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Title: An introduction and basics of the 'mole' and molar equations for G.C.S.E and AS level Chemistry
Description: Provides a comprehensive but easy to understand explanation of: - What is the mole? - Why do we use it? - How did we arrive at the mole? - What is the molar equation? - How to work out the relative atomic/formula mass for elements and compounds. - How do we use the mole in questions? Also included - Advice on how to go about answering a question in practice. - Practice molar equation questions on both elements and compounds. This powerpoint is 118 slides long but is very unlikely to take longer than around 20 minutes (excluding practice questions). The powerpoint is focused on introducing the mole and molar equations and showing the learner how to apply them upto, but not including titration which will be discussed in a later powerpoint and likely not applicable to G.C.S.E students. This powerpoint is therefore applicable for G.C.S.E Chemistry students and AS level Chemistry students.
Description: Provides a comprehensive but easy to understand explanation of: - What is the mole? - Why do we use it? - How did we arrive at the mole? - What is the molar equation? - How to work out the relative atomic/formula mass for elements and compounds. - How do we use the mole in questions? Also included - Advice on how to go about answering a question in practice. - Practice molar equation questions on both elements and compounds. This powerpoint is 118 slides long but is very unlikely to take longer than around 20 minutes (excluding practice questions). The powerpoint is focused on introducing the mole and molar equations and showing the learner how to apply them upto, but not including titration which will be discussed in a later powerpoint and likely not applicable to G.C.S.E students. This powerpoint is therefore applicable for G.C.S.E Chemistry students and AS level Chemistry students.
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THE ‘MOLE’
A REVISION POWERPOINT OF THE MOLE IN CHEMISTRY
WHAT IS A ‘MOLE’
• The ‘mole’ (pronounced ‘mole’ like the animal) is a unit used in chemistry to define how much ‘stuff’
there is in a sample of a chemical
...
• E
...
; “There is 6
...
”
WHAT IS A ‘MOLE’
• The ‘mole’ (pronounced ‘mole’ like the animal) is a unit used in chemistry to define how much ‘stuff’
there is in a sample of a chemical
...
g
...
2 mole of carbon in that sample
...
4 grams of carbon?”
WHY USE MOLES ?
• In order to answer that question let’s relate it to something that almost everybody has heard at some
point;
“Which weighs more? A tonne of feathers or a tonne of bricks?”
WHY USE MOLES ?
• In order to answer that question let’s relate it to something that almost everybody has heard at some
point;
“Which weighs more? A tonne of feathers or a tonne of bricks?”
• The answer (hopefully) is pretty obvious
...
WHY USE MOLES ?
• In order to answer that question let’s relate it to something that almost everybody has heard at some
point;
“Which weighs more? A tonne of feathers or a tonne of bricks?”
• The answer (hopefully) is pretty obvious
...
WHY USE MOLES ?
• In order to answer that question let’s relate it to something that almost everybody has heard at some
point;
“Which weighs more? A tonne of feathers or a tonne of bricks?”
• The answer (hopefully) is pretty obvious
...
• The material doesn’t matter because they weigh the same amount
...
Both weigh exactly one tonne so neither weighs more than
the other
...
• However, let’s see what happens if we switch the question around a little bit
...
Both weigh exactly one tonne so neither weighs more than
the other
...
• However, let’s see what happens if we switch the question around a little bit
...
Both weigh exactly one tonne so neither weighs more than the
other
...
• However, let’s see what happens if we switch the question around a little bit
...
This time the answer is the brick
...
WHY USE MOLES ?
• In order to answer that question let’s relate it to something that almost everybody has heard at some point;
“Which weighs more? A tonne of feathers or a tonne of bricks?”
• The answer (hopefully) is pretty obvious
...
• The material doesn’t matter because they weigh the same amount
...
“Which weighs more? A feather or a brick?”
• Again, this should be pretty simple
...
One feather is clearly lighter than one
brick
...
Both weigh exactly one tonne so neither weighs more than the other
...
•
However, let’s see what happens if we switch the question around a little bit
...
This time the answer is the brick
...
