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Dr. Mohamed Abdel Salam

1Dr. Mohamed Abdel Salam

A mixture is a combination of two or more substances in which the substances retain their distinct identities.

1. Homogenous mixture – composition of the mixture is the same throughout.

2. Heterogeneous mixture – composition is not uniform throughout.

soft drink, milk, solder

cement, iron filings in sand

1 2Dr. Mohamed Abdel Salam

Classifications of Matter

2


A physical change does not alter the composition or identity of a substance.

A chemical change alters the composition or identity of the substance(s) involved.

ice meltingsugar dissolving

in water

hydrogen burns in air to form water

Physical or Chemical?

34


International System of Units (SI)

4




Density – SI derived unit for density is kg/m3

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

density = mass

volume d =mV

A piece of platinum metal with a density of 21.5 g/cm3 has a volume of 4.49 cm3. What is its mass?

d =mV

m = d x V = 21.5 g/cm3 x 4.49 cm3 = 96.5 g

6


K = 0C + 273.15

0F = x 0C + 3295

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

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


Convert 172.9 0F to degrees Celsius.

0F = x 0C + 3295

0F – 32 = x 0C95

x (0F – 32) = 0C95

0C = x (0F – 32)95

0C = x (172.9 – 32) = 78.395

8


9Dr. Mohamed Abdel Salam 1.8

Scientific Notation

The number of atoms in 12 g of carbon:

602,200,000,000,000,000,000,000

6.022 x 1023

The mass of a single carbon atom in grams:

0.0000000000000000000000199

1.99 x 10-23

N x 10n

N is a number between 1 and 10

n is a positive or negative integer


Scientific Notation

1.8

568.762

n > 0

568.762 = 5.68762 x 102

move decimal left

0.00000772

n < 0

0.00000772 = 7.72 x 10-6

move decimal right

Addition or Subtraction

1. Write each quantity with the same exponent n

2. Combine N1 and N2

3. The exponent, n, remains the same

4.31 x 104 + 3.9 x 103 =

4.31 x 104 + 0.39 x 104 =

4.70 x 104

10


Scientific Notation

1.8

Multiplication

1. Multiply N1 and N2

2. Add exponents n1 and n2

(4.0 x 10-5) x (7.0 x 103) =(4.0 x 7.0) x (10-5+3) =

28 x 10-2 =2.8 x 10-1

Division

1. Divide N1 and N2

2. Subtract exponents n1 and n2

8.5 x 104 ÷ 5.0 x 109 =(8.5 ÷ 5.0) x 104-9 =

1.7 x 10-5


Significant Figures

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

12



How many significant figures are in each of the following measurements?

24 mL 2 significant figures

3001 g 4 significant figures

0.0320 m3 3 significant figures

6.4 x 104 molecules 2 significant figures

560 kg 2 significant figures


Significant Figures

Addition or Subtraction

The 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

14


Significant Figures

Multiplication or Division

The number of significant figures in the result is set by the original number that has the smallest number of 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


Significant Figures

Exact Numbers

Numbers from definitions or numbers of objects are consideredto have an infinite number of significant figures

The average of three measured lengths; 6.64, 6.68 and 6.70?

6.64 + 6.68 + 6.70

3= 6.67333 = 6.67

Because 3 is an exact number

= 7

16



Accuracy – how close a measurement is to the true value

Precision – how close a set of measurements are to each other

accurate&

precise

precisebut

not accurate

not accurate&

not precise


Dimensional Analysis Method of Solving Problems

Conversion Unit 1 L = 1000 mL

1L

1000 mL1.63 L x = 1630 mL

1L

1000 mL1.63 L x = 0.001630

L2

mL

How many mL are in 1.63 L?

18


The speed of sound in air is about 343 m/s. What is this speed in miles per hour?

1 mi = 1609 m 1 min = 60 s 1 hour = 60 min

343ms

x1 mi

1609 m

60 s

1 minx

60 min

1 hourx = 767

mi

hour

meters to miles

seconds to hours

conversion units


mass p = mass n = 1840 x mass e-

2.220



Atomic number (Z) = number of protons in nucleus

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

= atomic number (Z) + number of neutrons

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

XAZ

H11 H (D)

21 H (T)

31

U23592 U238

92

Mass Number

Atomic NumberElement Symbol

2.3

Atomic number, Mass number and Isotopes

2122


6 protons, 8 (14 - 6) neutrons, 6 electrons

6 protons, 5 (11 - 6) neutrons, 6 electrons

Do You Understand Isotopes?

2.3

How many protons, neutrons, and electrons are in C14

6 ?

How many protons, neutrons, and electrons are in C11

6 ?

