Compounds & Compounds & Molecules Molecules NaCl, salt NaCl, salt Buckyball, C Buckyball, C 60 60 Ethanol, C Ethanol, C 2 H H 6 O O
Dec 21, 2015
Compounds & Compounds & MoleculesMoleculesCompounds & Compounds & MoleculesMolecules
NaCl, saltNaCl, salt
Buckyball, CBuckyball, C6060
Ethanol, CEthanol, C22HH66OO
Compounds & Compounds & MoleculesMoleculesCompounds & Compounds & MoleculesMolecules
•COMPOUNDSCOMPOUNDS are a combination of 2 are a combination of 2
or more elements in definite ratios by mass. or more elements in definite ratios by mass.
• The character of each element is lost when The character of each element is lost when
forming a compound.forming a compound.
•MOLECULESMOLECULES are the smallest unit of a are the smallest unit of a
compound that retains the characteristics of compound that retains the characteristics of
the compound.the compound.
MOLECULAR FORMULASMOLECULAR FORMULASMOLECULAR FORMULASMOLECULAR FORMULAS
• Formula for glycine is Formula for glycine is CC22HH55NONO22• In one molecule there areIn one molecule there are
– 2 C atoms2 C atoms
– 5 H atoms5 H atoms
– 1 N atom1 N atom
– 2 O atoms2 O atoms
WRITING FORMULASWRITING FORMULASWRITING FORMULASWRITING FORMULAS
• Can also write glycine formula as
–H2NCH2COOHto show atom ordering
• or in the form of a structuralstructural formula formula
C
H
H C
H
H
O
O HN
MOLECULAR MOLECULAR MODELINGMODELING
MOLECULAR MOLECULAR MODELINGMODELING
C
H
H C
H
H
O
O HN
Ball & stickBall & stick Space-fillingSpace-filling
Drawing of glycineDrawing of glycine
Resources for Resources for Molecular ModelingMolecular Modeling
Resources for Resources for Molecular ModelingMolecular Modeling
• Modeling software on the Modeling software on the General General Chemistry Interactive CD-ROMChemistry Interactive CD-ROM
–CACheCAChe — see the folder — see the folder labeled CAChe on the CD-ROMlabeled CAChe on the CD-ROM
–RasmolRasmol
MOLECULAR WEIGHT MOLECULAR WEIGHT AND MOLAR MASSAND MOLAR MASS
Molecular weightMolecular weight = sum of the = sum of the
atomic weights of all atoms in the atomic weights of all atoms in the
molecule.molecule.
Molar massMolar mass = molecular weight in = molecular weight in
gramsgrams
What is the molar What is the molar mass of ethanol, mass of ethanol, CC22HH66O?O?
1 mol contains1 mol contains
2 mol C (12.01 g C/1 mol) = 24.02 g C2 mol C (12.01 g C/1 mol) = 24.02 g C
6 mol H (1.01 g H/1 mol) = 6.06 g H6 mol H (1.01 g H/1 mol) = 6.06 g H
1 mol O (16.00 g O/1 mol) = 16.00 g O1 mol O (16.00 g O/1 mol) = 16.00 g O
TOTAL = TOTAL = molar mass = 46.08 g/molmolar mass = 46.08 g/mol
How many How many molesmoles of alcohol of alcohol are there in a “standard” can are there in a “standard” can of beer if there are 21.3 g of of beer if there are 21.3 g of CC22HH66O?O?
(a) Molar mass of C2H6O = 46.08 g/mol
(b) Calc. moles of alcohol
21.3 g • 1 mol
46.08 g = 0.462 mol
How many How many moleculesmolecules 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?
= 2.78 x 1023 molecules
We know there are 0.462 mol of C2H6O.
0.462 mol • 6.022 x 1023 molecules
1 mol
How many How many atoms of Catoms of C are there are there in a “standard” can of beer if there in a “standard” can of beer if there are 21.3 g of Care 21.3 g of C22HH66O?O?
= 5.57 x 1023 C atoms
There are 2.78 x 1023 molecules.
Each molecule contains 2 C atoms.
