142
AP Chemistry Introductory Material
AP Chemistry
Introductory
Material
Chemical FoundationsChapter 1
• Scientific MethodObservations
Hypotheses
PredictionsTheoryOr Model
Predictions
ExperimentModifyTheory
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 decimal places as the least
precise measurement.
12.1118.0 1.01331.12331.1
• 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
corrected6.4
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 ExperimentsDescribe 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
F19
9
-1How many protons?
How many neutrons?
How many electrons?
Periodic TableDescribe the following:
• Period• Metals• Non-metals• Semi-metals• Alkali Metals• Alkali Earth Metals• Transition Metals• Halogens• Noble gases
Modern periodic table
H
Li
Na
Cs
Rb
K
TlHgAuHfLuBa
Fr
PtIrOsReWTa
He
RnAtPoBiPb
Be
Mg
Sr
Ca
CdAgZrY PdRhRuTcMoNb
LrRa
ZnCuTiSc NiCoFeMnCrV
In XeITeSbSn
Ga KrBrSeAsGe
Al ArClSPSi
B NeFONC
1 2 13 14 15 16 17 18I A II A III A IV A V A VI A VIIA 0
3 4 5 6 7 8 9 10 11 12III B IVB V B VIB VIIB VIII B IB IIB
1
2
3
4
5
6
7 Gd
Cm
Tb
Bk
Sm
Pu
Eu
Am
Nd
U
Pm
Np
Ce
Th
Pr
Pa
Yb
No
La
Ac
Er
Fm
Tm
Md
Dy
Cf
Ho
Es
*+
+
*
Metals
Li
Na
Cs
Rb
K
TlHgAuHfLuBa
Fr
PtIrOsReWTa PoBiPb
Be
Mg
Sr
Ca
CdAgZrY PdRhRuTcMoNb
LrRa
ZnCuTiSc NiCoFeMnCrV
In Sn
Ga
Al
Gd
Cm
Tb
Bk
Sm
Pu
Eu
Am
Nd
U
Pm
Np
Ce
Th
Pr
Pa
Yb
No
Er
Fm
Tm
Md
Dy
Cf
Ho
Es
At Rn
B
He
NeFONC
Te XeISb
As
Si
KrBrSe
ArClSP
Ge
La
Ac
*+
*+
H
NonmetalsH
Li
Na
Cs
Rb
K
TlHgAuHfLyBa
Fr
PtIrOsReWTa PoBiPb
Be
Mg
Sr
Ca
CdAgZrY PdRhRuTcMoNb
LrRa
ZnCuTiSc NiCoFeMnCrV
In SbSn
Ga Ge
Al
Gd
Cm
Tb
Bk
Sm
Pu
Eu
Am
Nd
U
Pm
Np
Ce
Th
Pr
Pa
Yb
No
Er
Fm
Tm
Md
Dy
Cf
Ho
Es
At
Te
As
Si
B
He
Rn
XeI
KrBrSe
ArClS
NeFO
P
NC
La
Ac
*+
*+
Semimetals or Metalloids
He
Rn
XeI
KrBrSe
ArClS
NeFO
P
NC
H
Li
Na
Cs
Rb
K
TlHgAuHfLuBa
Fr
PtIrOsReWTa PoBiPb
Be
Mg
Sr
Ca
CdAgZrY PdRhRuTcMoNb
LrRa
ZnCuTiSc NiCoFeMnCrV
In SbSn
Ga Ge
Al
Gd
Cm
Tb
Bk
Sm
Pu
Eu
Am
Nd
U
Pm
Np
Ce
Th
Pr
Pa
Yb
No
La
Ac
Er
Fm
Tm
Md
Dy
Cf
Ho
Es
At
Te
As
Si
B
*+
*+
Bonds and Stuff
• Ion
• Cation and anion
• Ionic Bond
• Covalent Bond
• Molecule
• Formula Unit
• Chemical Formula
• Structural Formula
Explain the following
Important: note that there are no easily identified NaCl molecules in the ionic lattice. Therefore, we cannot use molecular formulas to describe ionic substances.
