1 molecules and compounds read chapter 3. study all examples and complete all exercises. complete...
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Molecules and CompoundsMolecules and CompoundsRead Chapter 3.
Study all examples and complete all exercises.
Complete all bold numbered problems.
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Chapter 3 OutlineChapter 3 Outline
• Molecular FormulaMolecular Formula
• Molar MassMolar Mass
• Empirical and Molecular Empirical and Molecular FormulaFormula
• NomenclatureNomenclature
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Compounds & MoleculesCompounds & Molecules
• COMPOUNDS COMPOUNDS are a combination of 2 or
more elements in definite ratios by mass.
• The character of each element is lost when
forming a compound.
• MOLECULES MOLECULES are the smallest unit of a smallest unit of a
compound that retains the characteristics of compound that retains the characteristics of
the compound. the compound.
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MOLECULAR FORMULASMOLECULAR FORMULAS• Formula for glycine isFormula for glycine is CC22HH55NNOO22• In one molecule there areIn one molecule there are
– 2 C 2 C atomsatoms
– 5 H5 H atomsatoms
– 1 N1 N atomatom
– 2 O2 O atomsatoms
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WRITING FORMULASWRITING FORMULAS
• FormulaFormula
HOCHHOCH22CHCH22OHOH
to show atom orderingto show atom ordering
• or in the form of a or in the form of a structural formulastructural formula
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Molecular ModelingMolecular Modeling
Ball & stickBall & stick Space-fillingSpace-filling
Drawing of glycineDrawing of glycineCC
HH
HH CC
HH
HH
OO
OO HHNN
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Resources for Resources for Molecular ModelingMolecular Modeling
• Oxford Molecular/CAChe Scientific Oxford Molecular/CAChe Scientific
software on Saunders General software on Saunders General
Chemistry CD-ROMChemistry CD-ROM
• Rasmol and Chime on the InternetRasmol and Chime on the Internet
• See http://www.saundercollege.comSee http://www.saundercollege.com
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ELEMENTS THAT EXIST ELEMENTS THAT EXIST AS MOLECULESAS MOLECULES
Allotrope of Allotrope of CC
BuckyballBuckyball, CC6060
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IONS AND IONIC COMPOUNDSIONS AND IONIC COMPOUNDS
• IONS IONS are atoms or groups of atoms are atoms or groups of atoms
with a positive or negative charge. with a positive or negative charge.
• Taking away an electron from an atom Taking away an electron from an atom
gives a gives a CATION with a positive CATION with a positive
charge.charge.
• Adding an electron to an atom gives an Adding an electron to an atom gives an
ANION with a negative charge.ANION with a negative charge.
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Formation of Formation of Cations & AnionsCations & Anions
A cation forms when an atom loses one or more electrons.
An anion forms when an atom gains one or more electrons
Mg --> MgMg --> Mg2+2+ + 2 e- + 2 e- F + eF + e-- --> F --> F--
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PREDICTING ION CHARGESPREDICTING ION CHARGES
In generalIn general
• metals (Mg) metals (Mg) lose electrons lose electrons ---> ---> cationscations
• nonmetals (F) nonmetals (F) gain electronsgain electrons ---> ---> anionsanions
LiLi1+1+ F F1-1- LiF LiF
MONATOMIC IONSMONATOMIC IONS
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METALSMETALSM ---> n e- + MM ---> n e- + Mn+n+
where n = periodic groupwhere n = periodic group
NaNa++
MgMg2+2+
AlAl3+3+
Transition metals --> MTransition metals --> M2+2+ or M or M3+3+
are most commonare most common
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NONMETALSNONMETALS
NONMETAL + n e- ------> XNONMETAL + n e- ------> Xn-n-
where n = 8 where n = 8 - - Group numberGroup number
CC4-4- carbidecarbide
NN3-3- nitridenitride
OO2-2- oxideoxide
FF-- fluoridefluoride
BromineBromine
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Some Common Some Common Polyatomic IonsPolyatomic Ions
HHNNOO33
nitric acidnitric acid
NNOO33--
nitrate ionnitrate ion
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Some Common Some Common Polyatomic IonsPolyatomic Ions
NNHH44++
ammonium ionammonium ion
One of the few common One of the few common polyatomic cationspolyatomic cations
22Some Common Some Common Polyatomic IonsPolyatomic Ions
COCO332-2-
carbonate ion
HCOHCO33--
bicarbonate ion
hydrogen carbonate
23Some Common Some Common Polyatomic IonsPolyatomic Ions
POPO443-3-
phosphate ion
CHCH33COCO22--
acetate ion
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SOSO442-2-
Sulfate ion
SOSO332-2-
Sulfite ion
Some Common Some Common Polyatomic IonsPolyatomic Ions
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COMPOUNDS FORMED FROM COMPOUNDS FORMED FROM IONSIONS
CATION + ANION CATION + ANION COMPOUNDCOMPOUND
NaNa++ + Cl + Cl-- NaClNaCl
A neutral compound requires equal number of + and - charges.
