1. Chemical reactions and chemical equation and mole concept
Lecture 3
Chemical reactions and chemical equation and mole concept
101 CHEM
Done by:
L. Amal Abu-Mostafa

2. Session Objectives: Mole concept Determining chemical formulas
Mass percentages from the formula
Chemical Reactions
Indicators of chemical reactions
Symbols used in equations
Reaction Energy
Types of chemical Reactions

3. The mole concept:
A mole of a substance (symbol mol) is that amount of the substance which contains the same number of particles (atoms, molecules or ions) that are present in 12g of Carbon–12.
The number of particles (atoms) present in 12g of Carbon –12 is x
This number is called Avagadro’s Number or Avagadro’s Constant.

4. The mole concept:
For example : One mole of ethanol, contains the same number of ethanol molecules as there are carbon atoms in 12 g of carbon-12.
Another example: a mole of oxygen atoms (with the formula O) contains x O atoms.
A mole of oxygen molecules (formula O2) contains x 1023 O2 molecules,
that is, 2 × x 1023 O atoms.

5. The mole concept:
The molar mass of a substance is the mass of one mole of the substance.
Carbon-12 has a molar mass of exactly 12 g/mol, by definition.
For all substances, the molar mass in grams per mole (g/mol) is numerically equal to the formula mass in atomic mass units (amu).

6. A mole represents two things :-
i) It represents a definite number of particles (atoms, molecules or ions) equal to x
ii) It represents a definite mass of a substance equal to the gram atomic mass of an element or the gram molecular mass of a compound.

7. Gram atomic mass of an element :- is, its atomic mass expressed in grams.
Eg :-
Gram atomic mass of Oxygen O2 =16 x 2 =32g.
Gram molecular mass of a compound :- is, its molecular mass expressed in grams.
Eg :- Gram molecular mass of water
H2O = (1 x 2) + 16 = = 18g.

8. Exercises: a. What is the mass in grams of a one chlorine atom, Cl?
b. What is the mass in grams of a hydrogen chloride
molecule, HCl?
Solution strategy:
We need to consider using molar mass
And the relationship between the number of atoms or molecules and the molar mass.
a-The atomic mass of Cl = 35.5 amu,
So the molar mass of Cl = 35.5 g/mol.
Dividing 35.5 g/mole by x 1023 (Avogadro’s number) gives the mass of one atom.

9. Solution Mass of one Cl atom= 35.45 g = 5.90 x 10-23 g 6.022 x 1023
b.The molecular mass MM of HCl = AM of H + AM of Cl, =
= amu
The molar mass of HCl =36.45 g/mol
Therefore, 1 mol of HCl contains → x 1023 HCl molecules.
So, the mass of a one HCl molecule =
= 6.06 x g

10. Mole Calculations:
Relationship between number of moles (n), mass (m), molar mass (M), and Avagadro number (NA).
m g
n =
M g. mol-1
Mass of one (atom, molecule, or unit) = M m= NA

11. Converting Moles of Substance to Grams:
Example:
Zinc iodide, ZnI2, can be prepared by the direct combination of elements. A chemist determines from the amounts of elements that mol ZnI2 can form. How many grams of zinc iodide in this? m
Solution: n = , M
m = n× M = mol × g . mol-1
= g

12. Solution:
m = n× M First we will count the molar mass (M) for ZnI2 M = AM of Zn + (AM of I × 2) = ( × 2) = g. mol-1 So: = mol × g. mol-1 = g = 20.9 g

13. Homework
Lead(II) chromate, PbCrO4, is a yellow paint pigment (called chrome yellow) prepared by a precipitation reaction. In preparation, 45.6 g of lead(II) chromate is obtained as a precipitate. How many moles of PbCrO4 in this?

14. Calculating the Number of Molecules in a Given Mass:
Example: How many molecules are there in a 3.46 g sample of hydrogen chloride, HCl?
Problem Strategy :
The number of molecules in a sample is related to moles of compound
(1 mol HCl x 1023 HCl molecules). Therefore, if you first convert grams HCl to moles, then you can convert moles to number of molecules.

15. Solution: The molar mass of HCl (M) = 36.5 g/mol m 3.46 g
n= = = mol
M g . mol-1
1 mol of HCl x 1023 HCl molecules
0.095 mol of HCl ??? HCl molecules
HCl molecules = x 1023 x 0.095 1
= 5.71 x HCl molecules

16. Determining chemical formulas
Mass percentages from the formula:
Suppose that A is an element in a compound or one substance in a mixture. We define the mass percentage of A as the parts of A per hundred parts of the total, by mass. That is
Mass % A = Mass of A in the whole × 100
Mass of the whole
You can look at the mass percentage of A as the number of grams of A in 100 g of the whole.

