AP Chemistry

1
AP Chemistry

• Introductory
• Material

2
Chemical FoundationsChapter 1
Observations Hypotheses Predictions

• Scientific Method

Theory Or Model
Predictions
Experiment
Modify Theory
3
Scientific Method

• You are given a computer and asked to make a
  graph. After booting the computer, opening excel
  and entering data, the screen goes blank.
• Oh Gees! Now What!

4
Units of Measurement

• Expect you to know
• pico to giga
• And be able to convert
• Units used in science
• Kilograms, meters, seconds, kelvins, amps, moles

5
Significant Figures

• There is more than one convention!
• AP Chemistry allows for some variation
• If you are within one sig fig, it is OK
• We will follow this
• Rules are on Pg 23 of your book
• We will use these for every calculation
• You lose a point for incorrect sig figs on test

6
Calculations

• Adding and subtraction
• Answer has the same number of decimal places
  as the least precise measurement.

12.11 18.0 1.013 31.123 31.1

• Multiplication and Division
• Answer has the same number of significant
  figures
• as the least precise measurement
• 4.56 x 1.4 6.38

• pH
• The number to the left of the decimal is the
  exponent
• The number to the right of the decimal contains
  the
• correct number of sig figs.
• pH 7.07 has 2 sig figs

7
Dimensional Analysis

• Do I really have to?
• No, but it will cost you extra work explaining
  yourself
• Units written out in Dim Analysis are self
  explanatory
• Its way easier!
• Way, way easier!!
• Just Do it!

8
Mercury poisoning is a debilitating disease that
is often fatal. In the human body, mercury
reacts with essential enzymes leading to
irreversible inactivity of these enzymes. If the
amount of mercury in a polluted lake is 00.4000
micrograms Hg per milliliter, what is the total
mass in kilograms of mercury in the lake. The
lake has a surface area of 0100. mi2 and an
average depth of 20.1 ft. (5280. ft in a mile,
12 in in a foot, 2.54 cm in an inch, 106
micrograms in a gram)
9
Classification of Matter

• What is a mixture?
• Name two types
• How can we separate hetero?
• Homo?
• If I say something is a pure substance, what does
  that mean?
• What is the difference between an element and a
  compound?
• What is an element made up of?

10
Its the Law

• Explain the following laws
• Conservation of Mass
• Definite proportion
• Multiple proportion
• Name four parts of Daltons Atomic Theory
• Atoms
• All atoms of same element are identical
• Same compound always has same elements in same
  proportions
• Atoms themselves do not change in chemical
  reactions

11
Famous Atomic ExperimentsDescribe the Experiment

• JJ Thompson and CRTs
• Used CRT to determine charge to mass ratio
• Discovered electron
• Rutherfords Gold Foil
• Used alpha particles and gold foil
• Discovered a dense, positive nucleus
• Millakans Oil Droplet
• Discovered the charge of an electron
• Calculated the mass of an electron with JJs
  reults

12
Modern Theory

• Subatomic particles are?
• Electron, neutron and proton
• Nucleus is composed of ?
• Neutron and proton
• Electrons are in clouds
• What does that mean?

13
Symbol
How many protons?
How many neutrons?
How many electrons?
14
Periodic TableDescribe the following

• Period
• Metals
• Non-metals
• Semi-metals
• Alkali Metals
• Alkali Earth Metals
• Transition Metals
• Halogens
• Noble gases

15
Modern periodic table
1 2 13 14 15 16 17
18 I A II A III A IV A
V A VI A VIIA 0
He
H
1 2 3 4 5 6 7
Li
Be
B
Ne
F
O
N
C
3 4 5 6 7 8 9
10 11 12 III B IVB V B VIB VIIB
VIII B IB IIB
Na
Mg
Al
Ar
Cl
S
P
Si
K
Ca
Zn
Cu
Ti
Sc
Ni
Co
Fe
Mn
Cr
V
Ga
Kr
Br
Se
As
Ge
Rb
Sr
Cd
Ag
Zr
Y
Pd
Rh
Ru
Tc
Mo
Nb
In
Xe
I
Te
Sb
Sn

Cs
Tl
Hg
Au
Hf
Lu
Ba
Pt
Ir
Os
Re
W
Ta
Rn
At
Po
Bi
Pb

Fr
Lr
Ra


16
Metals
H
Li
Be
Na
Mg
Al
K
Ca
Zn
Cu
Ti
Sc
Ni
Co
Fe
Mn
Cr
V
Ga
Rb
Sr
Cd
Ag
Zr
Y
Pd
Rh
Ru
Tc
Mo
Nb
In
Sn
Te
Xe
Sb
I

Cs
Tl
Hg
Au
Hf
Lu
Ba
Pt
Ir
Os
Re
W
Ta
Po
Bi
Pb

Fr
Lr
Ra
La
Gd
Tb
Sm
Eu
Nd
Pm
Ce
Pr
Yb
Er
Tm
Dy
Ho

Cm
Bk
Pu
Am
U
Np
Th
Pa
No
Fm
Md
Cf
Es
Ac

17
Nonmetals
He
H
Li
Be
B
Ne
F
O
N
C
Na
Mg
Al
Si
Ar
Cl
S
P
K
Ca
Zn
Cu
Ti
Sc
Ni
Co
Fe
Mn
Cr
V
Ga
Ge
As
Kr
Br
Se
Rb
Sr
Cd
Ag
Zr
Y
Pd
Rh
Ru
Tc
Mo
Nb
In
Sb
Sn
Te
Xe
I

