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ATOMIC STRUCTURE

ATOMIC STRUCTURE. HISTORY OF THE ATOM. Democritus develops the idea of atoms. 460 BC. he pounded up materials in his pestle and mortar until he had reduced them to smaller and smaller particles which he called. ATOMA ( greek for indivisible ). HISTORY OF THE ATOM. John Dalton. 1808.

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ATOMIC STRUCTURE

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  1. ATOMIC STRUCTURE

  2. HISTORY OF THE ATOM Democritus develops the idea of atoms 460 BC he pounded up materials in his pestle and mortar until he had reduced them to smaller and smaller particles which he called ATOMA (greek for indivisible)

  3. HISTORY OF THE ATOM John Dalton 1808 suggested that all matter was made up of tiny spheres that were able to bounce around with perfect elasticity and called them ATOMS

  4. HISTORY OF THE ATOM Joseph John Thompson 1898 found that atoms could sometimes eject a far smaller negative particle which he called an ELECTRON

  5. HISTORY OF THE ATOM 1904 Thompson develops the idea that an atom was made up of electrons scattered unevenly within an elastic sphere surrounded by a soup of positive charge to balance the electron's charge like plums surrounded by pudding. PLUM PUDDING MODEL

  6. HISTORY OF THE ATOM Ernest Rutherford 1910 oversaw Geiger and Marsden carrying out his famous experiment. they fired Helium nuclei at a piece of gold foil which was only a few atoms thick. they found that although most of them passed through. About 1 in 10,000 hit

  7. HISTORY OF THE ATOM gold foil helium nuclei helium nuclei They found that while most of the helium nuclei passed through the foil, a small number were deflected and, to their surprise, some helium nuclei bounced straight back.

  8. HISTORY OF THE ATOM Rutherford’s new evidence allowed him to propose a more detailed model with a central nucleus. He suggested that the positive charge was all in a central nucleus. With this holding the electrons in place by electrical attraction However, this was not the end of the story.

  9. HISTORY OF THE ATOM Niels Bohr 1913 studied under Rutherford at the Victoria University in Manchester. Bohr refined Rutherford's idea by adding that the electrons were in orbits. Rather like planets orbiting the sun. With each orbit only able to contain a set number of electrons.

  10. Bohr’s Atom electrons in orbits nucleus

  11. HELIUM ATOM Shell proton N + - + N - neutron electron What do these particles consist of?

  12. ATOMIC STRUCTURE Particle Charge Mass proton + ve charge 1 neutron No charge 1 electron -ve charge nil

  13. ATOMIC STRUCTURE He 2 Atomic number the number of protons in an atom 4 Atomic mass the number of protons and neutrons in an atom number of electrons = number of protons

  14. ATOMIC STRUCTURE Electrons are arranged in Energy Levels or Shells around the nucleus of an atom. • first shell a maximum of 2 electrons • second shell a maximum of 8 electrons • third shell a maximum of 8 electrons

  15. ATOMIC STRUCTURE There are two ways to represent the atomic structure of an element or compound; 1. Electronic Configuration 2. Dot & Cross Diagrams

  16. ELECTRONIC CONFIGURATION With electronic configuration elements are represented numerically by the number of electrons in their shells and number of shells. For example; Nitrogen configuration = 2 , 5 7 2 in 1st shell 5 in 2nd shell N 2+5 = 7 14

  17. ELECTRONIC CONFIGURATION Write the electronic configuration for the following elements; 20 11 8 Na O Ca a) b) c) 16 23 40 2,8,8,2 2,8,1 2,6 17 14 5 Cl Si B d) e) f) 11 35 28 2,8,7 2,8,4 2,3

  18. DOT & CROSS DIAGRAMS With Dot & Cross diagrams elements and compounds are represented by Dots or Crosses to show electrons, and circles to show the shells. For example; X Nitrogen N 7 X X N X X 14 X X

  19. DOT & CROSS DIAGRAMS Draw the Dot & Cross diagrams for the following elements; X 8 17 X O Cl a) b) X 35 X 16 X X X X X Cl X X X X X X X O X X X X X X X X X X

  20. SUMMARY • The Atomic Number of an atom = number of • protons in the nucleus. • The Atomic Mass of an atom = number of • Protons + Neutrons in the nucleus. • The number of Protons = Number of Electrons. • Electrons orbit the nucleus in shells. • Each shell can only carry a set number of electrons.

