Mole Cambridge chemistry IGCSEc04.pptxc06.pptx

Mole Cambridge chemistry IGCSEc04.pptxc06.pptx

1. 1 THE MOLE CHAPTER 6 The following content has not been through the Cambridge Assessment International Education endorsement process.

2. 2 What are some units we use to represent different quantities of items? Why do we need to use units to represent different quantities of items? Do you think we can count the number of atoms in a substance like how we count the number of rice grains in a sack of rice? Why? Questions

3. 3 In this section, you will learn the following: Describe relative atomic mass ( A r ). Define relative molecular mass ( M r ). 6.1 Relative Atomic Mass, Relative Molecular Mass and Relative Formula Mass

4. 4 What is relative atomic mass? The relative atomic mass ( A r ) is the average mass of the isotopes of an element compared to of the mass of an atom of 12 C. Relative atomic mass is a ratio and therefore has no unit .   6.1 Relative Atomic Mass, Relative Molecular Mass and Relative Formula Mass Comparing the masses of different atoms relative to one another, using A r values

5. 5 What is relative molecular mass? The relative molecular mass ( M r ) of a molecular substance is the sum of the relative atomic masses of its constituent elements. Relative molecular mass is a ratio and therefore has no unit . Calculating the relative molecular mass of a molecular substance The relative molecular mass of a molecular substance is calculated by adding together the relative atomic masses of all the atoms in its chemical formula.

6. 6 Ionic compounds like sodium chloride do not exist as molecules. The relative molecular mass of an ionic compound is more accurately known as the relative formula mass . Like relative molecular mass, relative formula mass is given the symbol M r and has no unit. What is relative formula mass? Calculating the relative formula mass of an ionic compound  The relative formula mass is calculated in exactly the same way as relative molecular mass.

7. 7 In this section, you will learn the following: State that the mole, mol, is the unit of amount of substance and that one mole contains 6.02 × 10 23 particles. Use the relationship, amount of substance (mol) = in calculations.   6.2 The Mole and Molar Mass

8. 8 The mole is the unit of amount of substance. The symbol for the mole is mol . Number of particles in a mole: One mole of any substance contains 6.02 × 10 23 particles. The particles could be atoms, molecules, ions and even subatomic particles such as electrons and protons. The value 6.02 × 10 23 is called the Avogadro constant or Avogadro number. Equal numbers of moles of all substances contain the same number of particles. Calculations involving the mole and the number of particles: What is the mole? 6.2 The Mole and Molar Mass  

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10. 10 The molar mass of an element is the mass of one mole of atoms of the element. The molar mass of an element is equal to its relative atomic mass ( A r ) in grams. What is molar mass? The molar mass of a molecular substance is equal to its relative molecular mass ( M r ) in grams . Molar mass of a molecular substance Molar mass of an element Molar masses of some elements or molecules  Element / Substance Formula A r M r Molar mass (g/mol) Number of atoms in one mole of element / substance Neon Ne 20 20 6.02 x 10 23 Carbon C 12 12 Sodium Na 23 23 Oxygen O 2 2 x 16 = 32 32 Ammonia NH 3 (1 x 14) + (3 x 1) = 17 17 Water H 2 O (2 x 1) + (1 x 16) = 18 18

11. 11 The molar mass of an ionic compound is equal to its relative formula mass ( M r ) in grams. Molar mass of an ionic compound  Molar masses of some ionic compounds  One mole of different substances have different masses

12. 12 One mole of calcium bromide contains 18.06 × 10 23 ions The number of ions in one mole of an ionic compound is not equal to Avogadro constant. It depends on the number of ions that make up one formula unit. The number of moles of a substance can be calculated using the following formula:    Calculations involving the mole and molar mass

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14. 14 6.3 Molar Volume of Gases In this section, you will learn the following: Use the molar gas volume in calculations involving gases.

15. 15 Avogadro’s Law Volumes of oxygen gas and carbon dioxide gas at r.t.p. One mole of any gas occupies 24 dm 3 (24 000 cm 3 ) at room temperature and pressure ( r.t.p .). This volume is called the molar volume of a gas. Avogadro’s Law states that equal volumes of gases, under the same conditions of temperature and pressure, contain the same number of particles. 6.3 Molar Volume of Gases

16. 16 How do we calculate the number of moles of a gas? The number of moles of a gas can be determined in two ways. This formula can be rearranged to give: Volume of gas in cm 3 = number of moles × 24 000 cm 3 Volume of gas in d m 3 = number of moles × 24 d m 3  

17. 17 Do balloons of the same volume contain the same number of particles? According to Avogadro’s Law, the balloons contain the same number of gaseous particles since they have the same volume.  Do balloons of the same mass contain the same number of particles? The equal masses of different gases do not contain the same number of particles. Calculating the number of moles of different gases with equal mass

18. 18 6.4 Chemical Calculations In this section, you will learn the following: Calculate stoichiometric reacting masses, volumes of gases at r.t.p. and limiting reactants.

