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Prentice-Hall ©General Chemistry:Slide 1 of 37
Philip Dutton
University of Windsor, Canada
Prentice-Hall © 2002
Chapter 3: Chemical Compounds
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th
Edition
Prentice-Hall ©General Chemistry:Slide 2 of 37
Contents
3-1 Molecular and Ionic Compounds
3-2 Molecular Mass
3-3 Composition
3-4 Oxidation States
3-5 Names and formulas
Focus on Mass Spectrometry
Prentice-Hall ©General Chemistry:Slide 3 of 37
Molecular compounds
1 /inch
0.4 /cm
Prentice-Hall ©General Chemistry:Slide 4 of 37
Standard color scheme
Prentice-Hall ©General Chemistry:Slide 5 of 37
Some molecules
H2O2 CH3CH2Cl P4O10
CH3CH(OH)CH3 HCO2H
Prentice-Hall ©General Chemistry:Slide 6 of 37
Ionic compounds
 Atoms of almost all elements can gain or lose electrons to
form charged species called ions.
 Compounds composed of ions are known as ionic
compounds.
 Metals tend to lose electrons to form positively
charged ions called cations.
 Non-metals tend to gain electrons to form
negatively charged ions called anions.
Prentice-Hall ©General Chemistry:Slide 7 of 37
Sodium chloride
Extended array of Na+
and Cl-
ions Simplest formula unit is NaCl
Prentice-Hall ©General Chemistry:Slide 8 of 37
Inorganic molecules
S8
P4
Prentice-Hall ©General Chemistry:Slide 9 of 37
Molecular mass
HO
O
H
H
HO
H
OH
OHH
H
OH
Molecular formula C6H12O6
Empirical formula CH2O
Glucose
6 x 12.01 + 12 x 1.01 + 6 x 16.00
Molecular Mass: Use the naturally occurring mixture of isotopes,
= 180.18
Exact Mass: Use the most abundant isotopes,
6 x 12.000000 + 12 x 1.007825 + 6 x 15.994915
= 180.06339
Prentice-Hall ©General Chemistry:Slide 10 of 37
Halothane C2HBrClF3
M(C2HBrClF3
) = 2MC + MH + MBr + MCl + 3MF
= (2  12.01) + 1.01 + 79.90 + 35.45 + (3  19.00)
= 197.38 g/mol
Chemical Composition
Mole ratio nC/nhalothane
Mass ratio mC/mhalothane
Prentice-Hall ©General Chemistry:Slide 11 of 37
Example 3.4
Calculating the Mass Percent Composition of a Compound
Calculate the molecular mass
M(C2HBrClF3
) = 197.38 g/mol
For one mole of compound, formulate the mass
ratio and convert to percent:
%17.12%100
38.197
)01.122(
% =×
×
=
g
g
C
Prentice-Hall ©General Chemistry:Slide 12 of 37
Example 3-4
%88.28%100
38.197
)00.193(
%
%96.17%100
38.197
45.35
%
%48.40%100
38.197
90.79
%
%51.0%100
38.197
01.1
%
%17.12%100
38.197
)01.122(
%
=×
×
=
=×=
=×=
=×=
=×
×
=
g
g
F
g
g
Cl
g
g
Br
g
g
H
g
g
C
Prentice-Hall ©General Chemistry:Slide 13 of 37
Empirical formula
1. Choose an arbitrary sample size (100g).
2. Convert masses to amounts in moles.
3. Write a formula.
4. Convert formula to small whole numbers.
5. Multiply all subscripts by a small whole number
to make the subscripts integral.
5 Step approach:
Prentice-Hall ©General Chemistry:Slide 14 of 37
Determining the Empirical and Molecular Formulas of a
Compound from Its Mass Percent Composition.
Dibutyl succinate is an insect repellent used against household
ants and roaches. Its composition is 62.58% C, 9.63% H and
27.79% O. Its experimentally determined molecular mass is
230 u. What are the empirical and molecular formulas of
dibutyl succinate?
