ca and % Cl in CaCl2
Molar mass Ca = g/mol

72. 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 CCl2F2 is 58.64% Cl by mass means that 100 g of CCl2F2 contains g Cl
This can be used as a conversion factor
100 g CCl2F2 : g Cl

73. Given that there are 39 g Na in 100 g NaCl and you have 2. 4 g Na…
Given that there are 39 g Na in 100 g NaCl and you have 2.4 g Na…. to find the number of grams of NaCl, you have, which conversion factor would you use?
39 g Na / 100 g NaCl
100 g NaCl / 39 g Na

74. Example 3.14: Find the mass of table salt containing 2.4 g of Na
Given:
Find:
2.4 g Na, 39% Na
g NaCl
Conceptual Plan:
Relationships:
100 g NaCl : 39 g Na
g Na
g NaCl
Solution:
Check:
because the mass of NaCl is more than 2x the mass of Na, the number makes sense

75. Benzaldehyde is 79. 2% carbon. What mass of benzaldehyde contains 19
Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C?
Given:
Find:
19.8 g C, 79.2% C
g benzaldehyde
Conceptual Plan:
Relationships:
100 g benzaldehyde : 79.2 g C
g C
g benzaldehyde
Solution:
Check:
because the mass of benzaldehyde is more than the mass of C, the number makes sense

76. Conversion Factors in Chemical Formulas
Chemical formulas show inherent relationships between the number of atoms and molecules
or between the moles of atoms and molecules
These relationships can be used to convert between amounts of constituent elements and molecules
like percent composition

77. Example 3.15: Find the mass of hydrogen in 1.00 gal of water
Given:
Find:
1.00 gal H2O, dH2O = 1.00 g/ml
g H
Conceptual Plan:
Relationships:
3.785 L = 1 gal, 1 L = 1000 mL, 1.00 g H2O = 1 mL,
1 mol H2O = g, 1 mol H = g, 2 mol H : 1 mol H2O
gal H2O
L H2O
mL H2O
g H2O
g H2O
mol H2O
moL H
g H
Solution:
Check:
because 1 gallon weighs about 3800 g, and H is light, the number makes sense

78. Find the mass of sodium in 6.2 g of NaCl
Given:
Find:
6.2 g NaCl
g Na
Conceptual Plan:
Relationships:
1 mol NaCl = g, 1 mol Na = g,
1 mol Na : 1 mol NaCl
g NaCl
mol NaCl
mol Na
g Na
Solution:
Check:
because the amount of Na is less than the amount of NaCl, the answer makes sense

79. Empirical Formulas via elemental analysis
Determined by combustion analysis or % composition
If given the percentages of each element, convert them to grams by assuming you have 100 g of total compound
If given grams, proceed to next step.
Example 3.17: A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
CONVERT TO A BASIS OF 100 TOTAL GRAMS
 C = g
 H = 4.48 g
 O = g

80. How much aspirin do you assume you have at the start?
50 g
1 mol
100 g
1/3 mole 15

81. Find: the empirical formula,
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: the empirical formula,

82. Apply the conceptual plan
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
CP: g C,H,O  mol C,H,O 
pseudo form.  mol ratio  emp. form.
Rel: 1 mol C = g; 1 mol H = g; 1 mol O = g
Apply the conceptual plan
calculate the moles of each element

83. Given 60.00 g C / 4.48 g H / 35.53 g O; convert the grams of each element into moles
1 mole C = g C mole H = g H mole O = g O
5.00 mol C / 4.44 mol H / 2.22 mol O
.200 mol C / mol H / mol O
721 mol C / 4.52 mol H / 568 mol O

84. Given 4.996 mol C, 4.44 mol H, 2.220 mol O; what is the pseudo formula
C4.996H4.44O2.220
C5H4O2

85. 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
CP: g C,H,O  mol C,H,O 
pseudo form.  mol ratio  emp. form.
Rel: 1 mol C = g; 1 mol H = g; 1 mol O = g
Apply the conceptual plan
write the pseudo formula
C4.996H4.44O2.220