17. Example
Example 1: Calculate the percentage composition of elements in CHCl3?
Solution:
% C= × 100 = %
119.5
% H = × 100 = 0.84 %
% Cl = × 100 = %
Total is 100%

18. Example Example 2: Calculate the mass of Fe in 10 g of Fe2O3?
First we will calculate the molar mass of Fe2O3 from the periodic table, to see how many grams of iron (Fe) in it
M of Fe2O3 = (2 × 55.85)+( 16 × 3)
= = g/mol
So, g of Fe2O3 contain → g Fe
When the mass of Fe2O3 is 10 g then → ?? g Fe
mfe = 10 × = 6.99 g
159.7

19. CHEMICAL FORMULA: 1) Empirical formula: 2) Molecular formula:
Empirical formula (or simplest formula) for a compound is the formula of a substance written with the smallest integer (whole number) subscripts.
Eg: H2O, CH2
2) Molecular formula:
The molecular formula, you may recall, tells you the precise number of atoms of different elements in a molecule of the substance.
Eg: C2H4

20. CHEMICAL FORMULA:
Compounds with different molecular formulas can have the same empirical formula(CH), and such substances will have the same percentage composition. An example is acetylene, C2H2, and benzene, C6H6.
3) Structural formula :
Example: CH3CH2OH and CH3OCH3
Gives information about the way in which the atoms in the molecule are linked

21. Examples: 1) A sample contain 2.3 g of N and 5.34 g of O
What is the empirical formula of the compound?
First we should calculate Number of moles of each element in the compound.
nN = 2.3 = mol nO = 5.34 = mol
N : O
0.164 : (Divide the highest answer by the smallest answer)
0.164 : 0.333
1: 2
The empirical formula of the compound is NO2

22. Examples:
2) What is the empirical formula of a compound of 43.7% P and 56.3% O by weight?
Solution: Suppose we have 100 g compound, then:
The mass of P = 43.7 g, and the mass of O= 56.3 g
nP = 43.7 = 1.41 mol nO = 56.3 = 3.52 mol
FORMULA : P : O
1.41: 3.52
1 : (× 2)
2 : 5
The empirical formula of the compound is P2O5

23. Examples:
3) A sample contains only C and H was burned to give CO2 and H2O , If the weight of CO2 is g and of H2O is g, then calculate the empirical formula of the compound
The Final Answer:
The empirical formula of the compound is CH3

24. Example:
4) If an empirical formula for a substance is NO2, and the Mwt of that compound is 92 g/mol. Determine the molecular formula?
Solution:
The empirical formula is NO2
Molar mass or Mwt of NO2
= ×16= 46 g/mol
92/46 = 2
Molecular formula is (NO2) N2O4

25. Example:
5) A 0.1 g sample contain C, H and O. if it is burned and produced g CO2 and g H2O. What is the empirical formula of the compound?

26. Solution: =C2H6O Mass of C in the sample = (12/44) × 0.191 = 0.052 g
Mass of H in the sample = (2/18) × = g
Mass of O in the sample = 0.1-( ) = g
Moles of C = 0.052/12= mol
Moles of H = /1= mol
Moles of O =0.035/16= mol
FORMULA:
C / H 0.013/ O /
=C2H6O

27. Chemical
Reactions

28. Chemical Reactions: Chemical reactions are the heart of chemistry.
All chemical reactions must involve detectable change.
A chemical reaction involves a change from: reactant substances to product substances.
The product substances will have physical and chemical properties different from those of the reactants.

30. A chemical reaction usually involves a physical and chemical changes:
1) Physical Properties:
color
melting point
boiling point
electrical conductivity
specific heat
density
state (solid, liquid, or gas)

31. 2)Physical Changes: Changes in physical properties melting boiling
condensation
No change occurs in the identity of the substance
Example:
Ice , rain, and steam are all water

32. 3)Chemical Changes:
Atoms in the reactants are rearranged to form one or more different substances
Old bonds are broken; new bonds form
Examples:
Fe and O2 form rust (Fe2O3)
Ag and S form tarnish (Ag2S)

33. Example:
Classify each of the following as a 1) physical change or 2) chemical change A. ____ a burning candle B. ____ melting ice C. ____ toasting a marshmallow D. ____ cutting a pizza E. ____ polishing silver

34. Solution: Classify each of the following as a
1) physical change or ) chemical change
A. __2__ a burning candle
B. __1_ melting ice
C. __2__ toasting a marshmallow
D. __1__ cutting a pizza
E. __2__ polishing silver

35. Chemical Reaction:
A process in which at least one new substance is produced as a result of chemical change.

36. Indicators of chemical reactions:
Emission of light or heat
Formation of a gas
Formation of a precipitate
Color change
Emission of odor