Step 1: Determine the mass of each element in a 100g sample.
C 62.58 g H 9.63 g O 27.79 g
Dibutyl succinate is an insect repellent used against household
ants and roaches. Its composition is 62.58% C, 9.63% H and
27.79% O. Its experimentally determined molecular mass is
230 u. What are the empirical and molecular formulas of
dibutyl succinate?
Example 3-5
Prentice-Hall ©General Chemistry:Slide 15 of 37
Step 2: Convert masses to amounts in moles.
Omol
Og
Omol
Ogn
Hmol
Hg
Hmol
Hgn
Cmol
Cg
Cmol
Cgn
O
H
C
737.1
999.15
1
79.27
55.9
008.1
1
63.9
210.5
011.12
1
58.62
=×=
=×=
=×=
Step 3: Write a tentative formula.
Step 4: Convert to small whole numbers.
C5.21H9.55O1.74
C2.99H5.49O
Example 3-5
Prentice-Hall ©General Chemistry:Slide 16 of 37
Step 5: Convert to a small whole number ratio.
Multiply 2 to get C5.98H10.98O2
The empirical formula is C6H11O2
Step 6: Determine the molecular formula.
Empirical formula mass is 115 u.
Molecular formula mass is 230 u.
The molecular formula is C12H22O4
Example 3-5
Prentice-Hall ©General Chemistry:Slide 17 of 37
Combustion analysis
Prentice-Hall ©General Chemistry:Slide 18 of 37
Oxidation States
Metals tend to
lose electrons.
Na  Na+
+ e-
Non-metals tend
to gain electrons.
Cl + e-
 Cl-
Reducing agents Oxidizing agents
We use the Oxidation State to keep track of the number of
electrons that have been gained or lost by an element.
Prentice-Hall ©General Chemistry:Slide 19 of 37
Rules for Oxidation States
1. The oxidation state (OS) of an individual atom in a free
element is 0.
2. The total of the OS in all atoms in:
i. Neutral species is 0.
ii. Ionic species is equal to the charge on the ion.
3. In their compounds, the alkali metals and the alkaline
earths have OS of +1 and +2 respectively.
4. In compounds the OS of fluorine is always –1
Prentice-Hall ©General Chemistry:Slide 20 of 37
Rules for Oxidation States
5. In compounds, the OS of hydrogen is usually +1
6. In compounds, the OS of oxygen is usually –2.
7. In binary (two-element) compounds with metals:
i. Halogens have OS of –1,
ii. Group 16 have OS of –2 and
iii. Group 15 have OS of –3.
Prentice-Hall ©General Chemistry:Slide 21 of 37
Assigning Oxidation States.
What is the oxidation state of the underlined element in each
of the following? a) P4; b) Al2O3; c) MnO4
-
; d) NaH
a) P4 is an element. P OS = 0
b) Al2O3: O is –2. O3 is –6. Since (+6)/2=(+3), Al OS = +3.
c) MnO4
-
: net OS = -1, O4 is –8. Mn OS = +7.
d) NaH: net OS = 0, rule 3 beats rule 5, Na OS = +1 and
H OS = -1.
Example 3-7
Prentice-Hall ©General Chemistry:Slide 22 of 37
Naming Compounds
Trivial names are used for common compounds.
A systematic method of naming compounds is
known as a system of nomenclature.
Organic compounds
Inorganic compounds
Prentice-Hall ©General Chemistry:Slide 23 of 37
Inorganic Nomenclature
Binary Compounds of Metals and Nonmetals
NaCl = sodium chloride
name is unchanged “ide” endingelectrically neutral
MgI2 = magnesium iodide
Al2O3 = aluminum oxide
Na2S = sodium sulfide
Prentice-Hall ©General Chemistry:Slide 24 of 37
Prentice-Hall ©General Chemistry:Slide 25 of 37
Binary Compounds of Two Non-metals
Molecular compounds
usually write the positive OS element first.
