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Week 5: Lectures 13 – 15

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]

Jensen Chem 110 Chap3 Page: 2

Chemical Formulas

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.


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


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 + O2 ! 2 H2O

! Changing the subscripts changes the _________

of the molecule.

! Changing the coefficient changes the _________

of that molecule, not the identity.


Working with Formulas

Example: How many moles of ammonium ions

are present in 30.9 g of ammonium sulfate?

A. 0.468 mol

B. 0.288 mol

C. 0.234 mol

D. 2.14 mol

E. 3.47 mol


Working with Formulas

Practice Example:

Which of the following has the greatest

number of total atoms present?

A. 6.022 x 1023 O2 molecules

B. 0.1 mol C6H12O6

C. 56 g Fe

D. 0.4 mol CH4

E. 4.4 g of H2


Percent Composition: Mass Percent

contributed by each element in compound

(Note: percentage must add to 100)

1. Given chemical formula,

!

% element =(# of element) (AW of element)

FW of compound"100

Example: What is the mass percent of C in CO2?

C= _____ amu; O = _____ amu; CO2 = _______ amu

%C =

2. If we have 100 g of sample,

!

% element =mass of element

100 g compound"100

Example: I have 2g of a sample that is 27.3% C by mass, how many grams of C are in the sample?


Practice Example:

How many grams of oxygen are in 65 g of

C2H2O2?

A) 18 g

B) 29 g

C) 9.0 g

D) 36 g

E) 130 g


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.


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. C4H8O2

B. C2H4O

C. C5H4O

D. C4H2O2

E. C4H16O


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

smallest)

Empirical Formula:

4. Get molecular formula from empirical formula


Combustion Analysis

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

1. Use grams of CO2 and H2O to find moles of C- and H- elements in the products and in the sample;

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)

Fuel C, H, O + O2 CO2 + H2O


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?

a) C4H9O

b) C2H6O

c) C3H6O2

d) CH3O

e) CH2O


Solve:

1. Use grams of CO2 and H2O to calculate moles of C, H elements in products and in the sample

C = _____________________________________ mol

H = _____________________________________ mol

2. Calculate mass of C, H, and O elements and moles of O elements in the sample

C = ______________________________________ g

H = ______________________________________ g O = ______________________________________ g

O = _____________________________________ mol

3. Find ratio of moles of elements (divide by

smallest)

Empirical Formula:

What’s Next?

Chemical Formulas tell us how atoms are connected to each other

Lewis structures tell us where the electrons are.

Valence Shell Electron Pair Repulsion (VSEPR): Predict molecular shape (spatial arrangement of atoms) from Lewis structure

Lewis Structures (2D)

Bonding e— pairs Lone e— pairs

Molecular Shape (3D):

Arrangement of Atoms in the Molecule

Jensen Chem 110 Chap 9 Page: 1

VSEPR Model

Electron Domain Geometry (EDG)

Electron pairs " Electron domains

One lone pair = one electron domain

One bond (single, double or triple bond)

= one electron domain.

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


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


1. Linear EDG (2 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)


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)

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= _______________


Lone Pairs and Bond Angles

! Nonbonding pairs are physically larger

than bonding pairs

! Therefore, their repulsions are greater;

this tends to _________ bond angels

on the other side of the molecule

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

the molecule are ___________


4. Trigonal bipyramidal EDG (5 electron domains)

• Two types of sites –_______and ________ • Angles between sites are different for

these two types.

Lone pairs prefer to be equatorial.

The Trigonal Bipyramidal Case

CH110 FA11 SAS 6

• Two types of sites – _____ and ________

• Angles between sites are different for

these two types.

• Lone pairs prefer to be equatorial. Why?

Ax

Ax

A

Eq

Eq

Eq 1200

900

900


4. Trigonal bipyramidal EDG (5 electron domains)

Four possible MG for trigonal bipyramidal EDG:

! Trigonal bipyramidal

! Seesaw

! T-shaped

! Linear


5. Octahedral EDG (6 electron domains)

• All positions are equivalent

• Two lone pairs go on ___________ sides of the

molecule

Three possible MG for octahedral EDG:

! Octahedral

! Square pyramidal

! Square planar


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?

