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Week 5: Lectures 13 – 15
Chemical Formulas
Lecture 13: W 9/21
Lecture 14: F 9/23
Lecture 15: M 9/26
Reading:
BLB Ch 2.6, 2.9, 3.3 – 3.5; 9.1 – 9.3
Homework:
BLB 3:25, 29, 35, 37, 51, 53; Supp 3:1 – 9
BLB 9:23, 25, 27, 38; Sup 9:1 – 7
Reminder:
ALEKS Objective 5 due on Tue, 9/27
“No Score” from Exam 1?? Not a problem—Go see
Mike Joyce in 210 Whitmore or email him at
[email protected]
Molecular Formula: chemical formula with the total
number of atoms and their relative proportions in
each molecule.
Empirical Formula: chemical formula with the
smallest integer subscripts for a given molecule.
Structural Formula: chemical formula with
structural information about connectivity and
angles.
Jensen
Chem 110 Chap3
Page: 2
Connecting microscopic (atoms) to macroscopic
(weight) using Important Connections
• Avogadro!s number (NA):
Connects molecules (or atoms) to moles
• Molar Mass (in g/mol) connects mass to moles;
connects experimentally measured property to
moles (or molecules) of substance; the molar
mass (in g/mol) of any substance is always
numerically equal to its formula weight (in amu).
• Empirical Formula: chemical formula with the
smallest integer subscripts; tells relative number
of moles (molar ratio) of each element in a
compound; obtained from mass percent or
molecular formula;
• Balanced chemical equation:
connects moles (molecules) of reactants with
moles (molecules) of products; related to
conservation of mass
• Conservation of mass and energy:
mass of products = mass of reactants;
energy is conserved but can change its form
Jensen
Chem 110 Chap3
Page: 3
Coefficients and Subscripts
• In a molecule, subscripts denote the number of
________ per molecule or formula unit
C6H12O6 : one C6H12O6 molecule has __ C, __ O,
and __ H atoms.
Mg(NO3)2 : one Mg(NO3)2 formula unit has __ Mg,
__ N, and __ O atoms
• In a reaction, coefficients denote the relative
number of ________ of each compound
2 H2
+
O 2 ! 2 H 2O
! Changing the subscripts changes the _________
of the molecule.
! Changing the coefficient changes the _________
of that molecule, not the identity.
Jensen
Chem 110 Chap3
Page: 4
Working with Formulas
Working with Formulas
Example: How many moles of ammonium ions
are present in 30.9 g of ammonium sulfate?
A.
B.
C.
D.
E.
0.468 mol
0.288 mol
0.234 mol
2.14 mol
3.47 mol
Jensen
Practice Example:
Which of the following has the greatest
number of total atoms present?
A.
B.
C.
D.
E.
Chem 110 Chap3
Page: 5
6.022 x 1023 O2 molecules
0.1 mol C6H12O6
56 g Fe
0.4 mol CH4
4.4 g of H2
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Chem 110 Chap3
Page: 6
Percent Composition: Mass Percent
contributed by each element in compound
(Note: percentage must add to 100)
A) 18 g
1. Given chemical formula,
% element =
!
Practice Example:
How many grams of oxygen are in 65 g of
C 2H 2O 2?
(# of element) (AW of element)
"100
FW of compound
B) 29 g
C) 9.0 g
Example: What is the mass percent of C in CO2?
D) 36 g
C= _____ amu; O = _____ amu; CO2 = _______ amu
E) 130 g
%C =
2. If we have 100 g of sample,
% element =
!
mass of element
"100
100 g compound
Example: I have 2g of a sample that is 27.3% C
by mass, how many grams of C are in the sample?
Jensen
Chem 110 Chap3
Page: 7
Jensen
Chem 110 Chap3
Page: 8
Connect Mass Percent to Chemical Formula
Given one quantity, we can calculate the other:
• Start with a chemical formula and calculate
the mass % elements;
or
• Start with mass % of elements (i.e. empirical
data) and calculate a chemical formula.
Jensen
Chem 110 Chap3
Page: 9
Practice Example:
Butyric acid, which has the smell of rancid
butter, is 54.53% carbon by mass and 9.15%
hydrogen by mass, and is composed only of
carbon, hydrogen, and oxygen. It has a
molar mass of 88.10 g/mol. What is the
molecular formula of butyric acid?
A.
B.
C.
D.
E.
C4H 8O 2
C2H 4O
C5H 4O
C 4H 2O 2
C4H16O
Jensen
Chem 110 Chap3
Page: 10
Combustion Analysis
Solve:
1. Assume 100 g of sample
C = _______ g
H = _______ g
O = ______ g
2. Find moles of each element
C = _____________________________________ mol
H = _____________________________________ mol
O = _____________________________________ mol
3. Find ratio of moles of elements (divide by
Fuel
C, H, O
+ O2
CO2 + H2O
Given grams of sample in excess O2, grams of CO2
and H2O produced; may have N or O in formula
(non-C,H atoms); determine empirical formula or
molecular formula of the sample.
