1. Chapter 7 Writing Formulas & Naming Compounds
Unit 4
Chapter 7
Writing Formulas & Naming Compounds

2. Chemical Formulas and Chemical Compounds
Chapter 7

3. Chapter 7 – Section 1: Chemical Names and Formulas
Chemical Formulas
A chemical formula indicates the number of each kind of atom in a chemical compound.
when there is no subscript next to an atom, the subscript is understood to be 1.
Examples:
octane — C8H18
aluminum sulfate — Al2(SO4)3
there are 18 hydrogen atoms in the molecule.
there are 8 carbon atoms in the molecule.
Parentheses surround the polyatomic ion to identify it as a group. There are 3 SO4- groups.
there are 2 aluminum atoms in the formula unit.

4. Chemical Formulas Sample Problem
Chapter 7 – Section 1: Chemical Names and Formulas
Chemical Formulas Sample Problem
Count the number of atoms in the following chemical formulas:
Solution:
Ca(OH)2
KClO3
NH4OH
Fe2(CrO4)3
1 Calcium,
2 Oxygens,
and 2 Hydrogens
1 Potassium,
1 Chlorine,
and 3 Oxygens
1 Nitrogen,
5 Hydrogens,
and 1 Oxygen
2 Irons,
3 Chromiums,
and 12 Oxygens

5. Chapter 7 – Section 1: Chemical Names and Formulas
Cations
Atoms with 1, 2, or 3 valence electrons tend to lose them to form positive ions, which are called a cations.

6. Chapter 7 – Section 1: Chemical Names and Formulas
Anions
Atoms with 5, 6, or 7 valence electrons tend to gain more in order to have an octet (8 electrons ) in their outer shell.
Gaining extra electrons forms negative ions, called anions.

7. Chapter 7 – Section 1: Chemical Names and Formulas
Monoatomic Ions
Monoatomic Ions are ions formed from a single atom.
Some main-group elements tend to form covalent bonds instead of ions (ex. C and Si.) +1 +2 +3 -3 -2 -1

8. Nomenclature Flowchart
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Compounds
Ionic
Ionic
Molecular
Acids
Prefix System
Polyatomic Ions
Stock System
Binary Acids
Oxyacids
Binary
Binary
Simple (Main Group Elements)
Simple (Main Group Elements)
Stock System (d-Block Elements)

9. Writing Binary Ionic Compounds
Chapter 7 – Section 1: Chemical Names and Formulas
Writing Binary Ionic Compounds
Binary Compounds are composed of two elements.
Rules for writing binary ionic compounds:
Write the symbols for the ions, and their charges. Note: The cation is always written first.
Cross over the charges (use the absolute value of each ion’s charge as the subscript for the other ion.)
Simplify the numbers and remove the 1’s.
Example: aluminum oxide
The correct formula for aluminum oxide is
Al3+
O2– 2 3
Al2O3

10. Naming Binary Ionic Compounds
Chapter 7 – Section 1: Chemical Names and Formulas
Naming Binary Ionic Compounds
The name of the cation is given first, followed by the name of the anion.
Monatomic cations are identified simply by the element’s name.
For monatomic anions, the ending of the element’s name is dropped, and the ending -ide is added.
Examples:
cation
anion
Al2O3
aluminum
oxide KF
potassium
fluoride

11. Binary Ionic Compounds Sample Problem
Chapter 7 – Section 1: Chemical Names and Formulas
Binary Ionic Compounds Sample Problem
Write chemical formulas for :
Magnesium Iodide
Calcium Oxide
Write the correct names for:
Li2S
ZnCl2
Solution:
Mg2+
I –
MgI2 1 2
Ca2+
O 2–
CaO 2 2
Hint: Always divide subscripts by their largest common factor .
Lithium Sulfide
Lithium
Zinc Chloride
Zinc

12. Nomenclature Flowchart
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Compounds
Ionic
Ionic
Molecular
Acids
Prefix System
Polyatomic Ions
Stock System
Binary Acids
Oxyacids
Binary
Binary
Simple (Main Group Elements)
Stock System (d-Block Elements)
Stock System (d-Block Elements)

13. Chapter 7 – Section 1: Chemical Names and Formulas
The Stock System
Most d-block elements (transition metals) can form 2 or more ions with different charges.
To name ions of these elements, scientists use the Stock system, designed by Alfred Stock in 1919.
The system uses Roman numerals to indicate an ion’s charge.
Example: Fe2+
Fe3+
iron(II)
iron(III)
Visual Concept

14. Stock System Naming Sample Problem A
Chapter 7 – Section 1: Chemical Names and Formulas
Stock System Naming Sample Problem A
Write the formula and give the name for the compound formed by the ions Cr3+ and F–. Solution: Write the ions side by side, cation first. Cross over the charges to give subscripts. Chromium forms more than one ion, so its name must include the charge as a Roman numeral.
Cr3+
F – 1 3
CrF3
Chromium (III)
Fluoride

15. Stock System Naming Sample Problem B
Chapter 7 – Section 1: Chemical Names and Formulas
Stock System Naming Sample Problem B
Write chemical formulas for :
Tin (IV) Iodide
Iron (III) Oxide
Write the correct names for:
VF3
CuO
Solution:
Sn4+
I –
SnI4 1 4
Fe3+
O 2–
Fe2O3 2 3 3+ - V F3
Vanadium (III)
Fluoride - 2 + 2 Cu O
Copper (II)
Oxide
Hint: “Uncross” subscripts to get the charges of the ions.
Be sure to verify the charge of the anion.