THE REASON FOR THIS IS THAT THERE IS SIMPLY MORE ‘STUFF’ IN ONE BRICK THAN IN ONE FEATHER
...
Both weigh exactly one tonne so neither weighs more than the other
...
•
However, let’s see what happens if we switch the question around a little bit
...
This time the answer is the brick
...
THE REASON FOR THIS IS THAT THERE IS SIMPLY MORE ‘STUFF’ IN ONE BRICK THAN IN ONE FEATHER
...
ATOMS AND MOLES
• In the previous example the brick weighs more than the feather
...
• Put simply; this is because the brick has more atoms than the feather
...
• Put simply; this is because the brick has more atoms than the feather
...
ATOMS AND MOLES
• In the previous example the brick weighs more than the feather
...
• Atoms are the building block of molecules which are in turn the building blocks of the feather and brick
...
ATOMS AND MOLES
• In the previous example the brick weighs more than the feather
...
• Atoms are the building block of molecules which are in turn the building blocks of the feather and brick
...
• Mass is simply a way for us to say that an object has more ‘stuff’ or atoms than another object
...
ATOMS AND MOLES
• So what does this have to do with the difference between grams and moles ?
• moles can be seen as a way of measuring how many atoms there are in an object (it’s mass)
...
e
...
ATOMS AND MOLES
• So lets forget about the brick and feather and apply this to Carbon for an example:
ATOMS AND MOLES
• So lets forget about the brick and feather and apply this to Carbon for an example:
ATOMS AND MOLES
• So lets forget about the brick and feather and apply this to Carbon for an example:
• Here is a lump of Carbon
...
• It has a weight of 74
...
ATOMS AND MOLES
• So lets forget about the brick and feather and apply this to Carbon for an example:
• Here is a lump of Carbon
...
4 grams
...
73364 x 10^24
atoms
...
• It has a weight of 74
...
• It has a mass of 3
...
A number followed by ‘x 10^’ means that you
add 0 however many times to the end of the first
number and get rid of the ‘
...
ATOMS AND MOLES
• So lets forget about the brick and feather and apply this to Carbon for an example:
• Here is a lump of Carbon
...
4 grams
...
73364 x 10^24
atoms
...
’ to show the real
number
...
2 x 10^2 = 120
...
• It has a weight of 74
...
• It has a mass of 3
...
A number followed by ‘x 10^’ means that you
add 0 however many times to the end of the last
number and get rid of the ‘
...
For example; 1
...
If you were to count up in ones, once every
second, it would take you
118,392,947,700,000,000,000,000,000 years to
count to 3
...
• It has a weight of 74
...
• It has a mass of 3
...
• This is actually the Carbon from the
beginning and has a molar mass of
6
...
A number followed by ‘x 10^’ means that you
add 0 however many times to the end of the first
number and get rid of the ‘
...
For example; 1
...
If you were to count up in ones, once every
second, it would take you
118,392,947,700,000,000,000,000,000 years to
count to 3
...
ATOMS AND MOLES
• Now, this is why moles matter
...
• Here is a lump of Sulphur
...
• Here is a lump of Carbon
...
73364 x
10^24 (6
...
• Here is a lump of Sulphur
...
73364
x 10^24 (6
...
ATOMS AND MOLES
• Now, this is why moles matter
...
• It has a mass of 3
...
2 moles)
• It has a weight of 74
...
• Here is a lump of Sulphur
...
73364
x 10^24 (6
...
4
grams
...
• Here is a lump of Carbon
...
73364 x
10^24 (6
...
4
grams
...
• It has a mass of 3
...
2 moles)
• It has a weight of 198
...
DISPITE HAVING THE EXACT SAME NUMBER OF ATOMS, THESE TWO
SAMPLES WEIGH COMPLETELY DIFFERENT AMOUNTS
...
4 grams of Carbon = 74
...
73364 x 10^24 atoms of Carbon > 1
...
022 x 10^23 atoms of Carbon = 6
...
022 x 10^23 atoms of Carbon = 6
...
022 x 10^23 atoms = one mole
(We’ll get on to that shortly)
ATOMS AND MOLES
• Conclusion: The atoms of Sulphur weigh more than the atoms of Carbon
...