Combustion Analysis Procedures
smallest)
1. Use grams of CO2 and H2O to find moles of Cand H- elements in the products and in the sample;
Empirical Formula:
4. Get molecular formula from empirical formula
2. Use massnon-C,H = masssample - massC - massH to
find mass and moles of non-C,H elements in the
sample;
3. Calculate the ratio of C, H, and non-C,H elements
(divide by smallest) in the sample;
4. Write empirical formula using ratios; and convert
to molecular formula using molar mass of sample
(if known)
Jensen
Chem 110 Chap3
Page: 11
Jensen
Chem 110 Chap3
Page: 12
Practice Example:
11.0 g sample of a compound containing only
C, H, and O was burned completely in air to
produce 15.6 g of CO2 and 9.60 g of water.
What is the empirical formula of this
compound?
Solve:
a) C4H9O
2. Calculate mass of C, H, and O elements and
moles of O elements in the sample
b) C2H6O
1. Use grams of CO2 and H2O to calculate moles of
C, H elements in products and in the sample
C = _____________________________________ mol
H = _____________________________________ mol
C = ______________________________________ g
c) C3H6O2
d) CH3O
H = ______________________________________ g
e) CH2O
O = ______________________________________ g
O = _____________________________________ mol
3. Find ratio of moles of elements (divide by
smallest)
Empirical Formula:
Jensen
Chem 110 Chap3
Page: 13
Jensen
Chem 110 Chap3
Page: 14
What’s Next?
VSEPR Model
Chemical Formulas tell us how atoms are
connected to each other
Lewis structures tell us where the electrons are.
Electron Domain Geometry (EDG)
Electron pairs " Electron domains
One lone pair = one electron domain
Lewis Structures (2D)
One bond (single, double or triple bond)
= one electron domain.
—
Bonding e pairs
—
Lone e pairs
Molecular Shape (3D):
Arrangement of Atoms in the Molecule
Valence Shell Electron Pair Repulsion (VSEPR):
Predict molecular shape (spatial arrangement of
atoms) from Lewis structure
VSEPR: Electron domains around
a central atom repel each other.
They must stay near nucleus but
otherwise try to get as far apart
from each other as possible
Jensen
Chem 110 Chap 9
Page: 1
Electron Domain Geometries (EDG)
Determine Electron Domain Geometry (EDG)
and Molecular Geometry (MG)
1. Draw the Lewis structure of the molecule
2. Count the number of electron domains in the
Lewis structure: one bond (single, double, or triple)
counts as ______ domain! And one lone pair counts
as ______ domain!)
3. The electron domain geometry corresponds to
the number of electron domains
4. The molecular geometry is defined by the
positions of only ________________ in the
molecules, _______ the nonbonding pairs
5. Some electron domains have more than one
molecular geometry
Jensen
Chem 110 Chap 9
Page: 2
Jensen
Chem 110 Chap 9
Page: 3
1. Linear EDG (2 electron domains)
Two possible MG for trigonal planar EDG:
# If all bonding electron domains, MG = _________
# If there is a nonbonding (lone) pair, MG = ______
3. Tetrahedral EDG (4 electron domains)
# In this domain, only one possible MG: linear.
# NOTE: If there are only two atoms in the
molecule, the molecule will be linear no matter
what the electron domain is.
2. Trigonal planar EDG (3 electron domains)
Three possible MG for tetrahedral EDG:
# If all are bonding pairs, MG = ________________
# If there is one lone pair, MG= _________________
# If there are two lone pairs, MG= _______________
Jensen
Chem 110 Chap 9
Page: 4
Jensen
Chem 110 Chap 9
Page: 5
The Trigonal Bipyramidal Case
Lone Pairs and Bond Angles
4. Trigonal bipyramidal EDG (5 electron domains)
! Nonbonding pairs are physically larger
Two types
of sites ________
– _____ and ______
• Two types of• sites
–_______and
• Angles between
sitesbetween
are different
• Angles
sitesfor
are different for
these two types.
these two types.
than bonding pairs
! Therefore, their repulsions are greater;
this tends to _________ bond angels
900
on the other side of the molecule
Eq
Ax
Eq
1200
A
Eq
900
Ax
• Lone pairs prefer to be equatorial. Why?
Multiple Bond and Bond Angles
! greater electron density on
one side of the central atom
! Therefore, bond angles involving multiple bond
are _____________, while angles on other side of
Lone pairs prefer to be equatorial.
the molecule are ___________
CH110 FA11 SAS
Jensen
Chem 110 Chap 9
Page: 6
Jensen
Chem 110 Chap 9
Page: 7
4. Trigonal bipyramidal EDG (5 electron domains)
5. Octahedral EDG (6 electron domains)
• All positions are equivalent
• Two lone pairs go on ___________ sides of the
Four possible MG for trigonal bipyramidal EDG:
!
!
!
!
Trigonal bipyramidal
Seesaw
T-shaped
Linear
Jensen
Chem 110 Chap 9
molecule
Three possible MG for octahedral EDG:
! Octahedral
! Square pyramidal
! Square planar
Page: 8
Jensen
Chem 110 Chap 9
Page: 9
Determine the Geometry of a Molecule
1. Draw the Lewis structure of the molecule.
2. Determine the electron domain geometry (EDG:
count electron domains: multiple bonds count as
one domain).