Cs
Tl
Hg
Au
Hf
Ly
Ba
Pt
Ir
Os
Re
W
Ta
Po
Bi
Pb
At
Rn

Fr
Lr
Ra
Gd
Tb
Sm
Eu
Nd
Pm
Ce
Pr
Yb
Er
Tm
Dy
Ho
La

Cm
Bk
Pu
Am
U
Np
Th
Pa
No
Fm
Md
Cf
Es
Ac

18
Semimetals or Metalloids
He
H
Ne
F
O
N
C
Li
Be
B
Ar
Cl
S
P
Na
Mg
Al
Si
Kr
Br
Se
K
Ca
Zn
Cu
Ti
Sc
Ni
Co
Fe
Mn
Cr
V
Ga
Ge
As
Xe
Rb
Sr
Cd
Ag
Zr
Y
Pd
Rh
Ru
Tc
Mo
Nb
In
Sb
Sn
Te
I

Rn
Cs
Tl
Hg
Au
Hf
Lu
Ba
Pt
Ir
Os
Re
W
Ta
Po
Bi
Pb
At

Fr
Lr
Ra

Gd
Tb
Sm
Eu
Nd
Pm
Ce
Pr
Yb
La
Er
Tm
Dy
Ho
Cm
Bk
Pu
Am
U
Np
Th
Pa
No
Ac
Fm
Md
Cf
Es

19
Bonds and Stuff
Explain the following

• Ion
• Cation and anion
• Ionic Bond
• Covalent Bond
• Molecule
• Formula Unit
• Chemical Formula
• Structural Formula

20
Ions and Ionic Compounds
Important note that there are no easily
identified NaCl molecules in the ionic lattice.
Therefore, we cannot use molecular formulas to
describe ionic substances.
21
Naming Compounds

• Memorize all polyatomic ions pg 63 and how to
  determine the rest
• Memorize names of elements
• s block, p block all
• Transition metals
• Need to know common metals
• Charges on Al3, Zn2, Ag1, Cd2

22
Ions

• Ions are charged particles formed by the transfer
  of electrons between elements or combinations of
  elements.
• Cation - a positively charged ion.
• Mg Mg2 2e-
• Anion - a negatively charged ion.
• F2 2e- 2F-

23
Writing Formulas
All compounds are electrically neutral. The sum
of the positive and negative charges must add up
to zero.
Al3
O2-
3
2
Use subscripts to indicate how many of each ion
is used.
Al2O3
24
Naming inorganic compounds

• When an element forms only one compound with a
  given anion.
• name the cation
• name the anion using the ending (-ide) for
  monatomic ions

NaCl sodium chloride MgBr2 magnesium
bromide Al2O3 aluminum oxide K3N potassium
nitride
25
Naming ionic compounds

• Many metals form more than one compound with some
  anions.
• For these, Roman numerals are used in the name
  to indicate the charge on the metal.
• Cu1 O2- Cu2O
• copper(I) oxide copper(I) oxide
• Cu2 O2- CuO
• copper(II) oxide copper(II) oxide

26
Naming ionic compounds

• Since the charge of some metal ions can vary,
  look at everything else first.
• What ever is left is the charge on the metal!
• FeBr3
• The three bromides are each 1- so iron must be 3
  for the compound to have zero net charge.
• Iron (III) bromide

27
Examples
iron (II) chloride iron (III) chloride tin (II)
sulfide tin (IV) sulfide silver chloride zinc
sulfide
FeCl2 FeCl3 SnS SnS2 AgCl ZnS
Note Some transition metals have only one
oxidation state, so Roman numbers are omitted.
28
Metals with multiple charges

• Transition metals.
• Here it is easier to list some of the common
  elements that only have a single oxidation state.
• All Group 3B are 3
• Zn and Cd are 2
• Ag is 1

29
Oxidation numbers and the P.T.

• Some observed trends in compounds.
• Metals have positive oxidation numbers.
• Transition metals typically have more than one
  oxidation number.
• Nonmetals and semimetals have both positive and
  negative oxidation numbers.
• No element exists in a compound with an oxidation
  number greater than 8.
• The most negative oxidation numbers equals the
  group number - 8

30
B 3
Ne
F -1
O -1 -2
N 5 4 3 2 1 -3
C 4 -2 -4
H 1
He
Al 3
Ar
Cl 7 5 3 1 -1
S 6 4 2 -2
P 5 3 -3
Si 4 -4
Zn 2
Cu 2 1
Ti 4 3 2
Sc 3
Ni 2
Co 3 2
Fe 3 2
Mn 7 6 4 3 2
Cr 6 3 2
V 5 4 3 2
Ga 3
Kr 4 2
Br 5 1 -1
Se 6 4 2-
As 5 3 3-
Ge 4 -4
Cd 2
Ag 1
Zr 4
Y 3
Pd 4 2
Rh 4 3 2
Ru 8 6 4 3
Tc 7 6 4
Mo 6 4 3
Nb 5 4
In 3
Xe 6 4 2
I 7 5 1 -1
Te 6 4 -2
Sb 5 3 -3
Sn 4 2
Tl 3 1
Hg 2 1
Au 3 1
Hf 4
Lu 3
Pt 4 2
Ir 4 3
Os 8 6
Re 7 6 4
W 6 4
Ta 5
Rn
At -1
Po 2
Bi 5 3
Pb 4 2
Lr 3
Common oxidation numbers
31
Polyatomic ions

• A special class of ions where a group of atoms
  tend to stay together.