• But how is this possible?
ATOMS AND MOLES
• Conclusion: The atoms of Sulphur weigh more than the atoms of Carbon
...
ATOMIC NUMBERS AND RELATIVE ATOMIC MASS
•
Conclusion: The atoms of Sulphur weigh more than the atoms of Carbon
...
• Here we have the elemental notations for
Carbon and Sulphur
...
• This number represents the element’s calculated atomic weight (known as the
‘relative atomic mass’ or ‘mr’ for short, representing the average amount of
protons and neurons within the atom across all isotopes
...
g, Carbon-12 has 6 protons and 6 neutrons whereas Carbon-11 has 6 protons and 5 neutrons
...
+ 6
...
065
• An element’s atomic weight is calculated by adding up the amount of protons
(found in the ‘atomic number’) and adding them together with the average
number of neutrons found in their isotopes
...
ATOMIC NUMBERS AND RELATIVE ATOMIC MASS
DEFINING THE MOLE
AVOGADRO'S CONSTANT
DEFINING THE MOLE
AVOGADRO'S CONSTANT
• Previously we said that one mole was equal to 6
...
DEFINING THE MOLE
AVOGADRO'S CONSTANT
• The atomic mass unit was developed using the Carbon-12 atom (C-12); the most
common isotope of Carbon
...
• As Carbon-12 has exactly 6 protons, 6 neutrons and the weight of an electron is
almost immeasurably small, 1/12th of the mass of a Carbon-12 atom is therefore
very, very close to the mass of a single proton or neutron
...
• As Carbon-12 has exactly 6 protons, 6 neutrons and the weight of an electron is
almost immeasurably small, 1/12th of the mass of a Carbon-12 atom is therefore
very, very close to the mass of a single proton or neutron
...
0221415 x 10^23 grams
...
• As Carbon-12 has exactly 6 protons, 6 neutrons and the weight of an electron is
almost immeasurably small, 1/12th of the mass of a Carbon-12 atom is therefore
very, very close to the mass of a single proton or neutron
...
0221415 x 10^23 grams
...
K
...
– 6
...
DEFINING THE MOLE
AVOGADRO'S CONSTANT
• Using this, we know that if we had one mole of C-12 atoms it would weigh exactly 12
grams
...
USING MOLAR EQUATIONS
• What on Earth does that mean you are probably asking
...
USING MOLAR EQUATIONS
• What on Earth does that mean you are probably asking
...
• Moles = Mass / Relative Formula Mass
• Mass = Moles x Relative Formula Mass
• Relative Formula Mass = Mass / Moles
*THIS FORMULA IS ONE OF THE MOST VITAL FORMULAE
IN CHEMISTRY*
USING MOLAR EQUATIONS
CALCULATIONS
USING MOLAR EQUATIONS
CALCULATIONS
• How do we use this calculation in practice then?
USING MOLAR EQUATIONS
CALCULATIONS
• How do we use this calculation in practice then?
• Here is a completed example:
USING MOLAR EQUATIONS
CALCULATIONS
• How do we use this calculation in practice then?
• Here is a completed example:
Q) How many moles are present in 32 grams of Silicone?
A) 32g / 28
...
139 mole(3dp)*
* ‘dp’ stands for decimal places
...
0855Mr = 1
...
*
USING MOLAR EQUATIONS
CALCULATIONS
Q) How many moles are present in 32 grams of Silicone?
• First and foremost you should always establish what it is you know in the given question
...
• What do we know?:
USING MOLAR EQUATIONS
CALCULATIONS
Q) How many moles are present in 32 grams of Silicone?
• First and foremost you should always establish what it is you know in the given question
...
USING MOLAR EQUATIONS
CALCULATIONS
Q) How many moles are present in 32 grams of Silicone?
• First and foremost you should always establish what it is you know in the given question
...
- We know that the mass of the substance is 32 grams
...
•
-
What do we know?:
The chemical we are being asked about is Silicone
...
We are being asked to calculate the number of moles
...
•
-
What do we know?:
The chemical we are being asked about is Silicone
...