86. 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
CP: g C,H,O  mol C,H,O 
pseudo form.  mol ratio  emp. form.
Rel: 1 mol C = g; 1 mol H = g; 1 mol O = g
Apply the concept plan
find the mole ratio by dividing by the smallest number of moles (i.e. normalize the smallest value to 1)

87. Apply the conceptual plan
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
CP: g C,H,O  mol C,H,O 
pseudo form.  mol ratio  emp. form.
Rel: 1 mol C = g; 1 mol H = g; 1 mol O = g
Apply the conceptual plan
multiply subscripts by factor to give whole number
{C2.25H2O1} x 4
C9H8O4

88. Write a conceptual plan
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
CP: g C,H,O  mol C,H,O  pseudo form.  mol ratio  emp. form.
Write a conceptual plan
g C
mol C
whole
number
ratio
mole
ratio
empirical
formula
pseudo-
formula
g H
mol H
g O
mol O
Collect needed relationships:
1 mole C = g C mole H = g H mole O = g O

89. Finding an Empirical Formula
Convert grams to moles using each element’s molar mass
60.00 g C x 1 mol C/12.01 g C = mol C
4.48 g H x 1 mol H/1.008 g H = 4.44 mol H
35.53 g O x 1 mole O/16.00 g O = mol O
Write a pseudo-formula using moles as subscripts
CxHyOz  C4.996H4.44O2.220

90. Finding an Empirical Formula
5. Divide all subscripts by the smallest number of moles
if result is within 0.1 of whole number, round to the whole number
find the mole ratio by dividing by the smallest number of moles (i.e. normalize the smallest value to 1) 

91. Finding an Empirical Formula
6. Multiply all mole ratios by a common factor to make sure the subscripts are all whole numbers
a) if the mole ratio is 0.5, multiply all by 2; if the ratio is 0.33 or 0.67, multiply by 3; if the ratio is 0.25 or 0.75, multiply by 4; etc.
skip this step if the subscripts are already whole numbers
{C2.25H2O1} x 4
C9H8O4 

92. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00)
Given: 75.7% Sn, F must be (100 – 75.3) = 24.3%
 in 100 g stannous fluoride there are 75.7 g Sn and 24.3 g F
Find: the empirical formula SnxFy
Relationships: 1 mol Sn = g; 1 mol F = g
Conceptual plan:
whole
number
ratio
g Sn
mol Sn
mole
ratio
pseudo-
formula
empirical
formula
g F
mol F

93. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00)
Apply the conceptual plan
Sn0.638F1.28
SnF2

94. Determine the empirical formula of magnetite, which contains 72
Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)
Given: 72.4% Fe, and O must be (100 – 72.4) = 27.6%
 in 100 g magnetite there are 72.4 g Fe and 27.6 g O
Find: FexOy
Relationships: 1 mol Fe = g; 1 mol O = g
Conceptual plan:
whole
number
ratio
g Fe
mol Fe
mole
ratio
pseudo-
formula
empirical
formula
g O
mol O

95. Determine the empirical formula of magnetite, which contains 72
Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)
Apply the conceptual plan 
Fe1.30O1.73

96. 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

97. 3.18 – Given the empirical formula, C2H3O, Find the molecular formula of butanedione
butanedione MM = g/mol
molecular formula
Conceptual Plan: and
Relationships:
Solution:
Check:
the molar mass of the calculated formula is in agreement with the given molar mass

98. Molecular formula = {C5H3} x 4 = C20H12
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)
C5 = 5(12.01 g) = g
H3 = 3(1.01 g) = g
C5H = g
Molecular formula = {C5H3} x 4 = C20H12

99. Combustion Analysis Hydrogen  H2O, Oxygen = original mass – (C + H)
A common technique for analyzing compounds
burn a known mass of the compound
weigh the quantity of products made
used for organic compounds containing C,H,O
Carbon  CO2,
Hydrogen  H2O,
Oxygen = original mass – (C + H)
Knowing the masses of all the constituent elements in the original compound, the empirical formula can be found

100. Combustion Analysis

101. Example 3.20 Determine the empirical formula of the compound
Combustion of a g sample of an organic compound containing only carbon, hydrogen, and oxygen produced the following products:
CO2 = g
H2O = g
Determine the empirical formula of the compound

