Download All you need to know about Additional Science

Atomic Structure
LO
• To review the mass of atoms and their size, using the
term ‘Nano’ and what the standard form is.
Atomic structure
Rows = periods
Row number =
Columns = groups
Group number =
Atomic structure
Atomic
number:
…………………
…………………
…………………
…………………
…………………
…………………
…………………
…………………
…………………
Mass
number:
…………………
…………………
…………………
…………………
…………………
…………………
…………………
…………………
…………………
Exam Practice
• Describe in as much detail as possible the
structure of a fluorine atom
Mark Scheme
9 protons /Proton Number 9
mass / atomic number is neutral
1
10 neutrons
1
electron arrangement 2,7 / 9 electrons
incorrect configurations neutral
if no points scored, allow 1 mark for nucleus surrounded by electrons or nucleus contains neutrons
and protons
How small is an
atom?
• Atoms are very small, they have a radius
of about 0.1 nm
• That’s 1 x 10-10 m !
Standard Form
• https://www.youtube.c
om/watch?v=LR5o2ikiTM
• Looking at the size of an
atom………..
• Atoms are about a
tenth of a billion of a
meter!!!!!
• How can we express
this?
Standard Form
• When using very large or small numbers,
scientists use the ‘powers of ten’ to express a
number.
• Distance at an atomic level is measured in
nanometes(nm).
• 1nm = 1 x 10-9m
• Example (large number);
1 000 000m
Written as…………..
1 x 106m
• Example (small number);
One millionth of a meter
Written as……………
1 x 10-6m
Have a go
• Write 15 000 000 in standard index form.
= 1. 5 × 10 7
Write this 2 850 000 in standard index form.
= 2.85 × 1 000 000 = 2.85 × 106
Write this 0.000467 in standard index form
4.67 × 10 -4
• Convert this back to ordinary numbers 3 x 104
= 30 000
How small is the nucleus compared with
the whole atom?
The radius of a nucleus is less than
1/10,000 of the atom!
But the nucleus makes up
almost all the mass of an ato
nucleus
.
Therefore: which
particles
are responsible for the
mass
of the atom?
• So an atom has a radius
of 0.1nm (0.1 x 10-10)
• The radius of the nucleus
is about 1 x 10-14
So there is a lot of space
Questions
• The atomic radius of a boron atom is 9 x 10-11m
Give it’s atomic radius in nanometers
• Calculate the approximate radius of its nucleus (in nm), given
that it will be about one ten thousandth the radius of the
boron atom. Give your answer in standard form.
• 0.09nm 1x10-11 – 1 x 10-9 = 1 x 10-2
= 9 x 10-2
Isotopes
• Atoms of the same
element always have the
same number of PROTONS
BUT they can have
different numbers of
NEUTRONS.
• These are called
ISOTOPES.
• Always have the same
atomic numbers BUT
different mass numbers!
• Chlorine's Ar of 35.5 is an average of the masses of the
different isotopes of chlorine. This is calculated by working
out the relative abundance of each isotope. For example, in
any sample of Chlorine 25% will be 37 Cl and 75% 35 Cl. The
relative atomic mass is therefore calculated using the
equation:
• (% of isotope 1 × mass of isotope 1) + (% of isotope 2 ×
mass of isotope 2) ÷ 100
So in the case of chlorine:
(75 × 35) + (25 × 37) ∕ 100
= 2625 + 925 ∕ 100
= 35.3
What is the relative atomic mass of thallium which is
made up from 30 per cent thallium-203 and 70 per cent
thallium-205?
• 204.4
Properties of Isotopes
Isotopes an element can have different physical
properties for example they can different
densities and may or may not be radioactive
They will always have the same chemical
properties
3.2 Masses of atoms
Relative atomic masses (Ar)
Mass of atom compared to
H
Relative formula masses (Mr)
Mass of a compound found by
adding Ar of each element
e.g. Na = 23, Cl = 35.5
1) H =
2) Mg =
3) Ti =
4) O =
e.g. NaCl = 23 + 35.5 = 58.5
1) MgO
2) H2O
3) CH4
4) C6H12O6
3.3 Percentages and formulae
Percentage mass
%
=
mass of element
………………………………………………………
Percentage composition / empirical formula
Al
Cl
Mass
9
35.5
Ar
27
35.5
Moles
(9/27) = 0.33
(35.5/35.5) = 1
Simplest ratio
(divide by smallest
number of moles)
(0.33 / 0.33) = 1
(1 / 0.33) = 3
Formula
AlCl3
3.4 Balancing equations
H2 + O2  H2O2
Elements
(Right-hand side)
Elements
(Left-hand side)
H=
H=
O=
O=
3.4 Reacting masses
2NaOH + Cl2  NaOCl + NaCl + H2O
If we have a solution containing 100 g of sodium
hydroxide, how much chlorine gas should we pass
through the solution to make bleach? Too much, and
some chlorine will be wasted, too little and not all of the
sodium hydroxide will react.
3.4 Reacting masses
2NaOH + Cl2  NaOCl + NaCl + H2O
100 g
Ar / Mr
Ratio
Mass
?
