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AP Chemistry
Introductory
Material
Chemical Foundations
Chapter 1
Observations
• Scientific Method
Theory
Or Model
Hypotheses
Predictions
Predictions
Experiment
Modify
Theory
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!
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
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
Calculations
• Adding and subtraction
– Answer has the same number of 12.11
18.0
decimal places as the least
1.013
precise measurement.
31.123
• Multiplication and Division
31.1
– Answer has the same number of significant figures
as the least precise measurement
4.56 x 1.4 = 6.38 corrected
6.4
• 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
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
• It’s way easier!
• Way, way easier!!
• Just Do it!
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)
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?
It’s the Law
• Explain the following laws:
– Conservation of Mass
– Definite proportion
– Multiple proportion
• Name four parts of Dalton’s 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
Famous Atomic Experiments
Describe the Experiment
• JJ Thompson and CRT’s
– Used CRT to determine charge to mass ratio
– Discovered electron
• Rutherford’s Gold Foil
– Used alpha particles and gold foil
– Discovered a dense, positive nucleus
• Millakan’s Oil Droplet
– Discovered the charge of an electron
– Calculated the mass of an electron with JJ’s reults
Modern Theory
• Subatomic particles are?
– Electron, neutron and proton
• Nucleus is composed of ?
– Neutron and proton
• Electrons are in “clouds”
– What does that mean?
Symbol
How many protons?
19
How many neutrons?
9
How many electrons?
F
-1
Periodic Table
Describe the following:
•
•
•
•
•
•
•
•
•
Period
Metals
Non-metals
Semi-metals
Alkali Metals
Alkali Earth Metals
Transition Metals
Halogens
Noble gases
Modern periodic table
1
IA
2
II A
13 14 15 16 17 18
III A IV A V A VI A VIIA 0
He
1
H
2
Li Be
3
3
4
5
6 7 8
Na Mg III B IVB V B VIB VIIB
4
K Ca Sc Ti
Y
V
9 10
VIII B
C
N
O
F
Ne
Al
Si
P
S
Cl Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
5
Rb Sr
6
Cs Ba *Lu Hf Ta W Re Os Ir
7
Fr Ra Lr
+
11 12
IB IIB
B
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
I
Xe
Pt Au Hg Tl Pb Bi Po At Rn
* La
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
+ Ac
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No
Metals
He
H
Li Be
B
C
N
O
F
Na Mg
Al
Si
P
S
Cl Ar
K Ca Sc Ti
Rb Sr
Y
V
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
Cs Ba *
Lu Hf Ta W Re Os Ir
+
Fr Ra Lr
Ne
* La Ce
+ Ac
I
Xe
Pt Au Hg Tl Pb Bi Po At Rn
Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No
Nonmetals
He
H
Li Be
B
C
N
O
F
Na Mg
Al
Si
P
S
Cl Ar
K Ca Sc Ti
Rb Sr
Y
V
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
Cs Ba *Ly Hf Ta W Re Os Ir
+
Fr Ra Lr
Ne
I
Xe
Pt Au Hg Tl Pb Bi Po At Rn
* La Ce
Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
+ Ac Th
Pa U Np Pu Am Cm Bk Cf Es Fm Md No
Semimetals or Metalloids
H
He
Li Be
B
C
N
O
F
Na Mg
Al
Si
P
S
Cl Ar
K Ca Sc Ti
Rb Sr
Y
V
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
Cs Ba *Lu Hf Ta W Re Os Ir
+
Fr Ra Lr
Ne
I
Xe
Pt Au Hg Tl Pb Bi Po At Rn
*
La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
+
Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No
Bonds and Stuff
Explain the following
•
•
•
•
•
•
•
•
Ion
Cation and anion
Ionic Bond
Covalent Bond
Molecule
Formula Unit
Chemical Formula
Structural Formula
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.
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 Al+3, Zn+2, Ag+1, Cd+2
Ions
• Ions are charged particles formed by the
transfer of electrons between elements or
combinations of elements.
•
•
Cation - a positively charged ion.
•
•
Anion - a negatively charged ion.
