1. Chapter 7 – Chemical Formulas and Chemical Compounds
Taken from Modern Chemistry written by Davis, Metcalfe, Williams & Castka

2. Section 7.1 – Chemical Names and Formulas
HW – Notes on section 7.1 pgs
Objectives
Students will be able to :
Explain the significance of a chemical formula
Determine the formula of an ionic compound formed between two given ions
Name an ionic compound given its formula
Using prefixes, name a binary molecular compound from its formula.
Write the formula of a binary molecular compound given its name.

3. Section 7.1 – Chemical Names and Formulas
Significance of a chemical formula
The chemical formula indicates the relative number of atoms of each element in a chemical compound.
Al2(SO4)3
C12H22O11
Note how in this example parenthesis surround a polyatomic anion and the subscript refers to the entire unit
Elements’ subscripts indicate the number of atoms in the compound.

4. Section 7.1 – Chemical Names and Formulas
By gaining or losing electrons many main-group elements form ions with stable configurations.
Monatomic Ions
Group 1 metals lose one e- to give 1+ cations.
Group 2 metals lose two e- to give 2+ cations.
Ions formed from a single atom are known as monatomic ions

5. Section 7.1 – Chemical Names and Formulas
Not all main-group elements readily form ions, C and Si form covalent bonds where they share electrons.
Monatomic Ions
(continued)
The nonmetals in groups 15, 16 & 17 gain e- to form anions.

6. Section 7.1 – Chemical Names and Formulas
Monatomic Ions
(continued) K+
Mg2+
Elements which give up 1 or more e- and take a positive (+) charge are called cations.
Elements which gain 1 or more e- and take a negative (-) charge are called anions. F-
N3-

7. Section 7.1 – Chemical Names and Formulas
Mg2+ K+
Naming Monatomic Ions
(continued)
Potassium
cation
Magnesium
cation
Cation naming is simple, element name and the word cation.
For anions you drop the end of the element name and add –ide to the root.
N3- F-
Fluorine  fluoride
Nitrogen  Nitride

8. Section 7.1 – Chemical Names and Formulas
Binary Ionic Compounds
Compounds composed of two different elements are known as binary compounds
Cation goes first : Mg2+, Br-, Br-
Balance to become electrically neutral
And you get
MgBr2

9. Section 7.1 – Chemical Names and Formulas
Naming Binary Ionic Compounds
The naming system involves combining the names of the compound’s positive and negative ions
Al2O3
Aluminum cation & oxide
And you get
Aluminum Oxide

10. Section 7.1 – Chemical Names and Formulas
Naming Binary Ionic Compounds
The Stock system of nomenclature
Some elements form more than one cation, (no elements form more than one monoatomic anion) each with a different charge – add Roman Numerals
Fe2+
FeO
Iron(II) oxide
Fe3+
Fe2O3
Iron(III) oxide

11. Section 7.1 – Chemical Names and Formulas
Naming Binary Ionic Compounds
Compounds Containing Polyatomic Ions
Oxyanions each is a polyatomic ion that contains oxygen.
Most
Common
(-ate) ending
NO3-
Nitrate
One
Less O
(-ite) ending
NO2-
Nitrite

12. Section 7.1 – Chemical Names and Formulas
Naming Binary Molecular Compounds
Less electronegative element given first, prefix only for multiples
Second element named with prefix indicating # of atoms, with few exceptions ends with –ide (only 2 elements)
The o or a at the end of the prefix is dropped when the word following begins with another vowel.
mono-
di-
tri-
tetra-
penta-
hexa-
hepta-
octa-
nona-
deca-

13. Section 7.1 – Chemical Names and Formulas
Naming Binary Molecular Compounds
Example pg 212
P4O10
Prefix need if more than one
Less-electronegative element
Prefix indicating number
Root name +ide
tetraphosphorus decoxide

14. Section 7.1 – Chemical Names and Formulas
Covalent-Network Compounds
Similar to naming molecular compounds
SiC
Silicon
Carbide
SiO2
Silicon
dioxide
Si3N4
Trisilicon
tetranitride

15. Section 7.1 – Chemical Names and Formulas
Acids and Salts
Acids are a specific class of compound which usually refer to a solution of water of one of these special compounds.
HCl
Hydrochloric
acid
H2SO4
Sulfuric
acid

16. Section 7.1 – Chemical Names and Formulas
Acids and Salts - (continued)
An ionic compound composed of a cation and the anion from an acid is often referred to as a salt.
NaCl
Common
Table Salt

17. Section 7.1 – Chemical Names and Formulas
Practices
PRACTICE – pg 207 Q 1 & 2 all
Practice Naming
PRACTICE – pg 209 Q 1 & 2 all
PRACTICE – pg 211 Q 1 & 2 all
PRACTICE – pg 213 Q 1 & 2 all
SECTION REVIEW – pg 215 Q 2,3 & 4 all