Ions and Ionic CompoundsIons and Ionic Compounds
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.
• CationCation - a positively charged ion.
• Mg Mg2+ + 2e-
• AnionAnion - a negatively charged ion.
• F2 + 2e- 2F-
Writing Formulas
Al3+ O2-
All compounds are electrically All compounds are electrically neutral. The sum of the positive neutral. The sum of the positive and negative charges must add up and negative charges must add up to zero.to zero.
Al2O3
Use subscripts to indicate how Use subscripts to indicate how many of each ion is used.many of each ion is used.
32
Naming inorganic compoundsWhen an element forms only one compound with a given anion.
• name the cation
• name the anion using the ending (-ide-ide) for monatomic ions
NaCl sodium chloride
MgBr2 magnesium bromide
Al2O3 aluminum oxide
K3N 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) bromideIron (III) bromide
ExamplesFeCl2FeCl3
SnSSnS2
AgClZnS
Note: Note: Some transition metals have only one oxidation state, so Roman numbers are omitted.
iron (II) chlorideiron (III) chloride
tin (II) sulfidetin (IV) sulfide
silver chloridezinc sulfide
Metals with multiple charges• Transition metals.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.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
Tl
+3+1
Hg
+2+1
Au
+3+1
Hf
+4Lu
+3
Li
+1
Na
+1
Cs
+1
Rb
+1
K
+1
Fr
+1
Pt
+4+2
Ir
+4+3
Os
+8+6
Re+7+6+4
W
+6+4
Ta
+5
H
+1He
RnAt
-1Po
+2
Bi
+5+3
Pb
+4+2
Cd
+2Ag
+1Zr
+4Y
+3
Pd
+4+2
Rh+4+3+2
Ru+8 +6
+4+3
Tc+7+6+4
Mo+6+4+3
Nb
+5+4
Lr
+3
Ba
+2
Be
+2
Mg
+2
Sr
+2
Ca
+2
Ra
+2
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
In
+3
Xe+6+4+2
I+7 +5
+1-1
Te+6+4-2
Sb+5+3-3
Sn
+4+2
Ga
+3
Kr
+4+2
Br+5+1-1
Se6+4+2-
As5+3+3-
Ge
+4-4
Al
+3 Ar
Cl+7 +5+3 +1
-1
S+6 +4
+2-2
P+5+3-3
Si
+4-4
B
+3 NeF
-1
O
-1-2
N+5 +4+3 +2+1 -3
C+4-2-4
Common oxidation numbersCommon oxidation numbers
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.
NH4+ ammonium
NO3- nitrate
SO42- sulfate
OH- hydroxide
O22- 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
Names and Formulas of Ionic CompoundsNames 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-
Naming Inorganic CompoundsNaming Inorganic Compounds
Oxidation number and nomenclature
• Polyatomic anions containing oxygenPolyatomic 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
Incr
ease
d #
oxyg
en
an
dO
xid
ati
on
nu
mb
er
Incr
ease
d #
oxyg
en
an
dO
xid
ati
on
nu
mb
er
Oxidation number and nomenclature• ExamplesExamples
• 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.
