Here are the steps to write the formula for calcium bicarbonate:1. Write the ions involved: Ca2+ and HCO3-2. Work out the ratio of ions that will balance the charges: Ca2+ : HCO3- = 1 : 23. Write it out without charges: Ca HCO34. Put subscripts in place to reflect the ratio: Ca(HCO3)2Therefore, the formula for calcium bicarbonate is Ca(HCO3)2

CHEMICAL REACTIONS
1. Describe chemical change in terms of formulae, equations and valency.

2. Recognise oxidation and reduction in terms of electron transfer.

3. Recognise oxidation and reduction in terms of transfer of oxygen and hydrogen.

4. Explain the mechanism of dissolving and explain the formation of a precipitate.

5. Predict the identity of a precipitate in a precipitation reaction.

6. Describe thermal decomposition of metal hydroxides, carbonates and hydrogen
carbonates.

7. Show an understanding of atomic an molecular masses.

8. Calculate percentage composition of compounds.

9. Calculate empirical formulae from data relating to
percentage composition.

10.Calculate mass ratios of chemicals involved in chemical
reactions.

Thursday, 16 September 2010

LANGUAGE


Term Ans Definition GLOSSARY 1 Match the term with its definition > ans’s only in BOB

A. Valency

B. Ion

C. Reactants
D. Products

E. Oxidation

F. Reduction

G. Half -
equation


Term Definition GLOSSARY 1- HANDOUT
A. Ion
B. Precipitate
C. Species
D. Redox 4. a term that describes any oxidation-reduction reaction


Term Ans Definition GLOSSARY 2 Match the term with its definition > ans’s only in BOB


Term Definition GLOSSARY 2- HANDOUT


FORMULAE
AND
EQUATIONS

FORMULAE


VALENCY Copy

Valency is a term that allows us to predict how an atom will bond with other atoms.
It is given as a number.

Two ways of thinking about valency:
1. For ionic compounds it is the number of electrons that an atom will gain or
lose in order for it to gain an octet/duet of electrons.
2. For covalent compounds it is the number of bonds that it will form that will
allow it to gain an octet/duet.
Examples
Oxygen, O has a valency of 2. (it will gain two electrons to form an octet when it
forms an ionic compound or share two electrons to form an octet when it forms a
covalent compound)
• MgO (the Oxygen atom has a 2- charge in this compound)
• CO2 (Each O atom has formed 2 bonds with carbon in this compound)

Repeat the above illustration for Phosphorus. Use Phosphorus trichloride and Sodium
phosphide as the compounds

ESA: p61 - 63 Ex 8A


WRITING FORMULAE Copy
Examples
Use your table of ions to write formluae for the following ionic compounds:
Sodium Fluoride _____________ Magnesium Oxide __________
Sodium Oxide _______________ Zinc Chloride _____________
Lead Nitrate ______________ Potassium Sulphate ____________
Calcium Hydrogen Carbonate __________ Aluminium Oxide _______

THE PROCESS FOR WRITING FORMULAE


Examples

Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE

Example
Write the formula
for Calcium
bicarbonate


Examples


Example 1. Work out the ratio of ions
Write the formula that will cause a balance
for Calcium of positive & negative
bicarbonate charges


Examples


Write the formula
for Calcium
that will cause a balance 1 Ca2+ HCO3 -
of positive & negative
bicarbonate 1 : 2
charges


Examples


Write the formula
for Calcium
bicarbonate 1 : 2
charges

2. Writing it out without the charges


Examples


Write the formula
for Calcium
bicarbonate 1 : 2
charges

2. Writing it out without the charges 2 Ca HCO3


Examples


Write the formula
for Calcium
bicarbonate 1 : 2
charges


3. Put the subscripts in place to reflect the
ratio


Examples


Write the formula
for Calcium
bicarbonate 1 : 2
charges


3. Put the subscripts in place to reflect the 3 Ca(HCO3)2
ratio


Examples


Write the formula
for Calcium
bicarbonate 1 : 2
charges


ratio
Note
The “1” is never shown as a subscript
Ions that are made up of groups must be bracketed if the subscript is “2” or
more.