18. Section 7.1 – Chemical Names and Formulas
Quiz Break
Quiz Key

19. Ions Section 7.1 – Chemical Names and Formulas - POGIL
How are ions made from neutral atoms?
KEY

20. Naming Ionic Compounds
Section 7.1 – Chemical Names and Formulas - POGIL
Naming Ionic Compounds
What are the structural units that make up
Ionic compounds and how are they named?
KEY

21. Can a group of atoms have a charge?
Section 7.1 – Chemical Names and Formulas - POGIL
Polyatomic Ions
Can a group of atoms have a charge?
KEY

22. Naming Molecular Compounds
Section 7.1 – Chemical Names and Formulas - POGIL
Naming Molecular Compounds
How are the chemical formula and name of a molecular compound related?
KEY

23. Section 7.1 – V2 – Ions and Compounds
Quiz Break
Quiz Key

24. Section 7.1 – Chemical Names and Formulas - POGIL
Naming Acids
KEY

25. Section 7.2 - Oxidation Numbers
Objectives
HW – Notes on section 7.2 pgs
Students will be able to :
List the rules for assigning oxidation numbers.
Give the oxidation number for each element in the formula of a chemical compound.
Name the binary molecular compounds using oxidation numbers and the Stock sytem.

26. Section 7.2 - Oxidation Numbers
To indicate the general distribution of electrons among bonded atoms in molecular compounds , oxidation numbers (or states) are assigned to the atoms that compose the same.
Some are arbitrary, but they are useful in naming compounds, in writing formulas and in balancing equations.

27. Section 7.2 - Oxidation Numbers
Assigning Oxidation Numbers – the rules
The following are guidelines...
Atoms of pure elements have an oxidation number of zero. Na O2 P4 S8
All zero.

28. Section 7.2 - Oxidation Numbers
Assigning Oxidation Numbers – the rules
The more-electronegative element in a binary compound is assigned the number equal to the negative charge it would have as an anion. The less-electronegative is assigned the number equal to the positive charge it would have as a cation.

29. Section 7.2 - Oxidation Numbers
Assigning Oxidation Numbers – the rules
Fluorine has an oxidation number of -1 as it is the most electronegative element.

30. Section 7.2 - Oxidation Numbers
Assigning Oxidation Numbers – the rules
Oxygen has an oxidation number of -2 in almost all compounds. Exceptions peroxides is -1, compounds with halogens +2
Hydrogen has an oxidation number of +1 in all compounds containing elements that are more- electronegative; it has an oxidation number of -1 in compounds with metals.

31. Section 7.2 - Oxidation Numbers
Assigning Oxidation Numbers – the rules
The algebraic sum of the oxidation numbers of all atoms in a neutral compound = zero.
The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion = the charge of the ion.
Rules 1-7 apply to covalently bonded atoms, oxidation numbers can also be assigned to atoms in ionic compounds.

32. Section 7.2 - Oxidation Numbers
Using Oxidation Numbers for Formulas and Names
Many non-metals have more than one oxidation state. Recall the use of Roman numerals to denote charges.
Formula
Prefix system
Stock System
PCl3
phosphorus trichloride
phosphorus(III) chloride NO
nitrogen
monoxide
nitrogen(II)
oxide
PbO2
lead
dioxide
lead (IV) oxide

33. Practices Section 7.2 Ba(NO3)2 Is the substance elemental?
No, three elements are present.
Is the substance ionic?
Yes, metal + non-metal.
Are there any monoatomic ions?
Yes, barium ion is monoatomic.
Barium ion = Ba2+
Oxidation # for Ba = +2
Which elements have specific rules?
Oxygen has a rule in most compounds
Oxidation # for O = -2
Which element does not have a specific rule?
N does not have a specific rule.
Use rule 8 to find the oxidation # of N
Let N = Oxidation # for nitrogen
(# Ba) (Oxid. # of Ba) + (# N) (Oxid. # N) + (# of O) (Oxid. # of O) = 0
1(+2) + 2(N) + 6(-2) = 0
N = +5
Ba(NO3)2

34. Practices Section 7.2 NF3 Is the substance elemental?
No, two elements are present.
Is the substance ionic?
No, two non-metals.
Are there any monoatomic ions?
Since it is molecular, there are no ions present.
Which elements have specific rules?
F = -1
Which element does not have a specific rule?
N does not have a specific rule.
Use rule 8 to find the oxidation # of N
Let N = oxidation # of N
(# N) (Oxid. # N) + (# F) (Oxid. # F) = 0
1(N) + 3(-1) = 0
N = +3

35. Practices Section 7.2 (NH4)2SO4 Is the substance elemental?
No, four elements are present.
Is the substance ionic?
Yes, even though there are no metals present, the ammonium ion is a common polyatomic cation.
Are there any monoatomic ions?
No, the cation and anion are both polyatomic.
Which elements have specific rules?
H = +1 because it is attached to a non-metal (N)
O = -2
Which elements do not have a specific rule?
Neither N nor S has a specific rule.
You must break the compound into the individual ions that are present and then use rule 9 to find the oxidation numbers of N and S. Notice that if you try to use rule 8, you end up with one equation with two unknowns: 2N + 8(+1) + 1S + 4(-2) = 0
The two ions present are NH4+ and SO42-.
N + 4(+1) = +1 so N = -3
S + 4(-2) = -2 so S = +6

