Download Lectures 13 - Chemistry Courses

LEWIS STRUCTURES
CHEMICAL CONNECTIVITY
THIS LEADS US TO
CHEMICAL ANALYSIS
&
STRUCTURE DETERMINATION
KEY CONCEPTS
Conservation of Mass
mass of reactants = mass of products
Balanced Chemical Equation
connects moles of reactants with moles
of products; related to cons. of mass
Avogadro’s Number
connects molecules (or atoms) to moles;
connects microscopic quantities to
macroscopic quantities
Formula Weight
connects mass to moles; connects a property
that can be measured experimentally to moles
(or molecules) of a substance given in a
balanced reaction
Empirical Formula
tells relative number of atoms in a molecule;
obtained from percent mass or molecular formula
CHEMICAL EQUATIONS
N2 + 3 H2 → 2 NH3
2N
6H
reactants
2N
products
6H
balanced means equal numbers of
atoms of each element on each side
When reactants and products are given,
only need to change coefficients to
achieve balance
Law of Conservation of Mass
Atoms (or mass) are neither created nor
destroyed in a chemical reaction
The mass of the reactants must equal the
mass of the products in a reaction
The same number of atoms of each element
must be present among the reactants and
products of a reaction
balanced equation
AMU REVIEW
The masses of atoms, molecules in amu
1 amu = 1.66056 x 10−24 g
Defn: 12 amu = mass of one 12C atom
1
1 amu = 12 mass of 12C atom
Atomic Weight
Defn: weighted average mass of an element
Examples
Carbon
12C
98.89%
13C 1.11%
(0.9889)(12) + (0.0111)(13) = 12.011
Chlorine
35Cl
75.5%
37Cl 24.5%
(0.755)(35) + (0.245)(37) = 35.5
FORMULA WEIGHT
FW = sum of atomic weights of each
atom in the chemical formula
MgCO3
1 Mg 24.3
1 C
12.0
3 O 16.0x3
24.3
12.0
48.0
84.3 amu
THE MOLE AND
AVOGADRO’S NUMBER
One mole of (molecules, ions, green peas)
contains Avogadro’s number of them
1 mol 12C atoms = 6.02 x 1023 12C atoms
1 mol CO2 molecules = 6.02 x 1023 CO2 molecules
1 mol peas = 6.02 x 1023 peas
1 H2O molecule weighs 18.0 amu
1 mol of H2O weighs 18.0 g
MOLES OF COMPOUNDS
molar mass of molecular compound
CH4 12.011 + 4(1.01) = 16.05 g/mol
molar mass of ionic compound
CaCl2
40.08 + 2(35.45) = 110.98 g/mol
use molar mass to convert between
moles and grams in computations
How many moles of Mg(NO3)2 are
in 100.0 g of the compound?
EXAMPLE
molar mass:
24.31 + 2(14.01) + 6(16.00) = 148.33 g/mol
100.0 g x
1 mol
= 0.67 mol
148.33 g
Chemical Formula Review
How many atoms are in the formula Mg3(PO4)2
a)
b)
c)
d)
3
5
10
13
How many moles of Mg3(PO4)2 are in a 10.0 g
sample?
a)
b)
c)
0.0380 moles
0.0595 moles
0.0431 moles
How many atoms are in 10.0 g of Mg3(PO4)2?
a)
b)
c)
d)
2.59×1022 atoms
3.58×1022 atoms
2.29×1022 atoms
2.97×1023 atoms
PERCENT COMPOSITION
What is the mass percent of P in H3PO4 ?
H = 1.008 amu
P = 30.97 amu
O = 16.00 amu
FW = 3(1.008) + 30.97 + 4(16.00)
= 97.99
%P =
30.97
x 100 = 31.6%
97.99
4(16.00)
x 100 = 65.3%
%O =
97.99
%H = 100 – 31.6 – 65.3 = 3.1%
EMPIRICAL FORMULA
formula with smallest integer subscripts
can determine from percentage composition
EXAMPLE
butyric acid has % composition of
%C = 54.2% %O = 36.6% %H = 9.15%
Assume that you have 100 g…
1 mol C
= 4.52
moles C = (54.2 g)
12.0 g
1 mol H
moles H = (9.15 g)
= 9.08
1.008 g
moles O = (36.6 g)
1 mol O
= 2.29
16.0 g
Calculate ratios
C
O
4.52
= 1.97
2.29
C2H4O
H
O
9.08
= 3.97
2.29
To continue and calculate the molecular
formula, need more information
Need molecular weight
EXAMPLE cont.
MW butyric acid is 88.1 g/mol
empirical formula C2H4O
12 x 2 = 24
1 x4= 4
16 x 1 = 16
44
so we must double the subscripts
Molecular Formula C4H8O2
O
CH3–CH2–CH2–C–OH
EMPIRICAL FORMULA
FROM ANALYSIS
Percent
Composition
Empirical
Formula
Example: butyric acid
%C
%H
%O
C2H4O
C4H8O2
Combustion Analysis
Fuel
C, H, O
CO2 + H2O
Use combustion data to calc. g of C and H,
then find g of O. Then calc. percentages.