Therefore, the number of C atoms is
2.78 x 1023 molecules • 2 C atoms1 molecule
ELEMENTS THAT EXIST AS ELEMENTS THAT EXIST AS MOLECULESMOLECULES
ELEMENTS THAT EXIST AS ELEMENTS THAT EXIST AS MOLECULESMOLECULES
See SCREEN 3.2 See SCREEN 3.2 on the CD-ROMon the CD-ROM
Allotropes of CAllotropes of CAllotropes of CAllotropes of C
ELEMENTS THAT EXIST AS ELEMENTS THAT EXIST AS DIATOMICDIATOMIC MOLECULES MOLECULES
ELEMENTS THAT EXIST AS ELEMENTS THAT EXIST AS DIATOMICDIATOMIC MOLECULES MOLECULES
ELEMENTS THAT EXIST AS ELEMENTS THAT EXIST AS POLYATOMICPOLYATOMIC MOLECULES MOLECULESELEMENTS THAT EXIST AS ELEMENTS THAT EXIST AS POLYATOMICPOLYATOMIC MOLECULES MOLECULES
White P4 and polymeric red phosphorus
S8 sulfur molecules
IONS AND IONIC COMPOUNDSIONS AND IONIC COMPOUNDSsee Screen 3.5see Screen 3.5
• IONSIONS are atoms or groups of atoms with a positive are atoms or groups of atoms with a positive
or negative charge. or negative charge.
• Taking awayTaking away an electron from an atom gives a an electron from an atom gives a
CATIONCATION with a with a positive chargepositive charge
• AddingAdding an electron to an atom gives an an electron to an atom gives an ANIONANION
with a with a negative chargenegative charge..
Forming Cations & Forming Cations & AnionsAnions
Forming Cations & Forming Cations & AnionsAnions
A A CATIONCATION forms forms when an when an atom atom losesloses one or one or more electrons.more electrons.
An An ANIONANION forms forms when an when an atom atom gainsgains one or one or more electronsmore electrons
Mg --> Mg2+ + 2 e- F + e- --> F-
PREDICTING ION CHARGESPREDICTING ION CHARGESPREDICTING ION CHARGESPREDICTING ION CHARGES
In generalIn general
• metalsmetals (Mg) (Mg) lose electrons lose electrons ---> ---> cationscations
• nonmetalsnonmetals (F) (F) gain electronsgain electrons ---> ---> anionsanions
•See CD-ROM Screen 3.5 and book Figure 3.7See CD-ROM Screen 3.5 and book Figure 3.7
Charges on Common Charges on Common IonsIons
Charges on Common Charges on Common IonsIons
+3
-4 -1-2-3+1
+2
By losing or gaining e-, atom has same By losing or gaining e-, atom has same number of e-’s as nearest Group 8A atom.number of e-’s as nearest Group 8A atom.
METALSMETALSM ---> n e- + MM ---> n e- + Mn+n+
where n = periodic groupwhere n = periodic groupNaNa++ sodium ionsodium ionMgMg2+2+ magnesium ionmagnesium ionAlAl3+3+ aluminum ionaluminum ionTransition metals --> MTransition metals --> M2+2+ or M or M3+3+
are commonare commonFeFe2+2+ iron(II) ioniron(II) ionFeFe3+3+ iron(III) ioniron(III) ion
NONMETALSNONMETALSNONMETALSNONMETALS
NONMETAL + n e- ------> XNONMETAL + n e- ------> Xn-n-
where n = 8 - Group no.where n = 8 - Group no.
CC4-4-,carbide,carbide NN3-3-, nitride, nitride OO2-2-, oxide, oxide
SS2-2-, sulfide, sulfide
FF--, fluoride, fluoride
ClCl--, chloride, chloride
Group 7AGroup 6AGroup 4A Group 5A
BrBr--, bromide, bromide
II--, iodide, iodide
POLYATOMIC POLYATOMIC IONSIONS
CD Screen 3.6CD Screen 3.6
POLYATOMIC POLYATOMIC IONSIONS
CD Screen 3.6CD Screen 3.6
Groups of atoms with a charge.Groups of atoms with a charge.
MEMORIZEMEMORIZE the names and formulas in the names and formulas in
Table 3.1, page 89.Table 3.1, page 89.
