1. Naming inorganic Compounds and Writing Formulas
Part 1
Naming inorganic Compounds and Writing Formulas

2. Common and systematic names
IUPAC (international Union of Pure and Applied Chemistry)
Each compound has a name indicating its composition and its properties
A lot of chemicals have common names as well as the proper IUPAC name
Chemicals that should always be named by common name and never named by the IUPAC method are:
H2O water, not dihydrogen monoxide
NH3 ammonia, not nitrogen trihydrid

3. To keep in mind! 3

4. Rules to assign Oxidation Numbers (O.N.)
Classification of molecules
Rules to assign names and write formula
4. Chemical reactions  balancing equations

5. 1. Oxidation number
Oxidation number of an element represents the positive or negative character of an atom of that element in a particular bonding situation

6. Rules for Assigning Oxidation Numbers (O.N.)
The oxidation number of an atom in the elemental state is zero Example: Cl2 and Al both are 0 The oxidation number of a monatomic ion is equal to its charge Example: In the compound NaCl, the sodium has an oxidation number of 1+ and the chlorine is 1- The algebraic sum of the oxidation numbers in the formula of a compound is zero Example: the oxidation numbers in the NaCl add up to 0 The oxidation number of hydrogen in a compound is 1+, except when hydrogen forms compounds called hydrides where it is 1- Examples: H is 1+ in H2O, but 1- in NaH The oxidation number of oxygen in a compound is 2-, except in peroxides when it is 1- Example: In H2O the oxygen is 2-, in H2O2 it is 1- The algebraic sum of the oxidation numbers in the formula for a polyatomic ion is equal to the charge of that ion Example: in the sulfate ion, SO42-, the oxidation numbers of the sulfur and the oxygens add up to 2-. The oxygens are 2- each, and the sulfur is 6+ 6 6

7. Assigning Oxidation Numbers (O.N.)
The oxidation number is equal to the charge
Ionic compounds
monatomic ions
Barium cloride BaCl2 (Ba2+ 2Cl-) (Ba) (Cl)
Sodium solfate Na2SO4 (2Na+ SO42-) (Na) (SO42-)
+6(S) -8(O) = -2
polyatomic ion
The algebraic sum of the O.N. in the formula is equal to the charge of the ion  O.N. of the sulfur and the oxygens add up to 2-
The O.N in the formula add up to 0
Covalent compounds
Hydrogen oxide H2O (O) (H)
Hydrogen H (H)
The O.N. of an atom in the elemental state is zero 7

8. Potassium dichromate K2Cr2O7 (2K+ Cr2O72-)
Determining the oxidation state
The algebraic sum of the oxidation numbers in the formula of a compound is zero
Potassium dichromate K2Cr2O7 (2K+ Cr2O72-)
N.O. Cr = ? (Cr = +2, +3, +6)
0 = +1(K) + x(Cr) -2(O)
0 = +1(2) + x(2) -2(7)
0 = +2 + x(2) -14
x = /2
 Cr = + 6 8

9. Classifying the inorganic compounds
BINARY COMPOUNDS composed of just two elements
TERNARY COMPOUNDS containing three or more elements

10. Classifying the inorganic compounds
IONIC OXIDE
Me + nMe
IONIC compounds
SALTS
COVALENT OXIDE
BINARY COMPOUNDS
nMe + nMe
COVALENT compounds
SALTS
H+ nMe
HYDRACID

11. Classifying the inorganic compounds
Cation Anion
IONIC OXIDE
Me + nMe
IONIC compounds
SALTS
COVALENT OXIDE
BINARY COMPOUNDS
nMe + nMe
COVALENT compounds
SALTS
H+ nMe
HYDRACID

12. Cation name (Me) + Anion name (n-Me) ending in –IDE
Naming binary compounds
IONIC COMPOUNDS (Me + n-Me)
Cation name (Me) + Anion name (n-Me) ending in –IDE
Ionic OXIDE
K2O (2K+ O2-) Potassium oxide
Me + nMe
 IONIC COMPOUNDS
Binary SALTS
BaCl2
(Ba2+ 2Cl-) Barium cloride

13. Monatomic Ions

14. When the METALS can assume more than one oxidation state……
Naming binary compounds
IONIC COMPOUNDS
When the METALS can assume more than one oxidation state……