Mg
F2 + 2e-
Mg2+ + 2e-
2F-
Writing Formulas
All compounds are electrically
neutral. The sum of the positive and
negative charges must add up to
zero.
23+
Al2
O
3
Use subscripts to indicate how
many of each ion is used.
Al2O3
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
MgBr2
Al2O3
K3N
sodium chloride
magnesium bromide
aluminum oxide
potassium nitride
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
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
Examples
FeCl2
FeCl3
iron (II) chloride
iron (III) chloride
SnS
SnS2
tin (II) sulfide
tin (IV) sulfide
AgCl
ZnS
silver chloride
zinc sulfide
Note: Some transition metals have only
one oxidation state, so Roman numbers
are omitted.
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+
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
Li
B
Be
+1 +2
H
+1
+3
He
Al
Na Mg
Ca Sc
+1 +2 3+
Rb Sr
Si
+4
+3
-4
+1 +2
K
C
N
O
+4 +5 +4
-1
-2 +3 +2
-4 +1 -3 -2
Y
Ti
V
+4 +5 +4
+3 +3
+2 +2
Zr
Hf
+6 +7 +6 +3
+3 +4 +3
+2 +2 +2
Nb Mo Tc
+5
+1 +2 +3 +4
+4
Cs Ba Lu
Cr Mn Fe Co
Ta
+6
+4
+3
W
+6
+1 +2 +3 +4 +5
+4
Fr Ra Lr
+1 +2 +3
Os
Zn Ga
Rh
+4
+3
+2
Ir
Pd
Pt Au Hg
5+
3+
3-
Sn Sb
+4 Ag Cd In +4
+2 +1 +2 +3 +2
Tl
Pb
-1
Ne
P
S
Cl
+5 +6 +4+7 +5
Ar
+3 +2 +3 +1
-3
-2
-1
Ge As
+3
+2
+4
+2
+2 +3
+2
+1
-4
Ru
+7 +8 +6
+6 +4
+4 +3
Re
+7
+6
+4
Ni
Cu
F
+5
+3
-3
Bi
Se
6+
4+
2-
Br
+5
+1
-1
Te
I
+6 +7 +5
+4 +1
-2
-1
Po
At
+8 +4 +4 +3 +2 +3 +4 +5
+2 -1
+6 +3 +2 +1 +1 +1 +2 +3
Common oxidation numbers
Kr
+4
+2
Xe
+6
+4
+2
Rn
Polyatomic ions
• A special class of ions where a group of
atoms tend to stay together.
NH4+
NO3SO42OHO22-
ammonium
nitrate
sulfate
hydroxide
peroxide
Your book contains a more complete list.
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
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
…. -ate
…. -ite
hypo- …. -ite
ClO41ClO31ClO21ClO1-
Oxidation number and nomenclature
•
•
•
•
•
Anions
per ________ate
________ate
________ite
hypo ________ite
Increased #oxygen and
Oxidation number
• Polyatomic anions containing oxygen rely on a
modification of the name of the other element to
indicate the oxidation number.
Oxidation number and nomenclature
• Examples
• Cl oxidation
•
number
Formula
Name
•
•
•
•
•
NaClO4
NaClO3
NaClO2
NaClO
NaCl
sodium perchlorate
sodium chlorate
sodium chlorite
sodium hypochlorite
sodium chloride
+7
+5
+3
+1
-1
• 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.
