chapter 1. density – si derived unit for density is kg/m 3 1 g/cm 3 = 1 g/ml = 1000 kg/m 3 density...

Chapter 1

Density – SI derived unit for density is kg/m3

1 g/cm3 = 1 g/mL = 1000 kg/m3

density = mass

volumed = m

V

K = 0C + 273.15

0F = x 0C + 3295

1.7

273 K = 0 0C 373 K = 100 0C

32 0F = 0 0C 212 0F = 100 0C

0C = x (0F – 32)95

Kelvin is the SI Unit of temperature: absolute temperature scale. 0K is the lowest temperature that can be achieved theoretically.

Significant Figures- The meaningful digits in a measured or calculated quantity- The last digit is uncertain; 6.0±0.1ml

1.8

• Any digit that is not zero is significant

1.234 kg 4 significant figures• Zeros between nonzero digits are significant

606 m 3 significant figures• Zeros to the left of the first nonzero digit are not significant

0.08 L 1 significant figure• If a number is greater than 1, then all zeros to the right of the decimal point are significant

2.0 mg 2 significant figures• If a number is less than 1, then only the zeros that are at the end and in the middle of the number are significant

0.00420 g 3 significant figures

Significant Figures

1.8

Addition or SubtractionThe answer cannot have more digits to the right of the decimalpoint than any of the original numbers.

89.3321.1+

90.432 round off to 90.4

one significant figure after decimal point

3.70-2.9133

0.7867

two significant figures after decimal point

round off to 0.79

Answer can have only as many decimal places (DP)as the number in the operation with the least number of decimal places:

3DP1DP

1DP

2DP4DP2DP

1.8

Multiplication or DivisionThe number of significant figures in the result is set by the original number that has the smallest number of significant figures

Significant Figures

4.51 x 3.6666 = 16.536366 = 16.5

3 sig figs round to3 sig figs

6.8 ÷ 112.04 = 0.0606926

2 sig figs round to2 sig figs

= 0.061

3 sig figs

2 sig figs

Chapter 2

Atomic number (Z) = number of protons in nucleus

Mass number (A) = number of protons + number of neutrons

= atomic number (Z) + number of neutrons

number of protons = number of electrons

XAZ

Mass Number

Atomic NumberElement Symbol

2.3

Atomic number, Mass number and Isotopes

2.3

The Isotopes of Hydrogen

U23592 U238

92

Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei

2.7

Steps of naming ionic and binary molecular compounds

Naming acids and base

Chapter 3

Natural lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

7.42 x 6.015 + 92.58 x 7.016100 = 6.941 amu

3.1

Average atomic mass of lithium:

Average atomic mass- average mass of naturally occurring mixture of isotopes.

Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

SO2

1S 32.07 amu2O + 2 x 16.00 amu SO2 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2 3.3

1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu

1 12C atom12.00 amu

x 12.00 g6.022 x 1023 12C atoms

=1.66 x 10-24 g

1 amu

3.2

M = molar mass in g/mol

NA = Avogadro’s number

Relationship between amu and grams

Percent composition by mass is the percent of each element in a compound.

Percent composition of compounds

3.5

3.5

Percent Composition and Empirical Formulas

How to “Read” Chemical Equations

2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgOIS NOT

2 grams Mg + 1 gram O2 makes 2 g MgO

3.7

2 Mg(s) + O2 (g) 2 MgO(s)

2 HgO(s) O2 (g) + 2Hg(l)

Indicate physical state

Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

C2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

3.7

2C2H6 NOT C4H12

Balancing Chemical Equations3. Start by balancing those elements that appear only once on each side of the

equation and balance them.