37. All chemical reactions:
have two parts:
1) Reactants: the substances you start with
2) Products: the substances you end up with
The reactants turn into the products.
Reactants → Products
In a chemical reactions:
The way atoms are joined is changed
Atoms aren’t created or destroyed.
Can be described several ways

38. Symbols used in equations
(s) after the formula –solid Cu(s)
(g) after the formula –gas H2 (g)
(l) after the formula -liquid H2O(l)
(aq) after the formula - dissolved in water, an aqueous solution. CaCl2 (aq)
­ used after a product indicates a gas (same as (g)) O2 ­
¯ used after a product indicates a solid (same as (s)) CaCo3 ¯

39. Symbols used in equations
indicates a reversible reaction.
Heat , ∆ , shows that heat is supplied to the reaction.
Pt , Catalyst it indicates that a catalyst is used or supplied. In this example, platinum.
Pressure , 2 atm indicates a pressure other than STP

40. Summary of Symbols

41. What is a catalyst?
A substance that speeds up a reaction without being changed by the reaction. For example Pt, Pd.
Enzymes are biological or protein catalysts.

42. Reaction Energy:
All chemical reactions are accompanied by a change in energy. 
Exothermic - reactions that release energy to their surroundings (usually in the form of heat)
Endothermic - reactions that need to absorb heat from their surroundings to proceed.

43. Types of chemical Reactions
Synthesis Reactions
Decomposition Reactions
Single Replacement Reactions
Double Replacement Reactions
Combustion Reactions

44. Synthesis Reactions Synthesis reactions always yield one product.
Synthesis reactions occur when two substances (generally elements) combine and form a compound.
The general form of a synthesis reaction is:
A + B→AB.
Example: 2H2 + O2  2H2O
Example: C + O2  CO2

45. Here is another example of a synthesis reaction.

46. Decomposition Reactions
These are the opposite of synthesis reactions, "take things apart". Just as synthesis reactions can only form one product, decomposition reactions can only start with one reactant.
With the format :
AB → A + B
Example: 2 H2O  2H2 + O2

47. Example of a decomposition Reaction: 2 HgO  2Hg + O2

48. Single Replacement Reactions
Single replacement reactions, also called single displacement.
Occur when one element replaces another in a compound.
A metal can replace a metal (+) OR a nonmetal can replace a nonmetal (-).

49. Single Replacement Reactions
In the format : AB + C→AC + B
A + BC  AC + B (if A is a metal)OR
A + BC  BA + C (if A is a nonmetal)
(remember the cation always goes first!)

50. Example on Single Replacement Reactions:

51. Another Example on Single Replacement Reactions:
Sodium chloride solid reacts with fluorine gas.
NaCl(s) + F2(g)  NaF(s) + Cl2(g)
Note that fluorine replaces chlorine in the compound

52. Double Replacement Reactions
Double Replacement Reactions occur when a metal replaces a metal in a compound and a nonmetal replaces a nonmetal in a compound
In the format :
AB + CD  AD + CB

53. Double Replacement Reactions
first and last ions go together + inside ions go together. Examples:
AgNO3(aq) + NaCl(s)  AgCl(s) + NaNO3(aq)
K2SO4(aq) + Ba(NO3)2(aq)  KNO3(aq) + BaSO4(s)

54. Double Replacement Reactions
1)Precipitation and
2)Acid-Base Neutralization
Are kinds of Double Replacement Reactions.
1)Precipitation:
This can occur when two soluble salts (ionic compounds) are mixed and form an insoluble one—the precipitate.
Pb(NO3)2 (aq) + 2NaI(aq) PbI2 (s) + 2NaNO3 (aq)

55. 2)Acid-Base Neutralization:
In simple terms, an acid is a substance which can lose a H+ ion (a proton) and a base is a substance which can accept a proton. When equal amounts of an acid and base react, they neutralize each other, forming species which aren’t as acidic or basic.
Acid (H+) + Base (OH-) → Salt + H2O(l)
Examples:
HCl (aq)+ NaOH (aq)→ H2O(l)+ NaCl (aq)
NH4OH (aq) + H2SO4 (aq) → (NH4)2SO4 (aq) + H2O(l)

56. Combustion Reactions:
Combustion reactions occur when a hydrocarbon reacts with oxygen gas.
This is also called burning!!! In order to burn something you need the 3 things in the “fire triangle”: 1) A Fuel (hydrocarbon) 2) Oxygen to burn it with 3) Something to ignite the reaction(spark)

57. Combustion Reactions:
In general: CxHy + O2  CO2 + H2O
Products in combustion are ALWAYS carbon dioxide and water. (although incomplete burning does cause some by-products like carbon monoxide)
Combustion is used to heat homes and run automobiles (octane, as in gasoline, is C8H18)
Example:
C5H O2  CO2 + H2O

58. Thank you