All of the elements on the periodic table
make up everything in our world!
The periodic table consisting of most of
the known elements was developed by a
Russian man called Dimitri Mendelev
around 1860. He managed to put
all the elements in order of weight !
without sophisticated scientific
equipment.
Mendeleevs First Periodic Table
Relative atomic Mass (Ar) numbers
All atoms were measured against Carbon. The
carbon atom was assigned an atomic mass of 12
6
C
12
This tell us that each carbon atom
has a mass of 12
It also tells us the number of protons
and neutrons
We say the relative atomic mass (Ar) of Carbon is 12
Relative atomic Mass (Ar) numbers
1
H
1
hydrogen has a relative mass of 1
compared to carbon
We say the relative atomic mass (Ar) of hydrogen is 1
Relative atomic Mass (Ar) numbers
8
O
16
oxygen has a relative mass of 16
compared to carbon
We say the relative atomic mass (Ar) of oxygen is 16
All of the relative atomic masses for all the atoms
are found on the periodic table
Avogadro's Number
Through numerous experiments Avogadro found that one
mole (ie 6.02x10 23 atoms ,yes that is a very, very large
number)
of any element equalled the relative atomic mass (Ar) in
grams of each substance
This means:
6.02x10 23 oxygen (O) atoms = 16 grams
6.02x10 23 carbon (C) atoms = 12 grams
6.02x10 23 hydrogen (H) atoms = 1 gram
6.02x10 23 sulfur (S) atoms = 32 grams
Atomic Mass (Ar) and Molecular Mass (Mr)
We can find the molecular mass (Mr) of a
compound by adding the atomic mass (Ar)
numbers of the atoms in a compound.
Be aware that the terms atomic mass and
molecular mass are also known as molar
masses
Ar and Mr are very similiar
• Ar = the relative atomic mass and is used for single
atoms eg Ca, N, O etc in gmol-1
Eg The Ar of Ca = 40 gmol -1
Mr = the relative molecular mass of a substance and is the
sum of the Ar values
Eg 1 mole of H2 = 1 + 1 = 2 gmol -1
Eg What is the Mr of H2O? (use your PT to find out)
(2 x H = 2 ) + (1 x O = 16) = 18gmol -1
What is the Mr of H2SO4 ?
(2 x H = 2) + (1 x S = 32) + (4 x O = 64)
2
+
32 +
64 = 98 gmol-1
Measuring in Chemistry
A baker uses a dozen (12) of something
A chemist uses a mole 6.02 x 1023 of something
Units for Mr are gmol-1
Moles, Mass and Mr relationship
Moles/Mass and Mr are all related by the formula
m
n
Mr
Where:
m
n
Mr
X
m = mass in grams
Mr = molar mass (g mol-1)
n = number of moles
Moles,Mass and Mr relationship
m
n
Mr
Using this relationship we are able to work out:
1.Moles of a substance given the mass and Mr
2.Mass of a substance given the number of moles
and Mr
3.Even the Mr of a substance given the mass and
number of moles
Questions
• Find the number of moles of Ca in 50
grams of the metal.
m
n
Mr
50g
n 
40g mol
n = 1.25 moles
n = 1.25 moles (3sf)
-1
Questions
• Find the number of moles in 8 grams of
oxygen gas (O2).
m
n
Mr
8g
n
32g mol
-1
n = 0.250 moles (3sf)
Questions
• Find the number of moles in 12 grams of
carbon dioxide gas (CO2).
m
n
Mr
12g
n
44g mol
-1
n = 0.273 moles (3sf)
Starter Questions – Using your periodic table
Find the molar mass (Mr) for CO2
12 + (2 x 16) = 44gmol-1
(this means that one mole of CO2 has a mass of 44g)
Find the atomic mass (Ar) for Na
23 gmol-1
(this means that one mole of Na has a mass of 23g)
Find the atomic mass of 0.5 mole of Na metal 11.5 g
Find the mass of 0.1 mole of Na metal
2.3 g
Toughie? :How many Na atoms are there in 0.1 mol of Na metal?
0.1 x 6.02 x 1023 = 6.02 x 10 22 Na atoms
Find the molar mass (Mr) for Na2CO3
(2 x 23) + 12 + (3 x 16) = 106 gmol-1
Moles, Mass and Mr relationship
Moles/Mass and Mr are all related by
what formula?
m (g)
n
Mr(gmol -1 )
Where:
m = mass in grams
Mr = molar mass (gmol-1)
n = number of moles
m
n
Mr
X
We will use a grid to help in calculations. These become very
helpful with complicated questions - just fill what you know then
use this to work out your unknown
Question- iron oxide (Fe2O3) is used in the production of iron. How
many moles of iron oxide would there be in 200 grams?
Copy out the grid
Fe2O3
Mr
m
n
Mr
mass
(2 x 56) + (3 x 16) = 160 gmol
200g
= 1.25 moles
n ?
160g
200 grams
-1
In industrial operations large quantities are used
Question - How many moles of iron oxide would there be in 1
tonne (1000kgs) ?
Copy out the grid
Fe2O3
Mr
(2 x 56) + (3 x 16) = 160 gmol
m
n
Mr
1 x 10 6 g
? = 6,250 moles
n
160g
mass
-1
1000kg x 1000 = 1 x 106 grams
How to work out the % of an element in a
compound
What % of Fe is there in Fe2O3?
Mr 2Fe
%Fe 
Mr Fe2O3
112