C2H6 + O2 CO2 + H2O

3.7

start with C or H but not O

2 carbonon left

1 carbonon right multiply CO2 by 2

C2H6 + O2 2CO2 + H2O

6 hydrogenon left

2 hydrogenon right multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O

Balancing Chemical Equations

4. Balance those elements that appear in two or more reactants or products.

3.7

2 oxygenon left

4 oxygen(2x2)

C2H6 + O2 2CO2 + 3H2O

+ 3 oxygen(3x1)

multiply O2 by 72

= 7 oxygenon right

C2H6 + O2 2CO2 + 3H2O72

remove fractionmultiply both sides by 2

2C2H6 + 7O2 4CO2 + 6H2O

Balancing Chemical Equations

5. Check to make sure that you have the same number of each type of atom on both sides of the equation.

3.7

2C2H6 + 7O2 4CO2 + 6H2O

Reactants Products

4 C

12 H

14 O

4 C

12 H

14 O

4 C (2 x 2) 4 C

12 H (2 x 6) 12 H (6 x 2)

14 O (7 x 2) 14 O (4 x 2 + 6)

1. Write balanced chemical equation

2. Convert quantities of known substances into moles

3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity

4. Convert moles of sought quantity into desired units

Amounts of Reactants and Products

3.8

Limiting ReagentsThe reactant used up first in a reaction is called limiting reagent.

3.9

2NO + 2O2 2NO2

7 mole NO and 8 mole O2

NO is the limiting reagent

O2 is the excess reagent

The maximum amount of products formed depends on the amount of the limiting reagent.

Chapter 4

Molecular equation, Ionic equation and net Ionic equation

molecular equation

ionic equation

net ionic equation

Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3

-

Na+ and NO3- are spectator ions: ions that are not involved in the overall reaction

Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq)

precipitate

Pb2+ + 2I- PbI2 (s)

4.2

Writing Net Ionic Equations1. Write the balanced molecular equation.2. Write the ionic equation showing the strong electrolytes completely dissociated into

cations and anions. Weak and non electrolytes are written as molecules3. Cancel the spectator ions on both sides of the ionic equation4. Check that charges and number of atoms are balanced in the net ionic equation

AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)

Ag+ + NO3- + Na+ + Cl- AgCl (s) + Na+ + NO3

-

Ag+ + Cl- AgCl (s) 4.2

Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

Oxidation number

The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.

1. Free elements (uncombined state) have an oxidation number of zero.

Na, Be, K, Pb, H2, O2, P4 = 0

2. In monatomic ions, the oxidation number is equal to the charge on the ion.

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -23. The oxidation number of oxygen is usually –2. In H2O2 and O2

2- it is –1.

4.4

4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds (e.g. LiH, CaH2). In these cases, its oxidation number is –1.

6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. For example,NH4

+ , the sum of oxidation numbers is -3+4(+1)=+1

5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.

HCO3-

O = -2 H = +1

3x(-2) + 1 + ? = -1

C = +4

Oxidation numbers of all the elements in HCO3

- ?

4.4

7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O2

-, is -½.

5. Disproportionation reaction

The Activity Series for Metals

The Activity Series* Arranges metals according to their ease of oxidation*The higher the metal on the Activity Series, the more active that metal (the easier it is oxidized.)Any metal can be oxidized by the metal ions below it.*Any metal above hydrogen will displace it from water or from an acid.

3. Halogen Displacement

F2 > Cl2 > Br2 > I2

F2 is the greatest oxidizing halogenI2 is the least oxidizing halogen

Example: -1 0 -1 02 Br- + Cl2 --> 2 Cl- + Br2

Br2 + Cl- --> no reaction (NR)

Solution StoichiometryThe concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

4.5

M = molarity =moles of solute

liters of solution=

mole

liters

n

V =

• Read 2.5 M NaCl as 2.5 molar sodium chlorideOften use square bracket [ ] to indicate the concentration

Example:What is the molarity of a solution made from 4.00 g of NaOH diluted to a final volume of 250 mL?First find the number of moles of NaOH.Then divide by the volume in Liters.

Chapter 5

Ideal Gas Equation

5.4

Charles’ law: V T(at constant n and P)

Avogadro’s law: V n(at constant P and T)

Boyle’s law: V (at constant n and T)1P

V nTP

V = constant x = RnTP

nTP

R is the gas constant

PV = nRT

Ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal gas equation.

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PVnT

=(1 atm)(22.414L)(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)=8.314J/(K·mol) = 8.314 L·kPa/(K·mol)

5.4

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

Molar volume of gas• 1 mole of gas at STP = 22.4 Liters• Example:2 moles of gas at STP = 44.8 L

In calculation, the units of R must match those for P,V,T and n.