H
Li
Na
Cs
Rb
K
TlHgAuHfLuBa
Fr
PtIrOsReWTa
He
RnAtPoBiPb
Be
Mg
Sr
Ca
CdAgZrY PdRhRuTcMoNb
LrRa
ZnCuTiSc NiCoFeMnCrV
In XeITeSbSn
Ga KrBrSeAsGe
Al ArClSPSi
B NeFONC
1 2 13 14 15 16 17 18I A II A III A IV A VA VI A VIIA 0
3 4 5 6 7 8 9 10 11 12III B IVB V B VIB VIIB VIII B IB IIB
1
2
3
4
5
6
7 Gd
Cm
Tb
Bk
Sm
Pu
Eu
Am
Nd
U
Pm
Np
Ce
Th
Pr
Pa
Yb
No
La
Ac
Er
Fm
Tm
Md
Dy
Cf
Ho
Es
*+
+
*
Polyatomic Ions-ate has 3 Oxygens
-ate has 4 Oxygens
Slivka’s Slivka’s SquareSquare
-ate & -ite charges usually = -ide charge
Polyatomic Ions“ate” 4 Oxygens .. Inside Slivka’s Square
ex: SO42- sulfate
3 Oxygens .. Borders the outside of the square
ex: NO31- nitrate
Polyatomic Ions“ite” 1 less Oxygen compared to the -ate
ex: ClO21- chlorite
SO32- sulfite
Polyatomic Ions “per” root name “ate” has 1 more O than the “ate” ex: IO4
1-
periodate“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)
H
Li
Na
Cs
Rb
K
TlHgAuHfLuBa
Fr
PtIrOsReWTa
He
RnAtPoBiPb
Be
Mg
Sr
Ca
CdAgZrY PdRhRuTcMoNb
LrRa
ZnCuTiSc NiCoFeMnCrV
In XeITeSbSn
Ga KrBrSeAsGe
Al ArClSPSi
B NeFONC
1 2 18 I A II A VIIIA
3 4 5 6 7 8 9 10 11 12III B IVB V B VIB VIIB VIII B IB IIB
1
2
3
4
5
6
7 Gd
Cm
Tb
Bk
Sm
Pu
Eu
Am
Nd
U
Pm
Np
Ce
Th
Pr
Pa
Yb
No
La
Ac
Er
Fm
Tm
Md
Dy
Cf
Ho
Es
*+
+
*
Polyatomic Ions Group B Elements follow Group A patterns
CrO42- chromate
MnO41- permanganate
III A IV A VA VI A VIIA
13 14 15 16 17
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 Wordscuprous = 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
sulfite ionsulfite ion vs. sulfur (VI) oxidesulfur (VI) oxide
What is the difference between
SOSO332-2- and SOSO33
polyatomic ionpolyatomic ion vs. neutral compoundneutral compound
Naming Covalent Compounds
SCl4 sulfur (IV) chlorideSCl sulfur (VI) chloride
CO carbon (II) oxideCO2 carbon (IV) oxide
Some nonmetals can have more than one positive oxidation state when they share electrons to form molecules called covalent compoundscovalent compounds. Therefore Roman numbers must be used.
Names and Formulas of Binary Molecular Names and Formulas of Binary Molecular CompoundsCompounds 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 CompoundsNaming Inorganic Compounds
Names and Formulas of Binary Molecular Names and Formulas of Binary Molecular CompoundsCompounds
Naming Inorganic CompoundsNaming Inorganic 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.
S S
F
FF
FF
F
F
F
FF
Other Methods of Naming molecules• For binary molecular compounds composed of composed of
two nonmetalstwo nonmetals, prefixes are sometimes used to indicate the number of atoms of each element present.
Common prefixes mono = 1mono = 1 di = 2 di = 2 tri = 3 tri = 3 tetra = 4tetra = 4 penta = 5 penta = 5 hexa = 6 hexa = 6 hepta = 7hepta = 7 octa = 8 octa = 8 deca = 10 deca = 10
Other Methods of Naming molecules
The rule may be modified to improve how a name sounds.
ExampleExample - use monoxide not monooxide.
N2O5
CO2
COCCl4
dinitrogen pentoxidecarbon dioxidecarbon monoxidecarbon tetrachloride
•name elements in the formula.•use prefixes to indicate how many atoms there are of each type.
