Transcript Chemical Formulas 4-18-07

Chemistry Chapter 9
Chemical Formulas
and
Chemical Compounds
Heart cell rhythm depends on the opening and
closing of a complex series of valves on the cell
membrane, called ion channels. Some valves let
certain ions ike potassium (K+) flow out, others let
different ions like sodium (Na+) flow in. There
are also pumps that actively move ions one
direction or another.
Nomenclature - Humor
Fe2+
Fe2+
Fe2+
Fe2+
Fe2+
Fe2+
Fe2+
Fe2+
“Ferrous Wheel”
Fe = iron (Latin = ferrum)
Fe2+ = lower oxidation state = ferrous Iron II
Fe3+ = higher oxidation state = ferric Iron III
BaNa2
“BaNaNa”
What weapon can you make
from the elements nickel,
potassium and iron?
A KNiFe
Ions
• Cation: A positive ion
• Mg2+, NH4+
• Anion: A negative ion
• Cl-, SO42-
• Ionic Bonding: Force of attraction
between oppositely charged ions.
Predicting Ionic Charges
Group 1: Lose 1 electron to form 1+ ions
H+
Li+ Na+
K+
Predicting Ionic Charges
Group 2: Loses 2 electrons to form 2+ ions
Be2+
Mg2+
Ca2+
Sr2+
Ba2+
Predicting Ionic Charges
B3+
Al3+
Ga3+
Group 13: Loses 3
electrons to form
3+ ions
Predicting Ionic Charges
Group 14: Lose 4
electrons or gain
4 electrons?
Predicting Ionic Charges
N3- Nitride
P3- Phosphide
As3- Arsenide
Group 15: Gains 3
electrons to form
3- ions
Predicting Ionic Charges
O2- Oxide
S2- Sulfide
Se2- Selenide
Group 16: Gains 2
electrons to form
2- ions
Predicting Ionic Charges
F1- Fluoride
Br1- Bromide
Cl1-Chloride
I1- Iodide
Group 17: Gains 1
electron to form
1- ions
Predicting Ionic Charges
Group 18: Stable
Noble gases do not
form ions!
Predicting Ionic Charges
Groups 3 - 12: Many transition elements
have more than one possible oxidation state.
Iron(II) = Fe2+
Iron(III) = Fe3+
Metal Cations
• Type I Group A metals
– Metals that can only have one possible charge
– Determine charge by position on the Periodic Table
• Type II Transitional Metals Group B
– Metals that can have more than one possible charge
– Determine metal cation’s charge from the charge on
anion (algebraic sumation)
– Know the following:
Memorize
these
Pb+2
Pb+4
Cu+1
Cu+2
Lead II
Lead IV
Copper I
Copper II
Sn+2
Sn+4
Fe+2
Fe+3
Tin II
Tin IV
Iron II
Iron III
Predicting Ionic Charges
Groups 3 - 12: Some transition elements
have only one possible oxidation state.
Zinc = Zn2+ Silver = Ag+ Aluminium = Al+3
Remember these
Chemical Nomenclature
Formula to Name
Types of Compounds
Ionic
Acids
Covalent
Contains a metal or a
polyatomic ion
Binary compound with two
nonmetals
Name ions
Representative metals
(Group IA, IIA, IIIA, Zn, Cd and Ag)
Charge = group #
Name = element name
Ca2+ = calcium ion
Transition metals
(All others)
Charge given (in Roman
numerals) in name
Metal name(charge)
Fe3+ = iron (III)
Nonmetal Ions
Monatomic
Polyatomic ions
Charge = group # - 8
Name = stemide
MEMORIZE
S2- = sulfide
NO CHARGES
Name has number of each
element in the compound
Prefix-first element prefixsecondide
P2O5= diphosphorus
pentoxide
H____(aq)
Name has word, acid, in it
Like ionic compound where H is
the metal
Binary Acids
Oxyacids
H—polyatomic ion
HX(aq)
Hydro-x-ic acid
Name depends on name of
polyatomic ion
HI(aq) = hydroiodic acid
If ion ends in ate then acid
ends in ic acid
If ion ends in ite then acid
ends in ous acid
chlorate = ClO3choric acid = HClO3(aq)
chlorite = ClO2chlorous acid = HClO2(aq)
4
Figure 5.1: A flow chart for naming
binary compounds.
