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Solids, Liquids and Gases
• Solids (s):
• Liquids (l):
• Gases (s):
Molecules are held in place by intermolecular
interactions.
Molecules are held next to one another by
intermolecular interactions, however, these
interactions are not strong enough to prevent the
molecules from flowing
past one another.
The intermolecular
interactions are too weak
to hold the molecules
next to one another,
so the molecules wander
off on their own.
Diffusion
Diffusion is the spontaneous spreading out of a substance, and is
due to the natural movement of its particles.
Cotton wool
soaked with
hydrochloric acid
Ammonium chloride
Cotton wool
soaked with
ammonia solution
Diffusion of ammonia and hydrogen chloride gases
Ink being diffused
in water
The mass of atoms are very
small
• Therefore scientists cannot count
individual atoms or molecules directly
• Instead a scientist count there
particles by weighing a certain number
of them
• Same as a bank cashier, by knowing
the weight of one coin they can
calculate the number of coins by
dividing the total mass by the mass of
one coin.
A. What is the Mole?
• A counting number (like a dozen)
• Avogadro’s number (NA)
• 1 mol = 6.02  1023 items
A
large amount!!!!
The Mole
• The mole is simply a chemist’s measurement for the
amount of a substance.
• It is a very large number (6 x 1023) as it ‘counts’ very
small ‘objects’ such as atoms and molecules which
have a very small masses.
• This rather strange number was not selected
deliberately – instead, it is the number of atoms in a
sample of any element that has a mass in grams that is
numerically equal to the elements atomic mass.
• It allows chemists to work with masses in the lab which
are feasible.
The Mole Definition
A mole of any substance is defined as the
amount of the substance that contains as many
particles as 12 g of carbon – 12.
B. Molar Mass
• Mass of 1 mole of an element or compound.
• Atomic mass tells the...
–
–
–
–
That 6.01 *1023 atoms will weigh this amount
6.01 *1023 atoms of carbons will weigh 12 grams
grams per mole (g/mol)
1 mole of carbon is 12 grams per litre
• Round to 2 decimal places
B. Molar Mass Examples
• carbon
12.01 g/mol
• aluminum
26.98 g/mol
• zinc
65.39 g/mol
B. Molar Mass Examples
• water
– 2(1.01) + 16.00 = 18.02 g/mol
– H2O
• sodium chloride
– NaCl
– 22.99 + 35.45 = 58.44 g/mol
B. Molar Mass Examples
• sodium bicarbonate
– NaHCO3
– 22.99 + 1.01 + 12.01 + 3(16.00)
= 84.01 g/mol
• sucrose
– C12H22O11
– 12(12.01) + 22(1.01) + 11(16.00)
= 342.34 g/mol
If a magnesium atom is twice as
heavy as an atom of carbon
Then 100 moles of magnesium
Are twice as heavy as 100 moles
Of carbon
Therefore if we have a piece of
Magnesium that is twice as heavy
As a piece of carbon they have an
Equal number of atoms
• Atom of carbon
weighs
• Mr=12
• Atom of hydrogen
weighs
• Mr=1
One gram of hydrogen atoms and twelve grams of
Carbon atoms both have the same number of atoms
• Atom of silver
weighs
• Ar=108
• Atom of carbon
weighs
• Ar=12
If we have 12g of carbon how do we calulate the mass of silver if it has the
same number of atoms
12g
24g
27g
40g
carbon magnesium aluminium calcium
56g
63.5
iron
copper silver
All these masses contain the same
number of atoms
Avogadro`s number = 6 X 1023 Symbol L
It is the number of atoms of carbon in 12g of the carbon-12 isotope
108g
The amount of a substance which contains 6
X 1023 particles is called a mole of that
substance
Symbol
mol
• One mole of
carbon
weighs
12grams
• One mole of
carbon
contains 6 X
23
10 atoms
• 48g of Mg
(one mole
weighs 24 g)
• Is 2 moles
of Mg
Converting moles to grams
• Mass of one mole
of an element=
relative atomic
mass in grams
Element
H
C
N
Mg
Cu
Fe
I
Relative atomic mass
1
12
14
24
63.5
56
127
The mass
of one
mole of
an
element
=
Relative
Atomic
Mass
in grams
The relative molecular mass of a
compound is the sum of the relative
atomic masses of all the atoms in the
compound
H2O = 2(1) + (16) = 18 (relative molecular mass of H2O)
CuSO4.H2O =(63.5) + (32) +4(16) +5 (18)=249.5
Converting Grams to moles
• 12g of carbon = 1 mole
• 1g of carbon = 1/12 moles
• 36g of carbon= 36/12=3 moles
How many moles of sulfuric
acid (H2SO4) are there in
12.25 g of sulfuric acid
• Relative mol mass =2(1) + (32) + 4(16) = 98
• Total mass/mass of one mole
• 12.25/98=0.125 mole
Converting moles to number of atoms
or molecules
• How many atoms of sodium are present
in 0.25 moles of a metal
• 1 mole of Na contained 6x1023 atoms
• 0.25 x 6x1023 atoms= 1.5 x1023 atoms
(i)
How many molecules are there in 0.5 mols of chlorine gas?