x 100
160
 70% Fe in Fe2O3
How much iron would you get from 100kgs of
Fe2O3?
0.7 x 100 = 70kg of iron
Reading Equations using Moles
1 Ca
+ 1 H2SO4
1 CaSO4 + 1 H2
The above reaction reads:
1 amount of Ca reacts with 1 amount of H2SO4 to give 1
amount of CaSO4 and 1 amount of H2 gas.
(we don’t usually put in the ones – but if we did it would look
like this)
The amount we use is the mole
Now the reaction reads:
1 mole of Ca reacts with 1 mole of H2SO4 to give 1 mole of
CaSO4 and 1 mole of H2 gas.
Reading Equations using Moles
Ca
+
H2SO4
CaSO4 +
H2
Questions:
1. If we have 1 mole of Ca how many moles of H2SO4
are needed to fully react the Ca? Ans: 1mole
2. If we had 0.5 mole of Ca how many moles of H2SO4
would be required? Ans: 0.5 mole
3. If 0.5 mole of Ca reacted with 0.5 mole of H2SO4
how many moles of CaSO4 and how many moles
of H2 gas are formed?
Ans : 0.5 mole of H2 gas, 0.5 mole of CaSO4
Reading Equations using Moles
Ca
+
Questions:
H2SO4
CaSO4 +
H2
If 0.5 mole of Ca reacted with 0.5 mole of H2SO4 how many
moles of CaSO4 and how many moles of H2 gas are formed?
Ans : 0.5 mole of H2 gas, 0.5 mole of CaSO4
Given the Ar (atomic mass) values
H = 1gmol-1 Ca = 40gmol-1 S = 32gmol-1 O =16gmol-1
Find the molar masses (Mr) of H2SO4 , CaSO4 and H2
Then use these and the mole (n) formula to work out the
mass of H2 gas and the mass of CaSO4 formed
(Hint you will need to rearrange the formula)
Reading Equations using Moles
Fe2O3
Questions:
+
CO
2Fe
+
2CO2
If 200kg of Fe2O3 is reacted in the above equation what mass
of Fe is produced?
Mr 2Fe
%Fe 
Mr Fe2O3
112