  21. 11B 10B Isotopes • Atoms of the same element (same Z) but different mass number (A). • Boron-10 (10B) has 5 p and 5 n • Boron-11 (11B) has 5 p and 6 n

  22. Two isotopes of sodium.

  23. Learning Check – Counting Naturally occurring carbon consists of three isotopes, 12C, 13C, and 14C. State the number of protons, neutrons, and electrons in each of these carbon atoms. 12C 13C 14C 66 6 #p+ _______ _______ _______ #n _______ _______ _______ #e- _______ _______ _______

  24. Answers 12C 13C 14C 6 6 6 #p+666 #no678 #e- 6 66

  25. Relative atomic mass, Ar • Atoms are amazingly small. In order to get a gram of hydrogen, you would need to count out 602,204,500,000,000,000,000,000 atoms ( to the nearest 100,000,000,000,000,000). • It would be silly to measure the masses of atoms in conventional mass units like grams. Instead, their masses are compared with the mass of an atom of the carbon-12 isotope, taken as a standard. We call this the “carbon-12 scale”.

  26. On this scale one atom of the carbon-12 isotope weighs exactly 12 units. • An atom of the commonest isotope of magnesium weighs twice as much as this and is therefore said to have a relative isotopic mass of 24. • An atom of the commonest isotope of hydrogen weighs only one twelfth that of the carbon-12 isotope, and so has a relative atomic mass of 1.

  27. The basic unit on this scale is therefore 1/12 of the mass of a 12C atom. Everything else is measured relative to that. • The relative atomic mass (RAM) of an element ( as opposed to one of its isotopes) is given the symbol Ar and is defined by: The RAM is the weighted average of the masses of the isotopes relative to 1/12 of the mass of a carbon-12 atom.

  28. What is a “weighted average”? Finding the RAM of chlorine. • In any sample of chlorine, some atoms have a relative mass of 35; others a relative mass of 37. • A simple average of 35 and 37 is, of course, 36 – but this is not the relative atomic mass of chlorine. • The problem is that there aren’t equal numbers of 35Cl and 37Cl.

  29. A typical sample of chlorine has: 35Cl 75% 37Cl 25% • If you had 100 typical atoms of chlorine, 75 would be 35Cl and 25 would be 37Cl. • The total mass of the 100 atoms would be (75x35) + (25x37) = 3550 • The average mass of 1 atom would be 3550/100 = 35.5 • The weighted average is closer to 35 than 37 because there are more 35Cl atoms than 37Cl atoms. A weighted average allows for the unequal proportions. • We have just calculated the relative atomic mass (RAM) of chlorine as 35.5.

  30. Relative formula mass. • Relative formula mass tells you the relative mass (on the 12C scale) of the substance whose formula you have written. • Relative formula mass is sometimes called relative molecular mass (RMM). Avoid this term, because it can only properly be applied to substances which are actually molecules – in other words , to covalent structures. You should not use it for things like magnesium oxide or sodium chloride which are ionic. • Relative formula mass is given the symbol, Mr.

