Unit 4 Chapter 7 Writing Formulas & Naming Compounds.
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Transcript of Unit 4 Chapter 7 Writing Formulas & Naming Compounds.
Unit 4Chapter 7
Writing Formulas & Naming Compounds
CHAPTER 7Chemical Formulas and Chemical Compounds
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
Chapter 7 – Section 1: Chemical Names and Formulas
there are 8 carbon atoms in the molecule.
there are 18 hydrogen atoms in the molecule.
there are 2 aluminum atoms in the formula unit.
Parentheses surround the polyatomic ion to identify it as a group. There are 3 SO4
- groups.
Chemical FormulasSample Problem
Count the number of atoms in the following chemical formulas:
Solution:a.Ca(OH)2
b.KClO3
c.NH4OH
d.Fe2(CrO4)3
Chapter 7 – Section 1: Chemical Names and Formulas
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
Cations
• Atoms with 1, 2, or 3 valenceelectrons tend to lose them to form positive ions, which are called a cations.
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.
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.)
Chapter 7 – Section 1: Chemical Names and Formulas
+1+2 +3 -1-2-3
Nomenclature Flowchart
Compounds
Ionic Molecular
Binary PolyatomicIons
PrefixSystem
Acids
StockSystem
Simple (Main Group Elements)
Stock System (d-Block Elements)
BinaryAcids
Oxyacids
Chapter 7 – Section 1: Chemical Names and Formulas
Compounds
Ionic
Binary
Simple (Main Group Elements)
Writing Binary Ionic Compounds• Binary Compounds are composed of two elements.• Rules for writing binary ionic compounds:
1. Write the symbols for the ions, and their charges.Note: The cation is always written first.
2. Cross over the charges (use the absolute value of each ion’s charge as the subscript for the other ion.)
3. Simplify the numbers and remove the 1’s.Example: aluminum oxide
The correct formula for aluminum oxide is
Chapter 7 – Section 1: Chemical Names and Formulas
Al3+
Al2O3
3O2–2
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:
Chapter 7 – Section 1: Chemical Names and Formulas
Al2O3 KF
cation anion aluminum potassium
oxide fluoride
Binary Ionic CompoundsSample Problem
Write chemical formulas for :
a.Magnesium Iodide
b.Calcium Oxide
Write the correct names for:
a.Li2S
b.ZnCl2
Chapter 7 – Section 1: Chemical Names and Formulas
Solution:
Hint: Always divide subscripts by their largest common factor .
Lithium
Zinc
Mg2+2I –
1
Ca2+2O 2–
2
MgI2
CaO
Lithium Sulfide
Zinc Chloride
Nomenclature Flowchart
Compounds
Ionic Molecular
Binary PolyatomicIons
PrefixSystem
Acids
StockSystem
Simple (Main Group Elements)
Stock System (d-Block Elements)
BinaryAcids
Oxyacids
Chapter 7 – Section 1: Chemical Names and Formulas
Compounds
Ionic
Binary
Stock System (d-Block Elements)
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+
Chapter 7 – Section 1: Chemical Names and Formulas
iron(II)iron(III)
Visual Concept
Stock System NamingSample 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.
Chapter 7 – Section 1: Chemical Names and Formulas
Cr3+3F –
1
CrF3
Chromium (III) Fluoride
2
I –
O 2–
Stock System NamingSample Problem B
Write chemical formulas for :
a.Tin (IV) Iodide
b.Iron (III) Oxide
Write the correct names for:
a.VF3
b.CuO
Chapter 7 – Section 1: Chemical Names and Formulas
Solution:
Hint: “Uncross” subscripts to get the charges of the ions.
Vanadium (III)Copper (II)
Sn4+41
Fe3+32
SnI4
Fe2O3
Fluoride
Oxide
Be sure to verify the charge of the anion.