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Some Ionic CompoundsSome Ionic CompoundsCaCa2+2+ + 2 F + 2 F-- CaF CaF22
Calcium fluoride
MgMg2+2+ + 2 NO + 2 NO33-- Mg(NO Mg(NO33))22
Magnesium nitrate
3 Fe3 Fe2+2+ + 2 PO + 2 PO443-3- Fe Fe33(PO(PO44))22
Iron(II) phosphate
Calcium FluorideCalcium Fluoride
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Sample QuestionsSample Questions• Predict the charges for the ions formed
from:SeSe GaGa
PP SrSr
• Give the formula for each ion in Al2(SO4)3
• Give the formula for the ionic compound
that forms between Na and SNa and S
Ga and OGa and O
Ba and NBa and N
Answers
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Properties of Ionic CompoundsProperties of Ionic CompoundsForming NaCl from Na and ClForming NaCl from Na and Cl22
• A metal atom can transfer an electron to a nonmetal.
• The resulting cation and anion are attracted to each other by
electrostatic electrostatic forcesforces..
3030
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Electrostatic ForcesElectrostatic Forces
The oppositely charged ions in ionic compounds are attracted to one another by
ELECTROSTATIC FORCESELECTROSTATIC FORCES..
These forces are governed by
COULOMB’S LAWCOULOMB’S LAW..
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Electrostatic ForcesElectrostatic ForcesCOULOMB’S LAWCOULOMB’S LAW
As ion charge increases, the attractive force _______________.
As the distance between ions increases, the attractive force ________________.
This idea is important and will come up many times in future
discussions!
Force of attractionForce of attraction Force of attractionForce of attraction ==== (charge on(charge on++)(charge on)(charge on--))(charge on(charge on++)(charge on)(charge on--))
(distance between ions)(distance between ions)22(distance between ions)(distance between ions)22
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Importance of Coulomb’s LawImportance of Coulomb’s Law
NaNaClCl,, NaNa++ and and ClCl--,,m.p. 801 m.p. 801 ooCC
MgMgOO,, MgMg2+2+ andand OO2-2-
m.p. 2800 m.p. 2800 ooCC
AlN, AlAlN, Al3+3+ and N and N3-3- m.p. 2900 m.p. 2900 ooCC
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Names of CompoundsNames of Compounds
• Rules for nomenclature are Rules for nomenclature are found in section 3.5 found in section 3.5
• STUDY them carefully!!STUDY them carefully!!
• We will be studying We will be studying nomenclature in the nomenclature in the laboratory in Experiment MA.laboratory in Experiment MA.
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Counting AtomsCounting Atoms• Mg burns in air (Mg burns in air (OO22) to) to
produce white produce white magnesium oxide,magnesium oxide, MgMgOO.. • How can we figure out How can we figure out
how much oxide is how much oxide is produced from a produced from a given mass ofgiven mass of MgMg??