Then, it’s a problem of % composition to
empirical formula.
COMBUSTION ANALYSIS
Combustion of 3.14 g isopropyl alcohol (rubbing
alcohol) produced 6.90 g CO2 and 3.76 g H2O.
What is isopropyl alcohol’s empirical formula?
Need to find moles of C, H, O in sample
Consider C first:
moles CO2: 6.90 g x
1 mol
= 0.157 mol CO2
44.00 g
0.157 mol CO2 means 0.157 mol C in sample
Now consider H:
moles H2O: 3.76 g x
1 mol
= 0.209 mol H2O
18.01 g
0.209 mol H2O means 2 x 0.209 = 0.418 mol H
…cont...
COMBUSTION ANALYSIS
Now consider O:
mass O = mass sample – mass C – mass H
C
0.157 mol x 12.00 g/mol = 1.88 g C
H
0.418 mol x 1.008 g/mol = 0.421 g H
3.14 g – 1.88 g C – 0.421 g H = 0.839 g O
0.839 g x
1 mol
16.00 g
= 0.0524 mol O
Find ratios of moles of atoms in sample
C/O
0.157
= 3.00
0.0524
H/O
0.418
= 7.98
0.0524
So…. 3 C to 8 H to 1 O
C3H8O or C3H7OH
Sample Problem
• An analysis of nicotine, a poisonous
compound found in tobacco leaves,
shows that it is 74.0%C, 8.65% H and
17.35% N. Its molar mass is 162 g/mol.
What are the empirical and molecular
formulae of nicotine?
ORGANIC CHEMISTRY
The chemistry of compounds containing C
bonded to H.
Often contain O, N, S, and halogens
Millions of organic compounds are known
They are the main constituents of living matter.
DNA, RNA, proteins, enzymes, …
Reasons for the large number of organic
compounds
1)
Carbon atoms can form strong, stable bonds to other
C atoms (forming rings, chains, etc.) and to atoms
such as H, O, N, S, halogens (small size).
2)
Carbon atoms form up to 4 bonds simultaneously:
(valence of 4) therefore molecule can be branched.
3)
Carbon atoms form multiple bonds with C or with O,
N, S: further structural variations are possible.
CLASSES OF ORGANIC COMPOUNDS
Classification is necessary to manage large
number of compounds
Hydrocarbons
Simplest organic compounds
Contain only C and H
Functional Group Classes
Contain a representative group of
elements in a fixed pattern:
Have similarities in structure and reactivity
Examples:
C OH
alcohol
C Cl
halide
C NH2
amine
CLASSES OF
HYDROCARBONS
contain only C and H
9 Alkanes
all single bonds
9 Alkenes
double bond
C=C
H – C ≡ C – H triple bond
9 Alkynes
9 Aromatic hydrocarbons
HC
HC
H
C
C
H
CH
CH
benzene
ring
IMPORTANT THEME OF CHEM 110
structure affects reactivity
The classes of hydrocarbons have significant
differences in structural features: these
differences affect the properties and reactivity of
the molecules.
To aid in deducing the structure of molecules given the
formula, we use the rules of valence.
VALENCE of an atom is the number of
covalent bonds that it typically
forms
Rules of valence
Element
H
C
N
O, S
F, Cl, Br
# Bonds
1
4
3
2
1
Using the Rules of Valence, structural
formulas of simple molecules can be
deduced.
Molecular
formula
Condensed
molecular
formula
NH3
NH3
CO2
CO2
C2H6O
CH3CH2OH
C2H6O
or
CH3OCH3
structural
formula
ALKANES
Name
Molecular
formula
Condensed
molecular formula
methane
CH4
ethane
C2H6
CH3-CH3
propane
C3H8
CH3CH2CH3
butane
C4H10
CH3CH2CH2CH3
pentane
C5H12
CH3CH2CH2CH2CH3
In general: CnH2n+2
Lewis Structures
H
H H
H C H
H C
H
H
H H H
C H
H C
H H
H
H
H
H
C
C
C
C
H
H
H
H
H
C
C
H
H H H
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
STRUCTURE OF ALKANES
All carbon atoms have tetrahedral geometry:
the bonds point at the vertices of a tetrahedron.
Bond angle = 109.5o
Can represent as zig-zags
or
Both of these are octane (C8H18):
an 8 carbon chain
ALKENES
contain one or more double bonds
C
C
CnH2n
The C=C–H or C=C–C bond angle is 120o
H
H
C
C
C
H
ethene
ethylene
H
H
H
H
C
CH 3
propene
Rotation about double bond is restricted
EXAMPLE
butene
There are several possible structures for a
chain of 4 carbons with one double bond
1-butene
2-methylpropene
2-butene
cis-2-butene
trans-2-butene
ISOMERS
Isomers are compounds that have the same
molecular formula, but have different structures.
Structural isomers:
same molecular formula: different structure
1-butene
2-methylepropene
2-butene
Geometric isomers:
same molecular formula: different geometry
cis-2-butene
trans-2-butene
For geometric isomers: atoms have the same
structure (are joined in the same way), but
arrangement in space (geometry) is differene.
