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# C03 relative masses of atoms and molecules

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### C03 relative masses of atoms and molecules

1. 1. Chapter 3 Relative Masses of Atoms and Molecules LEARNING OUTCOMES Define relative atomic mass, Ar Define relative molecular mass, Mr
2. 2. Chapter 3 Relative Masses of Atoms and Molecules Relative Atomic Mass  The masses of atoms and molecules are very small and it very hard for us to compare them or use them in calculations.  For e.g. the mass of a hydrogen atom is 0.000 000 000 000 000 000 000 0014 or (1.4 x 10-24)g and the mass of a carbon atom is 1.68 x 10-23 g.
3. 3. Chapter 3 Relative Masses of Atoms and Molecules Relative Atomic Mass  Instead of using the true masses of atoms, scientists compare the masses of atoms with the mass of a hydrogen atom, which is assigned a mass of one unit.  If we take the mass of a hydrogen atom to be 1, then the mass of a carbon atom will be 12, since a carbon atom is 12 times as heavy as a hydrogen atom.
4. 4. Chapter 3 Relative Masses of Atoms and Molecules 12 H atoms Relative Atomic Mass  The mass of a carbon atom is 12 times as heavy as a hydrogen atom, so we say that the relative atomic mass of carbon is 12. C  Scientists prefer to use the carbon atom instead of the hydrogen atom as a standard unit, so if we take 1/12 of the mass of a carbon atom, we will still get 1 unit.  Hence we define: The relative atomic mass of an element is the average mass of an atom of the element compared to the mass of 1/12 of the mass of a carbon-12 atom.
5. 5. Chapter 3 Relative Masses of Atoms and Molecules Relative Atomic Mass Relative atomic mass (Ar) Mass of one atom of the element Mass of 1 of an atom of carbon-12 = 12  The symbol for Relative Atomic Mass is Ar  Relative atomic mass has no units.  The relative atomic mass of an element can be obtained from the mass number (nucleon number) of the element in the Periodic Table.  Examples : 2311Na, 5626Fe Hence, Ar of Na = 23, Ar of Fe = 56
6. 6. Chapter 3 Relative Masses of Atoms and Molecules Quick check 1 1. 2. 3. 4. 5. Define the relative atomic mass of an element. Using the Periodic Table, find the relative atomic mass of the following: Mg, Ca, Cl, O, S, Ne, Br. What is the mass of a calcium atom compared to the mass of a helium atom? How many hydrogen atoms have the same mass as one potassium atom? How many oxygen atoms have the same mass as one bromine atom? Solution
7. 7. Chapter 3 Relative Masses of Atoms and Molecules Solution to Quick check 1 1. 2. 3. 4. 5. The relative atomic mass of an element is the average mass of an atom of the element compared to the mass of 1/12 of a carbon-12 atom. Ar: Mg=24, Ca=40, Cl=35.5, O=16, S=32, Ne=20, Br=80. Mass of a Ca atom =10 x Mass of a He atom 39 hydrogen atoms 5 oxygen atoms Return
8. 8. Chapter 3 Relative Masses of Atoms and Molecules Relative Molecular Mass  The picture shows the mass of a molecule of water compared to the mass of hydrogen atoms.  It shows that one molecule of water is 18 times as heavy as one hydrogen atom.  Therefore, the relative molecular mass of water is 18. 8
9. 9. Chapter 3 Relative Masses of Atoms and Molecules Relative Molecular Mass  If we use the mass of a carbon-12 atom as the standard unit of comparison, then the relative molecular mass of a substance is defined as: The relative molecular mass of a substance is the average mass of one molecule of the substance compared to the mass of 1/12 of a carbon-12 atom. Relative molecular mass Mass of one molecule of a substance = 1 (Mr) Mass of 12 of a carbon-12 atom 9
10. 10. Chapter 3 Relative Masses of Atoms and Molecules Relative Molecular Mass  The symbol for Relative Molecular Mass is Mr  The relative molecular mass of a molecule can be found by adding up the relative atomic masses of all the atoms present in the molecule.  E.g. Mr of water, H2O = Mass of 2 H atoms + mass of 1 O atom = 2 x 1 + 16 = 18 10
11. 11. Chapter 3 Relative Masses of Atoms and Molecules Relative Formula Mass  Ionic compounds (e.g. sodium chloride) are not made up of single molecules; instead they are made up of a crystal lattice consisting of many oppositely charged ions.  Hence, instead of calculating the relative molecular mass of an ionic compound, we calculate the mass based on its formula (formula mass).  We can take the relative formula mass as equivalent to the relative molecular mass in our calculations. 11