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Counting AtomsCounting Atoms
• Chemistry is a quantitative science — we need a “counting unit.”
• The MOLEThe MOLE
• 1 mole is the amount of substance that contains as many particles (atoms, molecules) as there are in 12.0 g of 12C.
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Particles in a MoleParticles in a Mole
6.02 x 106.02 x 102323
Avogadro’s Number
Amedeo Avogadro1776-1856
There is Avogadro’s number of There is Avogadro’s number of particles in a mole of any substance.particles in a mole of any substance.
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MoleMole in Chemistry is NOT:in Chemistry is NOT:
•An informer / spyAn informer / spy
•Dark spot on Cindy Crawford’s upper lipDark spot on Cindy Crawford’s upper lip
•Rodent that burrows in the groundRodent that burrows in the ground
•A tunneling machineA tunneling machine
•Wave breakWave break
•Spicy Mexican sauceSpicy Mexican sauce
•An informer / spyAn informer / spy
•Dark spot on Cindy Crawford’s upper lipDark spot on Cindy Crawford’s upper lip
•Rodent that burrows in the groundRodent that burrows in the ground
•A tunneling machineA tunneling machine
•Wave breakWave break
•Spicy Mexican sauceSpicy Mexican sauce
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A mole is a convenient A mole is a convenient measuring tool.measuring tool.
Pair Dozen Ream
Baseballs 2 baseballs 12 baseballs 13 baseballs500 baseballs
Pineapples2 Pineapples
12 Pineapples
13 Pineapples500 Pineapples
Calculators2 Calculators
12 Calculators
13 Calculators500 Calculators
Planets 2 Planets 12 Planets 13 Planets 500 Planets
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Mole = mol = Mole = mol = 6.022 6.022 10 102323 “particles” “particles”
Mole
Baseballs 6.022 1023 baseballs
Pineapples 6.022 1023 Pineapples
Calculators 6.022 1023 Calculators
Planets 6.022 1023 Planets
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Just as
1 doz eggs = 12 eggs
or
1 mol of eggs = 6.022 1023 eggs
A mole is a “Number”
1 mol of H = 6.022 1023 atoms of H
1 mol of O = 6.022 1023 atoms of O
1 mol of Al = 6.022 1023 atoms of Al
1 mol of Cr = 6.022 1023 atoms of Cr
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In chemistry, the In chemistry, the molmol is is more then just a numbermore then just a number
1 mol 1 mol amu = 1.00 g amu = 1.00 g
12 2411 1.661 10
12amu mass C g
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1 mol 1 mol amu = 1.00 g amu = 1.00 gProof:
1 mol = 6.022 1023
amu = 1.661 10-24 g
1 mol amu = (6.022 1023) (1.661 10-24g) = 1.00 g
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Molar MassMolar Mass1 mol of 12C
= 12.00 g of C= 6.02 x 1023 atoms of C
12.00 g of 12C is its
MOLAR MASSTaking into account all
of the isotopes of C, the molar mass of C is 12.011 g/mol
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Molar MassMolar Mass1 mol of 12C = 12.00 g of C
= 6.02 x 1023 atoms of C
12.00 g of 12C is its MOLAR MASS
Taking into account all of the isotopes of
C, the molar mass of C is 12.011 g/mol
Find molar mass from periodic table
1313
AlAl
26.981526.9815
atomic numberatomic number
symbolsymbol
atomic weightatomic weight
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PROBLEM: How many moles PROBLEM: How many moles are represented by 0.200 g of are represented by 0.200 g of Mg?Mg?
How many atoms in this piece of Mg?
0.200 g
=
1 mole
24.3 g0.00823 mole
0.00823 mole
=
6.02 x 1023 atom
1 mole
4.95 x 1021 atom
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MOLECULAR WEIGHT MOLECULAR WEIGHT AND MOLAR MASSAND MOLAR MASS
Molecular weight Molecular weight is the sum
of the atomic weights of all atoms
in the molecule.