For alkenes, geometric isomers occur because of
the restricted rotation about the C=C bond.
ALKYNES
contain one or more triple bonds
⎯C≡C⎯
H⎯C≡C⎯H
ethyne
(acetylene)
CH3⎯C≡C⎯H
propyne
Bond angle is 180o
CH3 –CH2 –C≡C–H
CH3–C≡C–CH3
1-butyne
2-butyne
structural isomers
No geometric isomers are possible
FUNCTIONAL GROUPS
Contain a representative group of
elements in a fixed pattern
Have similarities in structure and
reactivity
Seemingly small changes in structure can
have big effects on molecular properties
Example: adding an oxygen changes a
hydrocarbon into an alcohol or an ether…
Functional Group Type
hydrocarbon
alcohol
ether
CH3⎯CH2⎯H
CH3⎯CH2⎯OH
CH3⎯O⎯CH3
ethane
ethanol
di-methyl ether
gas
liquid
gas
bp –89°C
bp 78°C
bp –24 °C
water soluble
FUNCTIONAL GROUP CLASSES
R≠H
Ethers
R–O–R’
R,R’ ≠ H
Amines
H–N–R
H
R–N–H
R’–N–R
H
H
–
–
R–OH
–
Alcohols
R,R’,R” = alkyl groups [–CH3, –CH2CH3, –CH2CH2CH3, etc.]
O
=
Groups having a carbonyl
C
O
=
=
O
C
C
R
H
Aldehydes
(R,R’ ≠ H)
Ketones
O
O
=
=
=
O
R
R’
R
C
C
C
OH
Carboxylic Acid
R
OR’
Ester
R
NHR’
Amide
FUNCTIONAL GROUPS DETERMINE
DRUG PROPERTIES
What functional groups are present in these
molecules?
O
C
OH
O C
CH3
O
Acetyl salicylic acid
H
HO
N
C
CH 3
O
Acetaminophen
CH3
CH3
CH CH2
CH3
O
CH C
OH
Ibuprofen
Which of these functional groups is an alcohol?
R–O–R’
R–O–H
R–N–H
A)
B)
H
H–O–H
R–N–H
–
A)
B)
C)
–
R’
Which of these functional groups is an ester?
=
O
A)
C
R’
R
=
O
B)
C
R
H
=
O
C)
C
OR’
R
O
=
D)
C
OH
R
=
O
E)
C
R
NHR’
ORGANIC ACIDS AND BASES
AMINES
weak bases, organic bases (like NH3)
(CH3)3N(aq) + H+(aq) U (CH3)3NH+(aq)
CARBOXYLIC ACIDS
weak acids, organic acids
O
=
=
O
C
R
OH
(aq) U
C
R
O
+
– (aq) + H (aq)
Polymers
High molecular weight materials formed from
many small molecules called monomers
Monomer = repeating unit
Polymer Synthesis
1. Addition polymers
monomer contains a double bond
polymer forms via addition reaction
2. Condensation polymers
polymers form via condensation reaction
a small molecule (H2O, HCl, CO2) is
eliminated in the reaction
3. Biological polymers
Proteins
DNA
Carbohydrates
ADDITION POLYMERS
polymers formed via addition reaction
Examples
Monomer
Polymer
CH2=CH2
polyethylene
(HDPE, LDPE)
–
CH3
CH2=CH
polypropylene
(PPP)
–
Cl
CH2=CH
polyvinylchoride
(PVC)
CF2=CF2
polytetrafluoroethylene
(teflon, PTFE)
CONDENSATION POLYMERS
polymers formed via condensation reaction
n
O
O
C
C
HO
OH
+n
CH 2 CH 2
OH
– H2O
HO
diacid
diol
O
O
C
CH2
C
O
O
CH2
O
Dacron (polyester)
H
n
HO
O
O
C
C
(CH 2 )6
+n
(CH 2 )6
OH
NH 2
diacid
O
– H2O
H N
diamine
O
O
C
C
(CH 2 )6
H
N
N
(CH 2 )6
H
Nylon 66 (polyamide)
BIOLOGICAL POLYMERS
Proteins
Monomer: amino acids (Figure 25.23)
React via condensation reaction
Proteins are polyamides
amide linkage
or peptide bond
condensation reaction (removal of H2O)
Protein backbone continues to grow by addition
of more amino acids. Backbone always has
terminal amino and carboxyl groups.
Proteins can fold into
complex secondary
structures such as
this a Helix.
These secondary structures
can fold into more complex
tertiary structures such as
this enzyme
BIOLOGICAL POLYMERS
DNA, RNA
Monomer: nucleotides (Figure 25.34)
React via condensation reaction
DNA and RNA are polyesters (of H3PO4)
Polynucleotide
Monomer nucleotide
has a phosphoric acid unit
a five-carbon sugar
and an organic base
[Adenine (A), Guanine (G),
Cytosine (C), Thymine (T), or
Uracil (U)]
Polynucleotide chain folds into a double helix
Hydrogen bonds