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5–10
Writing Binary Ionic Formulas
Ionic Nomenclature Tutorial
Example: Aluminum sulfide
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges
are balanced.
3. Balance charges , if necessary,
using subscripts. Use parentheses
if you need more than one of a
polyatomic ion.
Simulation
3+
Al
2
2S
3
Not balanced!
Al2S3
Naming Binary Ionic Compounds
• 1. Cation first, then anion
• 2. Monatomic cation = name of the
element
• Ca2+ = calcium ion
• 3. Monatomic anion = root + -ide
• Cl- = chloride
• CaCl2 = calcium chloride
Figure 5.2: Overall strategy for naming
chemical compounds.
Copyright © Houghton Mifflin Company.All rights reserved.
5–12
Writing Ionic Compound Formulas
Example: Barium nitrate
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges are
balanced.
2+
(
Ba NO3 ) 2
3. Balance charges , if necessary,
using subscripts. Use parentheses
if you need more than one of a
polyatomic ion.
Not balanced!
Ba(NO3)2
Writing Ionic Compound Formulas
Example: Ammonium sulfate
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges
are balanced.
( NH4+) SO422
3. Balance charges , if necessary,
using subscripts. Use parentheses
if you need more than one of a
polyatomic ion.
Not balanced!
(NH4)2SO4
Writing Ionic Compound Formulas
Example: Magnesium carbonate
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges
are balanced.
Mg2+ CO32They are balanced!
MgCO3
Writing Ionic Compound Formulas
Example: Zinc hydroxide
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges are
balanced.
2+
Zn
3. Balance charges , if necessary,
using subscripts. Use parentheses
if you need more than one of a
polyatomic ion.
( OH- )2
Not balanced!
Zn(OH)2
Writing Ionic Compound Formulas
Example: Aluminum phosphate
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges are
balanced.
3+
Al
PO4
3-
They ARE balanced!
AlPO4
Naming Ternary Ionic Compounds
(continued)
Metals with multiple oxidation states
some metal forms more than one cation
• - use Roman numeral in name
• -
• PbCl2
• Pb2+ is cation
• PbCl2 = lead(II) chloride
Determining names of transition
metal ions from formulas
Have to look at what it is combined with:
FeBr2
Ions:
Fe = transition metal:
iron( x ) (We need to determine the charge.)
Br = nonmetal ion:
We need to determine the charge on the iron. We can do this by looking at what it is combined with as
below where x represents the charge on the iron. We use the same method we used to determine formulas. We
find the least common multiple. Here we have the multipliers (subscripts)
Ion
subscript LCM
Fex
X 1
=
Br
X 2
= -2
Since the charge on the bromine is -1 and there are two of them, the LCM must be 2.
Ion
subscript
Fex
Br-
LCM
X 1
X 2
= 2
= -2
The total charge from the iron must be 2. Since there is only 1 iron ion, the charge on that ion (x) must be +2 so it
is
iron (II).
65
Assigning Oxidation Numbers
H +1 with nonmetals
O -2 with nonmetals
H -1 with metals (hydride)
O +2 with halogens
( most common)
O -1 in peroxide ion O2-2
Writing Ionic Binary Formulas w/
Transitional Elements
Example: ________
Iron(III) chloride
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges
are balanced.
3. Balance charges , if necessary,
using subscripts. Use parentheses
if you need more than one of a
polyatomic ion.
FeCl
Fe3+ Cl-
3
Not balanced!
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5–8
Naming Covalent Binary Compounds
•
•
•
•
•
Compounds between two nonmetals
- First element in the formula is named first.
- Second element is named as if it were an anion
(-ide).
- Use prefixes
- Only use mono on second element -
P2O5
CO2
CO
N2O
= diphosphorus pentoxide
= carbon dioxide
= carbon monoxide
= dinitrogen monoxide
Figure 5.3: A flow chart for naming acids. The acid is considered
as one or more H+ ions attached to an anion.
A piece of copper metal about to be placed in nitric acid (left). Copper reacts with
nitric acid to produce colorless NO which immediately reacts with the oxygen in
the air to form reddish-brown NO2 gas (right).