(ii)
How many atoms are there in 2 mols of water?
(iii)
How many electrons are there in 1.5 mols of Calcium?
(i)
1 mol of Cl2  6  1023 molecules
[Cl2 is composed of molecules]
 0.5 mols of Cl2  0.5  6  1023  3  1023 molecules
(ii)
1 mol of H2 0  6  1023 molecules
[H2 O is composed of molecules]
 2mols of H2 0  2  6  1023  1.2  1024 molecules
Each molecule of water contains 3 atoms
 1.2  1024 molecules contains 3  1.2  1024  3.6  1024 atoms.
(ii)
1 mol of Ca  6  1023 atoms [Ca is composed of atoms]
 1.5 mols of H2 0  1.5  6  1023  9  1023 atoms
Each atom of Calcium contains 20 electrons
 9  1023 atoms contain 23  9  1024  2.07  1025 electrons.
Molar Volume
• One mole of any gas should occupy the same volume
as one mole of any other gas at the same conditions
of temperature and pressure.
• Gases are often compared at STP, standard
temperature and pressure.
Standard pressure = 101325 Pa
Standard temperature = 273 K
• Molar Volume = 22.4 L = 22,400 cm3 = 2.24 x 10–2 m3
(i)
What is the volume in litres at STP of 0.01 mols of methane (CH4 ) gas?
(ii)
How many mols are there in 280 cm3 of fluorine gas at STP?
(i)
1 mol of CH4  22400 cm3
0.01mols of CH4  0.01  22400  224 cm3
(ii)
1 mol of F2  22400 cm3
1
mols of F2 = 1cm3
22400
1
 280 mols of F2 = 280cm3
22400
0.0125 molsof F2  280cm3
Relative Molecular Mass
The relative molecular mass is the average mass
of a molecule relative to one twelfth the mass of
the carbon 12 atom.
The relative molecular mass has no units as it is
the ratio of two masses.
(i)
Calculate the relative molecular mass of (a) sulfuric acid, (H2 SO4 ) (b) Oxygen, (O2 ).
(i)
(a)
H2 SO4  2(1)  32  4(16)  98
(b)
O2  2(16)  32
Molar Mass
• The molar mass of a substance is the mass in grams
of one mole of the substance.
• The molar mass has the same numerical value as its
relative molecular mass, but its units are grams (g).
Substance
Relative molecular mass Molar mass
Sulfuric aicd, H2 SO4
98
98 g
Glucose, C6H12 O6
180
180 g
Oxygen, O2
32
32 g
Sulfur dioxide, SO2
64
64 g
1 mol of carbon, 12 g
(i)
How many moles are there in 990 g of cabon dioxide (CO2 )
(ii)
The daily intake of calcium for an adult is 800 mg, how many moles of calcium is this?