HCl hydrogen chloride
mono 1 penta 5
di 2 hexa 6
tri 3 hepta 7
tetra 4 octa 8
Some pairs form more than one compound
Prentice-Hall ©General Chemistry:Slide 26 of 37
Prentice-Hall ©General Chemistry:Slide 27 of 37
Binary Acids
Emphasize the fact that a molecule is an acid by altering the
name.
HCl hydrogen chloride hydrochloric acid
HF hydrogen fluoride hydrofluoric acid
Acids produce H+
when dissolved in water.
They are compounds that ionize in water.
Prentice-Hall ©General Chemistry:Slide 28 of 37
Polyatomic Ions
Polyatomic ions are very common.
Table 3.3 gives a list of some of them. Here are a few:
ammonium ion NH4
+
acetate ion C2H3O2
-
carbonate ion CO3
2-
hydrogen carbonate HCO3
-
hypochlorite ClO-
phosphate PO4
3-
chlorite ClO2
-
hydrogen phosphate HPO4
2-
chlorate ClO3
-
sulfate SO4
2-
perchlorate ClO4
-
hydrogensulfate HSO4
-
Prentice-Hall ©General Chemistry:Slide 29 of 37
Prentice-Hall ©General Chemistry:Slide 30 of 37
Naming Organic Compounds
Organic compounds abound in nature
Fats, carbohydrates and proteins are foods.
Propane, gasoline, kerosene, oil.
Drugs and plastics
Carbon atoms form chains and rings and act as
the framework of molecules.
Prentice-Hall ©General Chemistry:Slide 31 of 37
Prentice-Hall ©General Chemistry:Slide 32 of 37
Visualizations of some hydrocarbons
Prentice-Hall ©General Chemistry:Slide 33 of 37
Visualizations of some hydrocarbons
Prentice-Hall ©General Chemistry:Slide 34 of 37
Isomers
Isomers have the same molecular formula but have different
arrangements of atoms in space.
H
(c)
Prentice-Hall ©General Chemistry:Slide 35 of 37
Functional Groups – carboxylic acid
Prentice-Hall ©General Chemistry:Slide 36 of 37
Functional Groups - alcohol
Prentice-Hall ©General Chemistry:Slide 37 of 37
Chapter 3 Questions
3, 5, 12, 24, 35,
46, 53, 61, 57, 73,
95, 97

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Ch03 121227103357-phpapp01

  • 1. Prentice-Hall ©General Chemistry:Slide 1 of 37 Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 Chapter 3: Chemical Compounds General Chemistry Principles and Modern Applications Petrucci • Harwood • Herring 8th Edition
  • 2. Prentice-Hall ©General Chemistry:Slide 2 of 37 Contents 3-1 Molecular and Ionic Compounds 3-2 Molecular Mass 3-3 Composition 3-4 Oxidation States 3-5 Names and formulas Focus on Mass Spectrometry
  • 3. Prentice-Hall ©General Chemistry:Slide 3 of 37 Molecular compounds 1 /inch 0.4 /cm
  • 4. Prentice-Hall ©General Chemistry:Slide 4 of 37 Standard color scheme
  • 5. Prentice-Hall ©General Chemistry:Slide 5 of 37 Some molecules H2O2 CH3CH2Cl P4O10 CH3CH(OH)CH3 HCO2H
  • 6. Prentice-Hall ©General Chemistry:Slide 6 of 37 Ionic compounds  Atoms of almost all elements can gain or lose electrons to form charged species called ions.  Compounds composed of ions are known as ionic compounds.  Metals tend to lose electrons to form positively charged ions called cations.  Non-metals tend to gain electrons to form negatively charged ions called anions.