V3+
F3-
Cu+
O -2
Nomenclature Flowchart
Compounds
Ionic Molecular
Binary PolyatomicIons
PrefixSystem
Acids
StockSystem
Simple (Main Group Elements)
Stock System (d-Block Elements)
BinaryAcids
Oxyacids
Chapter 7 – Section 1: Chemical Names and Formulas
Compounds
Ionic
PolyatomicIons
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
Chapter 7 – Section 1: Chemical Names and Formulas
There are 3 sulfate ions in this compound
Common Polyatomic Ions
Visual Concept
(NO3)2
Write chemical formulas for :
a.Calcium Hydroxide
b.Tin (IV) Sulfate
Write the correct names for:
a.(NH4)3 PO4
b.Cu(NO3)2
Sn4+ SO4 2–
OH –
Polyatomic IonsSample Problem
Chapter 7 – Section 1: Chemical Names and Formulas
Solution:
Hint: Remember to divide subscripts by their largest common factor .
Ammonium
Copper(II)
Ca2+21
42
Ca(OH)2
Sn(SO4)2
Phosphate
NitrateHint: “Uncross” subscripts to get the charges of the ions.
Cu2+ -
Nomenclature Flowchart
Compounds
Ionic Molecular
Binary PolyatomicIons
PrefixSystem
Acids
StockSystem
Simple (Main Group Elements)
Stock System (d-Block Elements)
BinaryAcids
Oxyacids
Chapter 7 – Section 1: Chemical Names and Formulas
Compounds
Molecular
PrefixSystem
The Prefix System
• Molecular compounds are composedof 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: P4O10
CO
Chapter 7 – Section 1: Chemical Names and Formulas
tetraphosphorus decoxidecarbon monoxide
The Prefix SystemSample Problem
Write chemical formulas for :
a.dinitrogen trioxide
b.carbon tetrabromide
Write the correct names for:
a.As2S3
b.PCl5
Chapter 7 – Section 1: Chemical Names and Formulas
Solution:
diarsenic
phosphorus
N2O3
CBr4
trisulfide
pentachloride
StockSystem
Nomenclature Flowchart
Compounds
Ionic Molecular
Binary PolyatomicIons
PrefixSystem
Acids
Simple (Main Group Elements)
Stock System (d-Block Elements)
BinaryAcids
Oxyacids
Chapter 7 – Section 1: Chemical Names and Formulas
Compounds
Acids
BinaryAcids
Oxyacids
Acids
• An acid is a certain typeof molecular compound.All acids start with H (e.g. HCl, H2SO4).
• Acids can be divided into two categories:1. Binary acids are acids that consist of H and a
non-metal. (e.g. HCl.)2. Oxyacids are acids that contain H and a
polyatomic ion that includes O (e.g. H2SO4.)
Chapter 7 – Section 1: Chemical Names and Formulas
Binary Acids
• General rules for naming a binary acid:1. Begin with the prefix hydro-.2. Name the anion, but change the ending
to –ic.3. Add acid to the name.Examples:HCl, hydrochloric acid.HBr, hydrobromic acid.H2S, hydrosulfuric acid.
Chapter 7 – Section 1: Chemical Names and Formulas
Oxyacids
• General rules for naming an oxyacid :1. Name the polyatomic ion.2. Replace -ate with -ic or -ite with -ous3. Add acid to the name.Examples:H2SO4, sulfuric acid.
H2SO3, sulfurous acid.
HNO3, nitric acid.
HNO2, nitrous acid.
Chapter 7 – Section 1: Chemical Names and Formulas
sulfate
Naming AcidsSample Problem
Write the correct name for each of the following:
a.HF
b.HNO2
c.H2S
d.H2SO4
e.H3PO4
Type of Acid:binary acid
oxyacid
Name:hydro nitrite
Chapter 7 – Section 1: Chemical Names and Formulas
fluorineic acid
binary acid hydro
sulfuric acid
ous acid
oxyacid uric acidoxyacid phosphateoric acid
StockSystem
Nomenclature Flowchart
Compounds
Ionic Molecular
Binary PolyatomicIons
PrefixSystem
Acids
StockSystem
Simple (Main Group Elements)
Stock System (d-Block Elements)
BinaryAcids
Oxyacids
Chapter 7 – Section 1: Chemical Names and Formulas
Compounds
Molecular
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 electronsgained or lost.