Polyatomic Ions
1
IA
-ate has 3
Oxygens
2
II A
H
-ate & -ite charges
usually = -ide charge
1
2
3
4
5
6
7
Li Be
3 4
5
6
7 8
9 10 11 12
Na Mg III B IVB V B VIB VIIB
VIII B
IB IIB
K Ca Sc Ti
Rb Sr
Y
V
Slivka’s
Square
+
13 14 15 16 17 18
III A IV A VA VI A VIIA 0
He
B
C
N
O
F
Ne
Al
Si
P
S
Cl Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
Cs Ba *Lu Hf Ta W Re Os Ir
Fr Ra Lr
-ate has 4
Oxygens
I
Xe
Pt Au Hg Tl Pb Bi Po At Rn
* La
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
+ Ac
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No
Polyatomic Ions
“ate”
4 Oxygens .. Inside
Slivka’s Square
2ex: SO4 sulfate
3 Oxygens .. Borders the
outside of the square
ex: NO31- nitrate
Polyatomic Ions
“ite”
1 less Oxygen
compared to the -ate
1ex: ClO2 chlorite
2SO3 sulfite
Polyatomic Ions
“per” root name “ate” has 1
more O than the “ate”
ex: IO4 periodate
1-
“hypo” root name “ite” has 2
less O than the “ate”
ex: ClO1- hypochlorite
Polyatomic Ions
“Per”-“ate” 1 more O
- “ate”
- “ite” 1 less O
“hypo”-“ite” 2 less O
(also notice oxidation # of
nonmetal changes)
Polyatomic Ions
1
IA
2
II A
18
VIIIA
13 14 15 16 17
CrO42- chromate
III A IV A VA VI A VIIA He
MnO41- permanganate
H
1
2
3
4
5
6
7
Group B Elements
follow Group A patterns
Li Be
3 4
5
6
7 8
9 10 11 12
Na Mg III B IVB V B VIB VIIB
VIII B
IB IIB
K Ca Sc Ti
Rb Sr
Y
V
+
C
N
O
F
Ne
Al
Si
P
S
Cl Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
Cs Ba *Lu Hf Ta W Re Os Ir
Fr Ra Lr
B
I
Xe
Pt Au Hg Tl Pb Bi Po At Rn
* La
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
+ Ac
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No
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+
Latin Root Words
cuprous = Cu1+
cupric = Cu2+
stannous = Sn2+
stannic = Sn4+
plumbous = Pb2+
plumbic = Pb4+
aurous = Au1+
auric = Au3+
Note that there is no pattern
between the -ous and -ic suffixes
and the actual charge of the ions.
Naming Covalent Compounds
What is the difference between
2SO3 and SO3
polyatomic ion vs. neutral compound
sulfite ion vs. sulfur (VI) oxide
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
SCl
sulfur (IV) chloride
sulfur (VI) chloride
CO carbon (II) oxide
CO2 carbon (IV) oxide
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
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
F
F
S
F
F
F
F
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
tetra = 4
penta = 5
hepta = 7
octa = 8
tri = 3
hexa = 6
deca = 10
Other Methods of Naming molecules
•name elements in the formula.
•use prefixes to indicate how many
atoms there are of each type.
N2 O 5
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.
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
HNO is nitrous acid
Naming Inorganic Acids
Naming Inorganic Acids
Salt Name
Formula
Acid Name
Formula
Acetic acid
HC2H3O2
Sodium acetate NaC2H3O2
Sodium chloride
NaCl
Hydrochloric acid HCl
Sodium hyponitrite NaNO
Hyponitrous acid HNO
Sodium phosphite Na3PO3
Phosphorous acid H3PO3
Sodium sulfate Na2SO4
Sulfuric acid
H2SO4
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
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.
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
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 compound’s
name.
copper (II) sulfate pentahydrate
CuSO45H2O
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
A quick review of nomenclature
Is a metal present
Is a nonmetal the
Is hydrogen
No
as the first element? No first element?
first element?
Yes
Yes
Yes
Can the metal have
more than one
oxidation state?
Use Roman numerals
Name as
or may use prefixes
an acid
(mono, di, tri ...)
-ides become
Yes
hydro- -ic acids
Roman numerals
-ates become
are not needed.
-ic acids
Use Roman numerals
-ites become
or may use Latin name
-ous acids
with -ous/-ic suffixes
A quick review of nomenclature
Look up the
name or formula
Is it a binary
compound?
Yes
Use the -ide
suffix for the
negative ion
Does it contain
one of the 8 No
common ions?
Yes
Name the common
polyatomic ion
Does it have more
or less O atoms than
one of the -ate ions?
Yes
Use per- -ate 1 more O
-ite 1 less O
hypo- -ite 2 less O
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 - don’t 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, don’t use a
number.
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%
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?
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.
Isotopes
• Isotopes Atoms of the same element but
having different masses.
» Each isotope has a different number of neutrons.