x 200
160
 140kg Fe in 200kg Fe2O3
Work out how many moles of HCl is used to completely react
60grams of Mg metal in the following reaction? Draw grid
Mg
+
2 HCl
MgCl2 +
Mg
Mr
mass
m
n
Mr
Reaction
ratio
24 gmol
H2
HCl
-1
36.5 gmol -1
60 grams
60g
n ?
= 2.5 moles
-1
24gmol
1
Therefore 2 x moles of Mg
2
= moles of HCl
Therefore 2 x 2.5 moles
= 5 moles of HCl
What was the mass of HCl used?
m
n
Mr
m = 5 x Mr
= 5 mol x 36.5 gmol-1
= 182.5 g
= 182 g (3sf)
Sherbet
Sherbet balanced reaction
NaHCO3 + C4 H6O6 + C6 H7 O7
Tartaric
Citric
acid
acid
Na tartarate + H2O + CO2
Sherbet
Given 1 gram of NaHCO3 work out how much of each ingredient
NaHCO3
Mr
m
n
Mr
84 gmol
1g
84gmol
1
Tartaric acid Citric acid
-1
= 0.0119 mol
Molar ratio
1
mass = n x Mr
1g
150 gmol -1
0.0119 mol
1
191 gmol
-1
0.0119 mol
1
0.0119 x 150 0.0119 x 191
= 1.785g
= 2.273g
Work out how many moles of carbon dioxide is created when
100grams of oxygen is reacted in the following reaction? Draw
grid
CH4
+
2O2
CO2 + 2H2O
O2
CO2
Mr
32 gmol
mass
100 grams
44 gmol -1
-1
m
n
Mr
100g
n ?
= 3.125 moles
-1
32gmol
Reaction
ratio
2
Therefore ½ x moles of O2
Therefore ½ x 3.125 moles
Would you give this answer?
1
= moles of CO2
= 1.5625 moles of CO2
1.56 moles of CO2 (3 sig fig)
Empirical Formula
This is the simplest ratio of atoms in a compound
For example the empirical formula for ethane C2H6
is CH3
the empirical formula for butane C4H10 is C2H5
the empirical formula for methane CH4 is CH4
the empirical formula for hydrogen peroxide H2O2 is HO
Molecular Formula
The molecular formula of a compound is the formula giving
the actual number of atoms in a compound
We will be asked to find the molecular formula given %
mass of a compound
For example find the molecular formula for a compound
which was found to contain 80% carbon and 20% hydrogen
Example: finding the empirical formula for a compound that
was found to contain 80% carbon and 20% hydrogen
Step one :
simply convert each % to
grams
Step two :
divide each element by
its Ar value
Step three :
What is the smaller number?
Now divide each of these by
this smaller number
Step four :
Now this is the empirical
formula for the compound just
write out the formula
C
H
80g
20g
80
12
20
 6.666
6.6666
6.6666
1
1
 20
20
6.66666
1xC
3xH
CH3
 3.00
Example : Now find the molecular formula for the compound
that was found to contain 80% carbon and 20% hydrogen.
Given that the compound was found to have a molar mass of
30 gmol-1
Step five: use the empirical formula you just found ie CH3 and
follow the method below
Step six : divide your given molar mass by the molar mass of
CH3
30gmol -1
given molar mass

2
empirical formula molar mass
15gmol -1
Step seven: now multiply your empirical formula by 2 to
get the molecular formula - 2 x CH3 = C2H6
A compound was found to have a % composition
of 2.0% Hydrogen, 32.7% Sulfur, 65.3% Oxygen
find the empirical formula
A compound was found to have a % composition
of 2.0% Hydrogen, 32.7% Sulfur, 65.3% Oxygen find the empirical formula
Step one :
simply convert each
% to grams
Step two :
H
S
2g
32.7g
2
divide each element
by its Ar value
2
1
32.7
32
O
65.3g
 1.02187
65.3
16
 4.08125
Step three :
What is the smaller
number? Now divide
each of these by
this smaller number
Step four :
Now round and just
write out the
empirical formula
2
1.02187
 1.957
1.02187
1.02187
1
4.08125
1.02187
 3.9939
2
1
4
H2
S
O4
A sample with composition 40% C, 6.7% H and
53.3% O had a molar mass 60 g mol-1 find the
Empirical formula and also the molecular formula
A sample with composition 40% C, 6.7% H and 53.3% O had a molar mass 60 g
mol-1 find the Empirical formula and also the molecular formula
Step one :
simply convert each
% to grams
Step two :
divide each element
by its Ar value
40
12
C
H
O
40g
6.7g
53.3g
 3.3333
6.7
1
 6.7
53.3
16
 3.3312
Step three :
What is the smaller
number? Now divide
each of these by
this smaller number
Step four :
Now round and just
write out the
empirical formula
3.3333
3.3312
 1.0006
6.7
3.3312
 2.011
3.3312
3.3312
1
2
1
C
H2
O
1
Step five: use the empirical formula you just found ie CH2O and
follow the method below
Step six : divide your given molar mass (60 gmol-1) by the molar
mass of CH2O (30 gmol-1)
given molar mass
empirical formula molar mass