NH4 ammonium NO3- nitrate SO42- sulfate OH-
hydroxide O22- peroxide Your book contains a
more complete list.
32
Polyatomic ions

• For compounds that contain 1 or 2 polyatomic
  ions, base the formulas upon the given ion
  name(s).
• ammonium chloride NH4Cl
• sodium hydroxide NaOH
• potassium permanganate KMnO4
• ammonium sulfate (NH4)2SO4

33
Naming Inorganic Compounds
Names and Formulas of Ionic Compounds Polyatomic
anions containing oxygen with more than two
members in the series are named as follows (in
order of decreasing oxygen) per- .
-ate ClO41- . -ate ClO31- .
-ite ClO21- hypo- . -ite ClO1-
34
Oxidation number and nomenclature

• Polyatomic anions containing oxygen rely on a
  modification of the name of the other element to
  indicate the oxidation number.
• Anions
• per ________ate
• ________ate
• ________ite
• hypo ________ite

Increased oxygen and Oxidation number
35
Oxidation number and nomenclature

• Examples
• Cl oxidation
• number Formula Name
• 7 NaClO4 sodium perchlorate
• 5 NaClO3 sodium chlorate
• 3 NaClO2 sodium chlorite
• 1 NaClO sodium hypochlorite
• -1 NaCl sodium chloride
• Usually, the overall charges of all ions for a
  nonmetal are the same. Sometimes the -ates and
  -ites have a different charge than the -ide ions.

36
-ate has 4 Oxygens
Polyatomic Ions
-ate has 3 Oxygens
1 2 13 14 15 16 17
18 I A II A III A IV A
VA VI A VIIA 0
H
He
-ate -ite charges usually -ide charge
1 2 3 4 5 6 7
Li
Be
B
Ne
F
O
N
C
3 4 5 6 7 8 9
10 11 12 III B IVB V B VIB VIIB
VIII B IB IIB
Na
Mg
Al
Ar
Cl
S
P
Si
Slivkas Square
K
Ca
Zn
Cu
Ti
Sc
Ni
Co
Fe
Mn
Cr
V
Ga
Kr
Br
Se
As
Ge
Rb
Sr
Cd
Ag
Zr
Y
Pd
Rh
Ru
Tc
Mo
Nb
In
Xe
I
Te
Sb
Sn

Cs
Tl
Hg
Au
Hf
Lu
Ba
Pt
Ir
Os
Re
W
Ta
Rn
At
Po
Bi
Pb

Fr
Lr
Ra


37
Polyatomic Ions
ate 4 Oxygens .. Inside Slivkas Square
ex SO42- sulfate 3 Oxygens .. Borders the
outside of the square ex NO31- nitrate

38
Polyatomic Ions
ite 1 less Oxygen compared to the -ate
ex ClO21- chlorite SO32- sulfite
39
Polyatomic Ions
per root name ate has 1 more O than the
ate ex IO41- periodate hypo root name
ite has 2 less O than the ate ex ClO1-
hypochlorite

40
Polyatomic Ions
Per-ate 1 more O - ate
- ite 1 less O hypo-ite 2 less
O (also notice oxidation of nonmetal changes)
41
Polyatomic Ions
Group B Elements follow Group A patterns
1 2 18 I A II A
VIIIA
CrO42- chromate MnO41- permanganate
13 14 15 16 17
H
He
III A IV A VA VI A VIIA
1 2 3 4 5 6 7
Li
Be
B
Ne
F
O
N
C
3 4 5 6 7 8 9
10 11 12 III B IVB V B VIB VIIB
VIII B IB IIB
Na
Mg
Al
Ar
Cl
S
P
Si
K
Ca
Zn
Cu
Ti
Sc
Ni
Co
Fe
Mn
Cr
V
Ga
Kr
Br
Se
As
Ge
Rb
Sr
Cd
Ag
Zr
Y
Pd
Rh
Ru
Tc
Mo
Nb
In
Xe
I
Te
Sb
Sn

Cs
Tl
Hg
Au
Hf
Lu
Ba
Pt
Ir
Os
Re
W
Ta
Rn
At
Po
Bi
Pb

Fr
Lr
Ra


42
Other Methods of Naming - Latin

• For ionic compounds containing a metal and a
  nonmetal, the Latin root word for the metal is
  sometimes used with an -ous or -ic suffix.
• The -ous suffix indicates a lower oxidation
  state, -ic a higher one.
• Ex. ferrous Fe2
• ferric Fe3

43
Latin Root Words
plumbous Pb2 plumbic Pb4 aurous
Au1 auric Au3
cuprous Cu1 cupric Cu2 stannous
Sn2 stannic Sn4
Note that there is no pattern between the -ous
and -ic suffixes and the actual charge of the
ions.
44
Naming Covalent Compounds
What is the difference between
SO32- and SO3
polyatomic ion vs. neutral compound
sulfite ion vs. sulfur (VI) oxide
45
Naming Covalent Compounds
Some nonmetals can have more than one positive
oxidation state when they share electrons to form
molecules called covalent compounds. Therefore
Roman numbers must be used.
SCl4 sulfur (IV) chloride SCl sulfur (VI)
chloride CO carbon (II) oxide CO2 carbon
(IV) oxide
46
Naming Inorganic Compounds
Names and Formulas of Binary Molecular Compounds
Binary molecular compounds are composed of two
nonmetallic elements. The element with the
positive oxidation number (the one closest to the
lower left corner on the periodic table) is
usually written first. Exception
NH3. Greek prefixes are used to indicate the
number of atoms in the molecule(subscripts). PCl5
is phosphorus pentachloride
47
Naming Inorganic Compounds
Names and Formulas of Binary Molecular Compounds
Roman numerals can be used to indicate the
positive oxidation number, but sometimes prefixes
more accurately describe the actual composition
of the molecule. Example sulfur (V) fluoride
exists as disulfur decafluoride molecules.
F
F
F
F
S
S
F
F
F
F
F
F
48
Other Methods of Naming molecules

• For binary molecular compounds composed of two
  nonmetals, prefixes are sometimes used to
  indicate the number of atoms of each element
  present.