Molar massMolar mass = molecular weight
in grams
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What is the molar mass of What is the molar mass of ethanol, Cethanol, C22HH66O?O?
1 mol contains
2 mol C (12.0 g C/1 mol) = 24.0 g C
6 mol H (1.0 g H/1 mol) = 6.0 g H
1 mol O (16.0 g O/1 mol) = 16.0 g O
TOTAL = molar mass = 46.0 g/mol
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How many moles of alcohol are How many moles of alcohol are there in a “standard” can of beer if there in a “standard” can of beer if there are 21.3 g of Cthere are 21.3 g of C22HH66O?O?
21.3 g=
1 mole
46.0 g0.463 mole0.463 mole
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How many How many molecules molecules of alcohol of alcohol are there in a “standard” can of are there in a “standard” can of beer if there are 21.3 g of Cbeer if there are 21.3 g of C22HH66O?O?
0.463 mole
=
6.02 x 1023 molecule
1 mole
2.79 x 102.79 x 102323 molecule molecule
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How many How many atoms of C atoms of C are there are there in a “standard” can of beer if in a “standard” can of beer if there are 21.3 g of Cthere are 21.3 g of C22HH66O?O?
2.79 x 1023 molecule
=
2 atom C
1 molecule
5.58 x 105.58 x 102323 atom C atom C
Sample problems
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Empirical and Molecular Empirical and Molecular FormulasFormulas
A pure compound always consists of the same elements combined in the same proportions by weight.
Therefore, we can express molecular composition as PERCENT BY WEIGHT
Ethanol, C2H6O
52.2% C, 13% H,
34.8% O
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Percent CompositionPercent CompositionConsider some of the family of
nitrogen-oxygen compounds:
NO2, nitrogen dioxide and closely related, NO, nitrogen monoxide (or nitric oxide)
Structure of NO2 Chemistry of NO, nitrogen monoxide (nitric oxide)
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Percent CompositionPercent CompositionConsider NO2, Molar mass = ?
What is the weight percent of N and of O?
Wt. % O = 2 (16 .0 g O per mole )
46 .0 g x 100% = 69 .6%
Wt. % N = 14.0 g N
46.0 g NO 2 • 100% = 30.4 %
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Percent CompositionPercent CompositionWhat are the weight percentages of N What are the weight percentages of N
and O in NO?and O in NO?
Wt. % O = 16.0 g O 30.0 g NO
100% = 53.3%•
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Percent CompositionPercent CompositionSamples ProblemsSamples Problems
1. Calculate the percent composition of H1. Calculate the percent composition of H22O.O.
Wt. % O = 16.0 g O 18.0 g H2O
• 100% = 88.9%
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Percent CompositionPercent CompositionSamples ProblemsSamples Problems
2. Calculate the percent O in NaOH.2. Calculate the percent O in NaOH.
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Percent CompositionPercent CompositionSamples ProblemsSamples Problems
3. Calculate the percent O and the 3. Calculate the percent O and the percent water in CuSOpercent water in CuSO44
..5H5H22O. O.
Wt. % O = 144.0 g O
249.6 g CuSO4 • 5H2O • 100% = 57.69 %
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Determining Determining FormulasFormulas
In chemical analysis we determine the In chemical analysis we determine the
% by weight of each element in a given % by weight of each element in a given
amount of pure compound and derive amount of pure compound and derive
thethe EMPIRICALEMPIRICAL or or SIMPLESTSIMPLEST formula.formula.
61A compound of B and H is A compound of B and H is 81.10% B. What is its empirical 81.10% B. What is its empirical formula?formula?
81.10 g 1 mole
10.8 g
18.90 g 1 mole
1.0 g
BB H H
1.00 mole B 2.5 mole H
7.51mole B 19 mole H
7.51 mole B 7.51 mole B
2.00 mole B 5.0 mole HB2H5
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A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. Its empirical formula is BIts empirical formula is B22HH55. What is . What is
its its molecular formulamolecular formula??