  31. Working out some relative formula masses • To find the RFM of magnesium carbonate, MgCO3. • Relative atomic masses: C=12; O=16; Mg=24 • All you have to do is to add up the relative atomic masses to give you the relative formula mass of the whole compound. • RFM = 24 +12 + (3X16) = 84

  32. Another example • To find the RFM of calcium hydroxide , Ca(OH)2. • Relative atomic masses: H= 1, O= 16 Ca= 40 • RFM = 40 + (16 +1) x 2 = 74

  33. Questions • Calculate the relative formula masses of the following compounds: • CO2 • (NH4)2SO4 • Na2CO3.10H2O

  34. Answers • A) RFM of CO2 = 12 + (2x16) =44 • B) RFM of (NH4)2SO4 = [14 + (4x1)] x 2 + 32 + (4x16) = 132 • C) RFM of Na2CO3.10H2O = (2 X 23) + 12 + (3X16) + 10 X [(2X1) + 16] = 286

  35. Using RFMs to find percentage compositions • To find the percentage of copper in copper(II) oxide, CuO • Relative atomic masses: O=16, Cu= 64 • RFM of CuO = 64 +16 = 80 • Of this, 64 is copper. • Percentage of copper + 64/80 x 100 = 80%

  36. Another example • To find the percentage of nitrogen in ammonium nitrate, NH4NO3 • Relative atomic masses: H=1 N=14 O=16 • RFM of NH4NO3 = 14 + (4X1) + 14 + (3X16) = 80 • Of this, (2 x14) is nitrogen • Percentage of nitrogen = (2 x14)/80 x100 = 35%

  37. Questions • Find the percentage of the named substances in each of the following examples: • Carbon in propane, C3H8 • Water in magnesium sulphate crystals, MgSO4.7H2O.

  38. Answers • Carbon in propane • RFM of C3H8 = (3x12) + (8x1) = 44 • Of this, (3x12) is carbon • Percentage of carbon = (3x12)/44 x100 = 81.8%

  39. Answers • RFM of MgSO4.7H2O = 24 + 32 + (4x16) + 7 x [(2x1)+ 16] = 246 • Of this, (7x18) is water • Percentage of water = (7x18)/246 x100 = 51.2%

  40. The mole

  41. The mole • In chemistry, the mole is a measure of amount of substance. You can use such expressions as; • A mole of copper(II) sulphate crystals, CuSO4.5H2O • A mole of oxygen gas , O2 • 0.1 mole of zinc oxide, ZnO • 3 moles of magnesium, Mg.

  42. A mole is a particular mass of that substance. You find the mass of 1 mole of a substance in the following way: Work out the relative formula mass, and attach the units “grams”.

  43. Working out the masses • 1 mole of iron(II) sulphate crystals, FeSO4.7H2O • Relative atomic masses H=1 O=16 S=32 Fe=56 • RFM of crystals = 56+32+ (4x16) +7 x [(2x1) + 16] = 278 • 1 mole of iron(II) sulphate crystals weighs 278g.

  44. You try • 1 mole of oxygen gas

  45. Answer • Relative atomic mass = o=16 • RFM of oxygen O2 = 2x16 • =32 • I mole of oxygen, O2 weighs 32g

  46. Simple calculations with moles • You need to be able to carry out interconversions like the following: • 1. What is the mass of 0.2 mol of calcium carbonate, CaCO3? • 2. How many moles is 54g of water, H2O? • There are 2 different approaches that you could use depending on your confidence with numbers. It doesn't matter in the least which you choose, as long as you get the right answer.

  47. If you are confident with numbers • 1. 1 mol of CaCO3 weighs 100g (Assume you have just worked that out) Therefore, 0.2 mol weighs 0.2 x 100g = 20g. • 2. 1 mol of H2O weighs 18g (again assume that you have just worked this out). 54g is 3 times 18g and so is 3 moles.

  48. If you aren’t confident with numbers • There is a simple formula that you can learn: Number of moles = mass (g)/mass of 1 mole (g)

  49. You can rearrange this to find whatever you need to find. E.g. • Mass(g) = number of moles x mass of 1 mole (g) • Mass of 1 mole (g) = mass (g)/number of moles

  50. Using the formula to do sums • What is the mass of 0.2 mol of calcium carbonate, CaCO3? • 1 mol of CaCO3 weighs 100G • Mass(g) = number of moles x mass of 1 mole (g) • = 0.2 x 100 g • = 20g

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