15. Cations formed from Transition Metals

16. Fe (+2, +3) (older name) (systematic-stock name)
+2  FeO ferrous oxide iron (II) oxide
+3 Fe2O ferric oxide iron (III) oxide
suffix OUS (lower)
suffix IC (higher)
Roman numerals

17. Classifying the inorganic compounds
Me + nMe
nMe + nMe
IONIC compounds
HYDRACID
IONIC OXIDE
SALTS
COVALENT compounds
COVALENT OXIDE
1H+ nMe
BINARY COMPOUNDS

18. Naming binary compounds
HYDRACID (1H + nMe)
prefix Hydro + anion name (nMe) ending in –IC + acid
-IDE  -IC
Anion Acid
1H+nMe  HYDRACID HF (H+ F- ) fluoride
hydrofluoric acid
hydrogen fluoride

19. Classifying the inorganic compounds
IONIC OXIDE
Me + nMe
IONIC compounds
SALTS
COVALENT OXIDE
BINARY COMPOUNDS
nMe + nMe
COVALENT compounds
SALTS
H+ nMe
HYDRACID

20. COVALENT COMPOUNDS (nMe + n-Me)
Naming binary compounds
COVALENT COMPOUNDS (nMe + n-Me)
1° n-Me name + 2° n-Me name ending in –IDE
PREFIX in case of two or more atoms
Prefix
(mono)- 1 di
tri
tetra
penta
hesa
(use the prefix!)
COVALENT COMPOUNDS
Covalent OXIDE CO2 carbon dioxide
N2O3 dinitrogen trioxide
SALTS CCl4 carbon tetracloride
PCl5 phosphorus pentachloride
nMe + nMe 

21. Si B P H C S I Br N Cl O F
The elemet that occurs first in this series is written and name first, the neme of the second element retains the -IDE ending

22. I prefissi valgono pure per gli ossidi ionici?!!!
Naming oxides
I prefissi valgono pure per gli ossidi ionici?!!!
stock
CrO3 chromium (VI) oxide chromium trioxide
MnO2 manganese (IV) oxide manganese dioxide
SO sulfur (VI) oxide sulfur trioxide
SO sulfur (IV) oxide sulfur dioxide
tradizionale

23. Classifying the inorganic compounds
OXYACID
HYDROXIDE
TERNARY SALTS
H + polyatomic anion
Me + OH-
Me + polyatomic anion
TERNARY
COMPOUNDS

24. TERNARY SALTS AND OXYACIDS CONTAIN POLYATOMIC IONS24

25. TERNARY SALTS Cation name (Me) + polyatomic anion ending in
Naming ternary compounds
TERNARY SALTS
Cation name (Me) polyatomic anion ending in
(higher) (lower) ox. State
–ATE or –ITE
Depending on the charge of the n-Me
Na3PO4 (3Na+ PO43-) sodium phosphate
MgSO4 (Mg2+ SO42-) magnesium sulfate S = -6
MgSO3 (Mg2+ SO32-) magnesium sulfite S = -4 25

26. Depending on the charge of the n-Me
Naming ternary compounds
OXYACID
Name of the polyatomic anion changing its ending as follows:
Anion Acid
-ATE  -IC
-ITE  -OUS
Depending on the charge of the n-Me
followed by –”acid”
H + polyatomic anion  OXYACID
H2SO4 (2H+ SO42-)  sulfate
sulfuric acid
HClO2 (H+ ClO2-)  clorite
chlorous acid 26

27. Naming ternary compounds
In case of multiple oxidation states
 Prefix and suffix
Anion name Acid name
for the lowest n.ox HYPO ………….–ITE HYPO….………OUS
2° n.ox ……………….-ITE ….…………….OUS
3° n.ox ………………-ATE ….…………………IC
for the higher n.ox PER ….………..-ATE PER….……………IC
Es. Cl n.ox. 1, 3, 5, 7
n.ox. +1 hypochlorite ClO-  HCIO hypochlorous acid
n.ox. +3  chlorite ClO  HClO2 chlorous acid
n.ox. +5  chlorate ClO  HClO chloric acid
n.ox. +7  Perchlorate ClO  HClO perchloric acid 27

