Download For H 2 O

Background terms
Oxidation: gain or loss of
electron from neutral atom
Oxidation state (valence):
"charge" on atom
General rules for determination
of oxidation states:
Group A elements
1. positive oxidation state is equal to column
number
2. negative oxidation state is equal to column
number minus 8 (beginning with column IVA)
3. if an atom is in an even numbered column, it
will have an even number oxidation state;
similarly with odd numbered columns
4. maximum change is going to be "8" (magic
number)
Group B elements
1. one possible positive oxidation
state is the column number except
for column VIIIB
2. another possible positive
oxidation state is "2+" except for
IIIB which will only have "3+"
More rules:
always
IA +1
IIA +2
Al, Sc +3
Zn, Cd +2
Metallic endings
new system
old system
iron (III)
Fe+3
ferric
iron (II)
Fe+2
ferrous
-ic suffix usually corresponds
to highest oxidation state
element usually takes
Example:
Cr (group VIB) maximum + charge=
group number
6+
3+ chromic (usually exists this way)
2+ chromous
Sn
tin (IV) or stannic SnF4 stannic fluoride
tin (II) or stannous SnF2 stannous fluoride
BINARY COMPOUNDS are compounds
made up of two elements
metal + non-metal
The metal always takes its given name; a common
suffix for a metal is -ium
non-metal suffixes are usually -on, -gen, and -ine
The metal takes its given name, the non-metal suffix
changes to -ide
ex.
NaCl
KF
AlN
sodium chloride
potassium fluoride
aluminum nitride
Cross-over process
write the symbols and charges for the ions next to
each other, always writing the cation first
2. cross over the charges by using the absolute value of
each ion’s charge as the subscript for the other ion
3. check the subscripts and divide them by their largest
common factor to give the smallest possible wholenumber ratio of ions.
1.
The purpose of this process is to ensure that the
compound is electrically neutral.
Binary two non-metals
Here we need the use of various prefixes to tell us
how many of each kind of atom we have present
mono- 1
hexa- 6
di- 2
hepta- 7
tri- 3
octa- 8
tetra- 4
nona- 9
penta- 5
deca- 10
When we name a binary compound composed of two
non-metals we state how many atoms of each kind
we have by using the above prefixes. If we only
have one of the first element listed, we do not need
to state that by using the prefix mono-. However, we
do need to state any other quantity of the elements.
Examples:
ex.
NO
N2O
NO2
N2O3
N2O4
mononitrogen monooxide
dinitrogen monoxide
nitrogen dioxide
dinitrogen trioxide
dinitrogen tetraoxide
When we have similar vowels together, such as two
O's or an A and an O we cancel out the first vowel
to make for more sensible spelling and pronunciation.
Ternary compounds
 Compounds made of 3 or more elements, one of which is
usually oxygen.
metal
(takes its
name)
+
radical (polyatomic ion)
(2 or more elements joined
together with a residual charge; when
written as the second component of a
compound it usually has a negative
charge)
The radical name is based on the amount of oxygen it
contains and is named according to this definition even if it
doesn't contain any oxygen.
RADICAL NAMING CHART
SUFFIX
FORMAT
NAME
1 more O
ClO4than normal
per-rootate
perchlorate
normal
ClO3amount of
O
1 less O
ClO2than normal
-ate
chlorate
-ite
chlorite
hyporoot-ite
hypochlorite
EXPLANATION
EXAMPLE
2 less O
ClOthan normal
More “radical” rules
In group VIIA: Cl, Br, and I follow the preceding
rules; F does not
In group VIA: S and those elements below it follow
the above rules (4 oxygens is the "normal" amount)
In group VA: P and As follow the above rules
(4 oxygens is "normal“ for P and As) [3 oxygens is the
“normal“ amount for N]
An important point to remember is that the oxidation
state of the radical does not change as the amount of
oxygen changes.
Examples that do not follow rules:
CNcyanide
CNOcyanate
SCNthiocyanate
(thio- tells us there is sulfur present)
Examples using polyatomic ions:
NaClO3
MgSO4
AlPO3
sodium chlorate
magnesium sulfate
aluminum phosphite
ACID NAMING
binary acids
To recognize a binary acid, it must be made up of
two elements (one of which is hydrogen) and
must have the word aqueous, abbreviated,
following the formula. The name must include
the prefix hydro- and the word acid, i.e.
Hydro -root-ic acid
ex.