22


An ion is an atom, or group of atoms, that has a net positive or negative charge.

cation – ion with a positive chargeIf a neutral atom loses one or more electronsit becomes a cation.

anion – ion with a negative chargeIf a neutral atom gains one or more electronsit becomes an anion.

Na11 protons

11 electrons Na+ 11 protons

10 electrons

Cl17 protons17 electrons Cl-

17 protons

18 electrons

2.52324


13 protons, 10 (13 – 3) electrons

34 protons, 36 (34 + 2) electrons

Do You Understand Ions?

2.5

How many protons and electrons are in ?Al2713

3+

How many protons and electrons are in ?Se7834

2-

24



A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance

An empirical formula shows the simplest whole-number ratio of the atoms in a substance

H2OH2O

molecular empirical

C6H12O6 CH2O

O3 O

N2H4 NH2

2.62526


Formula of Ionic Compounds

Al2O3

2.6

2 x +3 = +6 3 x -2 = -6

Al3+ O2-

CaBr2

1 x +2 = +2 2 x -1 = -2

Ca2+ Br-

Na2CO3

1 x +2 = +2 1 x -2 = -2

Na+ CO32-

26


Natural lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

7.42 x 6.015 + 92.58 x 7.016

100= 6.941 amu

3.1

Average atomic mass of lithium:


x6.022 x 1023 atoms K

1 mol K=

Do You Understand Molar Mass?

How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K

1 mol K = 6.022 x 1023 atoms K

0.551 g K1 mol K

39.10 g Kx

8.49 x 1021 atoms K

3.228



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

SO2

1S 32.07 amu

2O + 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

3.329 30


Do You Understand Molecular Mass?

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

3.3

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =

30


Do You Understand Formula Mass?

What is the formula mass of Ca3(PO4)2 ?

3.3

1 formula unit of Ca3(PO4)2

3 Ca 3 x 40.08

2 P 2 x 30.97

8 O + 8 x 16.00

310.18 amu


Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 moleof the compound

C2H6O

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

3.532



Percent Composition and Empirical Formulas

Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent.

nK = 24.75 g K x = 0.6330 mol K1 mol K

39.10 g K

nMn = 34.77 g Mn x = 0.6329 mol Mn1 mol Mn

54.94 g Mn

nO = 40.51 g O x = 2.532 mol O1 mol O

16.00 g O

33 34Dr. Mohamed Abdel Salam 3.5

Percent Composition and Empirical Formulas

K : ~~ 1.00.6330

0.6329

Mn : 0.6329

0.6329= 1.0

O : ~~ 4.02.532

0.6329

nK = 0.6330, nMn = 0.6329, nO = 2.532

KMnO4

34


g CO2 mol CO2 mol C g C

g H2O mol H2O mol H g H

g of O = g of sample – (g of C + g of H)

Combust 11.5 g ethanol

Collect 22.0 g CO2 and 13.5 g H2O

6.0 g C = 0.5 mol C

1.5 g H = 1.5 mol H

4.0 g O = 0.25 mol O

Empirical formula C0.5H1.5O0.25

Divide by smallest subscript (0.25)

Empirical formula C2H6O35 36


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 C4H1236




3. Start by balancing those elements that appear in only one reactant and one product.

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 + 3H2O37 38



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

38



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

3.840



Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH1 mol CH3OH

32.0 g CH3OHx

4 mol H2O

2 mol CH3OHx

18.0 g H2O

1 mol H2Ox =

235 g H2O

3.841 42


Limiting Reagents

3.9

2NO + 2O2 2NO2

NO is the limiting reagent

O2 is the excess reagent

42


Do You Understand Limiting Reagents?

In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

OR

g Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al

27.0 g Alx

1 mol Fe2O3

2 mol Alx

160. g Fe2O3

1 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent

3.943 44


Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102. g Al2O3

1 mol Al2O3

x = 234 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

3.944



Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtainedfrom a reaction.

% Yield = Actual Yield

Theoretical Yieldx 100

3.10

Reaction Yield


λ x ν = c

λ = c/ν

λ = 3.00 x 108 m/s / 6.0 x 104 Hz

λ = 5.0 x 103 m

Radio wave

A photon has a frequency of 6.0 x 104 Hz. Convertthis frequency into wavelength (nm). Does this frequency

fall in the visible region?

λ = 5.0 x 1012 nm

λλλλ

νννν

7.146


E = h x ν

E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)

E = 1.29 x 10 -15 J

E = h x c / λ

7.2

When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules)

associated with the photons if the wavelength of the X rays is 0.154 nm.