3. Focus on bonding electron pairs ONLY to
determine the molecular geometry (MG)
Example: What is the molecular geometry of
IF5?
Practice Example
The molecular geometry of which molecule is
square planar?
A.
B.
C.
D.
E.
CCl4
XeF4
PH3
XeF2
BrF3
A. see-saw
B. octahedral
C. square pyramidal
D. square bipyramidal
E. square planar
Jensen
Chem 110 Chap 9
Page: 10
Jensen
Chem 110 Chap 9
Page: 11
VSEPR can be applied to Larger Molecules
!
Describe geometry around
Determine Approximate Bond Angel
Practice Example: Determine the approximate
bond angles indicated.
a particular atom
Example: What is the approximate C—C—O
bond angel as indicated?
A. 90°
B. 105°
O
1
NCH2CH2CH2C 2
H
OH
H
Angle #1
Angle #2
A.
109°
109°
B.
180°
120°
C.
120°
90°
D.
109°
120°
E.
120°
109°
C. 109°
D. 120°
E. 180°
Jensen
Chem 110 Chap 9
Page: 12
Jensen
Chem 110 Chap 9
Page: 13
Molecular Geometries: Summary
# of
electron
domains
2
# of
bonded
atoms
2
3
molecular
geometry
Example
linear
CO2
3
trigonal planar
BF3
3
2
bent
O3
4
4
tetrahedral
CH4
4
3
4
2
5
5
5
4
trigonal
bipyramidal
seesaw
5
3
T-shaped
BrF3
5
2
linear
XeF2
6
6
octahedral
SF6
6
5
6
4
Jensen
trigonal
pyramidal
bent
square
pyramidal
square planar
Chem 110 Chap 9
NH3
Molecular Polarity
Molecular geometry and relative
electronegativity of elements determine
many properties of molecules
• How does microwave cooking work?
• Why do water and methane differ so much in
their boiling point (BP)?
H 2O
BP of H2O = 100 ˚C; BP of CH4 = -161 ˚C
PCl5
• Why do alcohols and water mix but oil and
water don!t?
SF4
BrF5
XeF4
Page: 14
Jensen
Chem 110 Chap 9
Page: 15
Bond Polarity
Molecular Polarity
The polarity of individual bond is determined
• Just because a molecule has polar bonds does
not mean the molecule as a whole will be polar.
by electronegativity difference between
elements; The greater the difference, the
Polarity of entire molecule: Dipole Moment (µ)
more polar the bond.
For a diatomic molecule: µ = Q r
H
r
Cl units = debye (D)
•%%%% •
• Most polar: ionic compound, e.g. NaF
+Q
• Polar covalent: (two different atoms)
&Q
• A molecule is polar if there is a NET charge
separation between two “ends” of the molecule: a
negative “end” and a positive “end”
A molecule is polar (has a net dipole) if it has:
a. Polar bonds
b. Geometry where bond dipoles do not cancel out
$+ and $— are partial charges for H and F
To determine the polarity of a molecule that has
more than two atoms:
• Nonpolar covalent: (two identical atoms);
e.g. H2, F2, Cl2, O2,
1. Find molecular shape using VSEPR
2. Find “bond” dipoles
3. Use vector analysis to find NET molecular dipole
Jensen
Jensen
Chem 110 Chap 9
Page: 16
Chem 110 Chap 9
Page: 17
Examples:
Examples:
CO2
NH3 (electronegativity: N > H)
(electronegativity: O > C)
EDG: tetrahedral
EDG: Linear
MG: trigonal pyrimidal
MG: Linear
Bond dipoles?
Bond dipoles?
Net dipole moment?
N
H
H
H
Net dipole moment?
CF4 (electronegativity: F > C)
H 2O
(electronegativity: O > H)
EDG: tetrahedral
EDG: tetrahedral
MG: tetrahedral
MG: bent
Bond dipoles?
Bond dipoles?
Net dipole moment?
F
Net dipole moment?
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Chem 110 Chap 9
Page: 18
Jensen
F
Chem 110 Chap 9
C F
F
Page: 19
Example: Which molecule below is polar?
Molecular Geometry & Molecular Polarity
Do not memorize this table!! Draw the generic
structures and think about them to verify!
Practice Example: Which molecule shown
below has non-zero dipole moment?
linear
bent
trigonal pyramidal
T-shaped
seesaw
square pyramidal
A. XeF2
B. BF3
C. SF4
D. IF2–
E. BeCl2
Jensen
MG
linear
trigonal planar
tetrahedral
trigonal bipyramidal
octahedral
square planar
Formula Example Polar?
AX2
CO2
AX3
BF3
AX4
CCl4
AX5
PCl5
AX6
SF6
AX4
XeF4
AX
AX2
AX3
AX3
AX4
AX5
HF, HCl
H2O, SO2
NH3
BrF3
SF4
BrF5
NOTE: X is the same atom on all positions
Chem 110 Chap 9
Page: 20
Jensen
Chem 110 Chap 9
Page: 21