Examples
NaF


Write the formula
for Calcium
bicarbonate 1 : 2
charges


ratio
Note
more.

Examples
NaF
Sodium Fluoride _____________ Magnesium Oxide __________ MgO



Write the formula
for Calcium
bicarbonate 1 : 2
charges


ratio
Note
more.

Examples
NaF

Sodium Oxide Na2O
_______________ Zinc Chloride _____________


Write the formula
for Calcium
bicarbonate 1 : 2
charges


ratio
Note
more.

Examples
NaF

Sodium Oxide Na2O
_______________ ZnCl2
Zinc Chloride _____________


Write the formula
for Calcium
bicarbonate 1 : 2
charges


ratio
Note
more.

Examples
NaF

Sodium Oxide Na2O ZnCl2
_______________ Zinc Chloride _____________
Pb(NO3)2


Write the formula
for Calcium
bicarbonate 1 : 2
charges


ratio
Note
more.

Examples
NaF

_______________ Zinc Chloride _____________
Pb(NO3)2
Lead Nitrate ______________ K2SO4
Potassium Sulphate ____________


Write the formula
for Calcium
bicarbonate 1 : 2
charges


ratio
Note
more.

Examples
NaF

_______________ Zinc Chloride _____________
Pb(NO3)2
Ca(HCO3)2


Write the formula
for Calcium
bicarbonate 1 : 2
charges


ratio
Note
more.

Examples
NaF

_______________ Zinc Chloride _____________
Pb(NO3)2
Ca(HCO3)2 Al2O3


Write the formula
for Calcium
bicarbonate 1 : 2
charges


ratio
Note
more.

Copy

Method 2: The swop and drop method

1. Write the ions down next to each other.
2. Drop each number down (without the charge) and
swop it to make the subscript of the other element

Example Write the formula for Aluminium Oxide

The valencies

If you are given the
formula then think
backwards to determine
the valencies.


Copy




The valencies

1
Al O 3+ 2-

If you are given the
formula then think
the valencies.


Copy




The valencies

1
Al O 3+ 2-

2
Al2O3 If you are given the
formula then think
the valencies.


FORMULAE FOR SIMPLE IONS

+1 +2 +3 _,,
-1
H* Mg2* Al3+ c1- gz-
hydrogen magnesium aluminium chloride oxide

Li* Ca2* Fe3* oH- COr'-
lithium calcium iron(III) hydroxide carbonate

Na+ Fe2* No,* Soo'-
sodium iron(II) nitrate sulfate

K+ Cu2* HCO3-
potasslum copper(II) hydrogen carbonate

PO43-
Zn2* Phosphate
zinc

Pb2+
lead


UTILIZING BONDING POTENTIAL

Valency can be shown using bonding symbols

Examples

Phosphorus has a valency of 3 and so has the potential to form 3 covalent bonds.
Sometimes written:

P


EQUATIONS
INTERACTIVE


Wri$ng
Chemical
Equa$ons
Chemical
reac$ons
are
the
chemists
shorthand
way

of
describing
chemical
reac$ons.


Wri$ng
Chemical
Equa$ons
Chemical
reac$ons
are
the
chemists
shorthand
way

of
describing
chemical
reac$ons.

The
equa$ons
can
be
either:


Wri$ng
Chemical
Equa$ons
Chemical
reac$ons
are
the
chemists
shorthand
way

of
describing
chemical
reac$ons.

The
equa$ons
can
be
either:

1.
Word
equa$ons


Wri$ng
Chemical
Equa$ons
Chemical
reac$ons
are
the
chemists
shorthand
way

of
describing
chemical
reac$ons.

The
equa$ons
can
be
either:

1.
Word
equa$ons

or

2.
Balanced
symbol
equa$ons


Wri$ng
Chemical
Equa$ons
Chemical
reac$ons
are
the
chemists
shorthand
way

of
describing
chemical
reac$ons.

The
equa$ons
can
be
either:

1.
Word
equa$ons

or

2.
Balanced
symbol
equa$ons

For
both
types
all
reactants
(star$ng
substances)
and

products
of
the
reac$on
(end
substances)
must
be

correctly

iden$ﬁed.


