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Transcript
```THE MOLE
A guide for A level students
KNOCKHARDY PUBLISHING
2008
SPECIFICATIONS
KNOCKHARDY PUBLISHING
THE MOLE
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it
may be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are
available from the KNOCKHARDY SCIENCE WEBSITE at...
www.knockhardy.org.uk/sci.htm
Navigation is achieved by using the left and right arrow keys on the keyboard
THE MOLE
CONTENTS
• What is a mole and why do we use it?
• Calculating the number of moles of a single substance
• Reacting mass calculations
• Solutions and moles
• Standard solutions
• Volumetric calculations
• Molar volume calculations
THE MOLE
Before you start it would be helpful to…
• know how to balance simple equations
• know how to re-arrange mathematical formulae
DON’T BE
LEFT IN THE
DARK!
THE MOLE
WHAT IS A MOLE ?
it is the standard unit of amount of a substance it is just a number, a very big number
it is a way of saying a number in words, just like...
DOZEN for 12
SCORE for 20
GROSS for 144
THE MOLE
WHAT IS A MOLE ?
it is the standard unit of amount of a substance it is just a number, a very big number
it is a way of saying a number in words, just like...
DOZEN for 12
SCORE for 20
GROSS for 144
HOW BIG IS IT ?
602200000000000000000000
(Approximately)... THAT’S
It is a lot easier to write it as...
BIG !!!
6.022 x 1023
THE MOLE
WHAT IS A MOLE ?
it is the standard unit of amount of a substance it is just a number, a very big number
it is a way of saying a number in words, just like...
DOZEN for 12
SCORE for 20
GROSS for 144
HOW BIG IS IT ?
602200000000000000000000
(Approximately)... THAT’S
It is a lot easier to write it as...
It is also known as...
BIG !!!
6.022 x 1023
It doesn’t matter what the number is as long as everybody sticks to the same value !
THE MOLE
WHY USE IT ?
Atoms and molecules don’t weigh much so it is easier to count large
numbers of them. In fact it is easier to weigh substances.
Using moles tells you...
how many particles you get in a certain mass
the mass of a certain number of particles
DO I NEED TO KNOW ANYTHING ELSE ?
Yes, it would help if you can balance equations
AND
Keep trying, you will get the idea ... EVENTUALLY!
THE MOLE – AN OVERVIEW
WHAT IS IT?
The standard unit of amount of a substance just as the standard unit of length is a METRE
It is just a number, a very big number
It is also a way of saying a number in words
like
DOZEN for 12
GROSS for 144
THE MOLE – AN OVERVIEW
WHAT IS IT?
The standard unit of amount of a substance just as the standard unit of length is a METRE
It is just a number, a very big number
It is also a way of saying a number in words
like
DOZEN for 12
GROSS for 144
HOW BIG IS IT ? 602200000000000000000000 (approx) - THAT’S BIG !!!
It is a lot easier to write it as 6.022 x 1023
And anyway it doesn’t matter what the number is
as long as everybody sticks to the same value !
THE MOLE – AN OVERVIEW
WHAT IS IT?
The standard unit of amount of a substance just as the standard unit of length is a METRE
It is just a number, a very big number
It is also a way of saying a number in words
like
DOZEN for 12
GROSS for 144
HOW BIG IS IT ? 602200000000000000000000 (approx) - THAT’S BIG !!!
It is a lot easier to write it as 6.022 x 1023
And anyway it doesn’t matter what the number is
as long as everybody sticks to the same value !
WHY USE IT ?
Atoms and molecules don’t weigh much so it is
easier to count large numbers of them.
In fact it is easier to weigh substances.
Using moles tells you :- how many particles you get in a certain mass
the mass of a certain number of particles
THE MOLE
CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE
MOLES
moles
mass
molar mass
=
=
=
=
MASS
MOLAR MASS
MASS
MOLES x MOLAR
MASS
mass / molar mass
moles x molar mass
mass / moles
COVER UP THE VALUE YOU
WANT AND THE METHOD
OF CALCULATION IS
REVEALED
UNITS
mass
molar mass
g
g mol-1
or
or
kg
kg mol-1
MOLES OF A SINGLE SUBSTANCE
1.