16. Nomenclature Flowchart
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Compounds
Ionic
Ionic
Molecular
Acids
Prefix System
Polyatomic Ions
Polyatomic Ions
Stock System
Binary Acids
Oxyacids
Binary
Simple (Main Group Elements)
Stock System (d-Block Elements)

17. Chapter 7 – Section 1: Chemical Names and Formulas
Polyatomic Ions
Common Polyatomic Ions
A polyatomic ion is a charged group of covalently bonded atoms.
Common endings are -ate or -ite, but there are exceptions.
For more than 1 polyatomic ion, use parentheses with the subscript on the outside.
Example: Al2(SO4)3
There are 3 sulfate ions in this compound
Visual Concept

18. Polyatomic Ions Sample Problem
Chapter 7 – Section 1: Chemical Names and Formulas
Polyatomic Ions Sample Problem
Write chemical formulas for :
Calcium Hydroxide
Tin (IV) Sulfate
Write the correct names for:
(NH4)3 PO4
Cu(NO3)2
Solution:
Ca2+
OH –
Ca(OH)2 1 2
Sn4+
SO4 2–
Sn(SO4)2 2 4
Hint: Remember to divide subscripts by their largest common factor .
Ammonium
Phosphate - 2+ Cu
(NO3)2
Copper(II)
Nitrate
Hint: “Uncross” subscripts to get the charges of the ions.

19. Nomenclature Flowchart
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Compounds
Ionic
Molecular
Molecular
Acids
Prefix System
Prefix System
Polyatomic Ions
Stock System
Binary Acids
Oxyacids
Binary
Simple (Main Group Elements)
Stock System (d-Block Elements)

20. Chapter 7 – Section 1: Chemical Names and Formulas
The Prefix System
Molecular compounds are composed of covalently-bonded molecules.
The old prefix system is still used for molecular compounds.
Name the prefix, then the element. Anions end in -ide.
The prefix mono- usually isn’t used for cations.
Examples: P4O CO
tetraphosphorus
decoxide
carbon
monoxide

21. The Prefix System Sample Problem
Chapter 7 – Section 1: Chemical Names and Formulas
The Prefix System Sample Problem
Write chemical formulas for :
dinitrogen trioxide
carbon tetrabromide
Write the correct names for:
As2S3
PCl5
Solution:
N2O3
CBr4
diarsenic
trisulfide
phosphorus
pentachloride

22. Nomenclature Flowchart
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Compounds
Ionic
Molecular
Acids
Acids
Prefix System
Stock System
Binary Acids
Binary Acids
Polyatomic Ions
Oxyacids
Oxyacids
Binary
Simple (Main Group Elements)
Stock System (d-Block Elements)

23. Chapter 7 – Section 1: Chemical Names and Formulas
Acids
An acid is a certain type of molecular compound. All acids start with H (e.g. HCl, H2SO4).
Acids can be divided into two categories:
Binary acids are acids that consist of H and a non-metal. (e.g. HCl.)
Oxyacids are acids that contain H and a polyatomic ion that includes O (e.g. H2SO4.)

24. Chapter 7 – Section 1: Chemical Names and Formulas
Binary Acids
General rules for naming a binary acid:
Begin with the prefix hydro-.
Name the anion, but change the ending to –ic.
Add acid to the name.
Examples:
HCl, hydrochloric acid.
HBr, hydrobromic acid.
H2S, hydrosulfuric acid.

25. Chapter 7 – Section 1: Chemical Names and Formulas
Oxyacids
General rules for naming an oxyacid :
Name the polyatomic ion.
Replace -ate with -ic or -ite with -ous
Add acid to the name.
Examples:
H2SO4, sulfuric acid.
H2SO3, sulfurous acid.
HNO3, nitric acid.
HNO2, nitrous acid.

26. Naming Acids Sample Problem
Chapter 7 – Section 1: Chemical Names and Formulas
Naming Acids Sample Problem
Write the correct name for each of the following: HF
HNO2
H2S
H2SO4
H3PO4
Type of Acid:
Name:
binary acid
hydro
fluorine
ic acid
oxyacid
nitrite
ous acid
binary acid
hydro
sulfur
ic acid
oxyacid
sulfate
uric acid
oxyacid
phosphate
oric acid

27. Nomenclature Flowchart
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Compounds
Ionic
Molecular
Molecular
Acids
Prefix System
Stock System
Stock System
Binary Acids
Polyatomic Ions
Oxyacids
Binary
Simple (Main Group Elements)
Stock System (d-Block Elements)

28. Chapter 7 – Section 2: Oxidation Numbers
In order to indicate the general distribution of electrons among covalently bonded atoms, oxidation numbers are assigned to the atoms.
Unlike ionic charges, oxidation #’s do not represent actual electrons gained or lost.
Many elements can have different oxidation #’s depending on what they’re combined with.
The 9 Oxidation #’s of Nitrogen

29. Rules for Assigning Oxidation Numbers
Chapter 7 – Section 2: Oxidation Numbers
Rules for Assigning Oxidation Numbers
The sum of the oxidation numbers for a neutral compound equals zero.
The sum of the oxidation numbers for an ion equals the charge of the ion.
Atoms in a pure element are zero.
The most electronegative element in a compound is assigned a negative number equal to the charge it would have as an anion.
Hydrogen is always either +1 or -1.

