1. References 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.

2. SI Units Units of Measurement Système International d’Unités
A different base unit is used for each quantity.

3. Metric System
Prefixes convert the base units into units that are appropriate for the item being measured.

4. 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.

5. Density is a physical property of a substance.
mass
Volume
d = m V

6. 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)

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

8. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Products appear on the right side of the equation.

9. Anatomy of a Chemical Equation
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.

10. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.

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

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

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

14. Combination & decomposition reactions
Reaction Types
Combination & decomposition reactions
Combustion in Air

15. 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)

16. 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)

17. 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)

18. 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

19. 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

20. 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

21. Percent Composition So the percentage of carbon in ethane is…
(2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu =
x 100
= 80.0%

22. Avogadro’s Number 6.02 x 1023 1 mole of 12C has a mass of 12 g.
1 mole of H2O has a mass of 18g.

23. 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).

24. Using Moles
Moles provide a bridge from the molecular scale to the real-world scale.

25. 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.

26. 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

27. 4- Calculate the mass, in grams, of 0.433 mol of Ca(NO3)2.
3- Calculate the number of moles of glucose (C6H12O6) in g of 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

28. 5- How many glucose molecules are in 5. 23 g of C6H12O6
5- How many glucose molecules are in 5.23 g of 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

29. Finding Empirical Formulas

30. Calculating Empirical Formulas
One can calculate the empirical formula from the percent composition.

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

32. Calculating Empirical Formulas
Assuming g of para-aminobenzoic acid,
C: g x = mol C
H: g x = 5.09 mol H
N: g x = mol N
O: g x = mol O
1 mol
12.01 g
14.01 g
1.01 g
16.00 g

33. Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of moles:
C: =  7
H: =  7
N: = 1.000
O: =  2
5.105 mol
mol
5.09 mol
1.458 mol

34. Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2

35. 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.

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

37. C3H8O 0.153 g C Moles of C = = 0.0128 mol 12 gmol-1 0.0343 g H
Moles of H = = mol
0.068 g O
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

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

39. Stoichiometric Calculations
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).

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

41. 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).

42. Limiting Reactants
In example below, O2 would be the excess reactant (excess reagent).

43. Theoretical 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.

44. Theoretical Yield
The actual yield is almost always less than the theoretical yield.
Why?
Part of the reactants may not react.
Side reaction.
Difficult recovery.

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

46. 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

47. 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

48. 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.

49. volume of solution in liters
Molarity
Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different.
Molarity is one way to measure the concentration of a solution.
moles of solute
volume of solution in liters
Molarity (M) =

50. 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.

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

52. 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

53. 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.

54. 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?
Ca2+ = M
NO3- = x 2 = 0.05 M

55. 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.

56. Dilution
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.

57. Dilution
Mc  Vc = Md  Vd
Moles solute before dilution = moles solute after dilution

58. 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

59. 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
Example:
36% HCl by mass contains 36 g of HCl for each 100 g of solution (64 g H2O)

60. 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

61. 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

62. 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

63. 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) =

64. 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.

65. 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

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