Calculate the number of moles of oxygen molecules in 4g
MOLES OF A SINGLE SUBSTANCE
1.
Calculate the number of moles of oxygen molecules in 4g
oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
MOLES OF A SINGLE SUBSTANCE
1.
Calculate the number of moles of oxygen molecules in 4g
oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles
=
mass
molar mass
=
4g
32g mol -1
= 0.125 mol
MOLES OF A SINGLE SUBSTANCE
1.
Calculate the number of moles of oxygen molecules in 4g
oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles
=
mass
molar mass
=
4g
32g mol -1
= 0.125 mol
MOLES OF A SINGLE SUBSTANCE
1.
Calculate the number of moles of oxygen molecules in 4g
oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles
=
mass
molar mass
=
4g
32g mol -1
2. What is the mass of 0.25 mol of Na2CO3 ?
= 0.125 mol
MOLES OF A SINGLE SUBSTANCE
1.
Calculate the number of moles of oxygen molecules in 4g
oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles
=
mass
molar mass
=
4g
32g mol -1
= 0.125 mol
2. What is the mass of 0.25 mol of Na2CO3 ?
Relative Molecular Mass of Na2CO3 = (2x23) + 12 + (3x16) = 106
Molar mass of Na2CO3
= 106g mol-1
MOLES OF A SINGLE SUBSTANCE
1.
Calculate the number of moles of oxygen molecules in 4g
oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles
=
mass
molar mass
=
4g
32g mol -1
= 0.125 mol
2. What is the mass of 0.25 mol of Na2CO3 ?
Relative Molecular Mass of Na2CO3 = (2x23) + 12 + (3x16) = 106
Molar mass of Na2CO3
= 106g mol-1
mass = moles x molar mass
= 0.25 x 106 = 26.5g
MOLES OF A SINGLE SUBSTANCE
1.
Calculate the number of moles of oxygen molecules in 4g
oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles
=
mass
molar mass
=
4g
32g mol -1
= 0.125 mol
2. What is the mass of 0.25 mol of Na2CO3 ?
Relative Molecular Mass of Na2CO3 = (2x23) + 12 + (3x16) = 106
Molar mass of Na2CO3
= 106g mol-1
mass = moles x molar mass
= 0.25 x 106 = 26.5g
REACTING MASS CALCULATIONS
CaCO3
+
2HCl
———>
CaCl2
1. What is the relative formula mass of CaCO3?
+
CO2
+
H2 O
40 + 12 + (3 x 16) = 100
2. What is the mass of 1 mole of CaCO3
100 g
3. How many moles of HCl react with 1 mole of CaCO3?
2 moles
4. What is the relative formula mass of HCl?
35.5 + 1 = 36.5
5. What is the mass of 1 mole of HCl?
36.5 g
6. What mass of HCl will react with 1 mole of CaCO3 ?
2 x 36.5g = 73g
7. What mass of CO2 is produced ?
moles of CO2 = moles of CaCO3
moles of CO2 = 0.001 moles
mass of CO2 = 0.001 x 44 = 0.044g
REACTING MASS CALCULATIONS
EQUATIONS
give you the ratio in which chemicals react and are formed
need to be balanced in order to do a calculation
CaCO3
+
2HCl
———>
CaCl2
+
CO2
+
H2O
1. What is the relative molecular mass of CaCO3?
40 + 12 + (3 x 16) = 100
2. What is the mass of 1 mole of CaCO3?
100 g
3. What does 0.1M HCl mean?
concentration is 0.1 mol dm-3
4. How many moles of HCl are in 20cm3 of 0.1M HCl?
0.1 x 20
1000
5. How many moles of CaCO3 will react ?
½ x 0.002 = 0.001 moles
6. What is the mass of 0.001 moles of CaCO3?
7. What mass of CO2 is produced ?
=
0.002 moles
mass = moles x molar mass
= 0.001 x 100 = 0.1 g
moles of CO2 = moles of CaCO3
moles of CO2 = 0.001 moles
mass of CO2 = 0.001 x 44 = 0.044g
THE MOLE
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
MOLES
UNITS
=
CONCENTRATION x VOLUME
concentration
volume
mol dm-3
dm3
MOLES
MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)
CONC x VOLUME
BUT IF...