(i)
1 mol of CO2  44 g
1
mols of CO2  1g
44
1
 990 mols of CO2  990 g
44
22.5 mols of CO2  990 g
(ii)
1 mol of Ca  40 g
1
mols of Ca  1g
40
1
 800 10 3 mols of Ca  800  10 3 g
40
0.02 mols of Ca  800  10 3 g
[800mg  800  10 3 g]
More Mole Calculations
Volume of X
Number of
particles of X
(if a gas at STP)
in litres at STP
Moles of X
x molar mass
÷ molar mass
Mass of X in g
Notice that there
is no direct link
from particles to
grams, you must
first convert to
moles. Likewise
going from
volume to mass.
What is the mass of one atom of calcium?
1 mol of Ca  40 g
1 mol of Ca  6  1023 atoms
 6  10 atoms  40 g
23
40
23
1 atom 

6.67

10
g
23
6  10
The Nissan Micra 1.5 Diesel SVE is quoted to have a CO2 emission figure of 120 g / km.
For a 40 km round trip to work calculate:
(i)
the mass of CO2 produced.
(ii)
the number of moles of CO2 produced.
(iii)
the volume of CO2 produced at room temperature and pressure.
[Molar volume at room temperature and pressure  24.0 litres]
(i)
120  40  4800 km
(ii)
no. of mols 
(iii)
1mol of CO2  24.0 L
actual mass
molecular mass
4800
no. of mols 
mols
44
no. of mols  109.09 mols
109.09 mols of CO2  109.09  24L
109.09 mols of CO2  2594.16 L
How many iron atoms should be consumed daily to meet the recommended daily intake
of iron in the diet of 0.014 g?
[H2007, Q4 (e)]
no. of mols 
acutalmass
molecular mass
0.014
56
no. of mols  2.5  10 4 mols
no. of mols 
1mol  6  10 3 atoms
2.5  10 4 mols  2.5  10 4  6  10 3 atoms
2.5  10 4 mols  1.5  10 0 atoms
Chemical Formulas
Empirical Formula
• The empirical formula of a compound is the
formula that gives that simplest whole
number ratio in which the atoms of the
elements in the compound are present.
• The empirical formula of glucose, C6H12O6
(ratio of atoms is 6:12:6) is CH2O, as this is the
lowest ratio (1:2:1) of the atoms C, H and O.
A sample of a brown coloured gas that is a major pollutant is found to contain 2.34 g of N
and 5.34 g of O. What is the simplest formula of the compound?
2.34
mols of N atoms 
 0.167
14
5.34
mols of O atoms 
 0.334
16
Ratio of Nitrogen atoms to Oxygen atoms  1: 2
 NO2
A dry - cleaning fluid composed of carbon and chlorine was found to have the
composition 14.5% C, 85.5% Cl. What is the empirical formula of this compound?
Element Percentage Percentage Simplest
Ar
Ratio
C
14.5%
Cl
85.5%
 CCl2
14.5
 1.208
12
85.5
 2.408
35.5
1
2
A 1.28 g sample of sulfur was allowed to react with an excess of chlorine to produce 4.12 g
of a product that contains only sulfur and chlorine. What is the empirical formula of the
compound?
Mass of sulfur consumed
=
1.28 g
Mass of chlorine consumed
=
4.12  1.28  2.84 g
Mols of sulfur consumed
=
Mols of chlorine comsumed
=
Ratio of sulfur atoms to chlorine atoms
=
 SCl2
1.28
 0.04
32
2.84
 0.08
35.5
1:2
Molecular Formula
The molecular formula of a compound is the
formula that gives the actual number of atoms
of each element present in a molecule of the
compound.
A colourless liquid used in rocket engines, whose empirical formula is NO2 ,has molecular
mass of 92. What is the molecular formula?
Formula mass
 46
Molecular mass
 92
92
2
46
 Molecular formula  NO2 2  N2 O4
No. of NO2 in formula =
Percentage Composition by Mass
If the empirical formula of a compound is
known, the percentage by mass of each element
present can be calculated. It can be useful to
calculate the percentage of a certain element in
a compound, for example the percentage of
nitrogen in fertilisers.
Calculate the percentage by mass of nitrogen present in urea, CO(NH2 )2 .
% of N 
mass of nitrogen
 100
Mr of Urea
2(14)
% of N 
 100
12  16  2[14  2(1)]
28
% of N   100  46.66%
60
Structural Formulas
The structural formula of a compound shows the arrangement of
the atoms within a molecule of the compound.