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5–9
Properties of Acids
 Acids taste sour
 Acids effect indicators
 Blue litmus turns red
 Methyl orange turns red
 Acids have a pH lower than 7
 Acids are proton (hydrogen ion, H+) donors
 Acids react with active metals, produce H2
 Acids react with carbonates
 Acids neutralize bases
Acids you must know:
Strong Acids
Weak Acids
Sulfuric acid, H2SO4
Phosphoric acid, H3PO4
Hydrochloric acid, HCl
Acetic acid, HC2H3O2
Nitric acid, HNO3
Naming Acids
Binary Acids:2 elements only
Contains Hydrogen plus a monatomic ion
HCl
H+ (Hydro)
+
Cl-(chlorine becomes –ic )
Hydrochloric Acid
Ternary (oxyacid) Acid: 3 or more elements
Contains Hydrogen plus a polyatomic ion
Phosphate (PO4-3) H3PO4 Phosphoric acid
Phosphite (PO3-3) H3PO3 Phosphorous acid
Copyright © Houghton Mifflin Company.All rights reserved.
5–13
Copyright © Houghton Mifflin Company.All rights reserved.
Sulfuric Acid
 Highest volume production of any
chemical
 Used in
 Used in
 Used in
in the U.S.
the production of paper
production of fertilizers
petroleum refining
.Sulfuric
acid is used as a drying agent to chemically remove water
from many substances. If poured on sugar crystals (C11H22O11), the
acid removes eleven molecules of water for every molecule of sucrose.
After dehydrating the sugar, all you have left is a brittle, spongy
black mass of carbon! If sulfuric acid gets on your skin, it immediately
begins to take water out of the molecules in the skin. This is what
causes an acid burn.
Nitric Acid
• Used in the production of
fertilizers
• Used in the production of
explosives
• Nitric acid is a volatile acid
– its reactive components
evaporate easily
• Stains proteins (including
skin!)
Hydrochloric Acid
• Used in the pickling of
steel
• Used to purify magnesium
from sea water
• Part of gastric juice, it
aids in the digestion of
protein
• Sold commercially as
“Muriatic acid”
Phosphoric Acid
o A flavoring agent in sodas
o Used in the manufacture
of detergents
o Used in the manufacture
of fertilizers
o Not a common laboratory
reagent
Acetic Acid
Used in the manufacture of
plastics
Used in making
pharmaceuticals
Acetic acid is the acid
present in vinegar
Acids are Proton Donors
Monoprotic acids
Diprotic acids
HCl
H2SO4
HC2H3O2
H2CO3
HNO3
Triprotic acids
H3PO4
Predicting type of substance
Nomenclature Practice
Calculating Formula Mass
Calculate the formula mass of magnesium carbonate,
MgCO3.
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
From previous slide:
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
 24.31 
Mg  
  100  28.83%
 84.32 
 12.01 
C 
  100  14.24%
 84.32 
 48.00 
O
  100  56.93%
 84.32 
100.00
Formulas
Empirical formula: the lowest whole number
ratio of atoms in a compound.
Molecular formula: the true number of
atoms of each element in the formula of a
compound.
 molecular formula = (empirical
formula)n [n = integer]
 molecular formula = C6H6 = (CH)6
 empirical formula = CH
Formulas
(continued)
Formulas for ionic compounds (formula units)
are ALWAYS empirical (lowest whole number
ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas
(continued)
Formulas for molecular compounds MIGHT
be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Empirical Formula Determination
1. Base calculation on 100 grams of compound.
2. Determine moles of each element in 100
grams of compound.
3. Divide each value of moles by the smallest of
the values.
4. Multiply each number by an integer to obtain
all whole numbers.
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and
6.85% H by mass. What is the empirical formula
of adipic acid?
 49.32 g C 1 mol C   4.107 mol C
12.01 g C 
 6.85g H 1 mol H   6.78 mol H
1.01 g H 
 43.84 g O 1 mol O   2.74 mol O
16.00 g O 
Empirical Formula Determination
(part 2)
Divide each value of moles by the smallest of the
values.
4.107
mol
C
Carbon:
 1.50
2.74 mol O
6.78 mol H
Hydrogen:
 2.47
2.74 mol O
2.74 mol O
Oxygen:
 1.00
2.74 mol O
Empirical Formula Determination
(part 3)
Multiply each number by an integer to obtain all
whole numbers.
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Oxygen: 1.00
x 2
2
Empirical formula: C3H5O2
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
2. Divide the molecular mass by the
mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
3. Multiply the empirical formula by this
number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
(C3H5O2) x 2
=
C6H10O4
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