R

P1V1

n1T1R=

The combined gas law

P2V2

n2T2

P1V1

n1T1=

P2V2

n2T2R=

If n1=n2

P2V2

T2

P1V1

T1=

Dalton’s Law of Partial PressuresPartial pressure is the pressure of the individual gas in the mixture.

V and T are constant

P1 P2 Ptotal = P1 + P2

5.6

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nBXB =

nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT

5.6

mole fraction (Xi) = ni

nT

Chapter 6

Another form of the first law for Esystem

6.3

E = q + w

E is the change in internal energy of a system

q is the heat exchange between the system and the surroundings

w is the work done on (or by) the system

w = -PV when a gas expands against a constant external pressure

H2O (s) H2O (l) H = 6.01 kJ/mol

• The physical states of all reactants and products must be specified in thermochemical equations.

Thermochemical Equations

6.4

H2O (l) H2O (g) H = 44.0 kJ/mol

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ/mol

266 g P4

1 mol P4

123.9 g P4

x 3013 kJ1 mol P4

x = 6470 kJ

Consider the reaction below: 2CH3OH(l) + 3O2(g) 4H2O(l) + 2CO2(g) ∆H = -1452.8 kJ/mol

What is the value of ∆H for the reaction of 8H2O(l) + 4CO2(g) 4CH3OH (l) + 6O2 (g)?

If you reverse a reaction, the sign of ΔH changes. If you multiply both sides of the equation by a factor n, then ∆H must change by the same factor n.∆H = 2*(+1452.8 kJ/mol) = +2905.6 KJ/mol

Practice problem of chap6:Q2

The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

C = m x s

Heat (q) absorbed or released:

q = m x s x t

q = C x t

t = tfinal - tinitial

6.5

Calorimetry- the measurement of heat.

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?

Establish an arbitrary scale with the standard enthalpy of formation (H0) as a reference point for all enthalpy expressions.

f

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. “0”-standard state at 1atm; “f”-formation

f

The standard enthalpy of formation of any element in its most stable form is zero.

H0 (O2) = 0f

H0 (O3) = 142 kJ/molf

H0 (C, graphite) = 0f

H0 (C, diamond) = 1.90 kJ/molf

6.6

Standard enthalpy of formation and reaction

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atm.rxn

aA + bB cC + dD

H0

rxndH0 (D)

fcH0 (C)

f= [ + ] - bH0 (B)

faH0 (A)

f[ + ]

H0

rxnnH0 (products)

f= mH0 (reactants)

f-

6.6

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

Chapter 7

Maxwell (1873), proposed that visible light consists of electromagnetic waves.– provides a mathematical description of the general behavior of light.

Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves, which has an electrical field component and a magnetic field component .

Speed of light (c) in vacuum = 3.00 x 108 m/s

All electromagnetic radiation x c

7.1

Energy of light:Quantum = packet of energyPhoton = packet of light• E = hPlanck’s constant (h)h = 6.63 x 10-34 J•s

Ephoton = E = Ef - Ei

Ef = -RH ( )1n2

f

Ei = -RH ( )1n2

i

i fE = RH ( )1

n2

1n2

nf = 1

ni = 2

nf = 1

ni = 3

nf = 2

ni = 3

7.3

Transition between energy states

if ni > nf, then emission occurs

= h x

Quantum Mechanical Description of the distribution of electrons in the atom:• 4 quantum numbers: n (principal quantum number)l (angular momentum quantum number)ml (magnetic quantum number)ms (spin quantum number)

Allowed values:• n = 1,2,3 …,n• l = 0,1,2,…,(n-1)• ml = (-l, …, 0. …+l)• ms = +1/2, -1/2

Electron Configurations

Periods 1, 2, and 3Three rules:1.Electrons fill orbitals starting with lowest n and moving upwards (Aufbau principle: Fill up electrons in lowest energy orbitals )2. No more than two electrons can be placed in each orbital.No two electrons can fill one orbital with the same spin (Pauli exclusion principle: no two electrons in an atomcan have the same four quantum numbers.)3. For degenerate orbitals, electrons fill each orbital singly before any orbital gets a second electron (Hund’s rule: The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins )

Period 4 and Beyondthe d orbitals begin to fill

Chapter 8

Electron Configurations of Cations and Anions

Na [Ne]3s1Na+ [Ne]

Ca [Ar]4s2Ca2+ [Ar]

Al [Ne]3s23p1Al3+ [Ne]

Atoms lose electrons so that cation has a noble-gas outer electron configuration.

