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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 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 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 concentration of solution in the graduated flask 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 concentration of solution in the graduated flask 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 concentration of solution in the graduated flask 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 concentration of solution in the graduated flask 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 concentration of solution in the graduated flask 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 concentration of solution in the graduated flask 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 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 clean beaker and dissolved in de-ionised water The solution was transferred 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 clean beaker and dissolved in de-ionised water The solution was transferred 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 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 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 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 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 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 © 2010 JONATHAN HOPTON & KNOCKHARDY PUBLISHING