Is the molecular formula B2H5, B4H10, B6H15, B8H20, etc.?
B2H6 is one example of this class of compounds.
BB22HH66
63A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. Its empirical formula is BIts empirical formula is B22HH55. What is . What is its molecular formula?its molecular formula?
We need to do an EXPERIMENTEXPERIMENT to find the MOLAR MASS.
Here experiment gives 53.3 g/mol.
Compare with the mass of B2H5 , 26.66 g/unit
Find the ratio of these masses.
mol 1
HB of units 2 =
HB ofg/unit 26.66
g/mol 3.53 52
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Molecular formula =Molecular formula = B B44HH1010
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Determine the formula of a Determine the formula of a compound of Sn and I using compound of Sn and I using the following datathe following data..
• Reaction of Sn and I2 is done using excess Sn.
• Mass of Sn in the beginning = 1.056 g
• Mass of iodine (I2) used = 1.947 g
• Mass of Sn remaining = 0.601 g
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Find the mass of Sn that combined with 1.947 g I2.
Mass of Sn initially = 1.056 g
Mass of Sn recovered = 0.601 g
Mass of Sn used = 0.455 g
Tin and Iodine Tin and Iodine CompoundCompound
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0.455 g 1 mole
118.7 g
1.947 g 1 mole
126.9 g
SnSn I I
1.00 mole Sn 4.01 mole I
0.0383 mole Sn 0.1534 mole I
0.0383 mole Sn 0.0383 mole Sn
SnISnI44
Tin and Iodine Tin and Iodine CompoundCompound
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More ProblemsMore Problems2. Calculate the formula for the iron sulfide
that forms when 53.73g Fe react with 46.27 g of sulfur.
53.73 g 1 mole
55.8 g
46.27 g 1 mole
32.1 g
FeFe S S
1.00 1.50
0.963 1.44
0.963 0.963
FeFe22SS33
2.00 3.00
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More ProblemsMore ProblemsCalculate the empirical formula a compound
containing 90.7% Pb and 9.33% O.
90.7 g 1 mole
207.2 g 9.33 g 1 mole
16.0 g
PbPb O O
1.00 1.33
0.438 0.583
0.438 0.438
PbPb33OO44
3.00 3.99
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More ProblemsMore ProblemsCalculate the empirical formula for a
compound containing 36.5% Na, 25.4% S and 38.1% O.
36.5 g 1 mole
23.0 g
25.4 g 1 mole
32.1 g
NaNa S O S O
2.01 1.00 3.01
NaNa22SOSO33
38.1 g 1 mole
16.0 g
1.59 0.791 2.38
0.791 0.791 0.791
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More ProblemsCalculate the empirical and molecular
formulas for nicotine, 74.0% C, 8.7% H and 17.3% N, with a molar mass of 160 g/mole.
74.0 g 1 mole
12.0 g
8.7 g 1 mole
1.0 g
C H N
4.98 7.0 1.00
17.3 g 1 mole
14.0 g
6.17 8.7 1.24
1.24 1.24 1.24
Empirical formula C5H7N
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More ProblemsMore Problems7. Calculate the empirical and molecular
formulas for nicotine, 74.0% C, 8.7% H and 17.3% N, with a molar mass of 160 g/mole.
Empirical formula C5H7N
Molecular formula C10H14N2
160= 2
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Practice ProblemsPractice Problems
Names/FormulasNames/Formulas
FeO
Pb(C2H3O2)2
magnesium bromide
sodium chromate
calcium phosphate
ammonium carbonate
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Practice ProblemsPractice Problems
As2S3
dinitrogen pentoxide
silicon tetrabromide
diphosphorus pentoxide
HBrO3
H3PO4(aq)
H2CO3
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Practice ProblemsPractice ProblemsCalculations
1. 26 g H2 is how many moles H2?
2. 4.25 x 1021 molecules NH3 is how many grams NH3?