  • 7. Prentice-Hall ©General Chemistry:Slide 7 of 37 Sodium chloride Extended array of Na+ and Cl- ions Simplest formula unit is NaCl
  • 8. Prentice-Hall ©General Chemistry:Slide 8 of 37 Inorganic molecules S8 P4
  • 9. Prentice-Hall ©General Chemistry:Slide 9 of 37 Molecular mass HO O H H HO H OH OHH H OH Molecular formula C6H12O6 Empirical formula CH2O Glucose 6 x 12.01 + 12 x 1.01 + 6 x 16.00 Molecular Mass: Use the naturally occurring mixture of isotopes, = 180.18 Exact Mass: Use the most abundant isotopes, 6 x 12.000000 + 12 x 1.007825 + 6 x 15.994915 = 180.06339
  • 10. Prentice-Hall ©General Chemistry:Slide 10 of 37 Halothane C2HBrClF3 M(C2HBrClF3 ) = 2MC + MH + MBr + MCl + 3MF = (2  12.01) + 1.01 + 79.90 + 35.45 + (3  19.00) = 197.38 g/mol Chemical Composition Mole ratio nC/nhalothane Mass ratio mC/mhalothane
  • 11. Prentice-Hall ©General Chemistry:Slide 11 of 37 Example 3.4 Calculating the Mass Percent Composition of a Compound Calculate the molecular mass M(C2HBrClF3 ) = 197.38 g/mol For one mole of compound, formulate the mass ratio and convert to percent: %17.12%100 38.197 )01.122( % =× × = g g C
  • 12. Prentice-Hall ©General Chemistry:Slide 12 of 37 Example 3-4 %88.28%100 38.197 )00.193( % %96.17%100 38.197 45.35 % %48.40%100 38.197 90.79 % %51.0%100 38.197 01.1 % %17.12%100 38.197 )01.122( % =× × = =×= =×= =×= =× × = g g F g g Cl g g Br g g H g g C
  • 13. Prentice-Hall ©General Chemistry:Slide 13 of 37 Empirical formula 1. Choose an arbitrary sample size (100g). 2. Convert masses to amounts in moles. 3. Write a formula. 4. Convert formula to small whole numbers. 5. Multiply all subscripts by a small whole number to make the subscripts integral. 5 Step approach:
  • 14. Prentice-Hall ©General Chemistry:Slide 14 of 37 Determining the Empirical and Molecular Formulas of a Compound from Its Mass Percent Composition. Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate? Step 1: Determine the mass of each element in a 100g sample. C 62.58 g H 9.63 g O 27.79 g Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate? Example 3-5
  • 15. Prentice-Hall ©General Chemistry:Slide 15 of 37 Step 2: Convert masses to amounts in moles. Omol Og Omol Ogn Hmol Hg Hmol Hgn Cmol Cg Cmol Cgn O H C 737.1 999.15 1 79.27 55.9 008.1 1 63.9 210.5 011.12 1 58.62 =×= =×= =×= Step 3: Write a tentative formula. Step 4: Convert to small whole numbers. C5.21H9.55O1.74 C2.99H5.49O Example 3-5
  • 16. Prentice-Hall ©General Chemistry:Slide 16 of 37 Step 5: Convert to a small whole number ratio. Multiply 2 to get C5.98H10.98O2 The empirical formula is C6H11O2 Step 6: Determine the molecular formula. Empirical formula mass is 115 u. Molecular formula mass is 230 u. The molecular formula is C12H22O4 Example 3-5
  • 17. Prentice-Hall ©General Chemistry:Slide 17 of 37 Combustion analysis
  • 18. Prentice-Hall ©General Chemistry:Slide 18 of 37 Oxidation States Metals tend to lose electrons. Na  Na+ + e- Non-metals tend to gain electrons. Cl + e-  Cl- Reducing agents Oxidizing agents We use the Oxidation State to keep track of the number of electrons that have been gained or lost by an element.