H 1s1H- 1s2 or [He]

F 1s22s22p5F- 1s22s22p6 or [Ne]

O 1s22s22p4O2- 1s22s22p6 or [Ne]

N 1s22s22p3N3- 1s22s22p6 or [Ne]

Atoms gain electrons so that anion has a noble-gas outer electron configuration.

Of Representative Elements

8.2

Remember: electrons are first removed from orbitals with the highest principal quantum number.

Electron Configurations of Cations of Transition Metals

8.2

When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals.

Fe: [Ar]4s23d6

Fe2+: [Ar]4s03d6 or [Ar]3d6

Fe3+: [Ar]4s03d5 or [Ar]3d5

Mn: [Ar]4s23d5

Mn2+: [Ar]4s03d5 or [Ar]3d5

Periodic Variation in Physical Properties

8.3

Cation is always smaller than atom from which it is formed. This is because the nuclear charge remains the same but the reduced electron repulsion resulting from removal of electrons make the electron clouds shrink.Anion is always larger than atom from which it is formed. This is because the nuclear charge remains the same but electron repulsion resulting from the additional electron enlarges the electron clouds. 8.3

Ionic Radius

•From top to bottom, both the atomic and ionic radius increase within group.•Across a period the anions are usually larger than cations.• For ions derived from different groups, size comparison is meaningful only if the ions are isoelectronic.E.g. Na+ (Z=11) is smaller than F- (Z=9).The larger effective charge results in a smaller radius.

Radius of tripositive ions < dipositive ions<unipositive ionsAl3+ <Mg2+ <Na+

Radius of uninegative ions < dinegative ionsO2->F-

General Trend in First Ionization Energies

8.4

Increasing First Ionization Energy

Incr

easi

ng F

irst I

oniz

ation

Ene

rgy

8.5

Variation of Electron Affinity With Atomic Number (H – Ba)

Overall trend: increases from left to right across the period. The values varied little in the group. The EA of metals are generally lower than those of nonmetals.

Chapter 9

The Octet RuleAtoms tend to gain, lose, or share electrons until theyare surrounded by 8 valence electrons.An octet = 8 electrons

= 4 pairs of electrons = noble gas arrangement = very stable

Exception: H = 2 electrons only

Example:Na [Ne]3s1 loses 1 e----> Na+ [Ne]Cl [Ne]3s23p5 gains 1 e- --> Cl- [Ar]

1. Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. Arrange atoms & connect with single bonds

– Single bond = 2 electrons

2. Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge.

3. Complete an octet for atoms bonded to the central atom except hydrogen (H=2). Then add leftover electrons to central atom. Electrons belonging to the central or surrounding atoms must be shown as lone pairs if they are not involved in bonding.

4. If the central atom has fewer than 8 electrons, try adding double and triple bonds on central atom as needed.

Writing Lewis Structures

9.6

9.7

An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.

formal charge on an atom in a Lewis structure

=1

2

total number of bonding electrons( )

total number of valence electrons in the free atom

-total number of nonbonding electrons

-

The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

FC = [ Val. e-] -[(nonbonded e-) + 1/2(bonded e-)]

• Remember: formal charges are NOT real charges

Formal Charge and Lewis Structures

9.7

1. For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.

2. Lewis structures with large formal charges are less plausible than those with small formal charges.

3. Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.

Which is the most likely Lewis structure for CH2O?

H C O H

-1 +1H

C OH

0 0

Exceptions to the Octet Rule

Odd-Electron Molecules

Result: unpaired electron

N – 5e-

O – 6e-

11e-

NO N O

The Expanded Octet (too many e- on central atom; central atom with principal quantum number n > 2)

SF6

S – 6e-

6F – 42e-

48e-

S

F

F

F

FF

F

6 single bonds (6x2) = 1218 lone pairs (18x2) = 36

Total = 48

9.9

Which of the following is correct concerning the lone pairs on the underlined atoms in compounds? (a)ICl, 3 (b) H2S, 6 (c) CH4, 4(d) CaH2, 2 (e) SCl2, 4

Practice problem of chap9:Q3,9

Which of the following obeys the octet rule?(a)BF3 (b) SF4 (c) NO(d) PF5 (e) NO3

-

F B F

F

Bond Energies (BE) and Enthalpy changes in reactions

H0 = total energy input – total energy released

= BE(reactants) – BE(products)

Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.