Ephoton = ∆E = Ef - Ei

Ef = -RH ( )1

n2f

Ei = -RH ( )1

n2i

i f

∆E = RH( )1

n2

1

n2

nf = 1

ni = 2

nf = 1

ni = 3

nf = 2

ni = 3

7.348



Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)

Ephoton = ∆E = -1.55 x 10-19 J

λ = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J

λ = 1280 nm

Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state.

Ephoton = h x c / λ

λ = h x c / Ephoton

i f

∆E = RH( )1

n2

1

n2Ephoton =

7.349 50


λ = h/mu

λ = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)

λ = 1.7 x 10-32 m = 1.7 x 10-23 nm

What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball

traveling at 15.6 m/s?

m in kgh in J•s u in (m/s)

7.450



How many 2p orbitals are there in an atom?

2p

n=2

l = 1

If l = 1, then ml = -1, 0, or +1

3 orbitals

How many electrons can be placed in the 3d subshell?

3d

n=3

l = 2

If l = 2, then ml = -2, -1, 0, +1, or +2

5 orbitals which can hold a total of 10 e-

7.652



Order of orbitals (filling) in multi-electron atom

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

7.753 54


Electron configuration is how the electrons are distributed among the various atomic orbitals in an

atom.

1s1

principal quantumnumber n

angular momentumquantum number l

number of electronsin the orbital or subshell

Orbital diagram

H

1s1

7.854


What is the electron configuration of Mg?

Mg 12 electrons

1s < 2s < 2p < 3s < 3p < 4s

1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons

7.8

Abbreviated as [Ne]3s2 [Ne] 1s22s22p6

What are the possible quantum numbers for the last (outermost) electron in Cl?

Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s

1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons

Last electron added to 3p orbital

n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½ 55 56Dr. Mohamed Abdel Salam 7.8

56



57

Write the electronic configuration, the number of unpaired electrons, the

magnetic property, and the four quantum numbers for the differentiating

electron of the element Osmium (76Os).

76Os: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d6

+2 +1 0 -1 -2

Number of unpaired electrons: 4 e−−−−

Paramagnetic

n = 5, llll = 2, mllll= 2, ms = -1/2


58

Write the electronic configuration, the number of unpaired electrons, the

magnetic property, and the four quantum numbers for the differentiating

electron of the element Chromium(24Cr).

24Cr: 1s2 2s2 2p6 3s2 3p6 4s2 3d4 (unstable)

3d

+2 +1 0 -1 -2

Number of unpaired electrons: 6 e−−−−

Paramagnetic

n = 3, llll = 2, mllll= -2, ms = +1/2

24Cr : 1s2 2s2 2p6 3s2 3p6 4s1 3d5 (stable)

4s

0


59

4s1 3d1 3d2 3d3 3d4 3d5

n 4 3 3 3 3 3

llll 0 2 2 2 2 2

mllll 0 2 1 0 -1 -2

ms 1/2 1/2 1/2 1/2 1/2 1/2

Give a set of four quantum numbers for each unpaired electron

in Cr.


60

a- F b- Ne c- Na d- Ar

The Ca2+ has the same electronic configuration as one of the

following elements (Ar, Na, Ne, F):

The ground state of nitrogen atom (7N) is:



61

The electronic configuration of antimony atom (51Sb) is:

a- [Kr] 5s2 4d3 b- [Kr] 5s2 4d9 5p4

c- [Kr] 5s2 4d10 5p3 d- [Kr] 5s1 5p5

How many electrons have (l = 0) antimony atom (51Sb) for ?

a- 10 b- 11 c- 12 d- 13

How many unpaired electrons are in antimony atom (51Sb)?

a- 2 b- 3 c- 4 d- 5

The maximum number of electrons the fourth level (n = 4) is:

a- 8 b- 16 c- 18 d- 32


ns

1

ns

2

ns

2n

p1

ns

2n

p2

ns

2n

p3

ns

2n

p4

ns

2n

p5

ns

2n

p6

d1

d5

d10

4f

5f

Ground State Electron Configurations of the Elements


Electron Configurations of Cations and Anions

Na [Ne]3s1 Na+ [Ne]

Ca [Ar]4s2 Ca2+ [Ar]

Al [Ne]3s23p1 Al3+ [Ne]

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

H 1s1 H- 1s2 or [He]

F 1s22s22p5 F- 1s22s22p6 or [Ne]

O 1s22s22p4 O2- 1s22s22p6 or [Ne]

N 1s22s22p3 N3- 1s22s22p6 or [Ne]

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

Of Representative Elements

8.264


Na+: [Ne] Al3+: [Ne] F-: 1s22s22p6 or [Ne]

O2-: 1s22s22p6 or [Ne] N3-: 1s22s22p6 or [Ne]

Na+, Al3+, F-, O2-, and N3- are all isoelectronic with Ne

What neutral atom is isoelectronic with H- ?