AcidsAcids 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 acidH2S is hydrosulfuric acid
-ate becomes -ic acidH3PO4 is phosphoric acid
-ite becomes -ous acidHNO2 is nitrous acid
Other Naming -Acids
Naming Inorganic AcidsNaming Inorganic Acids
Naming Inorganic AcidsNaming Inorganic AcidsSalt Name FormulaSalt Name Formula
Acid Name FormulaAcid Name FormulaSodium acetate NaC2H3O2
Acetic acid HC2H3O2 Sodium chloride NaCl
Hydrochloric acid HCl Sodium hyponitrite NaNO
Hyponitrous acid HNOSodium phosphite Na3PO3
Phosphorous acid H3PO3
Sodium sulfate Na2SO4 Sulfuric acid H2SO4
Names and Formulas of AcidsNames 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
Naming Inorganic CompoundsNaming Inorganic Compounds
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:
Other Naming -Double & Triple Salts
COCO332-2- is named carbonatecarbonate, but HCOHCO33
-- is hydrogen carbonatehydrogen carbonate (or bicarbonate)
HH22POPO44-- dihydrogen phosphatedihydrogen phosphate anion.
Note that these are not named as acids, since another cation is still needed to balance the charge.
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.
Other Naming -Double & Triple Salts
NaHCO3 sodium hydrogen carbonate (or sodium bicarbonate)
AlK(SO4)2 aluminum potassium sulfate
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.
Other Naming - Hydrates
copper (II) sulfate pentahydrate CuSO45H2O
pentahydrate 5H2O
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.
Other Naming - Historical Names
For example, H2O is called water, not dihydrogen monoxide
Check out http://www.dhmo.org
A quick review of nomenclatureIs a metal presentas the first element?
Can the metal havemore than oneoxidation state?
Use Roman numerals or may use prefixes (mono, di, tri ...)
Roman numeralsare not needed.
Use Roman numeralsor may use Latin name with -ous/-ic suffixes
NoNo
NoNo
YesYes
YesYes
Is hydrogenfirst element?
Name asan acid
YesYes
Is a nonmetal thefirst element? NoNo
YesYes
-ides become hydro- -ic acids-ates become -ic acids -ites become -ous acids
A quick review of nomenclatureIs it a binarycompound?
Use the -idesuffix for thenegative ion
Name the commonpolyatomic ion
YesYes
Look up thename or formula
Use per- -ate 1 more O -ite 1 less O
hypo- -ite 2 less O
YesYes
Does it containone of the 8common ions?
YesYes
NoNo
NoNoDoes it have moreor less O atoms thanone of the -ate ions?
NoNo
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
• IsotopesIsotopes 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
11
21
31
12 6
13 6
14 6
Isotopes• Most elements occur in nature as a mixture of
isotopes.
– ElementElement Number of stable isotopesNumber 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.
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 masses92.23100
(27.9769265 u) = 25.80 u
4.67100
(28.9764947 u) = 1.35 u
3.10100
(29.9737702 u) = 0.929 u
28Si
29Si
30Si
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 - thethe molemole.
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 uu.
• Using moles gives us a practical unit.
• We can then relate atoms, ions and molecules, using an easy to measure unit - thethe gramgram.
Masses of atomsand molecules
• Atomic massAtomic 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 massMolecular or formula mass
• The total mass for all atoms in a compound.
Masses of atomsand moleculesHH22OO - water
2 hydrogen 2 x 1.008 u1 oxygen 1 x 16.00 u
mass of molecule 18.02 u18.02 g / mol
Rounded off basedon significant figures
Rounded off basedon significant figures
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 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 methodFactor 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 massFormula mass - Add the masses of all the
atoms in formula; for molecular and ionic compounds.
• Molecular massMolecular mass - Calculated the same as formula mass; only valid for molecules.
• Both have units of either u or grams / mole.
