Download Unit 4 - WTPS.org

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

DNA-encoded chemical library wikipedia , lookup

Drug discovery wikipedia , lookup

Metalloprotein wikipedia , lookup

Transcript
Unit 4
Chapter 7
Writing Formulas & Naming Compounds
Chemical Formulas and Chemical Compounds
CHAPTER 7
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
there are 8 carbon
atoms in the molecule.
there are 18 hydrogen
atoms in the molecule.
aluminum sulfate — Al2(SO4)3
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.
Chapter 7 – Section 1: Chemical Names and Formulas
Chemical Formulas
Sample Problem
Count the number of atoms in the following
chemical formulas:
Solution:
1 Calcium, 2 Oxygens, and 2 Hydrogens
a. Ca(OH)2
1 Potassium, 1 Chlorine, and 3 Oxygens
b. KClO3
1 Nitrogen, 5 Hydrogens, and 1 Oxygen
c. NH4OH
2 Irons, 3 Chromiums, and 12 Oxygens
d. Fe2(CrO4)3
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.
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.)
+1
+2
+3
-3 -2 -1
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Molecular
Ionic
Binary
Polyatomic
Ions
Simple
(Main Group Elements)
Stock System
(d-Block Elements)
Prefix
System
Stock
System
Acids
Binary
Acids
Oxyacids
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:
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
Al23+ O32–
The correct formula for aluminum oxide is
Al2O3
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
Chapter 7 – Section 1: Chemical Names and Formulas
Binary Ionic Compounds
Sample Problem
Write chemical formulas for :
a. Magnesium Iodide
b. Calcium Oxide
–
Mg2+
I
1
2
Ca22+ O 2–
2
Solution:
MgI2
CaO
Hint: Always divide subscripts by their largest common factor .
Write the correct names for:
a. Li2S
b. ZnCl2
Lithium Sulfide
Zinc Chloride
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Molecular
Ionic
Binary
Polyatomic
Ions
Simple
(Main Group Elements)
Stock System
(d-Block Elements)
Prefix
System
Stock
System
Acids
Binary
Acids
Oxyacids
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+ iron(II)
Fe3+ iron(III)
Visual Concept
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:
3+ F –
Cr
Write the ions side by side, cation first.
1
3
Cross over the charges to give subscripts.
CrF3
Chromium forms more than one ion, so its name
must include the charge as a Roman numeral.
Chromium (III) Fluoride
Chapter 7 – Section 1: Chemical Names and Formulas
Stock System Naming
Sample Problem B
Write chemical formulas for :
a. Tin (IV) Iodide
b. Iron (III) Oxide
Sn4+1 I – 4
Fe23+ O 2–
3
Solution:
SnI4
Fe2O3
Write the correct names for:
3+
V F3
a. VF3
Vanadium (III) Fluoride
2+
2Copper (II) Oxide
Cu O
b. CuO
Hint: “Uncross” subscripts to get the charges of the ions.
Be sure to verify the charge of the anion.
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Molecular
Ionic
Binary
Polyatomic
Ions
Simple
(Main Group Elements)
Stock System
(d-Block Elements)
Prefix
System
Stock
System
Acids
Binary
Acids
Oxyacids
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.
There are 3 sulfate ions
Example: Al2(SO4)3
in this compound
Visual Concept
Chapter 7 – Section 1: Chemical Names and Formulas
Polyatomic Ions
Sample Problem
Write chemical formulas for :
a. Calcium Hydroxide
b. Tin (IV) Sulfate
Solution:
Ca(OH)2
Ca2+1 OH 2–
Sn4+ SO4 2– Sn(SO4)2
2
4
Hint: Remember to divide subscripts by their largest common factor .
Write the correct names for:
a. (NH4)3 PO4
2+
-
b. Cu(NO3)2 Cu (NO3)2
Ammonium Phosphate
Copper(II) Nitrate
Hint: “Uncross” subscripts to get the charges of the ions.