Common prefixes mono 1 di 2
tri 3 tetra 4 penta 5 hexa 6
hepta 7 octa 8 deca 10
49
Other Methods of Naming molecules

• name elements in the formula.
• use prefixes to indicate how many atoms there are
  of each type.

N2O5 CO2 CO CCl4
dinitrogen pentoxide carbon dioxide carbon
monoxide carbon tetrachloride
The rule may be modified to improve how a name
sounds. Example - use monoxide not monooxide.
50
Other Naming -Acids
Acids are substances that produce H ions in
water solutions (aqueous). The names of acids are
related to the names of the anions to which H is
bonded -ide becomes hydro-.-ic acid H2S
is hydrosulfuric acid -ate becomes -ic
acid H3PO4 is phosphoric acid -ite becomes
-ous acid HNO2 is nitrous acid
51
Naming Inorganic Acids
52
Naming Inorganic Acids
Salt Name Formula Acid
Name Formula Sodium acetate NaC2H3O2
Acetic acid HC2H3O2 Sodium chloride
NaCl Hydrochloric acid HCl Sodium
hyponitrite NaNO Hyponitrous acid
HNO Sodium phosphite Na3PO3
Phosphorous acid H3PO3 Sodium sulfate Na2SO4
Sulfuric acid H2SO4
53
Naming Inorganic Compounds
Names and Formulas of Acids Acids contain
hydrogen as the only cation. The names of acids
are related to the names of the anions -ide
becomes hydro-.-ic acid H2S is hydrosulfuric
acid -ate becomes -ic acid H3PO4 is phosphoric
acid -ite becomes -ous acid. HNO2 is nitrous
acid
54
Other Naming -Double Triple Salts
Polyatomic anions containing oxygen with
additional hydrogens are named by adding hydrogen
(or bi-) for one extra H, dihydrogen (for two
extra H), etc., to the name as follows
CO32- is named carbonate, but HCO3- is hydrogen
carbonate (or bicarbonate) H2PO4- dihydrogen
phosphate anion.
Note that these are not named as acids, since
another cation is still needed to balance the
charge.
55
Other Naming -Double Triple Salts
Two or three different positive ions can be
attracted to the same negative ion to form a
single compound. These are called double or
triple salts. Each ion is named as it appears.
NaHCO3 sodium hydrogen carbonate (or sodium
bicarbonate) AlK(SO4)2 aluminum potassium sulfate
56
pentahydrate 5H2O
Other Naming - Hydrates
Hydrated compounds physically trap water
molecules as part of their structure. A prefix
is used to indicate the relative number of water
molecules present with the word hydrate added
after the compounds name.
copper (II) sulfate pentahydrate CuSO4?5H2O
57
Other Naming - Historical Names
Sometimes the names of compounds are based upon
their historical significance or derivation.
There are no patterns or rules for determining
these names, so they would have to be memorized.
For example, H2O is called water, not dihydrogen
monoxide Check out http//www.dhmo.org
58
A quick review of nomenclature
No
Is a metal present as the first element?
No
Is a nonmetal the first element?
Is hydrogen first element?
Yes
Yes
Yes
Can the metal have more than one oxidation state?
Use Roman numerals or may use prefixes (mono,
di, tri ...)
Name as an acid
No
-ides become hydro- -ic acids -ates become
-ic acids -ites
become -ous acids
Yes
Roman numerals are not needed.
Use Roman numerals or may use Latin name with
-ous/-ic suffixes
59
A quick review of nomenclature
Look up the name or formula
Is it a binary compound?
No
No
Yes
Does it contain one of the 8 common ions?
Does it have more or less O atoms than one of the
-ate ions?
No
Use the -ide suffix for the negative ion
Yes
Yes
Name the common polyatomic ion
Use per- -ate 1 more O -ite 1 less O
hypo- -ite 2 less O
60
Naming Compounds Summary

• Simple rules that will keep you out of trouble
  most of the time.
• Groups IA, 2A, 3A (except Tl) only have a single
  oxidation state that is the same as the group
  number - dont use numbers.
• Most other metals and semimetals have multiple
  oxidation states - use numbers.
• If you are sure that a transition group element
  only has a single state, dont use a number.

61
Average Atomic Mass

• What is the average atomic mass of carbon-12?
• Why is this a bad question?
• If I traveled to alpha centauri, would the
  average atomic mass of chlorine be 35.45?
• Can you calculate the AAM of the following
• 1H 99
• 2H 1

62
Moles and Moles

• How many atoms in a mole?
• What does the mole do?
• How do you calculate molar mass?
• What is an empirical formula?
• What is a molecular formula?

63
Atomic masses

• Atoms are composed of protons, neutrons and
  electrons.
• Almost all of the mass of an atom comes from the
  protons and neutrons.
• All atoms of the same element will have the same
  number of protons. The number of neutrons may
  vary - isotopes.
• Most elements exist as a mixture of isotopes.

