1. Metals

2. Metal Reactivity
Metals display a wide range of reactivity with other substances, varying from very reactive to no reaction at all.
Question: Why is knowing the reactivity of a metal
useful to us?
The other substances that most influence the choice of metal for a particular purpose are oxygen, water and acids.
The order of metal reactivity is called the activity series of metals.

3. Reaction of metals with oxygen
Most metals react with oxygen to form metallic oxides.
All the oxides formed are ionic compounds. Why?
Metal + Oxygen Metal oxide

4. Reactions of metals with oxygen
For the reaction of iron with oxygen:
Step 1: Write the word equation
Iron + oxygen Iron oxide
Step 2: Write the forumla
Fe + O2 Fe 2O 3
Where do the subscripts come from?

5. Reactions of metals and oxygen
Step 3: Balance the equation
4Fe + 3O2 2Fe2O3
Step 4: Add states
4Fe (s) + 3O2 (g) 2Fe2O3 (s)

6. Reaction of metals with water
Most metals undergo no change when placed in cold water. Some exceptions to this are: lithium, potassium, sodium, calcium. These react with cold water to form hydrogen and a metal hydroxide.
Sodium + water sodium hydroxide + hydrogen
This reaction involves the transfer of electrons from sodium atoms to hydrogen atoms in the water.

7. Reaction of a metal and water
For the reaction of sodium with water:
Step 1: Write the word equation
Sodium + water sodium hydroxide + hydrogen
Step 2: Write the formula
Na + H2O NaOH + H2

8. Reaction of a metal and water
Step 3: Balance the equation
2Na + 2H2O 2NaOH + H2
Step 2: Add states
2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g)

9. Reaction of metals with water
Some less reactive metals (Al, Zn, Fe) will NOT react with cold water but will react with steam to produce steam to produce hydrogen and a metal oxide.
Zinc + steam zinc oxide + hydrogen

10. Reaction of metals with acids
Some metals will react with acids to produce salt and hydrogen.
Metal + acid salt + hydrogen
During the reaction between a metal and an acid the metal loses electrons and becomes positively charge ions. Hydrogen ions from the acid gain electrons to form hydrogen gas.

11. Reaction of metals and acids
For the reaction of zinc and hydrochloric acid:
Step 1: Write the word equation
Zinc + hydrochloric zinc hydrogen
acid chloride
Step 2: Write the formula
Zn + HCl ZnCl2 + H2

12. Reaction of metals with acids
Step 3: Balance the equation
Zn + 2HCl ZnCl2 + H2
Step 4: Add states
Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g)

13. A common feature
A common feature of all reactions of metals with oxygen,
water and dilute acids is that atoms of the metals lose
electrons to become positive ions.
Reactions with oxygen: ionic oxides are formed
MgO = ionic compound containing Mg 2+ and O 2-
Reactions with water: ionic hydroxides are formed
LiOH = ionic compound containing Li+ and H-

14. A common feature MgCl2 = ionic compound containing Mg2+ and Cl22-
Reactions with acids: ionic metallic chlorides and sulfates are formed
FeSo4 = ionic compound containing Fe2+ and SO42-
MgCl2 = ionic compound containing Mg2+ and Cl22-

15. Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g)
Ionic equations
Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g)
Species of the reactants are?
Zn atoms, H+ ions and Cl- ions
Species of the products are?
Zn 2+ ions, H2 molecules and Cl- ions
So we could write the complete ionic equation:
Zn(s) + 2H+(aq)+ 2Cl-(aq) Zn 2+(aq) + 2Cl-(aq) + H2(g)

16. Zn(s) + 2H+(aq)+ 2Cl-(aq) Zn 2+(aq)+ 2Cl-(aq)+ H2(g)
there are two Cl- ions on the right and two on the left.
Zn atoms have changed to Zn 2+ ions. For this to happen zinc atoms must have given up 2 electrons. We can write this as:
Zn Zn e-
Hydrogen ions have changed to hydrogen ions have changed to hydrogen molecules. Therefore each hydrogen ion must have gained one electron.
2H+ + 2e- H2

17. Zn (s) + 2H+ (aq) Zn 2+ (aq) + H2 (g)
Net ionic equations
Notice that this reaction is really between Zn and H because they are the only species that undergo a change.
A net ionic equation only shows the ionic species that undergo a CHANGE in the reaction.
Zn (s) + 2H+ (aq) Zn 2+ (aq) + H2 (g)

18. Spectator ions Note that in the reaction:
Zn(s) + 2H+(aq)+ 2Cl-(aq) Zn 2+(aq) + 2Cl-(aq) + H2(g)
the Cl- ions do not undergo a chemical change: there are
two Cl- ions on the right and two Cl- ions on the left. Ions
that do not undergo a chemical change during the reaction
are called spectator ions

19. Working out net ionic equations
Ionic equation for the reaction of Al and HCl
Step 1: Write the word equation
aluminium + hydrochloric Al + hydrogen
acid chloride
Step 2: Write the formula
Al + HCl AlCl3 + H2