  • 19. Prentice-Hall ©General Chemistry:Slide 19 of 37 Rules for Oxidation States 1. The oxidation state (OS) of an individual atom in a free element is 0. 2. The total of the OS in all atoms in: i. Neutral species is 0. ii. Ionic species is equal to the charge on the ion. 3. In their compounds, the alkali metals and the alkaline earths have OS of +1 and +2 respectively. 4. In compounds the OS of fluorine is always –1
  • 20. Prentice-Hall ©General Chemistry:Slide 20 of 37 Rules for Oxidation States 5. In compounds, the OS of hydrogen is usually +1 6. In compounds, the OS of oxygen is usually –2. 7. In binary (two-element) compounds with metals: i. Halogens have OS of –1, ii. Group 16 have OS of –2 and iii. Group 15 have OS of –3.
  • 21. Prentice-Hall ©General Chemistry:Slide 21 of 37 Assigning Oxidation States. What is the oxidation state of the underlined element in each of the following? a) P4; b) Al2O3; c) MnO4 - ; d) NaH a) P4 is an element. P OS = 0 b) Al2O3: O is –2. O3 is –6. Since (+6)/2=(+3), Al OS = +3. c) MnO4 - : net OS = -1, O4 is –8. Mn OS = +7. d) NaH: net OS = 0, rule 3 beats rule 5, Na OS = +1 and H OS = -1. Example 3-7
  • 22. Prentice-Hall ©General Chemistry:Slide 22 of 37 Naming Compounds Trivial names are used for common compounds. A systematic method of naming compounds is known as a system of nomenclature. Organic compounds Inorganic compounds
  • 23. Prentice-Hall ©General Chemistry:Slide 23 of 37 Inorganic Nomenclature Binary Compounds of Metals and Nonmetals NaCl = sodium chloride name is unchanged “ide” endingelectrically neutral MgI2 = magnesium iodide Al2O3 = aluminum oxide Na2S = sodium sulfide
  • 25. Prentice-Hall ©General Chemistry:Slide 25 of 37 Binary Compounds of Two Non-metals Molecular compounds usually write the positive OS element first. HCl hydrogen chloride mono 1 penta 5 di 2 hexa 6 tri 3 hepta 7 tetra 4 octa 8 Some pairs form more than one compound
  • 27. Prentice-Hall ©General Chemistry:Slide 27 of 37 Binary Acids Emphasize the fact that a molecule is an acid by altering the name. HCl hydrogen chloride hydrochloric acid HF hydrogen fluoride hydrofluoric acid Acids produce H+ when dissolved in water. They are compounds that ionize in water.
  • 28. Prentice-Hall ©General Chemistry:Slide 28 of 37 Polyatomic Ions Polyatomic ions are very common. Table 3.3 gives a list of some of them. Here are a few: ammonium ion NH4 + acetate ion C2H3O2 - carbonate ion CO3 2- hydrogen carbonate HCO3 - hypochlorite ClO- phosphate PO4 3- chlorite ClO2 - hydrogen phosphate HPO4 2- chlorate ClO3 - sulfate SO4 2- perchlorate ClO4 - hydrogensulfate HSO4 -
  • 30. Prentice-Hall ©General Chemistry:Slide 30 of 37 Naming Organic Compounds Organic compounds abound in nature Fats, carbohydrates and proteins are foods. Propane, gasoline, kerosene, oil. Drugs and plastics Carbon atoms form chains and rings and act as the framework of molecules.