64
Isotopes

• Isotopes Atoms of the same element but
  having different masses.
• Each isotope has a different number of
  neutrons.
• Isotopes of hydrogen H H H
• Isotopes of carbon C C C

1 1
2 1
3 1
12 6
13 6
14 6
65
Isotopes

• Most elements occur in nature as a mixture of
  isotopes.
• Element Number of stable isotopes
• H 2
• C 2
• O 3
• Fe 4
• Sn 10
• This is one reason why atomic masses(weights) are
  not whole numbers. They are based on averages.

66
Atomic masses

• As a reference point, we use the atomic mass unit
  (u), which is equal to 1/12th of the mass of a
  12C atom.
• (One atomic mass unit (u) 1.66 x 10-24 gram)
• Using this relative system, the mass of all
  other atoms can be assigned.
• Examples 7Li 7.016 004 u
• 14N 14.003 074 01 u
• 29Si 28.976 4947 u

67
Average atomic masses

• One can calculate the average atomic weight of an
  element if the abundance of each isotope for that
  element is known.
• Silicon exists as a mixture of three isotopes.
  Determine its average atomic mass based on the
  following data.
• Isotope Mass (u) Abundance
• 28Si 27.9769265 92.23
• 29Si 28.9764947 4.67
• 30Si 29.9737702 3.10

68
Average atomic masses
69
The mole

• The number of atoms in 12.000 grams of 12C can be
  calculated.
• One atom 12C 12.000 u 12 x (1.661 x 10-24 g)
• 1.993 x 10-23 g / atom
• atoms 12.000 g (1 atom / 1.993 x 10-23 g)
• 6.021 x 1023 atoms
• The number of atoms of any element needed to
  equal its atomic mass in grams will always be
  6.02 x 1023 atoms - the mole.

70
Moles and masses

• Atoms come in different sizes and masses.
• A mole of atoms of one type would have a
  different mass than a mole of atoms of another
  type.
• H - 1.008 grams / mol
• O - 16.00 grams / mol
• Mo - 95.94 grams / mol
• Pb - 207.2 grams / mol
• We rely on a straight forward system to relate
  mass and moles.

71
The mole

• 1 mole of any element 6.02 x 1023 atoms
• gram atomic mass
• Atoms, ions and molecules are too small to
  directly measure in u.
• Using moles gives us a practical unit.
• We can then relate atoms, ions and molecules,
  using an easy to measure unit - the gram.

72
Masses of atomsand molecules

• Atomic mass
• The average, relative mass of an atom in an
  element. Can be expressed in relative amtomic
  mass units (u) or grams / mole.
• Molecular or formula mass
• The total mass for all atoms in a compound.

73
Masses of atomsand molecules
H2O - water 2 hydrogen 2 x 1.008 u 1
oxygen 1 x 16.00 u mass of molecule
18.02 u 18.02 g / mol
Rounded off based on significant figures
74
The mole

• If we had one mole of water and one mole of
  hydrogen, we would have the same number of
  molecules of each.
• 1 mol H2O 6.022 x 1023 molecules
• 1 mol H2 6.022 x 1023 molecules
• We cant weigh out moles-we use grams.
• We would need to weigh out a different number of
  grams to have the same number of molecules

75
Converting units

• Factor label method
• Regardless of conversion, keeping track of units
  makes thing come out right
• Must use conversion factors
• - The relationship between two units
• Canceling out units is a way of checking that
  your calculation is set up right!

76
Molecular mass vs. formula mass

• Formula mass - Add the masses of all the atoms in
  formula for molecular and ionic compounds.
• Molecular mass - Calculated the same as formula
  mass only valid for molecules.
• Both have units of either u or grams / mole.
• Molar mass is the generic term for the mass of
  one mole of anything.

77
Formula mass, FM

• The sum of the atomic masses of all elements in a
  compound based on the chemical formula.
• You must use the atomic masses of the elements
  listed in the periodic table.
• CO2 1 atom of C and 2 atoms of O
• 1 atom C x 12.011 u 12.011 u
• 2 atoms O x 15.9994 u 31.9988 u
• Formula mass 44.010 u
• or g / mol

78
Another example
CH3CH2OH - ethyl alcohol 2 carbon 2 x 12.01
u 6 hydrogen 6 x 1.008 u 1 oxygen 1 x 16.00
u mass of molecule 46.07 u 46.07 g /mol
79
Molar masses

• Once you know the mass of an atom, ion, or
  molecule, just remember
• Mass of one unit - use amu
• Mass of one mole of units - use grams / mole
• The numbers DONT change -- just the units.

80
Example - (NH4)2SO4

• How many atoms are in 20.0 grams of ammonium
  sulfate?
• Formula weight 132.14 grams/ 1 mol
• Atoms in formula 15 atoms / 1 formula unit

atoms 0.151 mol x
x
atoms 1.36 x1024
81
Example - (NH4)2SO4

• Other information can be derived from the
  chemical formula of a compound.
• How many moles of ammonium ions are in 20.0 grams
  of ammonium sulfate?

Formula weight 132.14 g / 1 mol (NH4)2SO4 2
moles NH4 / 1 mol (NH4)2SO4
x moles NH4 0.151 mol (NH4)2SO4 x
moles NH4 0.302
82
Example - (NH4)2SO4

• How many grams of sulfate ions are in 20.0 grams
  of ammonium sulfate?