20. Working out net ionic equations
Step 3: Balance the equation
2Al + 6HCl 2AlCl3 + 3H2
Step 4: Add states
2Al (s) + 6HCl (aq) 2AlCl3 (aq) + 3H2 (g)

21. Working out net ionic equaitons
Step 5: Determine the species of the reactants and products
Reactants: Al atoms, H+ ions and Cl- ions
Products: Al 3+ atoms, H2 molecules and Cl- ions
Step 6: write the complete ionic equation
2Al (s) H+ (aq) + 6Cl- (aq)
2Al 3+ (aq) + 6Cl- (aq) + 3H2 (g)

22. Complete ionic equations
Step 7: Write the net ionic equation
2Al (s) H+ (aq) Al 3+ (aq) + 3H2 (g)

23. Ionisation energy
The reactivity of a metal is related to the ease with
which it loses valence electrons to form ions. Ionisation energy is a measure of the energy needed to remove the most loosely bound electron from an atom in the gaseous state.
In general reactive metals have low ionisation energies and less reactive metals have high ionisation energies

24. Oxidation – reduction reactions
Reactions which involve the transfer of electrons are called oxidation-reduction reactions.
When an atom LOSES one or more electrons we say it has been oxidised.
When an atom GAINS one or more electrons we say it has been reduced.
OXIDATION = LOSS OF ELECTRONS
REDUCTION = GAIN OF ELECTRONS

25. Oxidation – reduction reactions
In normal chemical reactions there can be no overall loss or gain of electrons. Hence oxidation and reduction occur simultaneously.
We call these reactions REDOX reactions.
Half equations can be used to describe the oxidation and reduction processes separately in terms of electrons lost or gained.

26. Half equations Reaction between magnesium and oxygen
Oxidation: Mg (s) Mg2+(s) + 2e-
Reduction: O2(g) + 4e O2-(s)

27. Relative atomic mass
Because atoms are so small it is difficult to measure
their actual individual masses.
Relative atomic mass is NOT the mass of an atom of that
element. It is just a relative mass – relative to the mass of
a carbon atom. It is not a mass at all – just a number with
no units.
Example: a titanium atom is 4x the mass of a carbon atom
so relative atomic mass of Ti is 48

28. Isotopes Most elements in nature consist of several isotopes with
slightly different masses. This is because isotopes have a
different number of neutrons in the nucleus.
Example: 75% of Cl atoms have 18 neutrons
25% of Cl atoms have 20 neutrons
There are two isotopes of Cl one with an atomic mass of
35 (Cl-35) and one with an atomic mass of 38 (Cl-37).
Remember all Cl atoms have 17 protons.

29. Isotopes Therefore strictly speaking when we say the ‘mass of an
atom’ we actually mean the ‘average mass of the atoms
in the naturally occurring element’.
Average mass = 75 x x 37 = 35.5
100
Note: average mass is closest to the atomic mass of the
most abundant isotope

30. Relative molecular mass
Relative atomic mass (Ar) is used to describe the mass
of atoms.
Relative molecular mass (Mr) is used to describe the
mass of molecules.
(Mr) = the mass of a molecule of a substance or compound relative to the mass of an atom of the carbon-12 isotope taken exactly as 12

31. Relative molecular mass
The relative molecular mass of a substance is found by
adding the relative atomic masses of the constituent
elements.
Example: molecular mass of H2O
= 2 x Ar(H) + 1 x Ar(O)
= 2 x
= 18

32. Relative formula mass
Many compounds, particularly ionic compounds (eg: NaCl)
exist as an array of ions or atoms bound to each other but
with no recognisable molecules. The formula NaCl instead
tells us that throughout a sample of NaCl sodium and
chlorine atoms are present in the ratio 1:1. Because ionic
compounds do not contain molecules the sum of the
relative atomic masses of the atoms in the formula is called
the relative formula mass (still given the symbol Mr).

33. The mole Chemists measure the amount of any substance in moles.
Mole: the quantity or amount of a substance that contains
the same number of particles as there are atoms in exactly
12 grams of carbon-12
Avogadro’s number (NA): the number of atoms in
exactly 12 grams of carbon-12 = x 1023
Therefore one mole of a substance contains x 1023
particles of that substance.

34. Working with moles When working in moles the particle or units being
counted should be stated as atoms, molecules or ions.
Eg:
1 mole Fe = x atoms of Fe
1 mole of Pb = x atoms of Pb
1 mole of H2SO4 = x molecules of H2SO
1 mole N2 gas = x 1023 molecules of nitrogen

35. Molar mass 6.022 x 1023 atoms of carbon has a mass of 12 grams
Since all relative atomic masses are measured against the
standard carbon 12 it follows that the atomic mass in
grams of an element (or the formula mass in grams of
any compound) is one mole of that substance and this
one mole contains avogadro’s number of particles.

36. Consider:
Titanium (atomic mass 48)
1 mole of titanium has x 1023 atoms (each of which have a mass 4x that of carbon)
Therefore mass of 1 mole of titanium = 4x12 =48
1 mole of titanium has a mass equal to its relative
atomic mass.