Polyatomic IonsPolyatomic IonsPolyatomic IonsPolyatomic Ions
HNOHNO33
nitric acidnitric acid
NONO33--
nitrate ionnitrate ion
Polyatomic IonsPolyatomic IonsPolyatomic IonsPolyatomic Ions
NHNH44++
ammonium ionammonium ion
One of the few common One of the few common polyatomic cationspolyatomic cations
Polyatomic Polyatomic IonsIonsPolyatomic Polyatomic IonsIonsCOCO33
2-2-
carbonate ioncarbonate ion
HCOHCO33--
bicarbonate ionbicarbonate ion
hydrogen carbonatehydrogen carbonate
POPO443-3-
phosphate ionphosphate ion
CHCH33COCO22--
acetate ionacetate ion
Polyatomic Polyatomic IonsIons
Polyatomic Polyatomic IonsIons
SOSO442-2-
sulfate ionsulfate ion
SOSO332-2-
sulfite ionsulfite ion
Polyatomic Polyatomic IonsIons
Polyatomic Polyatomic IonsIons
NONO33--
nitrate ionnitrate ion
NONO22--
nitrite ionnitrite ion
Polyatomic Polyatomic IonsIons
Polyatomic Polyatomic IonsIons
CATIONCATION + + ANIONANION ---> --->
COMPOUNDCOMPOUND
CATIONCATION + + ANIONANION ---> --->
COMPOUNDCOMPOUND
A neutral compd. A neutral compd. requiresrequires
equal number of + equal number of + and - charges.and - charges.
A neutral compd. A neutral compd. requiresrequires
equal number of + equal number of + and - charges.and - charges.
COMPOUNDCOMPOUNDS FORMED S FORMED FROM IONSFROM IONS
COMPOUNDCOMPOUNDS FORMED S FORMED FROM IONSFROM IONS
NaNa++ + Cl + Cl- - --> NaCl--> NaCl
IONIC COMPOUNDSIONIC COMPOUNDSIONIC COMPOUNDSIONIC COMPOUNDS
NH4+
Cl-
ammonium chloride, NHammonium chloride, NH44ClCl
Some Ionic CompoundsSome Ionic CompoundsSome Ionic CompoundsSome Ionic Compounds
MgMg2+2+ + NO + NO33-- ----> ---->
Mg(NOMg(NO33))22
magnesiummagnesium nitratenitrate
FeFe2+2+ + PO + PO443-3- ----> ---->
FeFe33(PO(PO44))22
iron(II) phosphateiron(II) phosphate
(See CD, Screen 3.11 for naming practice)(See CD, Screen 3.11 for naming practice)
calcium fluoridecalcium fluoride
CaCa2+2+ + 2 F + 2 F-- ---> ---> CaFCaF22
Properties of Ionic Properties of Ionic CompoundsCompounds
Forming NaCl from Na and ClForming NaCl from Na and Cl22
Properties of Ionic Properties of Ionic CompoundsCompounds
Forming NaCl from Na and ClForming NaCl from Na and Cl22
• A metal atom can A metal atom can transfer an transfer an electron to a electron to a nonmetal.nonmetal.
• The resulting The resulting cation and anion cation and anion are attracted to are attracted to each other by each other by
electrostatelectrostatic forcesic forces..
Electrostatic ForcesElectrostatic Forces
The oppositely charged ions in ionic compounds are The oppositely charged ions in ionic compounds are attracted to one another by attracted to one another by ELECTROSTATIC ELECTROSTATIC FORCESFORCES..
These forces are governed by These forces are governed by COULOMB’S COULOMB’S LAWLAW..
Electrostatic ForcesElectrostatic ForcesCOULOMB’S LAWCOULOMB’S LAW
Force of attraction = (charge on +)(charge on -)
(distance between ions)2
As ion charge increases, the attractive force As ion charge increases, the attractive force _______________._______________.As the distance between ions increases, the As the distance between ions increases, the attractive force ________________.attractive force ________________.This idea is important and will come up This idea is important and will come up many times in future discussions!many times in future discussions!
Importance of Coulomb’s Importance of Coulomb’s LawLaw
NaCl, NaNaCl, Na++ and Cl and Cl--,,m.p. 804 m.p. 804 ooCC
MgO, MgMgO, Mg2+2+ and O and O2-2-
m.p. 2800 m.p. 2800 ooCC
Molecular CompoundsMolecular CompoundsCompounds without IonsCompounds without Ions
CH4 methane
CO2 Carbon dioxide
BCl3 boron trichloride
Naming Molecular Naming Molecular CompoundsCompounds
CH4 methaneBCl3 boron trichloride
CO2 Carbon dioxide
All are formed from two or more nonmetals.