19. 19 Stoichiometry is the relationship between the number of moles of reactants and the number of moles of products involved in that chemical reaction. Consider the following chemical equation. 6.4 Chemical Calculations

20. 20 Calculating masses of reactants and products By using a balanced equation, we can calculate the mass of any reactant used up or the mass of any product formed in a reaction. 

21. 21 Calculating volumes of gaseous reactants and products Nitrogen dioxide is an air pollutant that is released from chemical plants. Consider the following reaction:

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23. 23 What are limiting reactants? A limiting reactant is the reactant that is completely used up in a reaction. It limits or determines the amount of products formed in the reaction. The reactants that are not used up in a reaction are said to be in excess.

24. 24 IGCSE Chemistry Student Book Results of experiment A In experiment A, the exact amounts of reactants required are used.  In experiments B and C, an excess of one reactant is used.  Effect of limiting reactants Consider the reaction below:

25. 25 Results of experiment B

26. 26 Results of experiment C The amount of a product formed in a reaction is always determined by the amount of the limiting reactant.

27. 6.5 The Concentration of a Solution 27 In this section, you will learn the following: State that concentration can be measured in g/dm 3 or mol/dm 3 . Calculate volumes and concentrations of solutions. Use experimental data from a titration to calculate the moles of solute, or the concentration or volume of a solution.

28. 28 The concentration of a solution is given by the amount of a solute dissolved in a unit volume of the solution. We can express concentration as either: Concentration in g/dm 3 (grams of solute per dm 3 ) Concentration in mol/dm 3 ( molar concentration ) 6.5 The Concentration of a Solution

29. Concentration in g/dm 3 The concentration of a solution can be determined using the following formula: By rearranging the formula, Mass of solute in g = concentration in g/dm 3 × volume of solution in dm 3   Concentration in mol/dm 3 (molar concentration) The concentration of a solution in mol/dm 3 can be calculated by using either of these formulae:   29

30. 30 Decrease the concentration of a solution by adding more solvent (see the figures below). Increase the concentration of a solution by increasing the amount of solute particles in the solution. This measuring cylinder also contains one mole of NaCl This measuring cylinder contains one mole of NaCl How can we change the concentration of a solution?

31. 31 What is volumetric analysis? Chemists can check the concentration of substances by carrying out volumetric analysis . Titration in volumetric analysis To do volumetric analysis, we use a method called titration. A titration involves a solution of unknown concentration (unknown); a solution of known concentration (titrant).

32. Let’s Investigate 6A In a titration, we find out the volume of titrant required to react with a fixed volume of the unknown. From these results, we can calculate the concentration of the unknown solution. Indicators such as methyl orange are used to determine whether a titration has reached the end-point 32

33. 33 In this section, you will learn the following: Calculate empirical formulae and molecular formulae. 6.6 Empirical and Molecular Formulae

34. 34 We can conduct experiments to find out the empirical formula of a compound. The steps are as follows: Find out the mass of the reactants taking part in the reaction. Work out the relative numbers of moles (mole ratio) of the reactants used. Find the formula of the compound using the mole ratio obtained. 6.6 Empirical and Molecular Formulae How can we work out the empirical formula of a compound? 

35. Let’s Investigate 6B Sample results Mass of crucible + lid = 26.52 g Mass of crucible + lid + magnesium = 27.72 g Mass of crucible + lid + magnesium oxide = 28.52 g Experimental set-up to find the empirical formula of magnesium oxide Finding the empirical formula of magnesium oxide  The empirical formula of magnesium oxide is thus MgO. The table shows the calculations for finding the empirical formula of magnesium oxide. Calculations Mass of magnesium = 27.72 − 26.52 = 1.20 g Mass of magnesium oxide produced = 28.52 − 26.52 = 2.00 g Mass of oxygen reacted = 2.00 − 1.20 = 0.80 g 35

36. 36 6.6 Empirical and Molecular Formulae How can we work out the molecular formula of a compound? We can find the molecular formula of a substance if we know two things about it: The empirical formula The relative molecular mass They are related as follows: If empirical formula = A x B y , molecular formula = ( A x B y ) n , where n = 1, 2, 3, etc. To find n,  

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38. 38 6.7 Percentage Yield, Percentage Composition and Percentage Purity In this section, you will learn the following: Calculate percentage yield, percentage composition by mass and percentage purity.

39. 39 6.7 Percentage Yield, Percentage Composition and Percentage Purity How to find percentage yield? The theoretical yield of a reaction is the calculated amount of product that would be obtained if the reaction were to be complete. The amount of pure product that is actually produced in the experiment is called the actual yield. The percentage yield shows the relationship between actual yield and theoretical yield.  

40. 40 How to find percentage composition of a compound?   How to find percentage purity?  

41. What have you learnt? ? ? ? ? ?

42. What have you learnt? ? ? ?

43. What have you learnt?

44. What have you learnt?

45. 45 Acknowledgements