THE MOLE A guide for A level students KNOCKHARDY PUBLISHING 2015 SPECIFICATIONS.

THE MOLETHE MOLEA guide for A level studentsA guide for A level students

KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING20152015

SPECIFICATIONSSPECIFICATIONS

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

KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING

THE MOLETHE 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 MOLETHE MOLE

Before you start it would be helpful to…

• know how to balance simple equations

• know how to re-arrange mathematical formulae

THE MOLETHE MOLE

DON’T BE LEFT IN THE

DARK!

WHAT IS A MOLE ?

it is the standard unit of amount of a substance - it is just a number, a very big numberit is a way of saying a number in words, just like...

DOZEN for 12 SCORE for 20 GROSS for 144

THE MOLETHE MOLE

HOW BIG IS IT ?

602200000000000000000000 (Approximately)... THAT’S BIG !!!

It is a lot easier to write it as... 6.022 x 1023

THE MOLETHE MOLE

WHAT IS A MOLE ?



HOW BIG IS IT ?

602200000000000000000000 (Approximately)... THAT’S BIG !!!

It is a lot easier to write it as... 6.022 x 1023

It is also known as... AVOGADRO’S NUMBER

It doesn’t matter what the number is as long as everybody sticks to the same value !

THE MOLETHE MOLE

WHAT IS A 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 MOLETHE MOLE

WHAT IS IT? The standard unit of amount of a substance - just as the standard unit of length is a METREIt is just a number, a very big numberIt is also a way of saying a number in wordslike DOZEN for 12

GROSS for 144

THE MOLE – AN OVERVIEWTHE MOLE – AN OVERVIEW


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 isas long as everybody sticks to the same value !



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 isas long as everybody sticks to the same value !

WHY USE IT ? Atoms and molecules don’t weigh much so it iseasier 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 massthe mass of a certain number of particles


CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE

moles = mass / molar mass mass = moles x molar mass

molar mass = mass / moles

UNITS

mass g or kgmolar mass g mol-1 or kg mol-1

THE MOLETHE MOLE

MOLES = MASS MOLAR MASS MOLES = MASS MOLAR MASS

MASS

MOLES x MOLAR MASS

COVER UP THE VALUE YOU WANT AND THE METHOD

OF CALCULATION IS REVEALED

1. Calculate the number of moles of oxygen molecules in 4g

MOLES OF A SINGLE SUBSTANCEMOLES OF A SINGLE SUBSTANCE

MASS

MOLES x MOLAR MASS

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 = 4g = 0.125 mol molar mass 32g mol -1


MASS

MOLES x MOLAR MASS








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


MASS

MOLES x MOLAR MASS







mass = moles x molar mass = 0.25 x 106 = 26.5g


CaCO3 + 2HCl ———> CaCl2 + CO2 + H2O

1. What is the relative formula mass of CaCO3? 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 CALCULATIONSREACTING MASS CALCULATIONS

EQUATIONS give you the ratio in which chemicals react and are formedneed 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 = 0.002 moles 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? mass = moles x molar mass

= 0.001 x 100 = 0.1 g

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 CALCULATIONSREACTING MASS CALCULATIONS

CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION

UNITS concentration mol dm-3

volume dm3

BUT IF... concentration mol dm-3

volume cm3

THE MOLETHE MOLE

MOLES

CONC x VOLUME

COVER UP THE VALUE YOU WANT AND THE

METHOD OF CALCULATION IS

REVEALED

MOLES = CONCENTRATION x VOLUME MOLES = CONCENTRATION x VOLUME

MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)

MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3) 1000

MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3) 1000

250cm3

25cm3

250cm3


concentration of solution in the graduated flask = 0.100 mol dm-3

volume pipetted out into the conical flask = 25.00 cm3

THE MOLETHE MOLE

The original solution has a concentration of 0.100 mol dm-3

250cm3

25cm3

250cm3




THE MOLETHE MOLE


This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solution

250cm3

25cm3

250cm3




THE MOLETHE MOLE


This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solutionTake out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles

250cm3

25cm3

250cm3




THE MOLETHE MOLE



moles in 1dm3 (1000cm3) = 0.100moles in 1cm3 = 0.100/1000moles in 25cm3 = 25 x 0.100/1000 = 2.5 x 10-3 mol

250cm3

25cm3

250cm3




THE MOLETHE MOLE



moles in 1dm3 (1000cm3) = 0.100moles in 1cm3 = 0.100/1000moles in 25cm3 = 25 x 0.100/1000 = 2.5 x 10-3 mol

250cm3

25cm3

250cm3




THE MOLETHE MOLE

1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH

MOLE OF SOLUTE IN A SOLUTIONMOLE OF SOLUTE IN A SOLUTION

MOLES = CONCENTRATION x VOLUMEMOLES = CONCENTRATION x VOLUME


moles = conc x volume in cm3

1000

= 2 mol dm-3 x 25cm3 = 0.05 moles 1000





1000

= 2 mol dm-3 x 25cm3 = 0.05 moles 1000

2 What volume of 0.1M H2SO4 contains 0.002 moles ?





1000

= 2 mol dm-3 x 25cm3 = 0.05 moles 1000

2 What volume of 0.1M H2SO4 contains 0.002 moles ?

volume = 1000 x moles (re-arrangement of above)

(in cm3) conc

= 1000 x 0.002 = 20 cm3

0.1 mol dm-3



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.