Symbol
Equa$ons

Each
reactant
and
product
is
represented
by
a

formula
that
represents
the
smallest
par$cle
that
can

take
part
in
the
reac$on.


Symbol
Equa$ons

Each
reactant
and
product
is
represented
by
a

formula
that
represents
the
smallest
par$cle
that
can

take
part
in
the
reac$on.

The
formula
of
any
solid
element
is
simply
the

symbol
(i.e.
for
one
atom)


Symbol
Equa$ons

Each
reactant
and
product
is
represented
by
a

formula
that
represents
the
smallest
par$cle
that
can

take
part
in
the
reac$on.

The
formula
of
any
solid
element
is
simply
the

symbol
(i.e.
for
one
atom)

e.g.
Magnesium
Metal

use
Mg


Balancing
Equa$ons

Equa$ons
must
be
“balanced”.
This
is
based
on
the

fact
that
during
a
reac$on
no
atoms
are
destroyed,

nor
are
any
new
atoms
created.

Therefore
the
numbers
of
each
type
of
atom
in
the

reactants
must
equal
the
number
of
each
type
of

atom
in
the
products.


Rules:


Rules:

1. Name
all
of
the
reactants
and
products


Rules:

1. Name
all
of
the
reactants
and
products
2. Write
the
correct
formula
for
each
substance

involved.


Rules:

1. Name
all
of
the
reactants
and
products
2. Write
the
correct
formula
for
each
substance

involved.
3.
Balance
so
the
number
of
each
type
of
atoms

remains
the
same
during
the
reac$on.
Formula
can

not
be
changed,
so
balancing
must
be
done
by

changing
the
number
of
par$cles
involved.


EQUATION 1

Potassium

+

Chlorine
(g)

Potassium
chloride

K

+

Cl2

KCl

2
K

+

Cl2

2
KCl

i.e.
2
atoms
of
potassium
react
with
1
molecule
of

chlorine
to
produce
2
units
of
potassium
chloride.

Why
“units”?


EQUATION 1

Potassium

+

Chlorine
(g)

Potassium
chloride

K

+

Cl2

KCl


EQUATION 1

Potassium

+

Chlorine
(g)

Potassium
chloride

K

+

Cl2

KCl

2
K

+

Cl2

2
KCl


EQUATION 1

Potassium

+

Chlorine
(g)

Potassium
chloride

K

+

Cl2

KCl

2
K

+

Cl2

2
KCl

i.e.
2
atoms
of
potassium
react
with
1
molecule
of

chlorine
to
produce
2
units
of
potassium
chloride.


EQUATION 2

Sodium

+

Oxygen
(g)

Sodium
oxide

Na

+

O2

Na2O

4
Na

+

O2

2
Na2O


EQUATION 2

Sodium

+

Oxygen
(g)

