Like this presentation? Why not share!

- Purification of substances by Ck Sal 960 views
- Atomic Structure Isotopes by Carolyn Khoo 10770 views
- Worksheet naming ions by gdelagdeg 1909 views
- 3 na calculating_mr by gueste21240a3 2033 views
- Goodness Gateway Exclusive Perks Ch... by Human Nature-Phils 257 views
- Motion, Velocity, and Acceleration ... by Miller_MMS 3012 views

743

-1

-1

Published on

No Downloads

Total Views

743

On Slideshare

0

From Embeds

0

Number of Embeds

1

Shares

0

Downloads

0

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Chapter 3 Relative Masses of Atoms and Molecules LEARNING OUTCOMES Define relative atomic mass, Ar Define relative molecular mass, Mr
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Il semblerait que vous avez déjà ajouté cette diapositive à .

Créer un clipboard

Be the first to comment