102. Example 3.19: Find the empirical formula of a compound with the given amounts of combustion products
Write down the given quantity and its units
Given: combustion of g of a compound
that contains only C,H,O; yields products:
2.445 g of CO2
g of H2O

103. Find: empirical formula, CxHyOz
Example: Find the empirical formula of a compound with the given amounts of combustion products
1. Write down the given information with units:
g compound,
2.445 g CO2
g H2O
2. Write down the quantity to find and/or its units
Find: empirical formula, CxHyOz
3. Write a conceptual plan g
CO2, H2O
mol
CO2, H2O
mol
C, H g
C, H g O
mol O
mol
C, H, O
pseudo
formula
mol
ratio
empirical
formula

104. Example: Find the empirical formula of a compound with the given amounts of combustion products
Information
Given: g compound,
2.445 g CO2, g H2O
Find: empirical formula, CxHyOz
CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula
Collect needed relationships 1 mole CO2 = g CO2 1 mole H2O = g H2O 1 mole C = g C 1 mole H = g H 1 mole O = g O 1 mole CO2 = 1 mole C 1 mole H2O = 2 mole H

105. Apply the conceptual plan calculate the moles of C and H
Example: Find the empirical formula of a compound with the given amounts of combustion products
Information
Given: g compound,
2.445 g CO2, g H2O
Find: empirical formula, CxHyOz
CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula
Rel: MM of CO2, H2O, C, H, O;
mol element : 1 mol compound
Apply the conceptual plan
calculate the moles of C and H

106. Apply the conceptual plan calculate the grams of C and H
Example: Find the empirical formula of a compound with the given amounts of combustion products
Information
Given: g compound,
2.445 g CO2, g H2O
Find: empirical formula, CxHyOz
CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula
Rel: MM of CO2, H2O, C, H, O;
mol element : 1 mol compound
Apply the conceptual plan
calculate the grams of C and H

107. 11am Apply the conceptual plan calculate the grams and moles of O
Example: Find the empirical formula of a compound with the given amounts of combustion products
Information
Given: g compound,
2.445 g CO2, g H2O
Find: empirical formula, CxHyOz
CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula
Rel: MM of CO2, H2O, C, H, O;
mol element : 1 mol compound
Apply the conceptual plan
calculate the grams and moles of O
11am

108. Example: Find the empirical formula of a compound with the given amounts of combustion products
Information
Given: g compound, g CO2, g H2O
mol C, g C, mol H, g H, g O, mol O
Find: empirical formula, CxHyOz
CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula
Rel: MM of CO2, H2O, C, H, O;
mol element : 1 mol compound
Apply the conceptual plan
write a pseudoformula
C H O

109. Apply the conceptual plan
Example: Find the empirical formula of a compound with the given amounts of combustion products
Information
Given: g compound, g CO2, g H2O
mol C, g C, mol H, g H, g O, mol O
Find: empirical formula, CxHyOz
CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula
Rel: MM of CO2, H2O, C, H, O;
mol element : 1 mol compound
Apply the conceptual plan
find the mole ratio by dividing by the smallest number of moles

110. Example: Find the empirical formula of a compound with the given amounts of combustion products
Information
Given: g compound, g CO2, g H2O
mol C, g C, mol H, g H, g O, mol O
Find: empirical formula, CxHyOz
CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula
Rel: MM of CO2, H2O, C, H, O;
mol element : 1 mol compound
Apply the conceptual plan
multiply subscripts by factor to give whole number, if necessary
write the empirical formula 1 12 10 O H C

111. The smell of dirty gym socks is caused by the compound caproic acid
The smell of dirty gym socks is caused by the compound caproic acid. Combustion of g of caproic acid produced g of H2O and 1.92 g of CO2. If the molar mass of caproic acid is g/mol, what is the molecular formula of caproic acid?
(MM C = 12.01, H = 1.008, O = 16.00)

112. C H O g
0.524
0.0877
0.232
moles
0.0436
0.0870
0.0145
C0.0436H0.0870O0.0145

113. Molecular formula = {C3H6O} x 2 = C6H12O2