28. Classifying the inorganic compounds
OXYACID
HYDROXIDE
TERNARY SALTS
H + polyatomic anion
Me + OH-
Me + polyatomic anion
TERNARY
COMPOUNDS 28

29. HYDROXIDE Cation name (Me) + “hydroxide” Naming ternary compounds
Me + OH-  HYDROXIDE NaOH (Na+ OH-) sodium hydroxide

30. Features and Balancing
Part 2
Chemical reaction:
Features and Balancing

31. One substance changes chemically into another substance
Chemical equation
One substance changes chemically into another substance
Reactants starting materials
Products resulting substance
Additional information in a chemical reaction:
Physical state of the substance (solid, liquid, or gas)
Identifies the solvent, if there is one (a solvent is the solution the materials are dissolved in, such as water)
Experimental conditions such as heat, light, or electrical energy added

32. Types of Chemical Reactions
There are four main types of chemical reactions
Precipitation reactions
Reactions with Oxygen
Acid-base reactions
Oxidation-reduction reactions

33. Precipitation reactions
A chemical change that produces an insoluble product that will form a solid.
Usually the solid can be seen “falling out” of the solution, hence, called precipitation. At other times the solid makes the solution turn from clear to cloudy.
2. Reactions with oxygen
Many substances react with oxygen.
If the substance contains carbon then carbon dioxide is usually produced.
If the substance contains hydrogen, then water is usually produced.

34. 4. Oxidation Reduction Reactions
3. Acid-base Reactions
This involves an acid combining with a base to form a salt
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(aq)
4. Oxidation Reduction Reactions
Involves the transfer of negative charge from one reactant to another
Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
An example would be

35. Features of a chemical reaction
CaCO3(s) →∆ CaO(s) + CO2(g)
The products are on the right of the arrow.
Reactants are on the left of the arrow.
The arrow indicates the reaction occurs in one direction.

36. Features of a chemical reaction
CaCO3(s) →∆ CaO(s) + CO2(g)
“l” would indicate the substance was a liquid.
“s” indicates the chemical is a solid substance
“g” indicates the substance is a gas

37. Features of a chemical reaction Law of Conservation of Mass
CaCO3(s) →∆ CaO(s) + CO2(g)
The main feature of a chemical equation is that it is balanced with the same number of elements in compounds on both sides of the arrow

38. Balancing a chemical equation
Apply the Law of Conservation of Mass to get the same number of atoms of every element on each side of the equation
CaCO3(s) → CaO(s) + CO2(g)
On the reactant side
On the product side
1 mol Ca
1 mol C
3 moles of O
The equation is balanced!

39. Stoichiometric Coefficients
Law of Conservation of Mass  In a balanced reaction, both sides of the equation have the same number of elements.
The stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation.
This stoichiometric coefficients establish the mole ratio between reactants and products. 

40. Steps to balancing a chemical equation
Step 1 Count the number of moles of atoms of each element on both product and reactant side
H2(g) + O2(g)  H2O(l)
On the reactant side:
2 moles of H
2 moles of O
On the product side:
2 moles of H
1 mole of O

41. Steps to balancing a chemical equation
Step 2 Determine which elements are not balanced
H2(g) + O2(g)  H2O(l)
The oxygen atoms are not balanced in this equation.

42. Steps to balancing a chemical equation
Step 3 Balance one element at a time
H2(g) + O2(g)  H2O(l)
First O: H2(g) + O2(g)  2H2O(l)
Then H: 2H2(g) + O2(g)  2H2O(l)

43. Steps to balancing a chemical equation
Step 4 After you believe you have successfully balanced the equation, check to make sure you have the same number of atoms on both sides of the equation.
2H2(g) + O2(g)  2H2O(l)
4 moles of H
2 moles of O
4 moles of H
2 moles of O

44. The equation is not balanced
Fe3O H Fe H2O
On the reactant side
On the product side
3 moles Fe
1 mole Fe
2 moles H
4 moles O
1 mole O
The equation is not balanced

45. Fe3O H Fe H2O
1) Fe: Fe3O H Fe H2O
2) O: Fe3O H Fe H2O
3) H: Fe3O H Fe H2O

46. The equation is balanced
Fe3O H Fe H2O
On the reactant side
On the product side
3 moles Fe
3 mole Fe
8 moles H
4 moles O
1 mole O
The equation is balanced