H-: 1s2 same electron configuration as He

8.2



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



Cation is always smaller than atom from which it is formed.Anion is always larger than atom from which it is formed.

8.368


Ionization energy is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state.

I1 + X (g) X+(g) + e-

I2 + X+(g) X2+

(g) + e-

I3 + X2+(g) X3+

(g) + e-

I1 first ionization energy

I2 second ionization energy

I3 third ionization energy

8.4

I1 < I2 < I3



General Trend in First Ionization Energies

8.4

Increasing First Ionization Energy

Incre

asin

g F

irst Io

niz

atio

n E

ne

rgy


Electron affinity is the negative of the energy change that occurs when an electron is accepted by an atom in the gaseous state to form an anion.

X (g) + e- X-(g)

8.5

F (g) + e- X-(g)

O (g) + e- O-(g)

∆H = -328 kJ/mol EA = +328 kJ/mol

∆H = -141 kJ/mol EA = +141 kJ/mol


Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond.

Electron Affinity - measurable, Cl is highest

Electronegativity - relative, F is highest

X (g) + e- X-(g)

9.571 72

Dr. Mohamed Abdel Salam 9.5

The Electronegativities of Common Elements

72



Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons thatparticipate in chemical bonding.

1A 1ns1

2A 2ns2

3A 3ns2np1

4A 4ns2np2

5A 5ns2np3

6A 6ns2np4

7A 7ns2np5

Group # of valence e-e- configuration

73 74Dr. Mohamed Abdel Salam 9.1

Lewis Dot Symbols for the Representative Elements &Noble Gases

74


Li + F Li+ F -

The Ionic Bond

1s22s11s22s22p5 1s21s22s22p6[He][Ne]

Li Li+ + e-

e- + F F -

F -Li+ + Li+ F -


A covalent bond is a chemical bond in which two or more electrons are shared by two atoms.

Why should two atoms share electrons?

F F+

7e- 7e-

F F

8e- 8e-

F F

F F

Lewis structure of F2

lone pairslone pairs

lone pairslone pairs

single covalent bond

single covalent bond

9.476



8e-

H HO+ + OH H O HHor

2e- 2e-

Lewis structure of water

Double bond – two atoms share two pairs of electrons

single covalent bonds

O C O or O C O

8e- 8e-8e-double bonds double bonds

Triple bond – two atoms share three pairs of electrons

N N

8e-8e-

N N

triple bondtriple bond

or

9.477 78


H F FH

Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms

electron richregion

electron poor

region e- riche- poor

δ+ δ-

9.578


1. Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center.

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

3. Complete an octet for all atoms excepthydrogen

4. If structure contains too many electrons, form double and triple bonds on central atom as needed.

Writing Lewis Structures

9.679 80


Writing Lewis Formula: The Octat Rule:

In most of their compounds, the representative elements achieve noble gas

configurations. This is because noble gas have 8 electrons in their outermost

shells (except He, which has 2 electrons).

S: is the total number of shared electrons

N: is the total number of valence shell electrons needed to achieve octat.

N = 8 x number of atoms + 2 x number of hydrogen atoms (if applicable).

A: is the number of valence electrons available.

Note: we add electrons to account for negative charges and subtract electrons to

account for positive charge

S = N - A



Write the Lewis structure of nitrogen trifluoride (NF3).

Step 1 – N is less electronegative than F, put N in center

F N F

F

Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)

5 + (3 x 7) = 26 valence electrons

Step 3 – Draw single bonds between N and F atoms and completeoctets on N and F atoms.

Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

9.681 82


Write the Lewis structure of the carbonate ion (CO32-).

Step 1 – C is less electronegative than O, put C in center

O C O

O

Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)

-2 charge – 2e-

4 + (3 x 6) + 2 = 24 valence electrons

Step 3 – Draw single bonds between C and O atoms and completeoctet on C and O atoms.

Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

9.6

Step 5 - Too many electrons, form double bond and re-check # of e-

2 single bonds (2x2) = 4

1 double bond = 4

8 lone pairs (8x2) = 16

Total = 24

82


Two possible skeletal structures of formaldehyde (CH2O)

H C O HH

C OH

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.


H C O H

C – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-


1 double bond = 4


Total = 12

formal charge on C

= 4 -2 -½ x 6 = -1

formal charge

on O= 6 -2 -½ x 6 = +1

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

-

-1 +1

9.784



C – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-


1 double bond = 4


Total = 12

HC O

H

formal charge on C

= 4 -0 -½ x 8 = 0

formal charge

on O= 6 -4 -½ x 4 = 0

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

-

0 0

9.785 86


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 +1 HC O

H

0 0

86


A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.