WRONGDissolve 1g of solute in 250cm3

of de-ionised water

SOLUTIONSSOLUTIONS

‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’

250cm3

1g

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.

WRONGDissolve 1g of solute in 250cm3

of de-ionised water

RIGHTDissolve 1g of solute in waterand then add enough water tomake 250cm3 of solution

SOLUTIONSSOLUTIONS

‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’

250cm3

1g

250cm3

1g

WATER WATER

STANDARD SOLUTIONSTANDARD SOLUTION

‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’

4.240g of Na2CO3 was placed in a clean The solution was transferred

beaker and dissolved in de-ionised water quantitatively to a 250 cm3

graduated flask and made upto 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 = 4.240g

molar mass of Na2CO3 = 106g mol -1

no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol










Concentration is normally expressed as moles per dm3 of solutionTherefore, as it is in 250cm3, the value is scaled up by a factor of 4










Concentration is normally expressed as moles per dm3 of solutionTherefore, 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


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? = 0.100 mol dm-3

How many moles will be in 1 dm3 ? = 0.100 molHow many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol






What is the formula of anhydrous sodium carbonate? = Na2CO3

What is the relative formula mass? = 106What is the molar mass? = 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 waterand then make the solution up to 250cm3 in a graduated flask.










ANS. The chemist will have to weigh out 2.650g, dissolve it in waterand then make the solution up to 250cm3 in a graduated flask.

VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS

Calculate the concentration of a solution of sodium hydroxide if 20cm3 isneutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.



1. Write out a BALANCED equation NaOH + HCl ——> NaCl + H2O




2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl





3. Calculate the number of moles of each substance HCl 0.100 x 25/1000 (i)

M is the concentration in mol dm-3 NaOH M x 20/1000 (ii)







4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HCl M x 20/1000 = 0.100 x 25/1000








5. Cancel the 1000’s M x 20 = 0.100 x 25

re-arrange the numbers to obtain M M = 0.100 x 25 20








5. Cancel the 1000’s M x 20 = 0.100 x 25

re-arrange the numbers to obtain M M = 0.100 x 25 20

6. Calculate the concentration of the NaOH = 0.125 mol dm-3

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 moles of NaOH = 2 x moles of H2SO4

or 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

REMEMBER... IT IS NOT A MATHEMATICAL EQUATION

2 moles of NaOH DO NOT EQUAL 1 mole of H2SO4

More examples follow




i.e moles of NaOH = 2 x moles of H2SO4

or 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 moles of HCl = 2 x moles of Na2CO3

or moles of Na2CO3 = moles of HCl

2



MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+

you need 5 moles of Fe2+ to react with every 1 mole of MnO4¯

i.e moles of Fe2+ = 5 x moles of MnO4¯

or 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


MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+

you need 5 moles of Fe2+ to react with every 1 mole of MnO4¯

i.e moles of Fe2+ = 5 x moles of MnO4¯

or moles of MnO4¯ = moles of Fe2+

5

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.



therefore moles of NaOH = 2 x moles of H2SO4

moles of H2SO4 = 0.120 x 20/1000 (i)

moles of NaOH = 0.100 x V/1000 (ii)

where V is the volume of alkali in cm3


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.



therefore moles of NaOH = 2 x moles of H2SO4

moles of H2SO4 = 0.120 x 20/1000 (i)

moles of NaOH = 0.100 x V/1000 (ii)

where V is the volume of alkali in cm3

substitute numbers moles of NaOH = 2 x moles of H2SO4

0.100 x V/1000 = 2 x 0.120 x 20/1000

cancel the 1000’s 0.100 x V = 2 x 0.120 x 20

re-arrange Volume of NaOH (V) = 2 x 0.120 x 20 = 48.00 cm3

0.100


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 stp0.25 mol of carbon dioxide will occupy a volume of 22.4 x 0.25 dm3 at stp0.25 mol of carbon dioxide will occupy a volume of 5.6 dm3 at stp

MOLAR VOLUMEMOLAR VOLUME

stp = standard temperature and pressure (273K and 105 Pa)ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure

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 stp0.25 mol of carbon dioxide will occupy a volume of 22.4 x 0.25 dm3 at stp0.25 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 = 12 + (4x1) = 16

Molar Mass of CH4 = 16g mol-1

Moles = mass/molar mass 0.08g / 16g mol-1 = 0.005 mols

1 mol of methane will occupy a volume of 22.4 dm3 at stp0.005 mol of carbon dioxide will occupy a volume of 22.4 x 0.005 dm3 at stp0.005 mol of carbon dioxide will occupy a volume of 0.112 dm3 at stp

MOLAR VOLUMEMOLAR VOLUME

stp = standard temperature and pressure (273K and 105 Pa)ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure

THE MOLETHE MOLETHE ENDTHE END

© 2015 JONATHAN HOPTON & KNOCKHARDY PUBLISHING© 2015 JONATHAN HOPTON & KNOCKHARDY PUBLISHING

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THE MOLE A guide for A level students KNOCKHARDY PUBLISHING 2015 SPECIFICATIONS.

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