Some Examples:
H
H
H
C
C
H
H
Methane, CH4
H
H
H
C
H
Ethene, C2H4
H
H
C
C
H
H
OH
Ethanol, C2H5OH
You will come across many more structural formulas in the
organic section of the course.
Chemical Equations
Chemical Equations
• A chemical equation is a way of representing a chemical change.
• It shows reactants and products.
• To balance an equation means to change the numbers of each
molecule involved, so that the same number of atoms of each
element appear on the reactants side and on the products side.
• Chemical equations balance on an atomic level, not molecular.
• You cannot change the formula of a substance, i.e. if the
equation has NH3 you cannot change this you can only put a
number in front of it, 2NH3, increasing the number of N’s and the
number of H’s.
• Never change the subscripts (small numbers).
• It is possible to write balanced equations for lots of reactions but
that does not mean that the reaction actually takes place.
Formation of Water
(Oxygen does not exist as single atoms)
(This reaction does not take place)
Balancing an Equation
Methane reacts with oxygen, forming carbon dioxide and water vapour only.
Write a balanced chemical equation for this reaction.
Solution
CH4  O2  CO2  H2 O
Examine each element in turn, to have the same no. of atoms of each on either side.
(Leave the O until last as it appears the most often)
Examining C: 1 on LHS and 1 on RHS, leave as they are.
Examining H: 4 on LHS and 2 on RHS, put a 2 in front of the H2 O, on the RHS.
CH4  O2  CO2  2H2 O
Examining O: 2 on LHS and 4 on RHS, put a 2 in front of the O2 , on the LHS.
CH4  2O2  CO2  2H2 O
Balance the following
1.
Fe  HCl  FeCl2  H2
Fe  2HCl  FeCl2  H2
2.
H2  O2  H2 O
H2  12 O2  H2 O
3.
C 4 H10  O2  CO2  H2 O
C 4 H10  132 O2  4CO2  5H2 O
4.
Al  O2  Al2 O3
2Al  32 O2  Al2 O3
Balancing Redox Equations
• In working out what is oxidised and what is
reduced in a reaction, it is important to
remember that oxidation numbers are not a
charge.
• Write the oxidation numbers below each atom
to which it applies, as shown in the examples.
H2 O
1 2
Cr2 O7 2
6
2
OH

2 1
• Oxidation is an increase in oxidation number.
• Reduction is a decrease in oxidation number.
Balance the following redox equation
Cr2 O72  Fe2  H  Cr 3  Fe3  H2O
1.
Cr2 O7 2  Fe2  H  Cr 3  Fe3  H2 O
2.
Cr  3e   Cr 3 
Cr2  6e   2Cr 3
Fe2  e   Fe3 
Fe2  e   Fe3
6
2
2
6
3
2
3.
Cr2
6Fe2
1
3
 6e
 6e
Cr2  6Fe2
3
3
1 2
3
6
2
Write down half equations of what is oxidised
and reduced.
Attach subscripts to atoms oxidised and reduced
and balance the half equations.
3
 2Cr 3
 6Fe3

Assign oxidation numbers to all
the atoms in the equation.
Balance the half equations.
2Cr 3  6Fe3
4.
Cr2 O72  6Fe2  H  2Cr 3  6Fe3  H2 O
Attach species that were attached to the oxidised
and reduced atoms.
5.
Cr2 O72  6Fe2  14H  2Cr 3  6Fe3  7H2 O
Include all the original species and complete the
balancing by inspection.
Calculations using Chemical Equations
• A balanced equation for a chemical reaction
gives the relative amounts of each reactant
and each product involved in the reaction.
• If the amount of one substance is known,
based on the molar ratios in the equation, the
amounts (masses, particles or volume)of other
substances can be calculated.
• Make sure to always work in moles.
A dry - cleaning fluid composed of carbon and chlorine was found to have the
composition 14.5% C, 85.5% Cl. What is the empirical formula of this compound?
Element Percentage Percentage Simplest
Ar
Ratio
C
14.5%
Cl
85.5%
 CCl2
14.5
 1.208
12
85.5
 2.408
35.5
1
2