Formula weight 132.14 g / 1 mol
(NH4)2SO4 96.06 grams SO4 / 1 mol (NH4)2SO4
x grams SO4 0.151 mol (NH4)2SO4 x
grams SO4 14.5
83
Masses of atoms and molecules
Law of Definite Composition - compounds always
have a definite proportion of the elements that
make it up. These proportions can be expressed
as ratios of atoms, equivalent mass values,
percentage by mass or volumes of gaseous
elements. Ex. Water always contains 2 H atoms
for every O atom, which is 2 g H for every 16 g O
or 11.1 H and 88.9 O by mass.
84
Percent Composition by Mass
Percent composition can also be determined from
experimental data.
Example When 2.47 g KClO3 is heated strongly,
0.96 g of O2 gas is driven off. What is the by
mass of oxygen in KClO3?

• 0 0.96 g O x 100 38.9 O
• 2.47 g KClO3
• Based upon the formula mass
• O 48.00 u O x 100 39.17 O
• 122.55 u KClO3

85
Gay-Lussacs Law

• Law of of Combining Volumes.
• At constant temperature and pressure, the
  volumes of gases involved in a chemical reaction
  are in the ratios of small whole numbers.
• Studies by Joseph Gay-Lussac led to a better
  understanding of molecules and their reactions.

86
Gay-Lussacs Law

• Example.
• Reaction of hydrogen and oxygen gases.
• Two volumes of hydrogen will combine with one
  volume of oxygen to produce two volumes of
  water.
• We now know that the equation is
• 2 H2 (g) O2 (g) 2 H2O (g)

87
Avogadros law

• Equal volumes of gas at the same temperature and
  pressure contain equal numbers of molecules (or
  moles of molecules).

Contain same number of moles of molecules
88
Standard conditions (STP)

• Remember the following standard conditions.
• Standard temperature 273.15 Kelvin
• (the normal freezing point of water, 0ºC)
• Standard pressure 1 atmosphere
• (the normal air pressure at sea level, 14.7 psi)

At these conditions One mole of any gas has a
volume of 22.4 liters at STP.
89
Applying Law of Definite Composition
In an expermiment, 10.0 grams of water is
decomposed by electrolysis.
Problem How many liters of O2 gas will be formed
at STP? How many grams of H2 gas will be formed?
X L O2 10.0g H2O x x
x
liters O2 6.22 Note that 12.4 L H2 will also
be formed.
X g H2 10.0g H2O x
1.12 g H2
90
Empirical formula

• This type of formula shows the ratios of the
  number of atoms of each kind in a compound.
• For organic compounds, the empirical formula can
  be determined by combustion analysis.
• Elemental analyzer
• An instrument in which an organic compound is
  quantitatively converted to carbon dioxide and
  water -- both of which are then measured.

91
Elemental analyzer
A sample is burned, completely converting it to
CO2 and H2O. Each is collected and measured as a
weight gain. By adding other traps elements like
oxygen, nitrogen, sulfur and halogens can also be
determined.
92
Elemental analysis

• Example A compound known to contain only
  carbon, hydrogen and nitrogen is examined by
  elemental analysis. The following information is
  obtained.
• Original sample mass 0.1156 g
• Mass of CO2 collected 0.1638 g
• Mass of H2O collected 0.1676 g
• Determine the of each element in the compound.

93
Elemental analysis

• Mass of carbon
• Mass of hydrogen
• Mass of nitrogen

0.1156 g sample - 0.04470 g C - 0.01875 g H
0.05215 g N
94
Elemental analysis

• Since we know the total mass of the original
  sample, we can calculate the of each element.

95
Empirical formula

• Empirical formula
• The simplest formula that shows the ratios of
  the number of atoms of each element in a
  compound.
• Example - the empirical formula for hydrogen
  peroxide (H2O2) is HO.
• We can use either our mass or our percent
  composition information from the earlier example
  to determine an empirical formula.

96
Empirical formula
97
Empirical formula

• The empirical formula is then found by looking
  for the smallest whole number mole ratio.
• C 0.003722 / 0.003722 1.000
• H 0.0186 / 0.003722 4.998
• N 0.003722 / 0.003722 1.000
• The empirical formula is CH5N

98
Empirical formula

• From experimental analysis, we found that a
  compound had a composition of
• If we assume that we have a 100.0 gram sample,
  then we can divide each percentage by the
  elements atomic mass and determine the relative
  number of moles of each.

C 38.67 H 16.22 N 45.11
99
Empirical formula
100
Empirical formula

• The empirical formula is found by looking for the
  smallest whole number ratio.
• C 3.220 / 3.220 1.000
• H 16.09 / 3.220 4.997
• N 3.220 / 3.220 1.000
• The empirical formula is determined to be the
  same, CH5N, whether using actual masses of the
  elements present in the sample or by using their
  composition by mass.

101
Molecular formula

• Molecular formula - shows the actual number of
  each type of atom in a molecule.
• They are multiples of the empirical formula.
• If you know the molecular mass, then the
  molecular formula can be found.
• For our earlier example, what would be the
  molecular formula if you knew that the molecular
  mass was 62.12?

102
Molecular formula

• Empirical formula CH5N
• Empirical formula mass 31.06 u
• Molecular mass 62.12
• Ratio 62.12 / 31.06 2
• The molecular formula is C2H10N2
• Note This does not tell you how the atoms are
  arranged in the compound!