HCl(aq) hydrochloric acid
H2S(aq) hydrosulfuric acid
Ternary acids
ORIGINAL
RADICAL
SUFFIX
per-rootate
-ate
-ite
Hypo-rootite
EXAMPLE
HClO4(aq)
HClO3(aq)
HClO2(aq)
HClO (aq)
ACID
SUFFIX
NAME
Per-root-ic Perchloric
acid
-ic
Chloric acid
-ous
Chlorous acid
Hypo-root- Hypochlorous
ous
acid
Bases
In general, the presence of the hydroxide ion and
the term aqueous are sufficient to denote a base.
ex. NaOH(aq)
sodium hydroxide
(present in lye, Drano, liquid plumber)
Bi(OH)3(aq) bismuth hydroxide
(present in pepto bismol)
Mg(OH) 2(aq) + Al(OH) 3(aq) magnesium hydroxide and
aluminum hydroxide
(present in mylanta)
Crossing over process
Ex: magnesium phosphate
We know that Mg has a +2 charge, and PO4 has a -3
charge. These two numbers do not add up to zero.
Thus, we find a least common denominator and
find out what we must multiply each number by to
get this result. Out LCD is 6, thus we multiply +2 by
3 and -3 by 2. This results in +6 and -6 cancelling
out to zero.
Mg3(PO4) 2
lead (II) nitrite = Pb(NO2) 2
A mole represents 6.02 x 1023 particles of a
substance
Type of matter
Type of particle
Elements
ionic compounds
atoms
formula units
molecular
compounds
molecules
Amadeo Avogadro di
Quarengo (Italian 1776-1856)
is credited with discovering
this number; therefore it is
called Avogadro's number
Examples:
1 mole NaCl
1 mole N2O4
1 mole Fe
1 mole Ca2+
= 6.02 x 1023 formula units
= 6.02 x 1023 molecules
= 6.02 x 1023 atoms
= 6.02 x 1023 ions
(A mole of molecules contains more than a
mole of atoms)
Calculation example: H2O
1 mole molecules = 6.02 x 1023 molecules
1 molecule = 3 atoms
6.02 x 1023 molecules x 3 atoms = 1.80 x 1024
1 mol H2O
molecule atoms/mole H2O
The mass of a mole varies with the
substance
1 mole C = 6.02 x 1023 atoms of C = 12 g C
Formula mass: the sum of the average
atomic masses of all atoms represented in
its formula (in a.m.u.)
CO2 (12.01 amu + 2(16.00 amu) = 44.01 amu
Molar mass: the mass of 1 mole of any
element or compound (in grams/mole)
[numerically equal to formula mass]
For an element the molar mass is the
atomic mass found on the periodic table.
H2O  2(1) + 16 = 18 g/mol
CO2  12 + 2(16) = 44 g/mol
Percent composition: % by mass of each
element in a compound
For H2O
%H = 2 g x 100 = 11.11%
18 g
%O = 16 g x 100 = 88.89%
18 g
For CO2
%C = 12 g x 100 = 27.27%
44 g
%O = 32 g x 100 = 73.73%
44 g
Empirical formula/simplest formula gives the
lowest whole-number ratio of the elements in a
compound (or the lowest whole-number ratio of
moles of atoms in a compound)
ex. Empirical formulas
CH
CH2O
Molecular formulas
C2H2 acetylene
C6H6 benzene
CH2O formaldehyde
C2H4O2 acetic acid
C6H12O6 glucose
Determining the empirical/simplest
formula
Using the % composition, determine
the mass of each of the elements in
100 g of that compound
Convert the masses in grams to
moles
Determine the simplest whole
number mole ratio between elements
and set that equal to the atoms ratio
in the simplest formula
Example: What is the empirical formula of a
compound containing 25.9% N and 74.1 % O?
25.9 g N x 1 mole N = 1.85 mole N ÷ 1.85 = 1 x 2 = 2
14 g N
74.1 g O x 1 mole O = 4.63 mole O÷1.85 = 2.5 x 2 = 5
16 g O
So, the simplest formula is N2O5
Alternate method
 If you are not provided with the percent composition of
the elements in the compound, you may follow this
method:
1. Divide the grams provided of each element
by its molar mass (atomic mass)
2. Divide all mole values by the smallest value
to obtain the smallest whole number ratio
possible.
3. Set those values equal to the subscripts in
the chemical formula.
To calculate the molecular formula from simplest
formula you need the molecular weight of the
compound and the simplest formula.
Ex. If the simplest formula for acetic acid is CH2O
and the molecular mass is 60, what is the
molecular formula?
CH2O
Simplest formula mass = 30 g/mole
30 X = 60
X=2
so multiply all subscripts in the empirical formula
by X to get C2H4O2
A chemical equation allows us to describe
in a concise manner, on paper, a chemical
reaction that has taken place. It represents,
with symbols and formulas, the identities
and relative molecular or molar amounts
of the reactants and products in a chemical
reaction.
Evidence for a chemical reaction:
Energy release as heat or light.