O O O+ -

OOO+-

O C O

O

- -O C O

O

-

-

OCO

O

-

- 9.8

What are the resonance structures of the carbonate (CO3

2-) ion?


Exceptions to the Octet Rule

The Incomplete Octet

H HBeBe – 2e-

2H – 2x1e-

4e-

BeH2

BF3

B – 3e-

3F – 3x7e-

24e-

F B F

F



Total = 24

9.988



Exceptions to the Octet Rule

Odd-Electron Molecules

N – 5e-

O – 6e-

11e-

NO N O

The Expanded Octet (central atom with principal quantum number n > 2)

SF6

S – 6e-

6F – 42e-

48e-

S

F

F

F

FF

F



Total = 48

9.989 90



Predicting Molecular Geometry

1. Draw Lewis structure for molecule.

2. Count number of lone pairs on the central atom and number of atoms bonded to the central atom.

3. Use VSEPR to predict the geometry of the molecule.

What are the molecular geometries of SO2 and SF4?

SO O

AB2E

bent

S

F

F

F F

AB4E

distortedtetrahedron

10.192


# 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

10.4




Sigma bond (σ) – electron density between the 2 atomsPi bond (π) – electron density above and below plane of nuclei

of the bonding atoms 10.5



Describe the bonding in CH2O.

CH

OH

C – 3 bonded atoms, 0 lone pairsC – sp2

10.5



Sigma (σ) and Pi Bonds (π)

Single bond 1 sigma bond

Double bond 1 sigma bond and 1 pi bond

Triple bond 1 sigma bond and 2 pi bonds

How many σ and π bonds are in the acetic acid(vinegar) molecule CH3COOH?

C

H

H

CH

O

O Hσ bonds = 6 + 1 = 7

π bonds = 1

10.598



Pressure of Gases and its Units

Pressure is defined as the force applied per unit area.

Common Units of PressureSI units of pressure is kg/m. s2

Pascal, Pa = Newton/meter2 = N/m2 (N = kg m/s2)Standard Atmospheric Pressure = 101.3 kPa

=1 atmosphere (atm)= 760 torr

= 760 mm Hg

Standard atmospheric pressure: the pressure that supports a column of

mercury exactly 760 mm high at 0oC at

sea level. A barometer is used to measure the pressure as shown.


Gas LawsBoyle’s Law:

(the relationship between volume and pressure):

“The volume of a sample of gas is inversely proportional to its

pressure, if temperature remains constant”

Pressure x Volume = Constant (T = constant)

P1 and V1 are initial conditions P2 and V2 are final conditions

P1V1 = P2V2 (T = constant)

ExampleA cylinder of oxygen has V = 50.0 L and P = 10 atm at 20oC. What will be the volume of the gas at atmospheric pressure (1 atm) and 20oC?

According to Boyle’s law P1V1 = P2V2

P1= 10 atm, V1 = 50.0 L, P2= 1 atm , V2 = ??

(10 atm x 50 L) = (1 atm x V2), and V2 = 500 L



Gas LawsCharles’s Law:

(the relationship between volume and temperature):

“The volume of a fixed amount of gas is directly proportional to the

absolute temperature of the gas at constant pressure.”

Volume = Temperature x Constant (P = constant)

ExampleA sample of gas at 1 atm pressure and 25oC has V = 1 L, The gas is

cooled to -196 oC. What is the new volume?

According to Charles’ law V1/T1 = V2/T2

V1= 1.0 L, T1 = 25oC = 25+ 273.15 = 298.15 K,

V2= ?? L, T2 = -196oC = -196 + 273.15 = 76.15 KV1/T1 = V2/T2 (1.0 L / 298.15 K) = (V2 L / 76.15 K)

V2= 0.26 L (temperature decreased, volume decreased)

2

2

1

1

T

V

T

V=


Gas LawsAvogadro’s Law:

(the relationship between volume and amount):

“The volume of a gas is directly proportional to the number of moles (n)

of gas at constant temperature and pressure.”

Volume = # moles x Constant

Example

In the following reaction: 3 H2 (g) + N2 (g) → 2 NH3 (g)

Before the reaction begins, the reactants occupy a volume of 1.32 L. If a constant

pressure is maintained in the reaction, what is the final volume occupied by the

product after the completion of the reaction (assume all reactants are consumed)?