30. Assigning Oxidation Numbers Sample Problem
Chapter 7 – Section 2: Oxidation Numbers
Assigning Oxidation Numbers Sample Problem
Assign oxidation numbers to each atom in the following compounds or ions:
UF6
H2SO4
c. ClO3-
Solution: 6+ 1-
-1 x 6 = -6
F = -1, U = +6 + 6 1+ 6+ 2-
-2 x 4 = -8
O = -2, H = +1, S = +6
+1 x 2 = +2 + 6 5+ 2-
O = -2, Cl = +5
-2 x 3 = -6 + 5 -1

31. Using Oxidation Numbers in Naming
Chapter 7 – Section 2: Oxidation Numbers
Using Oxidation Numbers in Naming
The Stock System is actually based on oxidation numbers.
It can be used as an alternative to the prefix system for naming molecular compounds.
Prefix system
Stock system
PCl3
phosphorus trichloride
phosphorus(III) chloride
PCl5
phosphorus pentachloride
phosphorus(V) chloride
N2O
dinitrogen monoxide
nitrogen(I) oxide NO
nitrogen monoxide
nitrogen(II) oxide
Mo2O3
dimolybdenum trioxide
molybdenum(III) oxide

32. Using Oxidation Numbers in Naming Sample Problem
Chapter 7 – Section 2: Oxidation Numbers
Using Oxidation Numbers in Naming Sample Problem
Write the correct prefix system name and the correct Stock system name for each of the following:
As2S3
SO3
Prefix System:
Stock System: 3+ 2-
diarsenic trisulfide
arsenic (III) sulfide
-2 x 3 = -6
+3 x 2 = +6 + 6+ 2-
sulfur trioxide
sulfur (VI) oxide
-2 x 3 = -6 +6 +

33. Chapter 7 – Section 3: Using Chemical Formulas
Formula Masses
The formula mass of any compound is the sum of the masses of all the atoms in its formula.
example: Formula mass of water, H2O: H2 = 1.0 amu x 2 = amu O = amu amu
A compound’s molar mass is numerically equal to its formula mass. Only the units are different. (Ex: Molar mass of H2O = 18.0 g.)

34. Molar Masses Sample Problem
Chapter 7 – Section 3: Using Chemical Formulas
Molar Masses Sample Problem
Determine the molar mass of each of the following compounds:
Al2S3
Ba(OH)2
Solution:
Al2 = 27.0 x 2 = g
S3 = 32.1 x 3 = g
150.3 g
Ba = g
O2 = 16.0 x 2 = g
H2 = x 2 = g
171.3 g

35. Molar Mass as a Conversion Factor
Chapter 7 – Section 3: Using Chemical Formulas
Molar Mass as a Conversion Factor
The molar mass of a compound can be used as a conversion factor to convert between moles and grams for a given substance.
Example:
What is the mass of 2.5 moles of H2O?
molar mass of H2O = 18.0 g/mol
conversion factor
given
18.0
g H2O
2.5 mol H2O x =
45 g H2O 1
mol H2O

36. Molar Mass as a Conversion Factor Sample Problem
Chapter 7 – Section 3: Using Chemical Formulas
Molar Mass as a Conversion Factor Sample Problem
Calculate the moles in 1170 g of copper (II) nitrate.
Solution: 2+ - Cu
(NO3)
Cu(NO3)2
1. Determine the correct formula: 2
2. Calculate the molar mass:
Cu = g
N2 = 14.0 x 2 = g
3. Convert from g to mol:
O6 = 16.0 x 6 = g
Conversion factor
187.5 g
Given 1
mol Cu(NO3)2
1170 g Cu(NO3)2 x =
6.24 mol Cu(NO3)2
187.5
g Cu(NO3)2

37. Chapter 7 – Section 3: Using Chemical Formulas
Percent Composition
The percentage by mass of each element in a compound is known as the percent composition of the compound.
mass of element in compound
% of element =
x 100
molar mass of compound
Visual Concept

38. Percent Composition Sample Problem
Chapter 7 – Section 3: Using Chemical Formulas
Percent Composition Sample Problem
Find the percentage composition of copper(I) sulfide, Cu2S.
Find the molar mass of Cu2S:
Find the percentage by mass of each element:
Cu2 = 63.5 x 2 = g
Solution:
S = g
159.1 g
127.0 g
% Cu =
x 100 =
79.8% Cu
159.1 g
32.1 g
% S =
x 100 =
20.2% S
159.1 g