We are being asked to calculate the number of moles
...
•
-
What do we know?:
The chemical we are being asked about is Silicone
...
We are being asked to calculate the number of moles
...
•
-
What do we know?:
The chemical we are being asked about is Silicone
...
We are being asked to calculate the number of moles
...
Moles = Mass / Relative Formula Mass
USING MOLAR EQUATIONS
CALCULATIONS
Q) How many moles are present in 32 grams of Silicone?
• First and foremost you should always establish what it is you know in the given question
...
We know that the mass of the substance is 32 grams
...
• And we have been given the mass
Moles = 32 grams / Relative Formula Mass
USING MOLAR EQUATIONS
CALCULATIONS
Q) How many moles are present in 32 grams of Silicone?
• First and foremost you should always establish what it is you know in the given question
...
We know that the mass of the substance is 32 grams
...
• And we have been given the mass and using the periodic table, we can work out the relative formula mass
...
• Silicon has an relative atomic weight / mass of 28
...
USING MOLAR EQUATIONS
CALCULATIONS
Q) How many moles are present in 32 grams of Silicone?
• Lets plug the relative formula mass into the equation
...
Moles = 32 grams / 28
...
Moles = 32 grams / 28
...
1
...
0086
USING MOLAR EQUATIONS
CALCULATIONS
Q) How many moles are present in 32 grams of Silicone?
• Now all that’s left to do is to calculate the answer using a calculator
...
392 (3dp) moles = 32 grams / 28
...
*
USING MOLAR EQUATIONS
CALCULATIONS
Q) How many moles are present in 32 grams of Silicone?
• Now all that’s left to do is to calculate the answer using a calculator
...
392 (3dp) moles = 32 grams / 28
...
*
AND THERE YOU GO!
MOLES!
USING MOLAR EQUATIONS
PRACTICE QUESTIONS
• Now you are ready for some practice questions before we move on to slightly more complicated
equations
...
• Make sure you have these things handy:
USING MOLAR EQUATIONS
PRACTICE QUESTIONS
• Now you are ready for some practice questions before we move on to slightly more complicated
equations
...
- Answers will be provided the slide immediately after the questions so be careful not to move forward
until you are ready
...
9 grams
...
1 grams
...
5 moles
...
• Q5) Calculate the total mass in a sample of 2
...
9 Moles of Nickle
...
*
USING MOLAR EQUATIONS
PRACTICE QUESTION ANSWERS
• Q1) 3
...
007 = 0
...
1 / 15
...
6313 moles
• Q3) 4
...
990 = 103
...
011 = 12
...
1 x 196
...
9 x 58
...
147 grams
USING MOLAR EQUATIONS
COMPOUNDS
USING MOLAR EQUATIONS
COMPOUNDS
• Exactly the same way in which the formula can be used to calculate the moles / mass in an element
sample, it can be applied to compounds
...
• The only additional work is addition itself
...
1 moles of Carbon-Dioxide (CO2)
...
1 moles of Carbon-Dioxide (CO2)
...
USING MOLAR EQUATIONS
COMPOUNDS
• Worked example:
- Q) Calculate the total mass of 2
...
- Firstly assess what we know:
- The chemical we are being asked about is Carbon-Dioxide (CO2)
USING MOLAR EQUATIONS
COMPOUNDS
• Worked example:
- Q) Calculate the total mass of 2
...
- Firstly assess what we know:
- The chemical we are being asked about is Carbon-Dioxide (CO2)
- We know that there is a total of 2
...
USING MOLAR EQUATIONS
COMPOUNDS
• Worked example:
- Q) Calculate the total mass of 2
...
- Firstly assess what we know:
- The chemical we are being asked about is Carbon-Dioxide (CO2)
- We know that there is a total of 2
...
- We are being asked to calculate the total mass
...
1 moles of Carbon-Dioxide (CO2)
...
1 moles in the sample
...
-
Let’s get up our equation:
USING MOLAR EQUATIONS
COMPOUNDS
• Worked example:
-
Q) Calculate the total mass of 2
...