According to Avogadro’s law V1/n1 = V2/n2

V1= 1.320 L, n1 = 3 mole (H2) + 1 mole (N2) = 4 moles,

V2= ?? L, n2 = 2 moles (NH3)V1/n1 = V2/n2 (1.32 L / 4 moles K) = (V2 L / 2 moles)

V2= 0.660 L (number of moles decreased, volume decreased)

2

2

1

1

n

V

n

V=


Ideal Gas Law

Boyle’s Law: V α 1/P (T = constant)

Charles’s Law: V α T (P = constant)

Avogadro’s Law: V α n T and P are constant

Therefore: V α nT/P or PV α nT

So, the Ideal Gas Law: PV = nRT

Where: R = proportionality constant called the universal gas constant and it is equal = 0.08206 L atm K-1 mol-1,

or 8.3145 J mol-1. K-1.

P = pressure in atm V = volume in Litres

n = moles T = temperature in Kelvin


Example What is the lung capacity (volume) of an average adult lung, if the number of moles of the air in the lungs is 0.15 moles? Assume that the person is at 1.00 atm pressure and has a normal body temperature of 37oC.

Answer According to the ideal gas law PV = nRTP = 1.00 atm, V= ??, n = 0.15 moles, R = 0.08206 L atm K-1

mol-1,T = 37 oC = 37 + 273.15 = 310.15 KV = nRT/P

= (0.15 mole x 0.08206 L atm K-1 mol-1 x 310.15 K)/1.00 atm

The lung capacity V = 3.8 L



Standard Temperature and Pressue “STP”The STP are considered when the pressure is equal to 1 atm and the temperature = 0 oC (273 K).

Example

What is the molar volume (volume of 1 mole) of an ideal gas at STP?

According to the ideal gas law: PV = nRT

• V = (nRT)/P = (1 mole x 0.08206 L atm K-1 mol-1 x 298.15 K)/ (1.00 atm)

• Molar volume Vm = 22.4 L

• The molar volume of ANY ideal gas is 22.4 liters at STP.


Dalton’s Law of Partial Pressure:

For a mixture of gases in a container, the overall pressure is the sum of

all the partial pressures of the individual components.PTotal = P1 + P2 + P3 + . . .

Example

A mixture of oxygen, hydrogen and nitrogen gases exerts a total pressure

of 278 kPa. If the partial pressures of the oxygen and the hydrogen are 112

kPa and 101 kPa respectively, what would be the partial pressure exerted by the nitrogen.

According to Dalton’s law of partial pressurePtotal = P1 + P2 + . . . Pn

278 kPa = 112 kPa + 101 kPa + Pnitrogen

Pnitrogen = 278 kPa - (112 kPa + 101 kPa)

Pnitrogen = 65 kPa


Mole Fraction (X)

• If a mixture contains two components, A and

B, then:

B molesA moles

A molesXA

+=

B molesA moles

B molesXB

+=

1XX BA =+

ExampleA gaseous mixture contains 0.38 moles of nitrogen gas and 0.45 moles

of oxygen gas. Determine the mole fractions of oxygen and nitrogen.

Total number of moles = 0.38 mol nitrogen + 0.45 mole oxygen = 0.83 moles

XOxygen = (moles of oxygen / total number of moles)

= (0.45 mole/ 0.83 moles) = 0.54

Xnitrogen = (moles of nitrogen / total number of moles)

= (0.38 mole/ 0.83 moles) = 0.46

Check: XOxygen + Xnitrogen = 0.54 + 0.46 =1.00 108Dr. Mohamed Abdel Salam

Determining Partial Pressure from Mole Fraction

samplein species all of moles

A molesXA =

V

RTnP A

A =V

RTnP total

total =

A

total

A

total

A

total

A Xn

n

V

RTnV

RTn

P

P==

=

AtotalA XPP =

Example The mole fraction of nitrogen in air is 0.7808. Calculate the partial

pressure of N2 in air when the atmospheric pressure is 760 torr.

Use the partial pressure formula:

==22 NtotalN XPP (760 torr)(0.7808) = 593 torr



Graham’s Law of diffusion and effusion of gases

Gas Diffusion is the gradual mixing of

molecules of different gases. A good

example is the diffusion of oxygen in the

lungs through the alveoli.

Gas Effusion is the escape of a gas from its

container through a small hole.2

2

H He

He H

rate MW

rate MW=

Graham’s Law stated that "The relative rates at which two gases under identical conditions of temperature and pressure will diffuse vary

inversely as the square roots of the molecular masses of the gases."

ExampleCalculate the relative rates of effusion of H2(g) and He(g)

B

B

B

A

MW

MW

Rate

Rate=

41.122

4

MW

MW

Rate

Rate

2

2

H

He

He

H====


Example

Write the equilibrium constant expression for Kc for the

following reactions:

(a) N2 (g) + 3H2 (g) 2NH3 (g)

(b) 2 O3 (g) 3O2 (g)

3

22

2

3

]][[

][

HN

NHK

c=

2

3

3

2

][

][

O

OKc =


Homogenous Equilibrium• In HOMOGENEOUS EQUILIBRIA all reactants and products are in the

same phase.