2NaOH
80
(80/80) = 1
1 x 100 = 100
100 g
Cl2
71
(71/80) = 0.8875
0.8875 x 100 = 88.75
88.75 g
3.5 Percentage yield
Very few chemical reactions have a yield of
100% because:
• Reaction is reversible
• Some reactants produce unexpected products
• Some products are left behind in apparatus
• Reactants may not be completely pure
• More than one product is produced and it may
be difficult to separate the product we want
3.5 Percentage yield
Percentage yield
% yield = amount of product produced (g) x 100%
max. amount of product possible (g)
3.5 Atom economy
The amount of the starting materials that end
up as useful products is called the atom
economy
% atom economy = Mr of useful product x 100%
………………………………………
3.6 Reversible reactions
A+B
C+D
…………… = reversible reaction
e.g. iodine monochloride and chlorine gas:
ICl + Cl2
ICl3
• increasing Cl2 increases ICl3
• decreasing Cl2 decreases ICl3
3.7 Haber process
• Fritz Haber invented
the Haber process
• A way of turning
nitrogen in the air
into ammonia
4.2 Collision theory
Collision theory
Chemical reactions
only occur when
reacting particles
collide with each other
with sufficient energy.
The minimum amount
of energy is called the
activation energy
Rate of reaction
increases if:
• temperature
increases
• concentration or
pressure increases
• surface area
increases
• catalyst used
4.2 Surface area
Why?
The inside of a large piece of solid is not in
contact with the solution it is reacting with, so
it cannot react
How?
………………………………………………………
………………………………………………………
………………………………………………………
4.3 Temperature
Why?
At lower temperatures, particles will collide:
a) ……………………………………
b) ……………………………………
How?
Put more energy into reaction
Increasing the temperature
by 10oC will double the rate
of reaction
4.4 Concentration
Why?
Concentration is a measure of how many
particles are in a solution. Units = mol/dm3
The lower the concentration, the fewer
reacting particles, the fewer successful
collisions
How?
……………………………………………………
……………………………………………………
4.4 Pressure
Why?
Pressure is used to describe particles in gases
The lower the pressure, the fewer successful
collisions
How?
Decrease the …………………………
Increase the ……………………………
4.5 Catalysts
Why?
Expensive to increase temperature or pressure
………………………………………………………………………………………
………………………………………………………………………………………
How?
Catalysts are made from transition metals, e.g.
iron, nickel, platinum
Provide surface area for reacting particles to
come together and lower activation energy
5.1 Energy changes
Exothermic reaction,
Endothermic reaction,
e.g. …………………………………
e.g. …………………………………
• Energy ‘exits’
reaction – heats
surroundings
• Energy ‘enters’
reaction – cools
surroundings
• Thermometer
readings rise
• Thermometer
readings fall
5.2 Energy and reversible reactions
………………………………………
Hydrated
copper sulphate
Anhydrous
copper sulphate + water
……………………………………….
5.3 Haber process (again!)
Exothermic
reaction
temperature
products
N2 + 3H2
Endothermic
reaction
temperature
products
temperature
products
2NH3
temperature
products
5.3 Haber process (again!)
Smaller vol. of
gas produced
pressure
products
N2 + 3H2
Larger vol. of
gas produced
pressure
products
2NH3
pressure
products
pressure
products
5.3 Haber process (again!)
Temperature:
- Forward reaction is exothermic, so low temperature is preferred
- But this makes reaction slow
- Compromise by using ……………
N2 + 3H2
Pressure:
- The higher the better
- High pressure is dangerous!
- Compromise by using ……………………….
2NH3
Catalyst:
- Iron
- Speeds up both
sides of reaction
6.1 Electrolysis
Electrolysis: splitting
up using electricity
Ionic substance
- molten ………
- dissolved ………
Non-metal ion
Metal ion
6.2 Changes at the electrodes
Solutions
Water contains the ions:
………………………………………
The less reactive element
will be given off at
electrode
Oxidation is loss
Reduction is gain
OIL
RIG
Molten (PbBr)
2Br-  Br2 + 2e-
Pb2+ + 2e-  Pb
Solution (KBr)
2Br-  Br2 + 2e-
2H+ + 2e-  H2
6.3 Electrolysing brine
At anode
At cathode
In solution
6.4 Purifying copper
At anode
At cathode
7.1 Acids and alkalis
Acids = H+ ions
Alkalis = OH- ions
Alkalis = soluble bases
7.2 + 7.3 Salts
Acid
Formula
Salt
Example
Hydrochloric HCl
Chloride Sodium chloride
Sulphuric
H2SO4
Sulphate Copper sulphate
Nitric
HNO3
Nitrate
Potassium nitrate
7.2 + 7.3 Salts – metals, bases
and alkalis
Metals:
Metal(s) + acid(aq)  salt(aq) + hydrogen(g)
Bases: Acid(aq) + base(aq)  salt(aq) + water(l)
Alkalis: Acid(aq) + alkali(aq)  salt(aq) + water(l)
Ionic equation (neutralisation): ……………………………
7.3 Salts – solutions
Solutions:
solution(aq) + solution(aq) 
precipitate(s) + solution(aq)
Solid precipitate is …………………………………………………