36. Practices Section 7.2 PRACTICE – pg 218 Q 1 all
SECTION REVIEW – pg 219 Q 1 & 2 all
Practice Sheet
Key

37. Section 7.2
Quiz
Key

38. Section 7.3 - Using Chemical Formulas
Objectives
HW – Notes on section 7.3 pgs
Students will be able to :
Calculate the formula mass or molar mass of any given compound.
Use molar mass to convert between mass in grams and amount in moles of a chemical compound.
Calculate the # of molecules, formula units, or ions in a given molar amount of a chemical compound.
Calculate the % composition of a given chemical compound.

39. Section 7.3 - Using Chemical Formulas
Formula Masses
The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all the atoms represented in the formula.
Book example
Average atomic mass of H : 1.01 amu
Average atomic mass of O : amu
H2O  (2 x 1.01 amu) + 16 amu = amu

40. Section 7.3 - Using Chemical Formulas
Molar Masses
A compound’s molar mass is numerically equal to its formula mass.
Book example
Formula mass of H2O = amu
Which is also the molar mass of water g/mol.

41. Section 7.3 – Using Chemical Formulas
Molar Mass as a Conversion Factor – Remember our old friends. . .
There are 3 mole equalities. They are:
1 mol = 6.02 x 1023 particles
1 mol = g-formula-mass (periodic table)
1 mol = 22.4 L for a gas at STP*
These become. . .
[ ] or [ ]
1 mol
6.02 x 1023 particles
[ ] or [ ]
1 mol
g-formula-mass (periodic table)
[----] OR [----]
1 mol
22.4 L

42. Section 7.3 - Using Chemical Formulas
Percentage Composition
The percentage by mass of each element in a compound is known as the percentage composition of the compound.
Mass of X in sample of compound
X 100 % =
Mass % X in compound
Mass of sample of compound

43. Section 7.3 - Using Chemical Formulas
Percentage Composition - example
Percent Composition Example:
Calculate the percent composition of Mg(NO3)2
1 Mg = 1 x 24 = 24
2 N = 2 x 14 = 28
6 O = 6 x 16 = 96
148g/mole
Double check -
do they total 100%?
% Mg = 24/148 x 100 = 16.2%
% N = 28/148 x 100 = 18.9%
% O = 96/148 x 100 = 64.0%

44. Section 7.3 - Using Chemical Formulas
Percentage Composition - practice
Quiz Break
Key

45. Section 7.4 - Determining Chemical Formulas
Objectives
HW – Notes on section 7.4 pgs
Students will be able to :
Define empirical formula, and explain how the terms applies to ionic and molecular compounds.
Determine an empirical formula from either a percentage or a mass composition.
Explain the relationship between the empirical formula and the molecular formula of a given compound.
Determine a molecular formula from an empirical formula

46. Section 7.4 - Determining Chemical Formulas
Calculation of Empirical Formulas
An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratios of the different atoms in the compound.

47. Section 7.4 - Determining Chemical Formulas
Calculation of Empirical Formulas - example
Let’s Determine the empirical formula for a compound which is 54.09% Ca, 43.18% O, and 2.73% H
1) Divide each percent by that element's atomic weight.
1.352/1.352 = 1
2.699/1.352 = 2
2.73/1.352 = 2
Ca = 54.09/40 = 1.352
O = 43.18/16 = 2.699
H = 2.73/1 = 2.73
2) To get the answers to whole numbers, divide through by the smallest one.
This gives us  CaO2H2
better yet  Ca(OH)2

48. Section 7.4 - Determining Chemical Formulas
Calculations of Molecular Formulas
An empirical formula may or may not be a correct molecular formula.
Book example , diborane1’s empirical formula is BH3, any multiple of that equals the same ratio – B2H6,B3H9,B4H12 etc
It is a colorless gas at room temperature with a repulsively sweet odor. Diborane mixes well with air, easily forming explosive mixtures. Diborane will ignite spontaneously in moist air at room temperature.

49. Section 7.4 - Determining Chemical Formulas
Calculations of Molecular Formulas – (continued)
The relationship between an empirical formula and a molecular formula is seen below:
X(empirical formula) = molecular formula
X is a whole-number multiple indicating the factor that you need to multiply the empirical formula by to get the molecular formula.

50. Section 7.4 - Determining Chemical Formulas
Calculations of Molecular Formulas – (continued)
X = molecular formula
empirical formula
Formula mass of diborane = amu
Empirical mass of diborane amu
The molecular formula of diborane is therefore B2H6
(BH3)2 = B2H6

51. Section 7.4 - Determining Chemical Formulas
PRACTICE
Empirical Formula Practice
Molecular Formula Practice