• For example; N2O4 (g) 2NO2 (g)

• Kc means concentrations are in mol/L (molar)

• Because concentration is proportional to gas pressure:

• Kp means equilibrium concentrations are in terms of gas pressure; Pressure must be in units of atm.

]O[N

][NO

k

k K

42

2

2

r

fc ==

42

22

ON

NO

pP

PK =


Heterogeneous Equilibrium• In HETEROGENEOUS EQUILIBRIA reactants and products

are in different phases.

For example:

CaCO3 (s) CaO (s) + CO2 (g)

• Species in solid or liquid phase DO NOT appear in an equilibrium expression because they are present in small quantities and could be neglected relative to the gas, which spread in the container.

• Kc = [CO2] or KP = PCO2

• Solid CaCO3 and CaO do not appear in the K expression.



KC and KP For Gases

• For the reaction aA bB

and

• From the ideal gas equation: PAVA = nART

RT]A[)RT(V

nP

A

AA ==

a

b

c]A[

]B[K = a

A

b

Bp

P

PK =

{ }{ }

)ab(

Ca

b

a

A

b

Bp )RT(K

)RT](A[

)RT](B[

P

PK −===

n

Cp )RT(KK ∆=

where ∆n is the difference in the stoichiometric coefficients of gaseousgaseousproducts and reactants.


The Equilibrium quotient (Q)For the reaction:

aA + bB cC + d D

At equilibrium, we can calculate the equilibrium constant KC, from the

equilibrium concentrations of both reactants and products.

But, any time we can calculate the equilibrium quotient (Q) using the

concentrations of both the reactants and products at any time.

[B][A]

[D][C]

Reactants

Products K

b a

d c

c ==

[B][A]

[D][C]

Reactants

Products Q

b a

d c

C ==

Kc is thus the special value that Q has when the reaction is at equilibrium.


• The value of Q in relation to K indicates the direction in

which any net reaction must proceed the ratio of Q/K

immediately tells us whether, and in which direction, a net

reaction will occur as the system moves toward its

equilibrium state:

Product concentration too low; net reaction

proceeds to right.

<1

System is at equilibrium; no net change will

occur.

=1

Product concentration too high; net reaction

proceeds to left.

>1

net reaction to reach equilibrium.Q/K


Example

For the synthesis of ammonia at 500oC, the equilibrium constant, KC is 6.0

x 10-2. Predict the direction in which the reaction will proceed to reach

equilibrium if the initial concentrations are:

[NH3] = 1.0 x 10-3 M, [N2] = 1.0 x 10-5, [H2] = 2.0 x 10-3 M.

N2 (g) + 3 H2 (g) 2 NH3 (g)

7

335

23

22

2

3 103.1)100.2)(100.1(

)100.1(

]][[

][x

xx

x

HN

NHQ ===

−−

−

AnswerCalculate Q and compare it with KC

But, Kc = 6.0 x 10-2 Thus Q >> KBecause Q ≠ K, the system is not in equilibrium.

The ratio of products to reactants is too high. In order to

achieve equilibrium, the reaction will proceed from right to left (consuming NH3 and producing N2 and H2).



Example

If 2.00 mol of NOCl is placed in an empty 1.00 L flask at

25oC. At equilibrium you find 0.66 mol/L of NO. Calculate

KC and KP.

2ClNO2NOCl2 +⇔

Answer:

1- Write balanced equation

2- Write the Equilibrium Expression [ ] [ ][ ]

( ) n

CP

C

RTKK

NOCl

ClNOK

∆

=

=2

2

2

3- Set up a table showing concentrations of each species.


4- Calculate equilibrium concentrations of each species using stoichiometric coefficients in balanced chemical equation:

5- Calculate KC from equilibrium concentrations.

[ ] [ ][ ]

( ) ( )( )

080.034.1

33.066.02

2

2

2

2

===NOCl

ClNOK

C

6- Calculate KP from KC


Factors Affecting Equilibrium

(Le Châtelier principle)"If a system at equilibrium is subjected to a change of pressure,

temperature, or the number of moles of a substance, there will be a tendency for a net reaction in the direction that tends to reduce the effect

of this change."

1) Concentration and Equilibrium:

The reaction goes to the opposite direction of the added substance.

2) Pressure and Equilibrium (only for Gases)

When pressure increased, the reaction goes to the less number of

moles side.