Sodium
oxide

Na

+

O2

Na2O


EQUATION 3

Sodium

+

sulfuric

sodium

+

water

hydroxide

acid

sulfate

NaOH

+

H2SO4

Na2SO4

+

H2O

2
NaOH

+

H2SO4

Na2SO4

+

2
H2O


EQUATION 3

Sodium

+

sulfuric

sodium

+

water

hydroxide

acid

sulfate

NaOH

+

H2SO4

Na2SO4

+

H2O


ACID/CARBONATE REACTION
EQUATION 4

Calcium

+

hydrochloric

calcium

+

carbon

+

water
carbonate

acid

chloride

dioxide

CaCO3

+

HCl

CaCl2

+

CO2

+

H2O

CaCO3

+

2
HCl

CaCl2

+

CO2

+

H2O


ACID/CARBONATE REACTION
EQUATION 4

Calcium

+

hydrochloric

calcium

+

carbon

+

water
carbonate

acid

chloride

dioxide

CaCO3

+

HCl

CaCl2

+

CO2

+

H2O


EQUATION 5

sodium

+

sulfuric

sodium

+

carbon

+

water
bicarbonate

acid

sulfate

dioxide

NaHCO3

+

H2SO4

Na2SO4

+

CO2

+

H2O

2
NaHCO3

+

H2SO4

Na2SO4
+
2
CO2
+

2
H2O


EQUATION 5

sodium

+

sulfuric

sodium

+

carbon

+

water
bicarbonate

acid

sulfate

dioxide

NaHCO3

+

H2SO4

Na2SO4

+

CO2

+

H2O


EQUATION 6

Ferric
oxide

+

carbon

iron
+
carbon
monoxide

Fe2O3

+

C

Fe

+

CO

Fe2O3

+

3
C

2
Fe

+

3
CO


EQUATION 6

Ferric
oxide

+

carbon

iron
+
carbon
monoxide

Fe2O3

+

C

Fe

+

CO


EQUATION 7

Magnesium

+

steam

magnesium
+

hydrogen

hydroxide

gas

Mg

+

H2O

Mg(OH)2

+

H2

Mg

+

2
H2O

Mg(OH)2

+

H2


EQUATION 7

Magnesium

+

steam

magnesium
+

hydrogen

hydroxide

gas

Mg

+

H2O

Mg(OH)2

+

H2


EQUATION 8

zinc

+

sulfuric

zinc

+

hydrogen

acid

sulfate

gas


EQUATION 8

zinc

+

sulfuric

zinc

+

hydrogen

acid

sulfate

gas

Zn

+

H2SO4

ZnSO4

+

H2


EQUATION 8

zinc

+

sulfuric

zinc

+

hydrogen

acid

sulfate

gas

Zn

+

H2SO4

ZnSO4

+

H2

(balanced,

no
modiﬁca2on
needed)


EQUATION 9

Aluminium

+

oxygen

aluminium
oxide

Al

+

O2

Al2O3

4
Al

+

3
O2

2
Al2O3


EQUATION 9

Aluminium

+

oxygen

aluminium
oxide

Al

+

O2

Al2O3


NOTES

OXIDATION
-
REDUCTION

METALS ARE REDUCING AGENTS


Copy
OXIDATION and REDUCTION

Definitions

“LEO the lion Loss of Electrons is Oxidation

goes GER” Gain of Electrons is Reduction

• Oxidation is also the addition of oxygen or the removal of hydrogen.

• Reduction is the removal of oxygen or the addition of hydrogen.

• Oxidation and reduction always occur together

Oxidants & Reductants

• An oxidant is a substance that causes oxidation and is itself reduced.
Example: In the reaction between Zinc and Oxygen, Oxygen is the oxidant. It has
caused Zinc to gain oxygen. Zinc Oxide is produced.

• A reductant is a substance that causes reduction and is itself oxidised.
Example: In the reaction between Hydrogen and Copper oxide, Hydrogen is the
reductant. It has caused Copper oxide to lose oxygen. Copper is produced.

Half equations Copy
REDOX
REACTION TYPES
Simple electron transfer

Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms
become ions.
Example 1
Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO


Half equations Copy
REDOX
REACTION TYPES

become ions.
Example 1

Magnesium loses electrons to form ions


Half equations Copy
REDOX
REACTION TYPES

become ions.
Example 1

Magnesium loses electrons to form ions (Mg --> Mg2+ + 2e-) x 2


Half equations Copy
REDOX
REACTION TYPES

become ions.
Example 1


Oxygen gains electrons to form ions


Half equations Copy
REDOX
REACTION TYPES

become ions.
Example 1


Oxygen gains electrons to form ions O2 + 4e- --> 2O2-


Half equations Copy
REDOX
REACTION TYPES

become ions.
Example 1



These equations can only be added once the first equation has
been doubled. This ensures that the number of electrons gained
in the reaction equals the number of electrons lost.


Half equations Copy
REDOX
REACTION TYPES

become ions.
Example 1




Adding the two equations and cancelling the charges gives the final equation.
The final equation describes both the reduction and oxidation together:


Half equations Copy
REDOX
REACTION TYPES

become ions.
Example 1





2Mg + O2 + 4e- --> 2 Mg2+ + 2O2- + 4e-


Half equations Copy
REDOX
REACTION TYPES

become ions.
Example 1





2Mg + O2 + 4e- --> 2 Mg2+ + 2O2- + 4e-

2Mg + O2 --> 2MgO


Spectator ions Copy

Example 2
Zinc powder added to Copper Sulphate solution. Sulphate ion is a spectator ion.
- final equation:
Zn + CuSO4 --> Cu + ZnSO4

Zn --> Zn2+ + 2e- Oxidation
- half equations:
Cu2+ + 2e- --> Cu Reduction

Zn + Cu2+ + 2e- --> Zn2+ + Cu + 2e-

- Adding the two half equations and cancelling out the charges gives the
final equation (above).