  • 32. Prentice-Hall ©General Chemistry:Slide 32 of 37 Visualizations of some hydrocarbons
  • 33. Prentice-Hall ©General Chemistry:Slide 33 of 37 Visualizations of some hydrocarbons
  • 34. Prentice-Hall ©General Chemistry:Slide 34 of 37 Isomers Isomers have the same molecular formula but have different arrangements of atoms in space. H (c)
  • 35. Prentice-Hall ©General Chemistry:Slide 35 of 37 Functional Groups – carboxylic acid
  • 36. Prentice-Hall ©General Chemistry:Slide 36 of 37 Functional Groups - alcohol
  • 37. Prentice-Hall ©General Chemistry:Slide 37 of 37 Chapter 3 Questions 3, 5, 12, 24, 35, 46, 53, 61, 57, 73, 95, 97

Editor's Notes

  1. Chemical formula – relative numbers of atoms of each element present Empirical formula – the simplest whole number formula Structural formula – the order and type of attachements – shows multiple bonds - may show lone pairs - hard to show 3-d
  2. Positive and negaive ions joined together by electrostatic forces Metals tend to lose electrons to form cations Non-metals tend to gain electrons to form anions Ionic solids formulae are reported as the formula unit – inappropriate to call it a molecular formula
  3. Na loses one electron to form the sodium ion Cl gains one electron to form the chloride ion Centers of ions are shown in the ball and stick model for clarity Space filling model shows how the ions are actually in contact with one another. We will discuss face centered cubic and other types of packing in chapter 13
  4. Some inorganic compounds for molecules Sulfur and phosporous for example. They come in various forms called allotropes – these are one allotrope of each
  5. Glucose Emprical formula leads us to the name “carbohydrate”
  6. Molecular formula tells us there are TWO moles of C per mole of halothane. We also know about the MASSES of the compound and its elemental components. Therefore we can talk about PERCENT COMPOSITION BY MASS
  7. These types of calculations can be carried out in reverse for the following reasons: Unknown compounds are analyzed for % composition. Relative proportion of elements present on a mass basis. Chemical formula requires mole basis, I.e. numbers of atoms.
  8. If you know the molecular wt it is beneficial to choose that number, then only first three steps are required.
  9. Read the problem carefully Pick out the critical information Think Follow the steps to solve the problem
  10. Step 5. You can multiply the rounded off one if you wish, but be careful of introducing an error If all the subscripts are within ±0.1 you are probably OK to round to the integer. Step 6: Simple multiplication is obvious here.
  11. Water vapour absorbed by magnesium perchlorate Carbon dioxide absorbed by sodium hydroxide. The differences in mass of the absorbers before and after yiled the masses of water and CO2 produced in the reaction Combustion takes place in an excess of oxygen so you cannot measure oxygen. Oxygen CAN be analyzed separately but is usually determined by difference.
  12. Metals are electron sources Non-metals are electron sinks Sodium goes to the +1 oxidation state Chlorine goes tot eh –1 oxidation state
  13. Rule 1 states OS of elements is 0 Rule 2 the total OS is 0, Rule 6 oxygen should be –2 to give a total of –6 for O, therefore 2 Al must be +6 or each Al is +3. Rule 2 the total OS is –1, Rule 6 oxygen should be –2 to give a total of –8 for O, therefore Mn must be +7. Rule 2 the total OS is –1, Rule 3 beats Rule 5, so Na OS = +1 and H OS = -1. There are other examples in the text and much more detail on the rules. Read this material carefully.
  14. Trivial names such as water, ammonia, sugar, acetone, ether.
  15. Write the unmodified name of the metal Then write the name of the nonmetal, modifed to end in ide. Ionic compounds must be electrically neutral
  16. We have already discussed simple anions such as hydride, fluoride, chloride, iodide etc.
  17. Most oxoacids are ternary compounds composed of hydrogen, oxygen and one other nonmental. Oxoacids are molecular compounds, salts are ionic compounds Ic and ate names are assigned to compounds (rather than ite and ate as in the oxoanions) in which the central nonmetal atom has an oxidation state equal to the periodic group number – 10 For halogens ic and ate names are assigned to compounds in which the halogen has an oxidation state of +5.
  18. These are structural isomers. The structures are different these molecules do not have the same formula, they are different c) Now these molecules have the same formula and ALSO the same connectivity. These are geometric isomers.