3. 1.5 x 102 formula units KClO3 is how many moles KClO3
4. 0.0042 mole Fe2O3 is how many formula units Fe2O3?
5. 2.15 moles MgSO4.7H2O is how many g
MgSO4.7H2O?
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Practice ProblemsPractice Problems6. 25 molecules HBr is how many mole
HBr?7. .00002 g Sn is how many atoms Sn?
8. 7.25 mole H2S is how many g H2S?
9. 5.2 g Sr(OH)2is how many formula units Sr(OH)2?
10. How many moles of CCl4 will contain 2.4 g of chlorine?
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Practice ProblemsPractice Problems
11. 19 g of HNO3 contains how many
a) molecules of HNO3? b) grams of O?
12. Calculate the percent composition of NaCl.
13. Calculate the empirical and molecular formulas for a compound containing 43.7g P and 56.3g O, with a molar mass of 140 g/mole.
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Practice Problems AnswersPractice Problems AnswersIron(II) oxide, ferrous oxide
lead(II) acetate, plumbous acetate
MgBr2 Na2CrO4 Ca3(PO4)2
(NH4)2CO3 ammonium sulfide
diarsenic trioxide sulfur dioxide
SiS2 As2S5 N2O
diarsenic trisulfide N2O5 SiBr4
P2O5 bromic acid phosphoric acid
carbonic acid
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Practice Problems AnswersPractice Problems Answers1. 13 mole 2. 0.120 g
3. 2.5 x 10-22 mole 4. 2.5 x 1021 atom
5. 530. g 6. 4.2 x 10-23 mole
7. 1 x 1017 atom 8. 247 g
9. 2.6 x 1022 formula units
10. 0.017 mole
11. a) 1.8 x 1023 molecule
b) 14 g
12. 39.3%, 60.7% 13. C5H7N
14. P2O5
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Sample QuestionsSample Questions• Predict the charges for the ions formed
from:Se -2 Ga +3 Al3+
P -3 Sr +2 SO42-
• Give the formula for each ion in Al2(SO4)3
• Give the formula for the ionic compound
that forms between Na and S Na2S
Ga and O Ga2O3
Ba and N Ba3N2
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Sample ProblemsSample Problems
1. 2.5 mole S = ? atom S 2.5 mole
=
6.02 x 1023 atom
1 mole
1.5 x 1024 atom S
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Sample ProblemsSample Problems
3. 1.42 g Mg = ? atom Mg3. 1.42 g Mg = ? atom Mg
1.42 g
=
1 mole
24.3 g
3.55 x 103.55 x 102222 atom atom
6.02 x 1023 atom
1 mole
84
Sample ProblemsSample Problems
4. 125.2 g O4. 125.2 g O22 = ? mole O = ? mole O22
125.2 g=
1 mole
32.0 g3.91 mole O3.91 mole O22
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Sample ProblemsSample Problems
1. 1.5 mole H1. 1.5 mole H22O = ? g HO = ? g H22OO
1.5 mole=
18.0 g
1 mole 27 g H27 g H2200
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Sample ProblemsSample Problems
2. 1.502. 1.50 mole CCl mole CCl44 = ? molecules CCl = ? molecules CCl44
1.5 mole
=
6.02 x 1023 molecule
1 mole
9.03 x 109.03 x 102323 molecule molecule
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Sample ProblemsSample Problems
3. 1.25 mole CaCl3. 1.25 mole CaCl22 = ? formula units CaCl = ? formula units CaCl22
1.25 mole
=
6.02 x 1023 formula unit
1 mole
7.52 x 107.52 x 102323 formula unit formula unit
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Sample ProblemsSample Problems
4. 2.5 g K = ? g KOH4. 2.5 g K = ? g KOH
2.5 g K
=
1 mole K
39.1 g K
3.6 g KOH3.6 g KOH
1 mole KOH
1 mole K
56.1 g KOH
1 mole KOH