In Summary
Half equations are used to show how electrons are lost and gained
in redox reactions (reactions involving reduction and oxidation)

Copy

Example 3 (class prac)
A bright steel nail is placed in 2 mL of copper sulphate solution (0.1 molL-1) in a test tube.

Observation:

- half equations:

- final equation:

A 2 cm strip of Magnesium ribbon is placed in 2 mL of Lead nitrate solution in a test tube.

Observation:

- half equations:

- final equation:


Copy

A 2 cm strip of Magnesium ribbon is placed in 2 mL of Silver nitrate solution in a test tube.

Observation:

- half equations:

- final equation:

A small (1 - 2 cm strip) of Copper foil is placed in 2 mL of Lead nitrate solution in a test
tube.

Observation:

- half equations:

- final equation:
More redox reaction pracs (ABA P34: “Reactions 4 & 5)

WHEN DISCUSSING CHEMISTRY OF
REDOX REACTION ALWAYS INCLUDE A
DISCUSSION OF ELECTRON TRANSFER IN
ADDITION TO APPLICATION OF OTHER
PRINCIPLES OF REDUCTION/OXIDATION


PRECIPITATION


Calcium Chloride solution (clear & colourless)
added to
Silver Nitrate solution (clear & colourless)

white precipitate.


Silver Nitrate solution (clear & colourless)
added to
Sodium Sulphide solution (clear & colourless)

black precipitate


Copy
WHAT IS A PRECIPITATE?

A precipitate may form when one ionic solution is added
to another. It is not possible to see through a solution
when a precipitate forms in it. The solution becomes
cloudy. The precipitate will eventually settle to the bottom
of the container.

Definition
A precipitation reaction has occurred when a solid forms as a result of a
reaction between two solutions. The solid forms because it is an insoluble
product of the reaction.


HOW PRECIPITATES FORM
Example
A solution of Copper Sulphate is added to a solution of Sodium Carbonate.
Copper ions spread through the water
The copper sulphate solution contains:
Sulphate ions spread through the water
Sodium ions spread through the water
The sodium carbonate solution contains:
Carbonate ions spread through the water

Forces of attraction between the Copper ions and Sulphate ions are weak enough for the
water molecules to get between the ions. The same is true for the sodium and carbonate
ions.
The same cannot be said for the Copper ions that collide with the Carbonate ions in this
mixture. A bond forms that is stronger than the attraction that water has for the ions and
so a precipitate is formed.
The Sodium and sulphate ions are dispersed and free to move amongst the water
molecules. They are spectator ions.

Copper Sulphate + Sodium Carbonate --> Copper Carbonate + Sodium Sulphate

Ion equation:

Complete formula equation:


SOLUBILITY RULES

“You can predict whether or not a compound is soluble by
using the following simple rules:”

1. All nitrates are soluble.

2. All group 1 metal compounds and ammonium compounds
are soluble.

3. All chlorides are soluble except for silver chloride and lead
chloride

4. All sulfates are soluble except for barium sulfate and lead
sulfate.

5. All carbonates are insoluble except group 1 and ammonium
compounds

6. All hydroxides and oxides are insoluble (exception Sodium
& Potassium)


THERMAL
DECOMPOSITION


EQUATIONS OF DECOMPOSITION REACTIONS

A decomposition reaction is one in which a compound breaks down to form simpler
compounds. A thermal decomposition requires heat.