12. 12. Chapter 3 Relative Masses of Atoms and Molecules Finding Relative Molecular Mass Worked examples 1. 2. 3. Find the relative molecular mass of carbon dioxide, CO 2. Mr of CO2 = 12 + 16 x 2 = 44 Find the relative molecular mass (formula mass) of copper(II) nitrate, Cu(NO3)2. Mr of Cu(NO3)2 = 64 + (14 + 16x3 )x2 = 64 + 62x2 = 188 Find the relative molecular mass (formula mass) of ammonium sulphate, (NH4)2SO4. Mr of (NH4)2SO4 = (14 + 1x4)x2 + 32 + 16x4 12
13. 13. Chapter 3 Relative Masses of Atoms and Molecules Quick check 2 Find the relative molecular mass (or formula mass) of each of the following: (a) Magnesium sulphate, MgSO4 (b) Calcium nitrate, Ca(NO3)2 (c) Ammonium carbonate, (NH4)2CO3 (d) Benzoic acid, C7H6O2 (e) Hydrated sodium carbonate, Na2CO3.10H2O 13 Solution
14. 14. Chapter 3 Relative Masses of Atoms and Molecules Solution to Quick check 2 a) Mr of MgSO4 = 24 + 32 + 16x4 = 120 b) Mr of Ca(NO3)2 = 40 + 2(14 + 16x3) = 164 c) Mr of (NH4)2CO3 = 2(14+4) + 12 + 16x3 = 96 d) Mr of C7H6O2 = 12x7 + 6 + 16x2 = 122 e) Mr of Na2CO3.10H2O = 23x2 + 12 + 16x3 + 10(2x1 + 16) = 286 14 Return
15. 15. Chapter 3 Relative Masses of Atoms and Molecules Percentage composition Worked example 1 What is the percentage composition of calcium carbonate, CaCO 3? Solution % of Ca = [Ca] x 100% = 40 x 100% [CaCO3] [40 + 12 + 16x3] = 40% % of C = [ C ] x 100% = 12 x 100% [CaCO3] 100 = 12% % of O = [ O3 ] x 100% = 16x3 x 100% [CaCO3] 100 = 48% Check 15 %Ca + %C + %O = 40 + 12 + 48 = 100%
16. 16. Chapter 3 Relative Masses of Atoms and Molecules Percentage composition Worked example 2 What is the percentage of sodium in sodium carbonate, Na2CO3? Solution % of Na = [Na2] x 100% = 23x2 x 100% [Na2CO3] [23x2 + 12 + 16x3] = 43.4% Worked example 3 Find the mass of oxygen in 90 g of water. Solution Mr of H2O = 1x2 + 16 = 18 Mass of oxygen = 16 x 90 g = 80 g 18 Worked example 4 What mass of magnesium oxide can be made from 6 g of magnesium? Solution Mr of MgO = 24 + 16 = 40 16 Mass of MgO = 6 x 40 g = 10 g 24
17. 17. Chapter 3 Relative Masses of Atoms and Molecules Percentage yield The percentage yield of a product is given by the formula: Percentage yield = Actual mass of product obtained x 100% Theoretical mass of product obtained Worked example 1 In an experiment to prepare magnesium sulphate, 2.4 g of magnesium was dissolved completely in dilute sulphuric acid. On crystallisation, 22.0 g of magnesium sulphate crystals, MgSO4.7H2O, were obtained. What is the percentage yield? 17
18. 18. Chapter 3 Relative Masses of Atoms and Molecules Percentage yield Solution Write the chemical equation: Mg + H2SO4  MgSO4 + H2 Number of mole of Mg reacted = 2.4 = 0.1 mol 24 From equation, 1 mol of Mg  1 mol of MgSO4 Therefore, 0.1 mol of Mg  0.1 mol of MgSO4.7H2O Mass of MgSO4.7H2O produced = 0.1 mol x 246 g/mol = 24.6 g Percentage yield = Actual mass of product obtained x 100% Theoretical mass of product obtained = 22.0 x 100 % 24.6 = 89.4 % 18
19. 19. Chapter 3 Relative Masses of Atoms and Molecules Percentage purity The percentage purity tells us how pure a prepared product is compared to the pure form of the substance.  For example, the percentage purity of a gold bar may be given as 99.99 %.  19
20. 20. Chapter 3 Relative Masses of Atoms and Molecules Percentage purity Worked example 2 On analysis, 5.00 g of a sample of marble (calcium carbonate) was found to contain only 4.26 g of pure calcium carbonate. What is the percentage purity of the marble? Solution Percentage purity = 4.26 x 100 % 5.00 = 85.2 % 20
21. 21. Chapter 3 Relative Masses of Atoms and Molecules Quick check 3 1. Find the percentage composition of each element in sulphuric acid, H2SO4. 2. Find the percentage of nitrogen in calcium nitrate, Ca(NO3)2. 3. Find the mass of calcium in 250 g of calcium carbonate, CaCO3. 4. What mass of iron can be obtained from 320 g of iron(III) oxide, Fe2O3? 5. 24 g of hydrogen combines with 192 g of oxygen to form water. What mass of hydrogen will combine with 24 g of oxygen? In the Haber process to manufacture ammonia, it was reported that in a certain factory, 2.8 tonnes of nitrogen gas produced 0.80 tonne of ammonia. What is the percentage yield of ammonia? The equation for the Haber process is: Solution N2(g) + 3H2(g)  2NH3(g) 21 6.
22. 22. Chapter 3 Relative Masses of Atoms and Molecules Solution to Quick check 3 1. 2. 3. 4. 5. 6. % composition of H2SO4 : H =2.04 %, S =32.7 %, O = 65.3 % % of nitrogen in Ca(NO3)2 = 17.1 % Mass of calcium = 40 x 250 = 100 g 100 Mass of iron = 112 x 320= 224 g 160 Mass of hydrogen = 24 x 24 = 3 g 192 Percentage yield = 0.8 x 100 % 3.4 = 23.5 % 22 Return
23. 23. Chapter 3 Relative Masses of Atoms and Molecules To learn more about Relative Atomic and Molecular Mass, click on the link below! http://www.ch.cam.ac.uk/magnus/MolWeight.html 23