A. see-saw

B. octahedral

C. square pyramidal

D. square bipyramidal

E. square planar


Practice Example

The molecular geometry of which molecule is

square planar?

A. CCl4

B. XeF4

C. PH3

D. XeF2

E. BrF3


VSEPR can be applied to Larger Molecules

Example: What is the approximate C—C—O

bond angel as indicated?

A. 90°

B. 105°

C. 109°

D. 120°

E. 180°

! Describe geometry around

a particular atom


Determine Approximate Bond Angel

Practice Example: Determine the approximate

bond angles indicated.

Angle #1 Angle #2

A. 109° 109°

B. 180° 120°

C. 120° 90°

D. 109° 120°

E. 120° 109°

NCH2CH2CH2C

H

H

O

OH

1 2


Molecular Geometries: Summary

# of electron domains

# of bonded atoms

molecular geometry

Example

2 2 linear CO2

3 3 trigonal planar BF3

3 2 bent O3

4 4 tetrahedral CH4

4 3 trigonal

pyramidal NH3

4 2 bent H2O

5 5 trigonal

bipyramidal PCl5

5 4 seesaw SF4

5 3 T-shaped BrF3

5 2 linear XeF2

6 6 octahedral SF6

6 5 square

pyramidal BrF5

6 4 square planar XeF4


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)?

BP of H2O = 100 ˚C; BP of CH4 = -161 ˚C

• Why do alcohols and water mix but oil and water don!t?


Bond Polarity

The polarity of individual bond is determined

by electronegativity difference between

elements; The greater the difference, the

more polar the bond.

• Most polar: ionic compound, e.g. NaF

• Polar covalent: (two different atoms)

$+ and $— are partial charges for H and F

• Nonpolar covalent: (two identical atoms); e.g. H2, F2, Cl2, O2,


Molecular Polarity

• Just because a molecule has polar bonds does

not mean the molecule as a whole will be polar.

Polarity of entire molecule: Dipole Moment (µ)

For a diatomic molecule: µ = Q r

H r Cl units = debye (D) •%%%% •

+Q &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

To determine the polarity of a molecule that has more than two atoms:

1. Find molecular shape using VSEPR

2. Find “bond” dipoles

3. Use vector analysis to find NET molecular dipole


Examples:

CO2

(electronegativity: O > C)

EDG: Linear

MG: Linear

Bond dipoles?

Net dipole moment?

H2O

(electronegativity: O > H)

EDG: tetrahedral

MG: bent

Bond dipoles?

Net dipole moment?


Examples:

NH3 (electronegativity: N > H)

EDG: tetrahedral

MG: trigonal pyrimidal

Bond dipoles?

Net dipole moment?

CF4 (electronegativity: F > C)

EDG: tetrahedral

MG: tetrahedral

Bond dipoles?

Net dipole moment?

F

CF

F

F

N

H

HH


Example: Which molecule below is polar?

Practice Example: Which molecule shown

below has non-zero dipole moment?

A. XeF2

B. BF3

C. SF4

D. IF2–

E. BeCl2


Molecular Geometry & Molecular Polarity

Do not memorize this table!! Draw the generic structures and think about them to verify!

MG Formula Example Polar? linear AX2 CO2

trigonal planar AX3 BF3

tetrahedral AX4 CCl4

trigonal bipyramidal AX5 PCl5

octahedral AX6 SF6

square planar AX4 XeF4

linear AX HF, HCl

bent AX2 H2O, SO2

trigonal pyramidal AX3 NH3

T-shaped AX3 BrF3

seesaw AX4 SF4

square pyramidal AX5 BrF5

NOTE: X is the same atom on all positions

“no score” from exam 1?? not a problem—go see mike ...courses.chem.psu.edu/chem110fall/lecture...

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