• Molar massMolar 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 = 12.011 u• 2 atoms O x 15.9994 u = 31.9988 u• Formula mass Formula mass == 44.010 u44.010 u
» or g / molor g / mol
Another example
CHCH33CHCH22OHOH - ethyl alcohol
2 carbon 2 x 12.01 u6 hydrogen 6 x 1.008 u1 oxygen 1 x 16.00 u
mass of molecule 46.07 u46.07 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’TDON’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 = 0.151 mol1 mol
132.14 g
atoms = 0.151 mol x x 15 atoms1 unit
6.02 x1023 units1 mol
atoms = 1.36 x1024
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?
x moles = 20.0 g x = 0.151 mol (NH4)2SO4
1 mol 132.14 g
x moles NH4 = 0.151 mol (NH4)2SO4 x 2 moles NH4
1 mol (NH4)2SO4
moles NH4 = 0.302
Formula weight = 132.14 g / 1 mol (NH4)2SO4
2 moles NH4 / 1 mol (NH4)2SO4
Example - (NH4)2SO4
• How many grams of sulfate ions are in 20.0 grams of ammonium sulfate?
x moles = 20.0 g x = 0.151 mol (NH4)2SO4
1 mol 132.14 g
x grams SO4 = 0.151 mol (NH4)2SO4 x 96.06 g SO4
1 mol (NH4)2SO4
grams SO4 = 14.5
Formula weight = 132.14 g / 1 mol (NH4)2SO4
96.06 grams SO4 / 1 mol (NH4)2SO4
Masses of atoms and molecules
Law of Definite CompositionLaw 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
– % 0 = 0.96 g O x 100 = 38.9 % O• 2.47 g KClO2.47 g KClO3
– Based upon the formula mass:– % O = 48.00 u O x 100 = 39.17 % O
• 122.55 u KClO122.55 u KClO3
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?
Gay-Lussac’s Law• Law of of Combining Volumes.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.
Gay-Lussac’s Law• Example.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)
+H2 H2 O2 H2O H2O
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 22.422.4 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 x x 1 mol H2O18.0 g H2O
liters O2 = 6.22 Note that 12.4 L H2 will also be formed.
0.5 mol O2
1 mol H2O22.4 L O2
1 mol O2
X g H2 =10.0g H2O x 2.02 g H2
18.02 g H2O= 1.12 g H2
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 analyzerElemental analyzer
• An instrument in which an organic compound is quantitatively converted to carbon dioxide and water -- both of which are then measured.
Elemental analyzer
furnace
CO2
trapH2Otrap
O2
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• ExampleExample: 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 carbonMass of carbon
• Mass of hydrogenMass of hydrogen
• Mass of nitrogenMass of nitrogen
0.1638 g CO2
12.01 g C 44.01 g CO2
= 0.04470 g C
0.1675 g H2O 2.016 g H
18.01 g H2O = 0.01875 g H
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.
% C = x 100% = 38.67 %
% H = x 100% = 16.22 %
% N = x 100% = 45.11 %
0.04470 g 0.1156 g
0.01875 g 0.1156 g
0.05215 g 0.1156 g
Empirical formula• Empirical formulaEmpirical formula
• The simplest formula that shows the ratios of the number of atoms of each element in a compound.
• ExampleExample - 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
0.01875 g H 1 mol H 1.008 g H
= 0.0186 mol H
0.04470 g C 1 mol C12.01 g C
= 0.003722 mol C
0.05215 g N 1 mol 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 = 4.998
• N 0.003722 / 0.003722 = 1.000
• The empirical formula is CHThe empirical formula is CH55NN
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 %
% C = 38.67 %% H = 16.22 %% N = 45.11 %
Empirical formula
16.22 g H 1 mol H 1.008 g H
= 16.09 mol H
38.67 g C 1 mol C 12.01 g C
= 3.220 mol C
45.11 g N 1 mol N 14.01 g N
= 3.220 mol 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 empirical formula is determined to be the same, CHthe same, CH55N, whether using actual masses of N, whether using actual masses of
the elements present in the sample or by using the elements present in the sample or by using their % composition by mass.their % composition by mass.
Molecular formula• Molecular formulaMolecular 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 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!
Hydrated CompoundsThe 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 Compounds4.262g 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 compound’s formula is BaCl22H2O
MolarityM =
moles solute molliters of solution L
=
MolarityMolarity• Recognizes that compounds have different
formula weights.
