1. General Chemistry 101 Chem
Dr. Fahad Ahmed Alharthi
Room: 2A104, Building No. 5

2. References ((المراجع
Chemistry the General Science 11E T. L. Brown, H. E. LeMay, B. E. Bursten and C. J. Murphy.
الكيمياء العامة. د.أحمد العويس، د.سليمان الخويطر، د.عبدالعزيز الواصل، د.عبدالعزيز السحيباني.

3. Matter
We define matter as anything that has mass and takes up space.

4. Three States of Matter

5. A Gas has neither a definite shape nor definite volume:
it adopts the volume and shape of the vessel containing it.
A Liquid has definite volume but not shape:
it adopts the shape of the vessel containing it.
A Solid has definite volume and shape:
which is independent of the vessel containing it.

6. Changes of States GAS LIQUID SOLID Vaporization
(heat or reduce pressure)
Condensation
(cool or increase pressure)
Liberates Energy
LIQUID
Requires Energy
Melting
(heat)
Freezing
(cool)
SOLID

7. Composition
Composition is the type and amount of substance that make up a sample of matter.
Atoms are the building blocks of matter.
Each element is made of the same kind of atom.
A compound is made of two or more different kinds of elements.

8. Properties
Properties are the characteristics that give each substance a unique identity.
Physical Properties
those which the substance
shows by itself without interacting with another substance, such as: colour, melting point, boiling
point, density
Chemical Properties
those which the substance shows as it interacts with, or transforms into, other substances, such as: flammability, corrosiveness

9. PROBLEM: Decide whether each of the following process is primarily a physical or a chemical change, and explain briefly.
Frost forms as the temperature drops on a humid winter night.
(b) Dynamite explodes to form a mixture of gases.
(c) Dissolving sugar and water.
(d) A silver fork tarnishes in air.
Criteria: “Does the substance change composition or just change form?”

10. SOLUTION:
physical change
(b) chemical change
(c) physical change
(d) chemical change

11. Units of Measurement Table 1.1 SI Base (or Fundamental) Units
Base Quantity
Name of Unit
Symbol
Length
meter m
Mass
kilogram kg
Time
second s
Current
ampere A
Temperature
kelvin K
Amount of substance
mole
mol
Luminous intensity
candela cd
Système International d’Unités
A different base unit is used for each quantity.

12. Metric System Table 1.2 Prefixes Used with SI Units Prefix Symbol
Meaning
Tera- T
1012
Giga- G
109
Mega- M
106
Kilo- k
103
Deci- d
10-1
Centi- c
10-2
Milli- m
10-3
Micro-
10-6
Nano- n
10-9
Pico- p
10-12

13. Volume
The most commonly used metric units for volume are the liter (L) and the milliliter (mL).
A liter is a cube 1 dm long on each side.
A milliliter is a cube 1 cm long on each side.
1 cm3 = (1 x 10-2 m)3 = 1 x 10-6 m3
1 dm3 = (1 x 10-1 m)3 = 1 x 10-3 m3
1 L = 1000 mL = 1000 cm3 = 1 dm3
1 mL = 1 cm3

14. How many mL are in 1.63 L? 1 cm3 = (1 x 10-2 m)3 = 1 x 10-6 m3
1 dm3 = (1 x 10-1 m)3 = 1 x 10-3 m3
1 L = 1000 mL = 1000 cm3 = 1 dm3
1 mL = 1 cm3
1 L = 1000 mL
How many mL are in 1.63 L? 1L
1000 mL
1.63 L x
= 1630 mL

15. Density – SI derived unit for density is kg/m3
1 g/cm3 = 1 g/mL = 1000 kg/m3
mass
volume
d = m V
Density =
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 = m V

16. Comparison of the three temperature scales
K = 0C
273 K = 0 0C
373 K = 100 0C
0F = x 0C + 32 9 5
32 0F = 0 0C
212 0F = 100 0C

17. Convert 172.9 0F to degrees Celsius and Kelvin
0F = x 0C + 32 9 5
0F – 32 = x 0C 9 5
x (0F – 32) = 0C 9 5
0C = x (0F – 32) 9 5
0C = x (172.9 – 32) = 9 5
K =

18. Atomic structure

19. Subatomic Particles (Table 1.3)
mass p = mass n = 1840 x mass e-

20. X H H (D) H (T) U 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 the nucleus
Mass Number X A Z
Element Symbol
Atomic Number H 1
H (D) 2
H (T) 3 U
235 92
238

21. Cation is an ion with a positive charge
If a neutral atom losses one or more electrons
it becomes a cation. Na
11 protons
11 electrons
Na+
11 protons
10 electrons
Anion is an ion with a negative charge
If a neutral atom gains one or more electrons
it becomes an anion. Cl
17 protons
17 electrons
Cl-
17 protons
18 electrons

22. How many protons and electrons are in
Monatomic ion contains only one atom
Na+, Cl-, Ca2+, O2-, Al3+, N3-
Polyatomic ion (molecular ion) contains more than one atom
OH-, CN-, NH4+, NO3-
Examples
How many protons and electrons are in Al 27 13 ? 3+
How many protons and electrons are in Se 78 34 2- ?

23. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Chemical Equations
Chemical equations are concise representations of chemical reactions.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

24. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Reactants appear on the left side of the equation.

25. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Products appear on the right side of the equation.

26. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
The states of the reactants and products are written in parentheses to the right of each compound.

27. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.

28. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule.

29. Subscripts and Coefficients Give Different Information
Coefficients tell the number of molecules.

30. Examples of balancing chemical equation
HCl + Zn ZnCl2 + H2 2
C3H O CO H2O
Zn + HNO Zn(NO3)2 + H2 2

31. Different Types of Chemical Reactions
Examples
Combination reactions
Decomposition reactions
Combustion in Air

32. Combination Reactions
In this type of reaction two or more substances react to form one product.
Examples:
2 Mg (s) + O2 (g)  2 MgO (s)
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)

33. Decomposition Reactions
In a decomposition one substance breaks down into two or more substances.
Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)

34. Combustion Reactions
These are generally rapid reactions that produce a flame.
Most often involve hydrocarbons reacting with oxygen in the air.
Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)

35. Formula Weights(FW)
A formula weight is the sum of the atomic weights for the atoms in a chemical formula.
So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1(40.1 amu*)
+ Cl: 2(35.5 amu)
111.1 amu
Formula weights are generally reported for ionic compounds.
*atomic mass unit

36. Molecular Weight (MW)
A molecular weight is the sum of the atomic weights of the atoms in a molecule.
For the molecule ethane, C2H6, the molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu

37. Percent Composition
One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:
% element =
(number of atoms)(atomic weight)
(FW of the compound)
x 100

38. The percentage of carbon in ethane (C2H6) is:
Example
The percentage of carbon in ethane (C2H6) is:
%C =
(2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu =
x 100
= 80.0%

39. Avogadro’s Number Avogadro’s number is 6.02 x 1023
The mole is a unit measurement express the amounts of a chemical substance.
One mole of chemical substance contains Avogadro’s number of elementary entities (e.g., atoms, molecules, ions, electrons).
Mass (m)
Molecular weight (Mwt)
Number of mole (n)=
Example
1 mole of H2O has a mass of 18g.

40. Molar Mass
By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol).
The molar mass of an element is the mass number for the element that we find on the periodic table.
The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).

41. Mole Relationships
One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.
One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.

42. Examples
1- Calculate how many atoms there are in moles of copper.
The number of atoms in one mole of Cu is equal to the Avogadro number = 6.02 x
Number of atoms in moles of Cu = (0.200 mol) x (6.02x1023 mol-1 ) = 1.20 x
2- Calculate how molecules of H2O there are in moles of water.
Number of water molecules = (12.10 mol)x(6.02 x 1023)
= x 1024

43. 3- Calculate the number of moles in 5.380 g of glucose (C6H12O6) .
Moles of C6H12O6 = = mol.
4- Calculate the mass, in grams, of mol of Ca(NO3)2.
Mass = o.433 mol x g/mol = 71.1 g.
5.380 g
180.0 gmol-1

44. 5- How many molecules are in 5. 23 g of glucose (C6H12O6)
5- How many molecules are in 5.23 g of glucose (C6H12O6)? How many oxygen atoms are in this sample?
Molecules of C6H12O6 = x (6.02x1023)
= 1.75 x 1022 molecules
Atoms O = 1.75 x 1022 x 6 = 1.05 x 1023
5.23 g
180.0 gmol-1

45. Calculating Empirical Formulas
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
H2O
molecular
empirical
H2O
C6H12O6
CH2O O3 O
N2H4
NH2

46. Example
The compound para-aminobenzoic acid (PABA) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%).
Find the empirical formula of PABA.

47. Assume that 100.00 g of para-aminobenzoic acid: C: = 5.105 mol C
H: = 5.09 mol H
N: = mol N
O: = mol O
61.31 g
12.01 g.mol -1
5.14g
1.01 g.mol
10.21 g
14.01 g.mol
Calculate the mole ratio by dividing by the smallest number of moles:

48. The empirical formula for para-aminobenzoic acid is: C7H7NO2
5.105 mol
mol
5.09 mol
1.458 mol
The empirical formula for para-aminobenzoic acid is:
C7H7NO2

49. Combustion Analysis
Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this.
C is determined from the mass of CO2 produced.
H is determined from the mass of H2O produced.
O is determined by difference after the C and H have been determined.

50. Determining Empirical Formula by Combustion Analysis
Example:
Combustion of g of isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol.
Grams of C = x 1 x 12 gmol-1 = g
0.561 g CO2
44 gmol-1
Grams of H = x 2 x 1 gmol-1 = g
0.306 g H2O
18 gmol-1
Grams of O = mass of sample – (mass of C + mass of H)
= g – ( g g) = g

51. C3H8O 0.153 g Moles of C = = 0.0128 mol 12 gmol-1 0.0343 g
Moles of H = = mol
0.068 g
16 gmol-1
Moles of O = = mol
Calculate the mole ratio by dividing by the smallest number of moles: C:H:O
2.98:7.91:1.00
C3H8O

52. Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products.

53. Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

54. C6H12O6 + 6 O2  6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6…
calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams.

55. Limiting Reactants
The reactant that is completely consumed in a reaction is called the limiting reactant (limiting reagent).
In other words, it’s the reactant that run out first (in this case, the H2).

56. In example below, O2 would be the excess reactant (excess reagent).

57. Theoretical (calculated) Yield
The theoretical yield is the maximum amount of product that can be made.
The amount of product actually obtained in a reaction is called the actual yield.
However, the actual yield is always less than the theoretical yield.
Why?
Part of the reactants may not react.
Side reaction.
Difficult recovery.

58. Percent yield calculation
The percent yield of a reaction relates to the actual yield to the theoretical (calculated) yield.
Actual Yield
Theoretical Yield
Percent Yield = x 100

59. Examples Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)
If we start with 150 g of Fe2O3 as the limiting reactant, and found actual yield of Fe was 87.9 g, what is the percent yield?
Actual Yield
Theoretical Yield
The percent yield = x 100

60. Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)
150 g Fe2O3
150 g
159 g mol-1
0.943 mol Fe2O3
2 mol Fe
1 mol Fe2O3 X
1.887 mol Fe
1.887 mol x g mol-1
105 g Fe
Theoretical yield = 105 g.
The percent yield = x 100 = 83.7 %
87.9 g
105 g

61. Solutions
Solutions are defined as homogeneous mixtures of two or more pure substances.
The solvent is present in greatest abundance.
All other substances are solutes.

62. volume of solution in liters
Molarity
Molarity is one way to measure the concentration of a solution.
The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution.
moles of solute
volume of solution in liters
Molarity (M) =
mol L M=

63. Mixing a Solution
To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute.
The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

64. Examples
Calculate the molarity of a solution made by dissolving g of sodium sulfate (Na2SO4) in 850 mL of water.
Moles of Na2SO4 = = mol
Molarity = = M
0.750 g
142 g mol-1
mol
0.850 L

65. How many moles of KMnO4 are present in 250 mL of a 0.0475 M solution?
Moles of KMnO4 = x 0.25 L
= mol
mol L
mol L

66. volume of solution in liters
How many milliliters of 11.6 M HCl solution are needed to obtain mol of HCl?
moles of solute
volume of solution in liters
Molarity (M) =
0.250 mol
volume in liters
11.6 M =
volume in liters = L = 22 mL.

67. What are the molar concentrations of each of the ions present in a 0
What are the molar concentrations of each of the ions present in a M aqueous solution of Ca(NO3)2?
In aqueous solution:
Ca(NO3) Ca NO3-
Ca2+ = M
NO3- = x 2 = 0.05 M
+ 2

68. Dilution One can also dilute a more concentrated solution by
Using a pipet to deliver a volume of the solution to a new volumetric flask, and
Adding solvent to the line on the neck of the new flask.

69. - The molarity of the new solution can be determined from the equation
Mc  Vc = Md  Vd
where Mc and Md are the molarity of the concentrated and dilute solutions, respectively, and Vc and Vd are the volumes of the two solutions. -
Mc  Vc = Md  Vd
Moles solute before dilution = moles solute after dilution

70. Examples
How many milliliters of 3.0 M H2SO4 are needed to make 450 mL of 0.10 M H2SO4 ?
Mc  Vc = Md  VV
3.0 M x Vc = 0.10 M x 450 mL
Vc = 15 mL

71. Ways of Expressing Concentrations of Solutions
Mass Percentage, ppm, and ppb
Mass Percentage
mass of A in solution
total mass of solution
Mass % of A =
 100
Note: use the same unit in both mass of A in solution and total mass of solution
Example:
36% HCl by mass contains 36 g of HCl for each 100 g of solution (64 g H2O)

72. Parts per Million (ppm)
mass of A in solution
total mass of solution
ppm =
 106
Parts per Billion (ppb)
mass of A in solution
total mass of solution
ppb =
 109
Note: use the same unit in both mass of A in solution and total mass of solution

73. Examples
Calculate the mass percentage of Na2SO4 in a solution containing 10.6 g Na2SO4 in 483 g water.
= 2.15 %
10.6 g
( ) g
Mass % of Na2SO4 =
 100

74. An ore contains 2. 86 g of silver per ton of ore
An ore contains 2.86 g of silver per ton of ore. What is the concentration of silver in ppm?
= 2.86 ppm
2.86 g
106 g
ppm =
 106

75. Mole Fraction, Molarity, and Molality Mole Fraction (X)
Molarity (M)
moles of A
total moles in solution
XA =
moles of solute
volume of solution in liters
Molarity (M) =

76. Since volume is temperature-dependent, molarity can change with temperature.
Molality (m)
moles of solute
Kilograms of solvent
m =
Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature-dependent.

77. Examples
An aqueous solution of hydrochloric acid contains 36 % HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution.
Moles HCl = = 0.99 mol
Moles H2O = = 3.6 mol
36 g
36.5 g mol-1
64 g
18 g mol-1

78. moles HCl
moles H2O + moles HCl
0.99
XHCl = =
= 0.22
Molality of HCl = = m
0.99 mol HCl
0.064 kg H2O