concentration
volume
mol dm-3
cm3
MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3)
1000
COVER UP THE VALUE
YOU WANT AND THE
METHOD OF
CALCULATION IS
REVEALED
THE MOLE
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
volume pipetted out into the conical flask
=
=
0.100 mol dm-3
25.00 cm3
25cm3
250cm3
250cm3
THE MOLE
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
volume pipetted out into the conical flask
=
=
0.100 mol dm-3
25.00 cm3
25cm3
250cm3
250cm3
The original solution has a concentration of 0.100 mol dm-3
THE MOLE
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
volume pipetted out into the conical flask
=
=
0.100 mol dm-3
25.00 cm3
25cm3
250cm3
250cm3
The original solution has a concentration of 0.100 mol dm-3
This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solution
THE MOLE
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
volume pipetted out into the conical flask
=
=
0.100 mol dm-3
25.00 cm3
25cm3
250cm3
250cm3
The original solution has a concentration of 0.100 mol dm-3
This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solution
Take out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles
THE MOLE
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
volume pipetted out into the conical flask
=
=
0.100 mol dm-3
25.00 cm3
25cm3
250cm3
250cm3
The original solution has a concentration of 0.100 mol dm-3
This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solution
Take out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles
moles in 1dm3 (1000cm3)
moles in 1cm3
moles in 25cm3
=
=
=
0.100
0.100/1000
25 x 0.100/1000
= 2.5 x 10-3 mol
THE MOLE
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
volume pipetted out into the conical flask
=
=
0.100 mol dm-3
25.00 cm3
25cm3
250cm3
250cm3
The original solution has a concentration of 0.100 mol dm-3
This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solution
Take out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles
moles in 1dm3 (1000cm3)
moles in 1cm3
moles in 25cm3
=
=
=
0.100
0.100/1000
25 x 0.100/1000
= 2.5 x 10-3 mol
MOLE OF SOLUTE IN A SOLUTION
MOLES
1
=
CONCENTRATION x VOLUME
Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH
MOLE OF SOLUTE IN A SOLUTION
MOLES
1
=
CONCENTRATION x VOLUME
Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH
moles =
=
conc x volume in cm3
1000
2 mol dm-3 x 25cm3
1000
= 0.05 moles
MOLE OF SOLUTE IN A SOLUTION
MOLES
1
CONCENTRATION x VOLUME
Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH
moles =
=
2
=
conc x volume in cm3
1000
2 mol dm-3 x 25cm3
1000
= 0.05 moles
What volume of 0.1M H2SO4 contains 0.002 moles ?
MOLE OF SOLUTE IN A SOLUTION
MOLES
1
CONCENTRATION x VOLUME
Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH
moles =
=
2
=
conc x volume in cm3
1000
2 mol dm-3 x 25cm3
1000
= 0.05 moles
What volume of 0.1M H2SO4 contains 0.002 moles ?
volume =
(in cm3)
=
1000 x moles
conc
1000 x 0.002
0.1 mol dm-3
(re-arrangement of above)
= 20 cm3
SOLUTIONS
‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’
Volumetric solutions are made by dissolving a known amount of solute in a solvent
(usually water) and then adding enough solvent to get the correct volume of solution.
WRONG
Dissolve 1g of solute in
of de-ionised water
250cm3
1g
250cm3
SOLUTIONS
‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’
Volumetric solutions are made by dissolving a known amount of solute in a solvent
(usually water) and then adding enough solvent to get the correct volume of solution.
WRONG
Dissolve 1g of solute in
of de-ionised water
1g
250cm3
250cm3
WATER
RIGHT
Dissolve 1g of solute in water
and then add enough water to
make 250cm3 of solution
WATER
1g
250cm3
STANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean
beaker and dissolved in de-ionised water
The solution was transferred
quantitatively to a 250 cm3
to the mark with de-ionised
(or distilled) water.
STANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean
beaker and dissolved in de-ionised water
The solution was transferred
quantitatively to a 250 cm3
to the mark with de-ionised
(or distilled) water.
What is the concentration of the solution in mol dm-3 ?
STANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean
beaker and dissolved in de-ionised water
The solution was transferred
quantitatively to a 250 cm3
to the mark with de-ionised
(or distilled) water.