• Isotopes of hydrogen
• Isotopes of carbon
1
1
12
6
H
2
1
C
13
6
H
3
1
H
C
14
6
C
Isotopes
• Most elements occur in nature as a mixture of
isotopes.
– Element
•
•
•
•
•
H
C
O
Fe
Sn
Number of stable isotopes
2
2
3
4
10
• This is one reason why atomic masses(weights)
are not whole numbers. They are based on
averages.
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
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 it’s 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 %
Average atomic masses
28Si
92.23
100
(27.9769265 u) = 25.80 u
29Si
4.67
100
(28.9764947 u) =
1.35 u
30Si
3.10
100
(29.9737702 u) =
0.929 u
Average atomic mass for silicon =
28.08 u
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.
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.
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.
Masses of atoms
and 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.
Masses of atoms
and
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
Theof mole
• If we had one mole
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 can’t weigh out moles-we use grams.
• We would need to weigh out a different number
of grams to have the same number of molecules
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!
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.
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
• 2 atoms O x 15.9994 u
•
Formula mass
»
=
=
=
12.011 u
31.9988 u
44.010 u
or g / mol
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
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 DON’T change -- just the units.
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
X moles = 20.0 g x
1 mol
= 0.151 mol
132.14 g
atoms = 0.151 mol x 6.02 x1023 units x
1 mol
atoms = 1.36 x1024
15 atoms
1 unit
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
1 mol
x moles = 20.0 g x
= 0.151 mol (NH4)2SO4
132.14 g
x moles NH4 = 0.151 mol (NH4)2SO4 x
moles NH4 = 0.302
2 moles NH4
1 mol (NH4)2SO4
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 moles = 20.0 g x
1 mol
= 0.151 mol (NH4)2SO4
132.14 g
x grams SO4 = 0.151 mol (NH4)2SO4 x
grams SO4 = 14.5
96.06 g SO4
1 mol (NH4)2SO4
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.
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
Gay-Lussac’s 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.
• Example. Gay-Lussac’s Law
•
Reaction of hydrogen and oxygen gases.
H2
H2
+
O2
H2O H2O
•
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)
Avogadro’s 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
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.
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 1 mol H2O x 0.5 mol O2 x 22.4 L O2
18.0 g H2O 1 mol H2O
1 mol O2
liters O2 = 6.22
X g H2 =10.0g H2O x
Note that 12.4 L H2 will also be formed.
2.02 g H2 = 1.12 g H2
18.02 g H2O
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.
O2
Elemental analyzer
CO
HO
2
trap
2
trap
furnace
sample
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.
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.
Elemental analysis
• Mass of carbon
12.01 g C
= 0.04470 g C
0.1638 g CO2
44.01 g CO2
• Mass of hydrogen
2.016 g H = 0.01875 g H
0.1675 g H2O
18.01 g H2O
• Mass of nitrogen
0.1156 g sample - 0.04470 g C - 0.01875 g H
= 0.05215 g N
Elemental analysis
• Since we know the total mass of the original
sample, we can calculate the % of each
element.
0.04470 g
%C=
x 100% = 38.67 %
0.1156 g
% H = 0.01875 g x 100% = 16.22 %
0.1156 g
% N = 0.05215 g x 100% = 45.11 %
0.1156 g
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.
Empirical formula
0.04470 g C
1 mol C = 0.003722 mol C
12.01 g C
1
mol
H
0.01875 g H
1.008 g H
= 0.0186 mol H
1
mol
N
0.05215 g N
14.01 g N
= 0.003722 mol N
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
•
N
0.003722 / 0.003722 = 1.000
•
The empirical formula is CH5N
= 4.998
Empirical formula
• From experimental analysis, we found that a
compound had a composition of:
% C = 38.67 %
% H = 16.22 %
% N = 45.11 %
• 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.
Empirical formula
1
mol
C
38.67 g C
= 3.220 mol C
12.01 g C
1
mol
H
16.22 g H
= 16.09 mol H
1.008 g H
1
mol
N
45.11 g N
= 3.220 mol N
14.01 g N
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.
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?