Ionic compounds generally involve a metal and nonmetal (NaCl)
Empirical & Molecular Empirical & Molecular FormulasFormulas
A pure compound always consists of the A pure compound always consists of the same elements combined in the same same elements combined in the same proportions by weight.proportions by weight.
Therefore, we can express molecular Therefore, we can express molecular composition as composition as PERCENT BY PERCENT BY WEIGHTWEIGHT
Ethanol, CEthanol, C22HH66OO
52.13% C52.13% C13.15% H 13.15% H 34.72% O34.72% O
Percent CompositionPercent CompositionPercent CompositionPercent CompositionConsider some of the family of nitrogen-Consider some of the family of nitrogen-
oxygen compounds:oxygen compounds:
NONO22, nitrogen dioxide and closely , nitrogen dioxide and closely related, NO, nitrogen monoxide (or related, NO, nitrogen monoxide (or nitric oxide) nitric oxide)
Structure of NOStructure of NO22
Chemistry of NO, Chemistry of NO, nitrogen monoxidenitrogen monoxide
Percent CompositionPercent CompositionPercent CompositionPercent CompositionConsider NOConsider NO22, Molar mass = ?, Molar mass = ?
What is the weight percent of N and of What is the weight percent of N and of O?O?
Wt. % O = 2 (16 .0 g O per mole)46 .0 g
x100 % = 69 .6%Wt. % O = 2 (16 .0 g O per mole)46 .0 g
x100 % = 69 .6%
Wt. % N = 14.0 g N
46.0 g NO2 • 100% = 30.4 %Wt. % N =
14.0 g N46.0 g NO2
• 100% = 30.4 %
What are the weight percentages of What are the weight percentages of N and O in NO?N and O in NO?
Mass Spectrum of EthanolMass Spectrum of EthanolMass Spectrum of EthanolMass Spectrum of Ethanol(from the NIST site)(from the NIST site)
46
45
CH3CH2OH+
CH3CH2O+31
CH2O+
Determining Determining FormulasFormulas
Determining Determining FormulasFormulas
In In chemical analysischemical analysis we determine we determine the % by weight of each element in a given the % by weight of each element in a given amount of pure compound and derive the amount of pure compound and derive the
EMPIRICALEMPIRICAL or or SIMPLESTSIMPLEST formula.formula.
PROBLEMPROBLEM: A compound of B : A compound of B and H is 81.10% B. What is its and H is 81.10% B. What is its empirical formula?empirical formula?
• Because it contains only B and H, it Because it contains only B and H, it must contain 18.90% H.must contain 18.90% H.
• In 100.0 g of the compound there are In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H.81.10 g of B and 18.90 g of H.
• Calculate the number of moles of each Calculate the number of moles of each constitutent.constitutent.
A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. What is its empirical formula?What is its empirical formula?A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. What is its empirical formula?What is its empirical formula?
Calculate the number of moles of each Calculate the number of moles of each element in 100.0 g of sample.element in 100.0 g of sample.
81.10 g B • 1 mol
10.81 g = 7.502 mol B81.10 g B •
1 mol10.81 g
= 7.502 mol B
18.90 g H • 1 mol
1.008 g = 18.75 mol H18.90 g H •
1 mol1.008 g
= 18.75 mol H
A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. What is its empirical formula?What is its empirical formula?A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. What is its empirical formula?What is its empirical formula?
Now, recognize that Now, recognize that atoms combine in atoms combine in the ratio of small whole numbers.the ratio of small whole numbers.
1 atom B + 3 atoms H --> 1 molecule BH1 atom B + 3 atoms H --> 1 molecule BH33
oror
1 mol B atoms + 3 mol H atoms ---> 1 mol B atoms + 3 mol H atoms ---> 1 mol BH1 mol BH33 molecules molecules
Find the ratio of moles of elements Find the ratio of moles of elements in the compound.in the compound.
A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. What is its empirical formula?What is its empirical formula?A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. What is its empirical formula?What is its empirical formula?
But we need a whole number ratio. But we need a whole number ratio.
2.5 mol H/1.0 mol B = 5 mol H to 2 mol B2.5 mol H/1.0 mol B = 5 mol H to 2 mol B
EMPIRICAL FORMULA = BEMPIRICAL FORMULA = B22HH55
Take the ratio of moles of B and H. Take the ratio of moles of B and H. AlwaysAlwaysdivide by the smaller number.divide by the smaller number.