37. Molar mass 1 mole of a substance has a mass equal to its:
relative atomic mass (expressed in grams)
relative molecular mass (expressed in grams)
relative formula mass (expressed in grams)
This is called the molar mass (M) of a substance. Molar
mass has units – usually grams per mole (g mol -1).

38. Calculating molar mass
Write down the symbol or formula of the substance
Add up the relative atomic masses of the elements involved
This is the symbol mass or formula mass
In grams this is one mole of the substance
This is made of avogadro’s number of particles

39. Summary There are therefore two ways of looking at a mole:
6.022 x = mole of Ti = 48 grams of Ti
atoms of Ti
A number of particles
(atoms, ions, molecules)
6.022 x 1023
A mass (the relative atomic, molecular or formula mass in grams)
MOLE

40. Moles and numbers of particles
Relationship between moles and number of particles (eg:
atoms, molecules):
Number of moles (n) = number of particles (N)
number of particles in one mole (NA)
n = N
6.022 x 1023
We often need to calculate the number of moles present in a given mass of a substance (either element or compound). At other times we need to calculate the number of atoms or molecules in a sample. Use the following formulas to do this.

41. Moles and numbers of particles
Using this formula it is possible to calculate:
Number of moles of a substance from the number of particles or basic units of a substance
Number of particles or basic units of a substance (atoms, formula units) from the number of moles

42. Moles and mass The relationship between the number of moles and mass
of a substances is:
Number of moles (n) = mass (m) in grams
mass of 1 mole (M)
n = m M

43. Using this formula it is possible to calculate:
Number of moles of any substance in a given mass
Mass of a substance in a given number of moles

44. Summary
n = m M
n = N
6.022 x 1023
Therefore:
N = m
6.022 x M

45. Percentage composition
Percentage composition of a compound is simply
the percentage by mass of each element present
in the compound.
To determine percentage composition you need two
things :
formula of the compound
Relative atomic masses of the elements present

46. Steps in determining percentage composition
Calculate the percentage composition of iron in Fe2O3
Step 1: determine mass of one mole of Fe2O3
2 x x 16 = grams
Step 2: One mole of Fe2O3 contains 2 moles of Fe.
2 moles of Fe : 2 x 55.9 = grams
Step 3: % composition of Fe = x 100 = 70%
159.8

47. Summary Therefore: % A in a compound
= mass of A in 1 mole of the compound x 100
mass of 1 mole of the compound

48. Molecular v’s empirical formula
Molecular formula: specifies the actual number of atoms of each element in a molecule
hydrogen peroxide: H2O2
Empirical formula: specifies the simplest whole number ratio of each element
hydrogen peroxide: HO

49. Determining empirical formula
The empirical formula of a substance can be established
by first determining the percentage composition of a
substance by chemical analysis.
Ex: Calculate the empirical formula of glucose with a chemical composition of 40% carbon, 6.6% hydrogen and 53.3% chlorine

50. Step 2: Determine the moles of each element in 100 grams
Step 1: List the elements and the mass of each element in 100grams of freon-12
Element: C H O
Mass in 100g:
Step 2: Determine the moles of each element in 100 grams
n = m M
= = = 3.33
Mole ratio 1 : 2 : 1
Step 3: Empirical formula CH2O

51. Determining molecular formula
If the molecular mass of glucose = what is its
molecular formula?
Step 1: Determine the relative empirical formula mass
CH2O = 1 x x x = 30.03
Step 2:
Relative molecular mass = molecular formula
Relative empirical formula mass empirical formula

52. molecular formula = 180.16 x CH2O 30.03
Relative molecular mass = molecular formula
Relative empirical formula mass empirical formula
= molecular formula
CH2O
molecular formula = x CH2O
30.03
molecular formula = 6 x CH2O
molecular formula = C6H12O6

53. Joseph Gay-Lussac
Gay Lussac’s law of combining gas volumes states that
when two gaseous elements combine:
‘the ratios of the volumes of gases involved, if measured
at the same temperature and pressure, are expressed by
small whole numbers’

54. Step 1: write the balanced equation for the reaction
Example: calculate the volume of hydrogen that will combine with 6 L of nitrogen to form ammonia
Step 1: write the balanced equation for the reaction
N2 (g) H2 (g) 2NH3 (g)
Step 2: determine the ratios y
1 mole N2 3 mole H mole NH3
1 volume 3 volume 2 volume
Step 3: calculate volume V (H2 ) = 3 x V(N2)
= 3 x 6.0L
= 18 L

55. Mass-mass calculations
Chemical equations show the number of moles of reactants
and products in a chemical reaction. They can also be used
to determine the relationship between the masses of the
reactants and products:
N2 (g) H2 (g) 2NH3 (g)
1 mole 3 mole 2 mole
28g + 6g 34g

56. Remember this! When carrying out mass-mass calculations
remember the following step by step method:
From the equation
N = m/N
N = m/N
Mass of the known
Moles of the known
Moles of the unknown