• Many elements can have differentoxidation #’s depending on whatthey’re combined with.
Chapter 7 – Section 2: Oxidation Numbers
The 9 Oxidation #’s of Nitrogen
Rules for Assigning Oxidation Numbers
1. The sum of the oxidation numbers for a neutral compound equals zero.
2. The sum of the oxidation numbers for an ion equals the charge of the ion.
3. Atoms in a pure element are zero.
4. The most electronegative element in a compound is assigned a negative number equal to the charge it would have as an anion.
5. Hydrogen is always either +1 or -1.
Chapter 7 – Section 2: Oxidation Numbers
Assigning Oxidation NumbersSample Problem
Assign oxidation numbers to each atom in the following compounds or ions:
a.UF6
b.H2SO4
c. ClO3-
Solution:F = -1, U = +6
O = -2, H = +1, S = +6
Chapter 7 – Section 2: Oxidation Numbers
O = -2, Cl = +5
1- -1 x 6 = -6 +
06
6+
-2 x 4 = -8
+ 06
2-+1 x 2 = +2
1+ 6+
-2 x 3 = -6 +
-15
2-5+
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.
Chapter 7 – Section 2: Oxidation Numbers
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
Using Oxidation Numbers in NamingSample Problem
Write the correct prefix system name and the correct Stock system name for each of the following:
a.As2S3
b.SO3
Prefix System:diarsenic trisulfide
sulfur trioxide
Chapter 7 – Section 2: Oxidation Numbers
Stock System:2-
-2 x 3 = -6 + 0
+3 x 2 = +6
3+arsenic (III) sulfide
-2 x 3 = -6 + 0
+6
2-6+sulfur (VI) oxide
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 = 2.0 amu. O = + 16.0 amu. 18.0 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.)
Chapter 7 – Section 3: Using Chemical Formulas
Molar MassesSample Problem
Determine the molar mass of each of the following compounds:
a.Al2S3
a.Ba(OH)2
Solution:Al2 = 27.0 x 2 = 54.0 g
Chapter 7 – Section 3: Using Chemical Formulas
S3 = 32.1 x 3 = + 96.3 g
150.3 g
Ba = 137.3 g O2 = 16.0 x 2 = 32.0 g H2 = 1.0 x 2 = + 2.0 g
171.3 g
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
Chapter 7 – Section 3: Using Chemical Formulas
given conversion factor
2.5 mol H2O g H2Omol H2O
45 g H2O=x118.0
2
Molar Mass as a Conversion FactorSample Problem
Calculate the moles in 1170 g of copper (II) nitrate.Solution:
Given Conversion factor
1170 g Cu(NO3)2mol Cu(NO3)2
g Cu(NO3)2
6.24 mol Cu(NO3)2
=x187.5
1
Chapter 7 – Section 3: Using Chemical Formulas
Cu = 63.5 g
O6 = 16.0 x 6 = + 96.0 g
187.5 g
(NO3)Cu2+
1. Determine the correct formula:2. Calculate the molar mass:
-Cu(NO3)2
N2 = 14.0 x 2 = 28.0 g 3. Convert from g to mol:
Percent Composition
• The percentage by mass of each elementin a compound is known as the percent composition of the compound.
Chapter 7 – Section 3: Using Chemical Formulas
% of element = mass of element in compoundmolar mass of compound
x 100
Visual Concept
Percent CompositionSample Problem
Find the percentage composition of copper(I) sulfide, Cu2S.
1.Find the molar mass of Cu2S:
2.Find the percentage by mass of each element:
Solution: Cu2 = 63.5 x 2 = 127.0 g
Chapter 7 – Section 3: Using Chemical Formulas
S = + 32.1 g
159.1 g
% Cu = 127.0 g159.1 g x 100 = 79.8% Cu
% S = 32.1 g159.1 g x 100 = 20.2% S