60gmol -1
30gmol -1
2
Step seven: now multiply your empirical formula by 2 to
get the molecular formula - 2 x CH2O = C2H4O2
Water of Crystallisation
• Some hydrated salts have H2O molecules in
their structure eg CuSO4 5H2O (hydrated means
with water)
• This means that each particle of CuSO4 is
surrounded by 5H2O molecules
• These H2O molecules are called the water of
crystallisation
• The water in the structure can be removed by
heating to form the anhydrous (without water)
salt ie CuSO4
Finding the Empirical Formula for a
Hydrated salt by Gravimetric
Analysis
• To find the mass of H2O in a hydrated
(with water) salt
• Firstly find the mass of the hydrated (with
water) salt
• Heat strongly for 10 minutes cool then
reweigh
• Then heat again and reweigh to check the
salt is at constant mass
• Then subtract the mass of the
anhydrous(dry) salt from the hydrated(wet)
salt
• This will give you the water mass
• Then use your anhydrous salt mass and
water mass to workout the empirical
formula of the hydrated salt using grid
Find the empirical formula for a hydrated salt Na2CO3XH2O
Given:
Mass of hydrated Na2CO3 salt = 2.58g
Mass of salt after heating to constant weight = 0.95g
Therefore: 2.58g – 0.95g = 1.63g (mass of water)
Calculation for finding the empirical formula for a hydrated
salt Na2CO3XH2O
Na2CO3
Step one :
106gmol-1
Find the Mr ‘s
Step two :
Given masses find 0.95
 0.0089622
moles of each
106
H2O
18gmol-1
1.63
 0.0905555
18
Step three :
What is the smaller
0.00896226
0.0905555
1
number?
 10.1
0.00896226
0.00896226
Now divide each of
these by
this smaller number
Step four :
1 x Na2CO3
10H2O
Make small adjustments
to give whole number
Na2CO310H2O
To give empirical formula
Concentration
Particles per volume
Can be in grams per litre but chemists usually
express concentration in moles per litre
This are related by the formula:
n
c 
V
Where:
n = moles
V = volume in litres
c = concentration in molL-1
Write the formula for:
1. Moles using mass and Molar mass
m (g)
n
-1
Mr(gmol )
2. Concentration using Volume and Moles
n
c
V
Concentration Example
15 grams of NaCl is dissolved in 1 litre of water, what is
the concentration of the solution in moles per litre?
(M NaCl = 58.5 gmol-1)
Find moles of NaCl
m
15
n

 0.2564 mol
M
58.5
Divide moles by volume in litres
n
0.2564mol
1
c

 0.256molL
V
1 litre
Concentration Example
20 grams of NaOH is dissolved in 500mls of water, what is
the concentration of the solution in moles per litre?
(M NaOH = 40gmol-1)
Find moles of NaOH
20g
m
n

 0.5 mol
-1
M
40gmol
Divide moles by volume in litres
n
0.5mol
c

 1 molL1
V
0.5litre
If solution A (25mls of 0.23 molL-1 AgNO3) is mixed with
solution B (82 mls of O.15molL-1 of AgNO3 )
What is the final concentration of the solution when they
are mixed
Find total moles by adding moles of A to moles of B
Solution A moles
n
V
n  c xV
Solution B moles
Total moles
c
n
V
n  c xV
moles A  moles B
n  0.23molL-1 x 0.025L
n  0.15 molL x 0.082L
n  5.75x10 -3 moles
n  0.0123 moles
c
5.75x10 3 moles  0.0123mole s
-1
 0.01805mol es
Divide total moles by total volume in litres
n
0.01805mol
c

 0.16869 molL1 0.169 molL-1 (3SF)
V
0.107litres
How many grams of water vapour H2O(g) will be produced in the
complete combustion of 500g methane, CH4(g)?
CH4
+
Mr
mass
2O2
CO2 +
2H2O
CH4
H2O
16 gmol -1
18 gmol -1
62.5x18 =1,125g
500 grams
= 1.13kg H2O
m
n
Mr
500g
?
n
= 31.25 moles
16gmol -1
Reaction
ratio
1
Therefore 2 x moles of CH4
Therefore 2 x 31.25 moles
2
= moles of H2O
= 62.5 moles of H2O
Homework for Friday - finish off all worksheet
pages in booklet except worksheet 4
Test preparation for Monday
Read Unit 11 page 41 in your pathfinder text
Complete all questions on page 42 and 43 before
Monday
See me before Monday if you are having any
trouble
How many grams of water vapour H2O(g) will be produced in the
complete combustion of 500g methane, CH4(g)?
Equation: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)