103
Hydrated Compounds
The formula for hydrated compounds are solved in
a similar fashion as empirical formulas. Example
When a 5.000 g sample of hydrated barium
chloride is heated to dryness, 0.738 g H2O is
lost. 5.000g hydrate - 0.738g H2O 4.262g
BaCl2
104
Hydrated Compounds
4.262g BaCl2 (1 mol BaCl2) 0.0205 mol
BaCl2 (208.2 g BaCl2) 0.738 g H2O ( 1
mol H2O ) 0.0410 mol H2O (18.0 g
H2O ) BaCl2 0.0205 / 0.0205 1.00 H2O
0.0410 / 0.0205 2.00 The compounds formula
is BaCl2?2H2O
105
Molarity

• Molarity
• Recognizes that compounds have different
• formula weights.
• A 1 M solution of sucrose contains the
• same number of molecules as 1 M ethanol.
• - special symbol which means molar
• ( mol/L )

106
Molarity

• Calculate the molarity of a 2.0 L solution that
  contains 10 moles of NaOH.
• M NaOH 10 mol NaOH / 2.0 L 5.0
  M

107
Solution preparation

• Solutions are typically prepared by
• Dissolving the proper amount of solute and
  diluting to volume.
• Dilution of a concentrated solution.
• Lets look at an example of the calculations
  required to prepare known molar solutions using
  both approaches.

108
Making a solution

• You are assigned the task of preparing 250.0 mL
  of a 1.00 M solution of sodium hydroxide.
• What do you do?
• First, you need to know how many moles of NaOH
  are in 250.0 mL of a 1.00 M solution.
• mol M x V (in liters)
• 1.00 M x 0.2500 liters
• 0.250 moles NaOH

109
Making a solution

• Next, we need to know how many grams of NaOH to
  weigh out.
• g NaOH mol x molar massNaOH
• 0.250 mol x 40.0 g/mol
• 10.0 grams NaOH

110
Making a solution

• Finally, youre ready to make the solution.
• Weigh out exactly 10.0 grams of dry, pure NaOH
  and transfer it to a volumetric flask, (or some
  other containing where the exact volume can be
  accurately measured.)
• Fill the flask about 1/2 of the way with
  distilled water and gently swirl until the solid
  dissolves.
• Now, dilute exactly to the mark, cap and mix.

111
Dilution

• Once you have a solution, it can be diluted by
  adding more solvent. This is also important for
  materials only available as solutions
• M1V1 M2V2
• 1 initial 2 final
• Any volume or concentration unit can be used as
  long as you use the same units on both sides of
  the equation.

112
Dilution

• How many ml of concentrated 12.0 M HCl must be
  diluted to produce 250.0 mL of 1.00 M HCl?
• M1V1 M2V2
• M1 12.0 M M2 1.00
• V1 ??? ml V2 250.0 mL
• V1 M2V2 / M1
• M2 (1.00 M) (250.0 mL) 20.8 mL
• (12.0 M)

113
Diluting an Acid

• When diluting concentrated acids, ALWAYS add the
  acid to water to help dissipate the heat
  released.
• Fill the volumetric flask (or some other
  containing where the exact volume can be
  accurately measured.) about 1/2 of the way with
  distilled water.
• Measure out exactly 20.8 mL of 12.0 M HCl and
  transfer it to a volumetric flask. Gently swirl
  to mix.
• Now, dilute exactly to the mark, cap and mix.

114
Other Methods of Expressing Concentration

• When making different solutions with a specific
  molarity, the number of milliliters of solvent
  needed to prepare 1 liter of solution will vary.
• Sometimes it is necessary to know the exact
  proportions of solute to solvent that are in a
  particular solution.
• Various methods have been devised to express
  these proportions.

115
Molality
moles solute
mol kilograms of solvent kg
Molality (m)

• Recognizes that the ratio between moles of solute
  and kg of solvent can vary.
• A 1 m solution of sucrose contains the same
  number of molecules as 1 m ethanol.
• The freezing point of water is lowered by
  1.86ºC/ m and the boiling point is raised by
  0.51ºC/ m.

116
Density
grams of solution g milliliters
of solution mL
Density (D)

• Focuses on the total solution and does not
  emphasize either the solute or solvent.
• g solution g solute g solvent
• Units may be expressed as other mass per volume
  ratios.

117
Percent Composition
value of the part Value of the whole
Percent Composition x 100

• by Mass g solute / g solution x 100
• by Volume mL solute / mL solution x100
• by Mass per Volume
• g solute/mL solution x 100
• Must specify which type of composition.

118
Mole Fraction
moles of solute or solvent total moles of
solute solvent
Mole fraction

• Often used to compare ratio between moles of
  gases in a mixture.
• The mole ratio of gases in a mixture is equal to
  their pressure ratio and their volume ratio.

119
Parts per Million or Billion
grams of solute 1,000,000 g solution
Parts per million (ppm)

• Parts per Million or Billion
• Used to express concentrations for very
• dilute solutions.
• For aqueous solutions, the mixture is
• mostly water. Therefore, the density of
• the solution 1 g/mL, and
• 1 ppm 1 g/1000 L.

120
Stoich Baby

• Given the following equation
• ___N2 ___H2 ? ___NH3

1 3
2
Given one mole of nitrogen gas, how many moles of
ammonia would form?
Assuming gases at STP, given one mole of
nitrogen, how many liters of ammonia would form?
Given 2 liters of nitrogen and 5 liters of
hydrogen, how may liters of ammonia are formed?
What is left over?
121
Stoichiometry

• Stoichiometry
• The study of quantitative relationships between
  substances undergoing chemical changes.
• Law of Conservation of Matter
• In chemical reactions, the quantity of matter
  does not change.
• The total mass of the products must equal that of
  the reactants.

