dr. mohamed abdel salam - kaumabdelsalam.kau.edu.sa/files/0053615/files/16374_summary for th… ·...
TRANSCRIPT
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
3Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
International System of Units (SI)
4
Dr. Mohamed Abdel Salam
5Dr. Mohamed Abdel Salam
5 6Dr. Mohamed Abdel Salam
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
7Dr. Mohamed Abdel Salam
K = 0C + 273.15
0F = x 0C + 3295
273 K = 0 0C 373 K = 100 0C
32 0F = 0 0C 212 0F = 100 0C
7 8Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
9 10Dr. Mohamed Abdel Salam
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
11Dr. Mohamed Abdel Salam
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
11 12Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
13Dr. Mohamed Abdel Salam
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
13 14Dr. Mohamed Abdel Salam
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
15Dr. Mohamed Abdel Salam
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
15 16Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
17Dr. Mohamed Abdel Salam
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
17 18Dr. Mohamed Abdel Salam
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
19Dr. Mohamed Abdel Salam
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
19 20Dr. Mohamed Abdel Salam
mass p = mass n = 1840 x mass e-
2.220
Dr. Mohamed Abdel Salam
21Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
23Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
25Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
27Dr. Mohamed Abdel Salam
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:
27 28Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
29Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
31Dr. Mohamed Abdel Salam
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
31 32Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
33Dr. Mohamed Abdel Salam 3.5
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
35Dr. Mohamed Abdel Salam 3.6
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
Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
37Dr. Mohamed Abdel Salam
Balancing Chemical Equations
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
Dr. Mohamed Abdel Salam
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
38
39Dr. Mohamed Abdel Salam
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)
39 40Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
41Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
Limiting Reagents
3.9
2NO + 2O2 2NO2
NO is the limiting reagent
O2 is the excess reagent
42
43Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
45Dr. Mohamed Abdel Salam
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
45 46Dr. Mohamed Abdel Salam
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
47Dr. Mohamed Abdel Salam
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.
47 48Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
49Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
51Dr. Mohamed Abdel Salam 7.6
51 52Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
53Dr. Mohamed Abdel Salam
Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
7.753 54
Dr. Mohamed Abdel Salam
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
55Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
57Dr. Mohamed Abdel Salam
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
58Dr. Mohamed Abdel Salam
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
59Dr. Mohamed Abdel Salam
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.
60Dr. Mohamed Abdel Salam
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:
Dr. Mohamed Abdel Salam
61Dr. Mohamed Abdel Salam
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
62Dr. Mohamed Abdel Salam 8.2
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
63Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
65Dr. Mohamed Abdel Salam
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
66Dr. Mohamed Abdel Salam 8.3
67Dr. Mohamed Abdel Salam
Cation is always smaller than atom from which it is formed.Anion is always larger than atom from which it is formed.
8.368
Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
69Dr. Mohamed Abdel Salam
General Trend in First Ionization Energies
8.4
Increasing First Ionization Energy
Incre
asin
g F
irst Io
niz
atio
n E
ne
rgy
70Dr. Mohamed Abdel Salam
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
71Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
73Dr. Mohamed Abdel Salam 9.1
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
75Dr. Mohamed Abdel Salam 9.2
Li + F Li+ F -
The Ionic Bond
1s22s11s22s22p5 1s21s22s22p6[He][Ne]
Li Li+ + e-
e- + F F -
F -Li+ + Li+ F -
75 76Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
77Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
79Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
81Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
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
83Dr. Mohamed Abdel Salam 9.7
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.
83 84Dr. Mohamed Abdel Salam
H C O H
C – 4 e-
O – 6 e-
2H – 2x1 e-
12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 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
Dr. Mohamed Abdel Salam
85Dr. Mohamed Abdel Salam
C – 4 e-
O – 6 e-
2H – 2x1 e-
12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 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
Dr. Mohamed Abdel Salam
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
87Dr. Mohamed Abdel Salam
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?
87 88Dr. Mohamed Abdel Salam
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
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24
9.988
Dr. Mohamed Abdel Salam
89Dr. Mohamed Abdel Salam
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
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
9.989 90
Dr. Mohamed Abdel Salam 10.1
91Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
# 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
Dr. Mohamed Abdel Salam
93Dr. Mohamed Abdel Salam 10.4
94Dr. Mohamed Abdel Salam
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
95Dr. Mohamed Abdel Salam 10.5
96Dr. Mohamed Abdel Salam
Describe the bonding in CH2O.
CH
OH
C – 3 bonded atoms, 0 lone pairsC – sp2
10.5
Dr. Mohamed Abdel Salam
97Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam 10.7
99Dr. Mohamed Abdel Salam
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.
100Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
101Dr. Mohamed Abdel Salam
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=
102Dr. Mohamed Abdel Salam
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=
103Dr. Mohamed Abdel Salam
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
104Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
105Dr. Mohamed Abdel Salam
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.
106Dr. Mohamed Abdel Salam
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
107Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
109Dr. Mohamed Abdel Salam
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====
110Dr. Mohamed Abdel Salam
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 =
111Dr. Mohamed Abdel Salam
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 =
112Dr. Mohamed Abdel Salam
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.
Dr. Mohamed Abdel Salam
113Dr. Mohamed Abdel Salam
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.
114Dr. Mohamed Abdel Salam
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.
115Dr. Mohamed Abdel Salam
• 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
116Dr. Mohamed Abdel Salam
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).
Dr. Mohamed Abdel Salam
117Dr. Mohamed Abdel Salam
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.
118Dr. Mohamed Abdel Salam
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
119Dr. Mohamed Abdel Salam
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)
120Dr. Mohamed Abdel Salam
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:
Dr. Mohamed Abdel Salam
121Dr. Mohamed Abdel Salam
• 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.
122Dr. Mohamed Abdel Salam
(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).
123Dr. Mohamed Abdel Salam
Organic Compounds
Acyclic
(open chain compounds)
Cyclic
(Closed chain compounds)
Alkane Alkene Alkyne
CnH2n+2 CnH2n CnH2n
aliphatic compounds
Aliphatic Aromatic
124Dr. Mohamed Abdel Salam
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.
Dr. Mohamed Abdel Salam
125Dr. Mohamed Abdel Salam
126Dr. Mohamed Abdel Salam
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
127Dr. Mohamed Abdel Salam
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:
128Dr. Mohamed Abdel Salam
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.
Dr. Mohamed Abdel Salam
129Dr. Mohamed Abdel Salam
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.
130Dr. Mohamed Abdel Salam
131Dr. Mohamed Abdel Salam
132Dr. Mohamed Abdel Salam
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
Dr. Mohamed Abdel Salam
133Dr. Mohamed Abdel Salam
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).
135Dr. Mohamed Abdel Salam
� � ا�� �ر و ا���ح � ��� ��� ����� � آ��ب �أUniversity Chemistry
� �%�$# خ!ارزم