• A 1 M solution of sucrose contains thesame 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 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)
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
• 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.
moles solute molkilograms of solvent kgMolality (m) = =
Density
• 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.
grams of solution gmilliliters of solution mLDensity (D) = =
Percent Composition
• % 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.
value of the partValue of the whole
Percent Composition = x 100
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.
moles of solute or solventtotal moles of solute & solvent
Mole fraction =
Parts per Million or Billion
Parts per Million or BillionParts per Million or Billion
•Used to express concentrations for verydilute 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.
# grams of solute1,000,000 g solutionParts per million (ppm) =
Stoich Baby
Given the following equation___N2 + ___H2 ___NH3
Assuming gases at STP, given one mole of nitrogen, how many liters of ammonia would form?
Given one mole of nitrogen gas, how many moles of ammonia would form?
1 3 2
Given 2 liters of nitrogen and 5 liters of hydrogen, how may liters of ammonia are formed? What is left over?
Stoichiometry• StoichiometryStoichiometry
• The study of quantitative relationships between substances undergoing chemical changes.
• Law of Conservation of MatterLaw 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»
• CaCO3 (s) CaO (s) + CO2 (g)
ReactantReactant Products Products
A balanced equation shows the relationshipbetween 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) CaO (s) + CO2 (g)
– Reactants Products» 1 Ca 1 Ca» 1 C 1 C» 3 O 3 O
Balancing chemical equations
Step 1Step 1 Count the number of atoms of each element on each side of the equation.
Step 2Step 2 Determine which atom numbers are not balanced.
Step 3Step 3 Balance one atom at a time by using coefficients in front of one or more substances.
Step 4Step 4 Repeat steps 1-3 until everything is balanced.
Example.Decomposition of urea
– (NH2)2CO + H2O ______> NH3 + CO2
• 2 N 1 N < < not balancednot balanced
• 6 H 3 H < not balanced< not balanced• 1 C 1 C• 2 O 2 O
• We need to double NH3 on the right.
– (NH2)2CO + H2O ______> 22NH3 + CO2
2 H2 + O2 -----> 2 H2O 2 H2 + O2 -----> 2 H2O
You need a balancedequation and you WILL
work with moles.
You need a balancedequation and you WILL
work with moles.
Mass relationshipsin chemical reactions
• StoichiometryStoichiometry - The calculation of quantities of reactants and products in a chemical reaction.
Stoichiometry,General steps.
11 Balance the chemical equation.
33 Convert masses to moles.
22 Calculate formula masses.
44Use chemical equation toget the needed answer.
Convert back to mass if needed.55
Mole calculations
• The balanced equation shows the reacting ratio between reactants and products.
• 2C2C22HH66 + 7O + 7O22 4CO 4CO22 + 6H + 6H22OO
– 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
• moles N2
– = 14 g x 1 mol /28.02 g – = 0.50 moles
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.
• 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 mol40.08 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.
• Limiting reactantLimiting 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 ReactionBalanced Chemical Reaction
• 2H2 + O2 ________> 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 mol1 mol32.0 g
1 mol2.0 g
Example• Balanced Chemical ReactionBalanced 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 yieldTheoretical yield
• The amount of product that should be formed according to the chemical reaction and stoichiometry.
• Actual yieldActual yield
• The amount of product actually formed.
• Percent yieldPercent yield
• Ratio of actual to theoretical yield, as a %.
• Quantitative reactionQuantitative reaction
• When the percent yield equals 100%.
Yield• Less product is often produced than expected.
• Possible reasonsPossible 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 = x 100
• In order to determine % yield, you must be able to recover and measure all of the product in a pure form.
Actual yieldTheoretical yield
StoichiometryStep 1 • Identify species present in solution and determine the reaction that
occurs
Step 2• Write the balanced net ionic equation
Step 3Calculate the moles of reaction
– solution = molarity x volume– heterogeneous = grams molar mass
Step 4Consider the limiting reactant
Step 5 Answer the question using stoich!