3) Temperature and Equilibrium

When temperature is raised (increased):If the heat energy (E) is in the reactant side, the reaction will go to the right

(products)

If the heat energy (E) is in the product side, the reaction will go to the left

(reactants)


ORGANIC CHEMISTRY• Organic chemistry is very important branch of chemistry and it study the

compounds which contain carbon (C) and hydrogen (H), in general, and may contains other atoms such as oxygen (O), nitrogen (N), sulfur (S),…etc.

•DRAWING ORGANIC MOLECULES

Molecular formulae: A molecular formula simply counts the numbers of

each sort of atom present in the molecule, but tells you nothing about the way they are joined together.

Example:the molecular formula of butane is C4H10, and the molecular formula of

ethanol is C2H6O.

Structural formulae: A structural formula shows how the various

atoms are bonded. Example:ethanoic acid would be shown in a fully displayed form and a simplified form as:



• Skeletal formulae: all the hydrogen atoms are removed from carbon

chains, leaving just a carbon skeleton with functional groups attached to it.

• Example:

• butan-2-ol Cyclohexane, C6H12

STRUCTURAL ISOMERISM:

Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space.

Types of structural isomerism

(A) Chain isomerism: due to possibility of branching in carbon chains.

Example:

there are two isomers of butane, C4H9.


(B) Position isomerism: the basic carbon skeleton remains unchanged, but important groups are moved around on that skeleton.

Example:there are two structural isomers

with the molecular formula C3H7Br.

Functional group isomerism

The isomers contain different functional groups, and they belong to

different families of compounds Examplemolecular formula C3H6O could be either propanal (an aldehyde)

or propanone (a ketone).


Organic Compounds

Acyclic

(open chain compounds)

Cyclic

(Closed chain compounds)

Alkane Alkene Alkyne

CnH2n+2 CnH2n CnH2n

aliphatic compounds

Aliphatic Aromatic


Aliphatic hydrocarbons• Alkanes:• they are saturated hydrocarbons, because all the carbon atoms are bonded

with 4 single covalent bonds.

• General formula: CnH2n+2

• Nomenclature:a chemical name has three parts:

prefix, parent, and suffix

• prefix: Tells #, types & where side groups attached

• parent: Tells # of Cs in longest continuous chain

• suffix: Tells which functional group is present

• Straight chain alkanes are named by counting the number of carbon atoms in the longest chain and adding the suffix -ane.




How to name open chain compounds?• To name alkanes and any other organic compounds:

• 1. Find the longest continuous chain of carbon atoms; this is a hexane (6 crabo atoms).

• 2. Alkane groups as substituent’s are named as follows:

• - CH3 methyl

• - CH2CH3 ethyl

• - CH2CH2CH3 propyl

• 3. Number the long chain so that the substituents are at the lowest numbers,and the substituent at carbon number 3.

The name for this alkane is: 3-ethylhexane


Draw the structures for n-butane and isobutene

n means normal chain (straight), and but means 4 carbon atoms.

put 4 carbon atoms beside each other and make each one connectedto 4 single bonds:

CH3CH2CH2CH3 or

iso means branched chain, and but means 4 carbon atoms.

n-butane

isobutane

put 3 carbon atoms beside each other and connect the forth C atom to the middle carbon, and make each one connected to 4 single

bonds:


Aliphatic hydrocarbons• Alkenes:

• Alkenes are unsaturated hydrocarbons (contain at least one

C=C double bond)

• General formula: CnH2n

•Naming Alkenes: count the number of the

carbon atoms in the longest chain, and add the suffix -ene at the end.

cis-trans isomerism

Alkenes have two different geometrical isomerism, cis and trans.



Aliphatic hydrocarbons• Alkynes:

• Alkynes are unsaturated hydrocarbons (contain at least one

C≡C triple bond)

• General formula: CnH2n-2

•Naming Alkynes: count the number of the carbon atoms in the longest

chain, and add the suffix -yne at the end.




Biochemistry• Biochemistry is the chemistry of living organisms.

• It is the application of chemistry to study the different biological processes at the cellular and molecular level.

• In general, Chemistry deals with objects at the molecular scale and the fundamental unit of living organisms is the cell (macroscopic scale), so, consequently, biochemistry tries to span both of these worlds (molecules and cells).

Basic Structures and MechanismsThe basic structures of biochemistry are biomolecules, which are

molecules created by living organisms. There are four main categories of biomolecules:

1) Proteins

2) Carbohydrates3) Lipids

4) Nucleic Acids



Proteins

• Proteins are macromolecules made up of amino acids. Amino acids: consist

of an amino group, a carboxyl group, a hydrogen atom and a distinctive R

group bonded to a carbon atom.

The type of bond in protein is peptide bond. 134Dr. Mohamed Abdel Salam

There are twenty different types of side chains (20 amino acids).


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