Color change
Evolution of gas (bubbles and/or odor)
Appearance of a solid (precipitate)
Parts of a chemical equation:
 The items on the left of the arrow are
called reactants;  (arrow) means
"yields; the right products
 States of matter are described:
s (↓), l, g (↑), aq
 Other symbols used in chemical
equations are on p. 266, Table 2
 Coefficients represent relative numbers
of particles that take part in the reaction
#atoms are conserved 
mass is conserved
Equations must be balanced; the number of
atoms on both sides of the arrow must be
equal.
Basics for balancing:
1. Start from left to right.
2. Balance polyatomic ions as a single
entity whenever possible.
3. Balance the H's and O's last, and
4. NEVER manipulate subscripts in a
formula.
5. Try and keep the coefficients to the
smallest whole numbers possible
(fractional coefficients are acceptable)
Combination/synthesis reactions: 2 or more
substances react to form a single substance.
Reactants are often elements and/or simple
compounds, often H2O
Products are compounds
Difficult to guess product of nonmetals (must
be told)
Often liberates energy
General format: A + X  AX
Examples:
8Ca (s) + S8 (s)  8CaS (s)
2Mg (s) + O2 (g)  2MgO (s)
2Fe (s) + O2 (g)  2FeO (s)
Na2O (s) + H2O (l)  2NaOH (aq)
Decomposition/analysis reactions: one
compound broken down into two or more
simpler products
Products are often elements and/or
compounds in any combination; difficult to
predict
Binary compounds break down into
their elements
Energy is required for the reaction to
take place
General format: AX  A + X
Decomposition/analysis reactions
Examples:
a. metallic carbonates metallic oxides + CO2 (g)
CaCO3 (s)  CaO (s) + CO2 (g)
(NH4) 2CO3(s)2NH3(g)+H2O(g) +CO2(g)
b. metallic hydroxides  metallic oxides +
H2O
Ca (OH) 2 (s)  CaO (s) + H2O (g)
NaOH and KOH are exceptions
c. metallic chlorates  metallic chlorides + O2
2KClO3 (s)  2KCl (s) + 3O2 (g)
d. some acids  nonmetallic oxides + H2O
H2CO3 (aq)  H2O(l) + CO2 (g)
H2SO3 (aq)  H2O(l) + SO2 (g)
e. some oxides decompose upon heating
2HgO (s)  2Hg (l) + O2 (g)
2PbO2 (s)  2PbO (s) + O2 (g)
f. electric current - electrolysis
2H2O(l)  2H2 (g) + O2 (g)
2NaCl (s)  2Na (s) + Cl2 (g)
2HI (g)  H2 (g) + I2 (g)
Single replacement reactions: atoms of one
element replace atoms of a 2nd similar
element in a compound
determined by relative reactivities of the 2
metals (Activity Series is a list of elements
arranged according to the ease with which
the elements undergo certain chemical
reactions.)
Halogens are nonmetals that are replaced
(activity decreases going down Group VIIA)
 General format:
A + BX  AX + B
Y + BX  BY + X
 Displacement of Hydrogen in water or
acid by a metal
Double replacement reactions: exchange of
positive ions between 2 compounds
Generally are reactions between ionic
compounds in aqueous solutions.
General format:
AX + BY  AY + BX
At least one statement below is usually
true of one of the products
Only slightly soluble and ppt. from
solution
is a gas and bubbles out of solution
is a molecular compound such as H2O
Examples:
2NaOH (aq) + H2SO4 (aq)  Na2SO4 (aq) + 2H2O (l)
(neutralization rxn)
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
(precipitation rxn)
ionic reactions:
Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl-(aq)  AgCl(s)
+ Na+ (aq) + NO3-(aq)
net ionic: Ag+ (aq) + Cl-(aq)  AgCl(s)
Net ionic reactions DO NOT include spectator ions!
Combustion reactions: oxygen reacts with
another substance often producing energy
in the form of heat and/or light
Commonly involve hydrocarbons
Complete combustion produces water
and carbon dioxide
Incomplete combustion produces CO and
C in addition to CO2 and H2O due to
decreased amounts of O2
Examples:
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
2C6H6 (l) + 15O2 (g)  12CO2 (g) + 6H2O(g)
Sub-category – Redox reactions
Oxidation-reduction reactions, typically
called “redox rxns” involve two simultaneous
processes
Oxidation – loss of electrons
Reduction – gain of electrons
Mnemonic device to remember this:
OILRIG – oxidation is loss, reduction is gain
Identifying Redox rxns
0
+2 -2
0
+1 -2
H2 (g) + CuO (s)  Cu (s) + H2O (l)
Hydrogen is oxidized; copper is
reduced
Disproportionation
This occurs when one substance is both
oxidized and reduced during the same
chemical reaction.
+1 -1
+1 -2
0
2H2O2 (aq)  2H2O (l) + O2 (g)
Oxygen is both oxidized and reduced in
this reaction.