What is the concentration of the solution in mol dm-3 ?
mass of Na2CO3 in a 250cm3 solution
molar mass of Na2CO3
no. of moles in a 250cm3 solution
= 4.240g
= 106g mol -1
= 4.240g / 106g mol -1
= 0.04 mol
STANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean
beaker and dissolved in de-ionised water
The solution was transferred
quantitatively to a 250 cm3
to the mark with de-ionised
(or distilled) water.
What is the concentration of the solution in mol dm-3 ?
mass of Na2CO3 in a 250cm3 solution
molar mass of Na2CO3
no. of moles in a 250cm3 solution
= 4.240g
= 106g mol -1
= 4.240g / 106g mol -1
Concentration is normally expressed as moles per dm3 of solution
Therefore, as it is in 250cm3, the value is scaled up by a factor of 4
= 0.04 mol
STANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean
beaker and dissolved in de-ionised water
The solution was transferred
quantitatively to a 250 cm3
to the mark with de-ionised
(or distilled) water.
What is the concentration of the solution in mol dm-3 ?
mass of Na2CO3 in a 250cm3 solution
molar mass of Na2CO3
no. of moles in a 250cm3 solution
= 4.240g
= 106g mol -1
= 4.240g / 106g mol -1
= 0.04 mol
Concentration is normally expressed as moles per dm3 of solution
Therefore, as it is in 250cm3, the value is scaled up by a factor of 4
no. of moles in 1000cm3 (1dm3)
= 4 x 0.04
= 0.16 mol
ANS.
0.16 mol dm-3
STANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate
from anhydrous sodium carbonate. How much will they need to weigh out?
STANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate
from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be?
How many moles will be in 1 dm3 ?
How many moles will be in 250cm3 ?
= 0.100 mol dm-3
= 0.100 mol
= 0.100/4
= 0.025 mol
STANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate
from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be?
How many moles will be in 1 dm3 ?
How many moles will be in 250cm3 ?
= 0.100 mol dm-3
= 0.100 mol
= 0.100/4
= 0.025 mol
What is the formula of anhydrous sodium carbonate?
What is the relative formula mass?
What is the molar mass?
= Na2CO3
= 106
= 106g mol -1
STANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate
from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be?
How many moles will be in 1 dm3 ?
How many moles will be in 250cm3 ?
= 0.100 mol dm-3
= 0.100 mol
= 0.100/4
= 0.025 mol
What is the formula of anhydrous sodium carbonate?
What is the relative formula mass?
What is the molar mass?
= Na2CO3
= 106
= 106g mol -1
What mass of Na2CO3 is in 0.025 moles
= 0.025 x 106 = 2.650g
of Na2CO3 ? (mass = moles x molar mass)
STANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate
from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be?
How many moles will be in 1 dm3 ?
How many moles will be in 250cm3 ?
= 0.100 mol dm-3
= 0.100 mol
= 0.100/4
= 0.025 mol
What is the formula of anhydrous sodium carbonate?
What is the relative formula mass?
What is the molar mass?
= Na2CO3
= 106
= 106g mol -1
What mass of Na2CO3 is in 0.025 moles
= 0.025 x 106 = 2.650g
of Na2CO3 ? (mass = moles x molar mass)
ANS.
The chemist will have to weigh out 2.650g, dissolve it in water
and then make the solution up to 250cm3 in a graduated flask.
STANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate
from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be?
How many moles will be in 1 dm3 ?
How many moles will be in 250cm3 ?
= 0.100 mol dm-3
= 0.100 mol
= 0.100/4
= 0.025 mol
What is the formula of anhydrous sodium carbonate?
What is the relative formula mass?
What is the molar mass?
= Na2CO3
= 106
= 106g mol -1
What mass of Na2CO3 is in 0.025 moles
= 0.025 x 106 = 2.650g
of Na2CO3 ? (mass = moles x molar mass)
ANS.
The chemist will have to weigh out 2.650g, dissolve it in water
and then make the solution up to 250cm3 in a graduated flask.
VOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 is
neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
VOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 is
neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation
NaOH
+
HCl
——>
NaCl
+
H2O
VOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 is
neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation
2. Get a molar relationship between the reactants
NaOH
+
HCl
——>
NaCl
+
H2O
moles of NaOH = moles of HCl
you need ONE NaOH for every ONE HCl
VOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 is
neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation
2. Get a molar relationship between the reactants
3. Calculate the number of
moles of each substance
M is the concentration in mol dm-3
NaOH
+
——>
HCl
NaCl
+
H2O
moles of NaOH = moles of HCl
you need ONE NaOH for every ONE HCl
HCl
0.100
x
NaOH
M
20/1000
x
25/1000
(i)
(ii)
VOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 is
neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation
2. Get a molar relationship between the reactants
3. Calculate the number of
moles of each substance
M is the concentration in mol dm-3
NaOH
+
——>
HCl
NaCl
+
H2O
moles of NaOH = moles of HCl
you need ONE NaOH for every ONE HCl
HCl
0.100
x
NaOH
M
20/1000
4. Look at the molar relationship and insert (i) and (ii)
x
25/1000
(i)
(ii)
moles of NaOH = moles of HCl
M x 20/1000 = 0.100 x 25/1000
VOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 is
neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation
2. Get a molar relationship between the reactants
3. Calculate the number of
moles of each substance
M is the concentration in mol dm-3
NaOH
re-arrange the numbers to obtain M
——>
HCl
NaCl
+
H2O
moles of NaOH = moles of HCl
you need ONE NaOH for every ONE HCl
HCl
0.100
x
NaOH
M
20/1000
4. Look at the molar relationship and insert (i) and (ii)
5. Cancel the 1000’s
+
x
25/1000
(i)
(ii)
moles of NaOH = moles of HCl
M x 20/1000 = 0.100 x 25/1000
M x 20
M
=
=
0.100 x 25
0.100 x 25
20
VOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 is
neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation
2. Get a molar relationship between the reactants
3. Calculate the number of
moles of each substance
M is the concentration in mol dm-3
NaOH
re-arrange the numbers to obtain M
6. Calculate the concentration of the NaOH
——>
HCl
NaCl
+
H2O
moles of NaOH = moles of HCl
you need ONE NaOH for every ONE HCl
HCl
0.100
x
NaOH
M
20/1000
4. Look at the molar relationship and insert (i) and (ii)
5. Cancel the 1000’s
+
x
25/1000
(i)
(ii)
moles of NaOH = moles of HCl
M x 20/1000 = 0.100 x 25/1000
M x 20
M
=
=
0.100 x 25
0.100 x 25
20
= 0.125 mol dm-3
VOLUMETRIC CALCULATIONS
Care must be taken when dealing with reactions that do not have a 1:1 molar ratio.
If you don’t understand what an equation tells you, it is easy to make a mistake.
2NaOH
+
——>
H2SO4
Na2SO4 +
2H2O
you need 2 moles of NaOH to react with every 1 mole of H2SO4
i.e
or
moles of NaOH
moles of H2SO4
=
=
2 x moles of H2SO4
moles of NaOH
2
REMEMBER... IT IS NOT A MATHEMATICAL EQUATION
2 moles of NaOH DO NOT EQUAL 1 mole of H2SO4
More examples follow
VOLUMETRIC CALCULATIONS
Care must be taken when dealing with reactions that do not have a 1:1 molar ratio.
If you don’t understand what an equation tells you, it is easy to make a mistake.
2NaOH
+
——>
H2SO4
Na2SO4 +
2H2O
you need 2 moles of NaOH to react with every 1 mole of H2SO4
moles of NaOH
moles of H2SO4
i.e
or
=
=
2 x moles of H2SO4
moles of NaOH
2
2HCl
+
Na2CO3
——>
2NaCl + CO2 + H2O
you need 2moles of HCl to react with every 1 mole of Na2CO3
i.e
or
moles of HCl
moles of Na2CO3
=
=
2 x moles of Na2CO3
moles of HCl
2
VOLUMETRIC CALCULATIONS
Care must be taken when dealing with reactions that do not have a 1:1 molar ratio.
If you don’t understand what an equation tells you, it is easy to make a mistake.