Molecular formula
•
•
Empirical formula
CH5N
Empirical formula mass
31.06 u
•
Molecular mass
•
Ratio:
62.12
62.12 / 31.06
=2
•
The molecular formula is C2H10N2
• Note: This does not tell you how the atoms
are arranged in the compound!
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
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
H 2O
0.0410 / 0.0205 = 2.00
The compound’s formula is BaCl22H2O
M =
Molarity
moles solute
liters of solution
=
mol
L
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 )
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
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.
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
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
Making a solution
• Finally, you’re 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.
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.
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 MM2 = 1.00
• V1 = ??? ml V2 = 250.0 mL
•
•
M2
V1 = M2V2 / M1
= (1.00 M) (250.0 mL) = 20.8 mL
»
(12.0 M)
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.
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.
Molality
Molality (m) =
moles solute
kilograms of solvent
=
mol
kg
•
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.
Density
Density (D) =
grams of solution
milliliters of solution
=
g
mL
•
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.
Percent Composition
Percent Composition =
value of the part
Value of the whole
•
•
•
% by Mass = g solute / g solution x 100
•
Must specify which type of % composition.
% by Volume = mL solute / mL solution x100
% by Mass per Volume =
g solute/mL solution x 100
x 100
Mole Fraction
Mole fraction =
moles of solute or solvent
total moles of solute & solvent
• 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.
Parts per Million or Billion
Parts per million (ppm) =
# grams of solute
1,000,000 g solution
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.
Stoich Baby
Given the following equation
1
3 2  ___NH
2
___N
2 + ___H
3
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?
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.
Chemical equations
• Chemist’s shorthand to describe a reaction.
• It shows:
• All reactants and products
• The state of all substances
• Any conditions used in the reaction
Reactant
Products
»

• CaCO3 (s)
CaO (s) + CO2 (g)
A balanced equation shows the relationship
between the quantities of all reactants and products.
Balancing chemical equations
• Each side of a chemical equation must have
the same number of each type of atom.
– CaCO3 (s)
– Reactants
» 1 Ca
»1C
»3O
CaO (s) + CO2 (g)
Products
1 Ca
1C
3O
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.
Example.
Decomposition
of
urea
______
– (NH2)2CO + H2O
•
> NH3 + CO2
2N
1 N < not balanced
• 6H
• 1C
• 2O
3 H < not balanced
1C
2O
• We need to double NH3 on the right.
– (NH2)2CO + H2O ______> 2NH3 + CO2
Mass relationships
in chemical reactions
• Stoichiometry - The calculation of quantities
of reactants and products in a chemical
reaction.
You need a balanced
equation and you WILL
work with moles.
2 H2 + O2 -----> 2 H2O
Stoichiometry,
General steps.
1
Balance the chemical equation.
2
Calculate formula masses.
3
Convert masses to moles.
4
Use chemical equation to
get the needed answer.
5
Convert back to mass if needed.
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.
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
=
=
14 g x 1 mol /28.02 g
0.50 moles
• moles N2
–
–
Mass calculations
• We don’t 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.
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
______
> CaCl2 + H2
– Note:
• We produce one H2 for each calcium.
• There is an excess of HCl so we have all we need.
Mass calculations
2HCl + Ca ____> 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
1 mol
40.08 g
Mass calculations
» 2HCl + Ca
_____
> 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
Mass calculations
• OK, so how many grams of CaCl2 were
made?
» 2HCl + Ca
»
_____
> 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
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.
Limiting reactant example
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 ________> 2H2O
• You need 2 moles of H2 for each mole of O2.
• Moles of H2
• Moles of O2
5g
1 mol
2.0 g
10g
= 2.5 mol
1 mol
32.0 g
= 0.31 mol
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 2:1
– but we have a ratio of 2.5 : 0.31 or 8.3 : 1.
– Hydrogen is in excess and oxygen is the
– limiting reactant.
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%.
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.
Percent yield
• The amount of product actually formed
divided by the amount of product calculated
to be formed, times 100.
•
% yield =
Actual yield
Theoretical yield
x 100
• In order to determine % yield, you must be
able to recover and measure all of the product
in a pure form.
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!