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Molecular
Ionic
Binary
Polyatomic
Ions
Simple
(Main Group Elements)
Stock System
(d-Block Elements)
Prefix
System
Stock
System
Acids
Binary
Acids
Oxyacids
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: P4O10
CO
tetraphosphorus decoxide
carbon monoxide
Chapter 7 – Section 1: Chemical Names and Formulas
The Prefix System
Sample Problem
Write chemical formulas for :
a. dinitrogen trioxide
b. carbon tetrabromide
Solution:
N2O3
CBr4
Write the correct names for:
a. As2S3
diarsenic trisulfide
b. PCl5
phosphorus pentachloride
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Molecular
Ionic
Binary
Polyatomic
Ions
Simple
(Main Group Elements)
Stock System
(d-Block Elements)
Prefix
System
Stock
System
Acids
Binary
Acids
Oxyacids
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:
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 -ous
3. 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
Naming Acids
Sample Problem
Write the correct name for each of the following:
Type of Acid:
Name:
binary acid
hydrofluorine
ic acid
a. HF
b. HNO2
c. H2S
oxyacid
binary acid
d. H2SO4
oxyacid
e. H3PO4
oxyacid
nitrite
ous acid
hydrosulfuric acid
sulfate
uric acid
oric acid
phosphate
Chapter 7 – Section 1: Chemical Names and Formulas
Nomenclature Flowchart
Compounds
Molecular
Ionic
Binary
Polyatomic
Ions
Simple
(Main Group Elements)
Stock System
(d-Block Elements)
Prefix
System
Stock
System
Acids
Binary
Acids
Oxyacids
Chapter 7 – Section 2: Oxidation Numbers
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
Chapter 7 – Section 2: Oxidation Numbers
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 Numbers
Sample Problem
Assign oxidation numbers to each atom in the
following compounds or ions:
Solution:
6+ 1-1 x 6 = -6
a. UF6
F = -1, U = +6
+ 6
0
1+ 6+ 2-2 x 4 = -8
O = -2, H = +1, S = +6
b. H2SO4
+1 x 2 = +2
+ 6
0
5+ 2O = -2, Cl = +5
c. ClO3-2 x 3 = -6
+ 5
-1
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
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:
Prefix System:
Stock System:
3+ 2a. As2S3 diarsenic trisulfide arsenic (III) sulfide
-2 x 3 = -6
+3 x 2 = +6
+
0
6+ 2sulfur (VI) oxide
sulfur trioxide
b. SO3
-2 x 3 = -6
+6
+
0
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 = 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 Masses
Sample Problem
Determine the molar mass of each of the
following compounds:
Solution:
Al2 = 27.0 x 2 =
54.0 g
a. Al2S3
S3 = 32.1 x 3 = + 96.3 g
150.3 g
=
a. Ba(OH)2 Ba
O2 = 16.0 x 2 =
H2 = 1.0 x 2 =
137.3 g
32.0 g
+ 2.0 g
171.3 g
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
given
conversion factor
2.5 mol H2O x 18.0 g H2O = 45 g H2O
1 mol H2O
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+
1. Determine the correct formula: Cu (NO3) 2 Cu(NO3)2
= 63.5 g
2. Calculate the molar mass: Cu
N2 = 14.0 x 2 = 28.0 g
3. Convert from g to mol:
O6 = 16.0 x 6 = + 96.0 g
Given
1170 g Cu(NO3)2 x
Conversion factor
187.5 g
1 mol Cu(NO3)2 = 6.24 mol
Cu(NO3)2
187.5 g Cu(NO3)2
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.
% of element = mass of element in compound x 100
molar mass of compound
Visual Concept
Chapter 7 – Section 3: Using Chemical Formulas
Percent Composition
Sample Problem
Find the percentage composition of copper(I)
sulfide, Cu2S.
Cu2 = 63.5 x 2 = 127.0 g
Solution:
S
= + 32.1 g
1. Find the molar
mass of Cu2S:
159.1 g
2. Find the percentage by mass of each element:
127.0 g
% Cu =
x 100 = 79.8% Cu
159.1 g
32.1 g
%S=
x 100 = 20.2% S
159.1 g