Word equations for simple decomposition reactions

1. The thermal decomposition of carbonates, Eg ...

Calcium Carbonate --> Calcium Oxide + Carbon Dioxide

Copper Carbonate --> Copper Oxide + Carbon Dioxide

2. The thermal decomposition of sodium bicarbonate, Eg...

Sodium Bicarbonate --> Sodium Carbonate + Water + Carbon Dioxide

3. The decomposition of hydroxides

Copper Hydroxide --> Copper Oxide + Water

Note
Not all metal carbonates, hydrogen carbonates and hydroxides will decompose on
heating. Position on the reactivity series is an important factor

REACTIVITY SERIES OF METALS

Most reactive K

Na

Li In a simple electron exchange reaction
(redox), a metal higher in the series will
Ca
donate electrons to one that is lower in
Mg the series
Al

Zn
Example
Fe

Sn Zn(s) + CuSO4(aq) ---> ZnSO4(aq) + Cu(s)

Pb
Grey Blue Colourless orange-brown

Cu

Hg

Ag
Least reactive
Au

CALCULATIONS


ATOMIC & MOLECULAR MASSES

The mole
Consider the formation of Carbon Monoxide:

C + O ---> CO
This is often understood as meaning
“1 atom of C” + “1 atom of O” ---> “1 molecule of CO”
But working with single atoms when you are performing calculations involving mass
means working with masses that are too small.
The chemist needs to find a larger unit so that masses in calculations are realistic.
This unit is called the mole.

A mole is a unit that consists of 6.023 x 1023 objects.

Atomic masses
“This means that 1 mole of hydrogen
atomic 1 (6.023 x 1023 atoms) has a mass of 1.0 g.”
number
H
Relative atomic
1.0
mass, Ar “The atomic mass of Carbon is
12 gmol-1.”


Mr = Relative molecular mass
= the number of grams per mole for the compound.

A formula gives us the mole ratio of each type of atom in a compound.

Molecular masses can be calculated from relative atomic mass values and
formulae.

Example
The formula for Methane is CH4.
This means that in a sample of Methane there are 4 moles of hydrogen atoms
for every mole of carbon.

Mr(CH4) = the number of grams per mole of methane molecules.

The periodic table tells us that:
C = 12.0g per mole
H = 1.0g per mole
So ......
Mr(CH4) = 12.0 + 4 x 1.0 = 16 gmol-1


Example: A question from the 2003 NCEA examination paper



Mr(FeCO3) = 55.9 + 12 + (3 x 16) = 115.9 gmol-1



Mr(FeCO3) = 55.9 + 12 + (3 x 16) = 115.9 gmol-1

Mr[Fe(NO3)3] = 55.9 + (3 x 14) + (9 x 16) = 241.9 gmol-1



Mr(FeCO3) = 55.9 + 12 + (3 x 16) = 115.9 gmol-1

Mr[Fe(NO3)3] = 55.9 + (3 x 14) + (9 x 16) = 241.9 gmol-1

Mr(Fe2O3) = 2 x 55.9 + 3 x 16 = 159.8 gmol-1


A question from the 2005 NCEA examination paper

“Try these questions out for yourself”

ESA: Ex 12A



159.6 gmol-1

ESA: Ex 12A



159.6 gmol-1

106 gmol-1

ESA: Ex 12A



159.6 gmol-1

106 gmol-1

261 gmol-1

ESA: Ex 12A

PERCENTAGE COMPOSITION

Percentage composition of an element/s in a compound is the ratio of the mass per
mole of the element/s to the mass per mole of the compound expressed as a
percentage.

Example
Consider that you need to calculate the percentage of Sulfur in Copper Sulphate. The
relative atomic masses are:

Cu = 63.5 gmol-1
S = 32.1 gmol-1
O = 16.0 gmol-1

Mr (CuSO4) = 159.6 gmol-1 (from previous question)
Ar (S) = 32.1 gmol-1

Percentage Sulfur = 32.1 = 20.1%
159.6


Here are the steps to write the formula for calcium bicarbonate:1. Write the ions involved: Ca2+ and HCO3-2. Work out the ratio of ions that will balance the charges: Ca2+ : HCO3- = 1 : 23. Write it out without charges: Ca HCO34. Put subscripts in place to reflect the ratio: Ca(HCO3)2Therefore, the formula for calcium bicarbonate is Ca(HCO3)2

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