MnO4¯ + 8H+ + 5Fe2+
——> Mn2+ + 4H2O + 5Fe3+
you need 5 moles of Fe2+ to react with every 1 mole of MnO4¯
i.e
or
moles of Fe2+
moles of MnO4¯
=
=
5 x moles of MnO4¯
moles of Fe2+
5
VOLUMETRIC CALCULATIONS
Care must be taken when dealing with reactions that do not have a 1:1 molar ratio.
If you don’t understand what an equation tells you, it is easy to make a mistake.
MnO4¯ + 8H+ + 5Fe2+
——> Mn2+ + 4H2O + 5Fe3+
you need 5 moles of Fe2+ to react with every 1 mole of MnO4¯
i.e
or
moles of Fe2+
moles of MnO4¯
=
=
5 x moles of MnO4¯
moles of Fe2+
5
2MnO4¯ + 5H2O2 + 6H+ ——> 2Mn2+ + 5O2 + 8H2O
you need 5 moles of H2O2 to react with every 2 moles of MnO4¯
i.e
moles of H2O2
=
5 x moles of MnO4¯
2
or
moles of MnO4¯
=
2 x moles of H2O2
5
VOLUMETRIC CALCULATIONS
Calculate the volume of sodium hydroxide (concentration 0.100 mol dm-3)
required to neutralise 20cm3 of sulphuric acid of concentration 0.120 mol dm-3.
2NaOH
+
H2SO4
——>
Na2SO4 +
2H2O
you need 2 moles of NaOH to react with every 1 mole of H2SO4
therefore moles of NaOH = 2 x moles of H2SO4
moles of H2SO4
moles of NaOH
= 0.120 x 20/1000
= 0.100 x V/1000
where V is the volume of alkali in cm3
(i)
(ii)
VOLUMETRIC CALCULATIONS
Calculate the volume of sodium hydroxide (concentration 0.100 mol dm-3)
required to neutralise 20cm3 of sulphuric acid of concentration 0.120 mol dm-3.
2NaOH
+
H2SO4
——>
Na2SO4 +
2H2O
you need 2 moles of NaOH to react with every 1 mole of H2SO4
therefore moles of NaOH = 2 x moles of H2SO4
moles of H2SO4
moles of NaOH
substitute numbers
(i)
(ii)
moles of NaOH
=
2 x moles of H2SO4
0.100 x V/1000
=
2 x 0.120 x 20/1000
0.100 x V
=
2 x 0.120 x 20
Volume of NaOH (V)
=
2 x 0.120 x 20 = 48.00 cm3
0.100
cancel the 1000’s
re-arrange
= 0.120 x 20/1000
= 0.100 x V/1000
where V is the volume of alkali in cm3
MOLAR VOLUME
ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 22.4dm3 at stp
1.
Calculate the volume occupied by 0.25 mols of carbon dioxide at stp
1 mol of carbon dioxide will occupy a volume of 22.4 dm3 at stp
0.25 mol of carbon dioxide will occupy a volume of 22.4 x 0.25 dm3 at stp
0.25 mol of carbon dioxide will occupy a volume of 5.6 dm3 at stp
stp = standard temperature and pressure (273K and 105 Pa)
ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure
MOLAR VOLUME
ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 22.4dm3 at stp
1.
Calculate the volume occupied by 0.25 mols of carbon dioxide at stp
1 mol of carbon dioxide will occupy a volume of 22.4 dm3 at stp
0.25 mol of carbon dioxide will occupy a volume of 22.4 x 0.25 dm3 at stp
0.25 mol of carbon dioxide will occupy a volume of 5.6 dm3 at stp
2.
Calculate the volume occupied by 0.08g of methane (CH4) at stp
Relative Molecular Mass of CH4
Molar Mass of CH4
Moles = mass/molar mass
0.08g / 16g mol-1
= 12 + (4x1)
= 16g mol-1
=
= 16
0.005 mols
1 mol of methane will occupy a volume of 22.4 dm3 at stp
0.005 mol of carbon dioxide will occupy a volume of 22.4 x 0.005 dm3 at stp
0.005 mol of carbon dioxide will occupy a volume of 0.112 dm3 at stp
stp = standard temperature and pressure (273K and 105 Pa)
ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure
THE MOLE
THE END
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