-
Firstly assess what we know:
-
The chemical we are being asked about is Carbon-Dioxide (CO2)
-
We know that there is a total of 2
...
-
We are being asked to calculate the total mass
...
USING MOLAR EQUATIONS
COMPOUNDS
• This time we are being asked to calculate the mass
...
• We must rearrange our equation:
Mass = Moles x Relative Formula Mass
USING MOLAR EQUATIONS
COMPOUNDS
• This time we are being asked to calculate the mass
...
1 moles x Relative Formula Mass
USING MOLAR EQUATIONS
COMPOUNDS
• This time we are being asked to calculate the mass
...
1 moles x Relative Formula Mass
• Total RFM: 0
USING MOLAR EQUATIONS
COMPOUNDS
• This time we are being asked to calculate the mass
...
1 moles x Relative Formula Mass
• Total RFM: 12
...
011
USING MOLAR EQUATIONS
COMPOUNDS
• This time we are being asked to calculate the mass
...
1 moles x Relative Formula Mass
• Total RFM: 28
...
011
• Oxygen: 15
...
USING MOLAR EQUATIONS
COMPOUNDS
• This time we are being asked to calculate the mass
...
1 moles x Relative Formula Mass
• Total RFM: 44
...
011
• Oxygen: 15
...
• We must rearrange our equation:
• Next lets plug in what we know:
Mass = 2
...
009 mr
• Now we can simply solve the equation
...
1 moles of Carbon-Dioxide (CO2)
...
1 x 44
...
419 grams (3dp)
USING MOLAR EQUATIONS
COMPOUND PRACTICE
• Now you are ready for some practice questions
...
• Make sure you have these things handy:
-
Periodic table
-
Calculator (A scientific calculator will come in very useful)
-
The molar conversion equation(s) [Do your best to learn it but have it on hand if you can’t]
-
You may also need a piece of paper for jotting notes
...
USING MOLAR EQUATIONS
COMPOUND PRACTICE
• Q1) Calculate the molar mass of a sample of Iron-Oxide (Fe2O3) weighing 6
...
• Q2) Calculate the molar mass of a sample of Sodium-Chloride (NaCl) weighing 2
...
• Q3) Calculate the mass of a 4
...
• Q4) Calculate the molar mass of a sample of Amonia (Nh3) weighing 10
...
• Q5) State which single element is bonded twice with Oxygen (O) when the overall mass is 41
...
3 and state the molecule
...
008 x 3) + 14
...
2g = 0
...
990 + 35
...
7g = 21
...
8 x (1
...
453) = 175
...
92 / ((1
...
007) = 0
Title: An introduction and basics of the 'mole' and molar equations for G.C.S.E and AS level Chemistry
Description: Provides a comprehensive but easy to understand explanation of: - What is the mole? - Why do we use it? - How did we arrive at the mole? - What is the molar equation? - How to work out the relative atomic/formula mass for elements and compounds. - How do we use the mole in questions? Also included - Advice on how to go about answering a question in practice. - Practice molar equation questions on both elements and compounds. This powerpoint is 118 slides long but is very unlikely to take longer than around 20 minutes (excluding practice questions). The powerpoint is focused on introducing the mole and molar equations and showing the learner how to apply them upto, but not including titration which will be discussed in a later powerpoint and likely not applicable to G.C.S.E students. This powerpoint is therefore applicable for G.C.S.E Chemistry students and AS level Chemistry students.
Description: Provides a comprehensive but easy to understand explanation of: - What is the mole? - Why do we use it? - How did we arrive at the mole? - What is the molar equation? - How to work out the relative atomic/formula mass for elements and compounds. - How do we use the mole in questions? Also included - Advice on how to go about answering a question in practice. - Practice molar equation questions on both elements and compounds. This powerpoint is 118 slides long but is very unlikely to take longer than around 20 minutes (excluding practice questions). The powerpoint is focused on introducing the mole and molar equations and showing the learner how to apply them upto, but not including titration which will be discussed in a later powerpoint and likely not applicable to G.C.S.E students. This powerpoint is therefore applicable for G.C.S.E Chemistry students and AS level Chemistry students.