122
Chemical equations

• Chemists shorthand to describe a reaction.
• It shows
• All reactants and products
• The state of all substances
• Any conditions used in the reaction
• CaCO3 (s) CaO (s) CO2 (g)

Reactant Products
?
A balanced equation shows the relationship between
the quantities of all reactants and products.
123
Balancing chemical equations

• Each side of a chemical equation must have the
  same number of each type of atom.
• CaCO3 (s) CaO (s) CO2 (g)
• Reactants Products
• 1 Ca 1 Ca
• 1 C 1 C
• 3 O 3 O

124
Balancing chemical equations

• Step 1 Count the number of atoms of each element
  on each side of the equation.
• Step 2 Determine which atom numbers are not
  balanced.
• Step 3 Balance one atom at a time by using
  coefficients in front of one or more substances.
• Step 4 Repeat steps 1-3 until everything is
  balanced.

125
Example.Decomposition of urea

• (NH2)2CO H2O ______gt NH3 CO2
• 2 N 1 N lt not balanced
• 6 H 3 H lt not balanced
• 1 C 1 C
• 2 O 2 O
• We need to double NH3 on the right.
• (NH2)2CO H2O ______gt 2NH3 CO2

126
Mass relationshipsin chemical reactions

• Stoichiometry - The calculation of quantities of
  reactants and products in a chemical reaction.

127
Stoichiometry,General steps.
128
Mole calculations

• The balanced equation shows the reacting ratio
  between reactants and products.
• 2C2H6 7O2 4CO2 6H2O
• For each chemical, you can determine the
• moles of each reactant consumed
• moles of each product made
• If you know the formula mass,
• mass quantities can be used.

129
Mole-gram conversion

• How many moles are in 14 grams of N2 ?
• Formula mass
• 2 N x 14.01 g/mol
• 28.02 g /mol
• moles N2
• 14 g x 1 mol /28.02 g
• 0.50 moles

130
Mass calculations

• We dont directly weigh out molar quantities.
• We can use measured masses like kilograms, grams
  or milligrams.
• The formula masses and the chemical equations
  allow us to use either mass or molar quantities.

131
Mass calculations

• How many grams of hydrogen will be produced if
  10.0 grams of calcium is added to an excess of
  hydrochloric acid?
• 2HCl Ca ______gt CaCl2 H2
• Note
• We produce one H2 for each calcium.
• There is an excess of HCl so we have all we need.

132
Mass calculations

• 2HCl Ca ____gt CaCl2 H2
• First - Determine the number of moles of
  calcium available for the reaction.
• Moles Ca grams Ca / FM Ca
• 10.0 g x
• 0.25 mol Ca

133
Mass calculations

• 2HCl Ca _____gt CaCl2 H2
• 10 g Ca 0.25 mol Ca
• According to the chemical equation, we get one
  mole of H2 for each mole of Ca.
• So we will make 0.25 moles of H2.
• grams H2 produced moles x FW H2
• 0.25 mol x 2.016 g/mol
• 0.504 grams

134
Mass calculations

• OK, so how many grams of CaCl2 were made?
• 2HCl Ca _____gt CaCl2 H2
• 10 g Ca 0.25 mol Ca
• We would also make 0.25 moles of CaCl2.
• g CaCl2 0.25 mol x FM CaCl2
• 0.25 mol x 110.98 g / mol CaCl2
• 27.75 g CaCl2

135
Limiting reactant

• In the last example, we had HCl in excess.
• Reaction stopped when we ran out of Ca.
• Ca is considered the limiting reactant.
• Limiting reactant - the material that is in the
  shortest supply based on a balanced chemical
  equation.

136
Limiting reactant example
137
Example

• For the following reaction, which is limiting if
  you have 5.0 g of hydrogen and 10 g oxygen?
• Balanced Chemical Reaction
• 2H2 O2 ________gt 2H2O
• You need 2 moles of H2 for each mole of O2.
• Moles of H2 5 g 2.5 mol
• Moles of O2 10g 0.31 mol

138
Example

• Balanced Chemical Reaction
• 2H2 O2 2H2O
• You need 2 moles of H2 for each mol of O2
• You have 2.5 moles of H2 and 0.31 mol of O2
• Need a ratio of 21
• but we have a ratio of 2.5 0.31 or 8.3 1.
• Hydrogen is in excess and oxygen is the
• limiting reactant.

139
Theoretical, actual, and percent yields

• Theoretical yield
• The amount of product that should be formed
  according to the chemical reaction and
  stoichiometry.
• Actual yield
• The amount of product actually formed.
• Percent yield
• Ratio of actual to theoretical yield, as a .
• Quantitative reaction
• When the percent yield equals 100.

140
Yield

• Less product is often produced than expected.
• Possible reasons
• A reactant may be impure.
• Some product is lost mechanically since the
  product must be handled to be measured.
• The reactants may undergo unexpected reactions -
  side reactions.
• No reaction truly has a 100 yield due to the
  limitations of equilibrium.

141
Percent yield

• The amount of product actually formed divided by
  the amount of product calculated to be formed,
  times 100.
• yield
  x 100
• In order to determine yield, you must be able
  to recover and measure all of the product in a
  pure form.

142
Stoichiometry

• Step 1 
• Identify species present in solution and
  determine the reaction that occurs
• Step 2
• Write the balanced net ionic equation
• Step 3
• Calculate the moles of reaction
• solution molarity x volume
• heterogeneous grams ? molar mass
• Step 4
• Consider the limiting reactant
• Step 5
• Answer the question using stoich!