THE MOLE A guide for A level students

THE MOLE A guide for A level students 2008 SPECIFICATIONS KNOCKHARDY PUBLISHING

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

DON’T BE LEFT IN THE DARK! THE MOLE • Before you start it would be helpful to… • know how to balance simple equations • know how to re-arrange mathematical formulae

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 BIG !!! It is a lot easier to write it as...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 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 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

MOLES = MASS MOLAR MASS THE MOLE 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 kg molar mass g mol-1or kg mol-1 MASS MOLES x MOLAR MASS COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED

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 = 4g = 0.125 mol molar mass 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 = 4g = 0.125 mol • molar mass 32g mol -1 • What is the mass of 0.25 mol of Na2CO3 ?

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

MOLES CONC x VOLUME 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 UNITSconcentration mol dm-3 volume dm3 BUT IF... concentration mol dm-3 volume cm3 MOLES = CONCENTRATION x VOLUME MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)

THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.100 mol dm-3 volume pipetted out into the conical flask= 25.00 cm3 25cm3 250cm3 250cm3

THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.100 mol dm-3 volume pipetted out into the conical flask= 25.00 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of0.100 mol dm-3

THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.100 mol dm-3 volume pipetted out into the conical flask= 25.00 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of0.100 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 concentration of solution in the graduated flask = 0.100 mol dm-3 volume pipetted out into the conical flask= 25.00 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of0.100 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 concentration of solution in the graduated flask = 0.100 mol dm-3 volume pipetted out into the conical flask= 25.00 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of0.100 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) = 0.100 moles in 1cm3 = 0.100/1000 moles in 25cm3 = 25 x 0.100/1000 = 2.5 x 10-3 mol

MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME 1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH

MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME 1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = conc x volume in cm3 1000 = 2 mol dm-3 x 25cm3= 0.05 moles 1000

MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME 1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = conc x volume in cm3 1000 = 2 mol dm-3 x 25cm3 = 0.05 moles 1000 2 What volume of 0.1M H2SO4 contains 0.002 moles ?

MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME 1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = conc x volume in cm3 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

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 250cm3 of de-ionised water

1g WATER WATER 1g 250cm3 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 250cm3 of de-ionised water RIGHT Dissolve 1g of solute in water and then add enough water to make 250cm3 of solution

STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up 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 cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up 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 cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up 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 = 4.240g molar mass of Na2CO3 = 106g mol -1 no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol

STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up 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 = 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 solution Therefore, as it is in 250cm3, the value is scaled up by a factor of 4

STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up 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 = 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 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 molANS.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? = 0.100 mol dm-3 How many moles will be in 1 dm3 ? = 0.100 mol How many moles will be in 250cm3 ? = 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? = 0.100 mol dm-3 How many moles will be in 1 dm3 ? = 0.100 mol How 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? = 106 What is the molar mass? = 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? = 0.100 mol dm-3 How many moles will be in 1 dm3 ? = 0.100 mol How 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? = 106 What 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)

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? = 0.100 mol dm-3 How many moles will be in 1 dm3 ? = 0.100 mol How 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? = 106 What 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 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 equationNaOH + 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 equationNaOH + HCl ——> NaCl + H2O 2. Get a molar relationship between the reactantsmoles 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 equationNaOH + HCl ——> NaCl + H2O 2. Get a molar relationship between the reactantsmoles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl 3. Calculate the number of moles of each substance HCl0.100 x 25/1000 (i) M is the concentration in mol dm-3NaOHM x 20/1000 (ii)

THE MOLE A guide for A level students