18.75 mol H7.502 mol B
= 2.499 mol H1.000 mol B
= 2.5 mol H1.0 mol B
18.75 mol H7.502 mol B
= 2.499 mol H1.000 mol B
= 2.5 mol H1.0 mol B
A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. What is its empirical formula?What is its empirical formula?A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. What is its empirical formula?What is its empirical formula?
A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. Its Its empirical formulaempirical formula is B is B22HH55. .
What is its What is its molecular molecular formulaformula??Is the molecular formula BIs the molecular formula B22HH55, B, B44HH1010, ,
BB66HH1515, B, B88HH2020, etc.? , etc.?
BB22HH66 is one example of this class of compounds. is one example of this class of compounds.
B2H6
A compound of B and H is 81.10% B. Its empirical A compound of B and H is 81.10% B. Its empirical
formula is Bformula is B22HH55. What is its molecular formula. What is its molecular formula??A compound of B and H is 81.10% B. Its empirical A compound of B and H is 81.10% B. Its empirical
formula is Bformula is B22HH55. What is its molecular formula. What is its molecular formula??
We need to do an We need to do an EXPERIMENTEXPERIMENT to to find the MOLAR MASS.find the MOLAR MASS.
Here experiment gives Here experiment gives 53.3 g/mol53.3 g/molCompare with the mass of BCompare with the mass of B22HH55
= = 26.66 g/unit26.66 g/unit
Find the ratio of these masses.Find the ratio of these masses.
53.3 g/mol26.66 g/unit of B2H5
= 2 units of B2H5
1 mol
53.3 g/mol26.66 g/unit of B2H5
= 2 units of B2H5
1 mol
Molecular formula = BMolecular formula = B44HH1010
Determine the formula of a Determine the formula of a compound of Sn and I using compound of Sn and I using the following data.the following data.
Determine the formula of a Determine the formula of a compound of Sn and I using compound of Sn and I using the following data.the following data.
• Reaction of Sn and IReaction of Sn and I22 is done using excess Sn. is done using excess Sn.• Mass of Sn in the beginning = 1.056 gMass of Sn in the beginning = 1.056 g• Mass of iodine (IMass of iodine (I22) used ) used
= 1.947 g = 1.947 g• Mass of Sn remaining Mass of Sn remaining
= 0.601 g= 0.601 g• See p. 104See p. 104
Find the mass of Sn that combined with Find the mass of Sn that combined with 1.947 g I1.947 g I22..
Mass of Sn initially = 1.056 gMass of Sn initially = 1.056 g
Mass of Sn recovered = 0.601 gMass of Sn recovered = 0.601 g
Mass of Sn used = 0.455 gMass of Sn used = 0.455 g
Find moles of Sn used:Find moles of Sn used:
0.455 g Sn • 1 mol
118.7 g = 3.83 x 10-3 mol Sn0.455 g Sn •
1 mol118.7 g
= 3.83 x 10-3 mol Sn
Tin and Iodine Tin and Iodine CompoundCompound
Tin and Iodine Tin and Iodine CompoundCompound
Now find the number of moles of INow find the number of moles of I22 that that combined with 3.83 x 10combined with 3.83 x 10-3-3 mol Sn. Mass mol Sn. Mass of Iof I22 used was 1.947 g. used was 1.947 g.
1.947 g I2 • 1 mol
253.81 g = 7.671 x 10-3 mol I21.947 g I2 •
1 mol253.81 g
= 7.671 x 10-3 mol I2
How many mol of How many mol of iodine atomsiodine atoms? ?
= 1.534 x 10-2 mol I atoms= 1.534 x 10-2 mol I atoms
7.671 x 10-3 mol I2 2 mol I atoms
1 mol I2
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 7.671 x 10-3 mol I2
2 mol I atoms1 mol I2
⎛ ⎝ ⎜
⎞ ⎠ ⎟
Tin and Iodine Tin and Iodine CompoundCompound
Now find the ratio of number of moles of moles Now find the ratio of number of moles of moles of I and Sn that combined.of I and Sn that combined.
1.534 x 10-2 mol I
3.83 x 10-3 mol Sn =
4.01 mol I1.00 mol Sn
1.534 x 10-2 mol I
3.83 x 10-3 mol Sn =
4.01 mol I1.00 mol Sn
Empirical formula is Empirical formula is
SnISnI44