9.10

Endothermic reaction Exothermic

reaction

Chapter 10

10.1

Molecules in which the central atom has no lone pairs

No Lone Pairs on Central Atom

• Arrangement of electron domains is very symmetrical• The molecular geometry is the same as the electron domain geometry

Steps to Assign Molecular Geometry

• Draw Lewis structure• Count # of electron domains around the Central atom(treat double and triple bonds as though there were single bond)• Assign electron domain geometry• Finally assign the molecular shape

– based on the arrangement of the atoms– not the arrangement of the domains

No Lone Pairs on Central Atom

• Arrangement of electron domains is very symmetrical• The molecular geometry is the same as the electron domain geometry

10.1

Molecules containing net dipole moments are called polar molecules. Otherwise they are called nonpolar molecules because they do not have net dipole moments.

Diatomic molecules: Determined by the polarity of bondPolar molecules:HCl, CO,NONonpolar molecules: H2,F2,O2

Molecules with three or more atoms: determined by the polarity of the bond and the molecular geometry

Dipole moment is a vector quantity, which has both magnitude and direction.

Chloroform, CCl3H, isasymmetric. The vectorsum of the bond dipolesis nonzero giving themolecule a net dipole.

• A molecule will be nonpolar if: (a) the bonds are nonpolar, or (b) there are no lone pairs in the valence shell of thecentral atom and all the atoms attached to the central atom are the same

10.4

# of Lone Pairs+

# of Bonded Atoms Hybridization Examples

2

3

4

5

6

sp

sp2

sp3

sp3d

sp3d2

BeCl2

BF3

CH4, NH3, H2O

PCl5

SF6

How do I predict the hybridization of the central atom?

1. Draw the Lewis structure of the molecule.2. Count the number of lone pairs AND the number of atoms

bonded to the central atom. Predict the overall arrangement of electron pairs using VSEPR model ( Table 10.1)

3. Deduce hybridization of central atom by matching the arrangement of the electron pairs with those of hybrid orbitals shown in Table 10.4.

10.4

1. The number of molecular orbitals (MOs) formed is always equal to the number of atomic orbitals combined.

2. The more stable the bonding MO, the less stable the corresponding antibonding MO.

3. The filling of MOs proceeds from low to high energies.

4. Each MO can accommodate up to two electrons.

5. Use Hund’s rule when adding electrons to MOs of the same energy.(Electrons spread out as much as possible, with spins unpaired, over orbitals that have the same energy)

6. The number of electrons in the MOs is equal to the sum of all the electrons on the bonding atoms.

10.7

Molecular Orbital (MO) Configurations-MOs follow the same filling rules as atomic orbitals

The approximate relative energies of molecular orbitals insecond period diatomic molecules. (a) Li2 through N2, (b) O2through Ne2.

10.7

bond order = 1

2

Number of electrons in bonding MOs

Number of electrons in antibonding MOs( - )

10.7

bond order½ 1 0½

Chapter 11

Types of Intermolecular Forces (IMF)• Should actually be called Interparticulate Forces (molecules, ions, and/or atoms)• Ion - ion forces• Ion-dipole forces• Dipole-dipole forces• Dispersion Forces• Hydrogen Bonds

Ion - ion forces: (lattice energy-ionic compound)• Remember Chapter 9• Force depends on the charge on the ions andthe distance separating the ions• (STRONG FORCES)

E = k Q+Q-r

11.4

1 atom/unit cell

(8 x 1/8 = 1)

2 atoms/unit cell

(8 x 1/8 + 1 = 2)

4 atoms/unit cell

(8 x 1/8 + 6 x 1/2 = 4)

11.4

A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas.

Phase Diagram of Water

11.9

phase diagram