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Chemistry SK016 CHAPTER 1 MATTER

Chemistry SK016 CHAPTER 1 MATTER. LECTURE 1. 1.1 Atoms & Molecules. LEARNING OUTCOMES: a) Describe proton, electron and neutron in terms of the relative mass and relative charge b) Define proton number, Z, nucleon number, A and isotope. c) Write isotope notation. Model of an Atom.

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Chemistry SK016 CHAPTER 1 MATTER

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  1. Chemistry SK016CHAPTER 1MATTER

  2. LECTURE 1 1.1 Atoms & Molecules LEARNING OUTCOMES: a) Describe proton, electron and neutron in terms of the relative mass and relative charge b) Define proton number, Z, nucleon number, A and isotope. c) Write isotope notation.

  3. Model of an Atom Nucleus atom: i. proton, p (+ve) ii. neutron, n (neutral) Electron, e (-ve) Surrounding nucleus atom

  4. Points to Ponder • All the elements, except hydrogen atom, possess neutrons in the atomic nuclei. What does a neutron do in an atomic nucleus? • Distinguish the nucleus, neutron and nucleon.

  5. The properties of the particles

  6. Symbol of element Nucleon number Charge on an ion c A X Z Proton number Example: Isotope Notation(notation for Nuclides) A = Z + neutron e = p  atom e > p  anion (-ve) e < p  cation (+ve)

  7. Proton number ,Z: The number of proton in the nucleus of an atom. • The particles that are found in the nucleus of an atom is termed as nucleon. It consists of proton and neutron. • Nucleon number, A: The total sum of protons and neutrons found in the nucleus of the atom. • Note: * Proton number is also known as Atomic Number * Nucleon Number is also known as Mass Number

  8. c A X Z Exercise 1) ____ is the proton number. It is the total number of _________ in an atomic nucleus. 2) A is the ______________ of the nuclide X. 3) Nucleon number is defined as the total number of _________ and _________ in an atomic nucleus.

  9. 2- 84 59 16 3+ Kr Co O 36 27 8 Exercise: • The nucleon number of Kr = _______ • The proton number of Co3+ = _______ • The number of neutrons in Kr= _____ • _____ contains 10 electrons. • Co3+ consists of ___ protons, ___ electrons and ______ neutrons.

  10. 18 2- T 8 Exercise: complete the following table 24 52 21 13 27 10 2- Write the notation for nuclide T.

  11. Points to Remember: • Writing the charge on an ion incorrect correct Na+ Na+1 Na1+ Ca+2 Ca2+ Cl Cl1 Cl1 S2 S2 • An atomlostelectrons – cation An atomgainedelectrons – anion An atomCANNOTdonate protons!!!

  12. Isotopes • Isotopes are two or more atoms of the same element that have the same number of protons in their nucleus but differ in number ofneutrons.

  13. H H H 1 1 1 tritium protium deuterium _ + H H Hydrogen ion @ proton Hydride ion Isotopes Hydrogen atom: 2 3 1 Proton number: 1 1 1 Number of neutrons: 0 1 2

  14. 13 14 15 T R Z 6 6 7 Exercise The nuclides are , , and : Write the nuclides that are isotopes. State the atoms that have the same number of neutrons.

  15. Compare and contrast between the isotopes of an element by referring to Ne-20, Ne-21 and Ne-22 Homework

  16. LECTURE 2 1.1 Atoms & Molecules

  17. LEARNING OUTCOMES: Define relative atomic mass, Ar and relative molecular mass, Mr based on the C-12 scale. Calculate the average atomic mass of an element given the relative abundances of isotopes or a mass spectrum.

  18. Definition Relative atomic mass, Ar of an element: mass of an atom in comparison to 1/12 of mass of carbon-12 atom Ar = average mass of one atom of the element 1 mass of one atom C-12 x 12

  19. Definition Relative molecular mass, Mr of a molecular substance: mass of a molecule in comparison to 1/12 of mass of a carbon-12 atom. Mr = average mass of one molecule of the substance 1 mass of one atom C-12 x 12

  20. Relative Abundance 18 7 m/e 85 87 Mass spectrum of Rubidium Interpretation • The mass spectrum of rubidium shows that naturally occurring rubidium consists of two isotopes (two peaks): 85Rb and 87Rb. • The height of each line is proportional to the abundance of each isotope. In this example, Rb-85 is more abundance than Rb-87.

  21. Average atomic mass Average atomic mass = ∑ (isotopic mass x abundance) ∑ abundance =  (mi x Qi)  Qi

  22. Relative Abundance 18 7 m/e 85 87 = 85.56 a.m.u Example: A sample of Rubidium is analyzed in a spectrometer. i) Calculate the relative atomic mass of Rubidium.ii)What is the percentage abundance of each of the isotopes? Calculation: i) Average atomic mass of Rb =  (mi x Qi)  Qi 85x18 + 87x7 = (18 + 7)

  23. 18 x 100 = 72 % 25 Relative atomic mass of Rb = 85.56 a.m.u 1/12 x 12.01 a.m.u = 85.56 no unit! ii) Percentage abundance of each of the isotopes: % Rb-85 = 100% - 72% = 28 % % Rb-85 =

  24. = 2.333 63 Cu 65 Cu Exercise: • The ratio of relative abundance of naturally occurring of copper isotopes is as follow: Based on the carbon-12 scale, the relative atomic mass of 63Cu=62.9396 and 65Cu=64.9278. Calculate the Ar of copper.

  25. 2.333 = 2.333 = 1 The abundance of Cu-63 = 2.333 and The abundance of Cu-65 = 1 63 63 Cu Cu 65 65 Cu Cu Solution: Ar Cu =

  26. Exercise: • The atomic masses of 6Li and 7Li are 6.0151u and 7.0160u respectively. What is the relative abundance of each isotope if the relative atomic mass of lithium is 6.941? Solution: Let the abundance of 6Li = w and the abundance of 7Li = 1-w 6.941 = w = (the abundance of Li-6) The abundance of Li-7 =

  27. LECTURE 3 1.2 Mole Concept LEARNING OUTCOMES: a) Define mole in terms of mass of carbon-12 and Avogadro’s constant, NA. b) Interconvert between moles, mass, number of particles, molar volume of gas at s.t.p. and room temperature. c) Define the terms empirical and molecular formulae d) Determine empirical and molecular formulae from mass composition or combustion

  28. Mole Concept Chemists have adopted the mole concept as a convenient way to deal with the enormous numbers of atoms, molecules or ions in the samples they work with. Mole (unit mol): The amount of substance that contains as many elementary particles as there are atoms in exactly 12.000 g of carbon-12.

  29. Avogadro’s constant (NA) The number of atoms in a 12 gram sample of carbon-12 is called Avogadro’s constant(symbol NAor L) The term mole, just like a dozen or gross, refers to a particular number of things. one dozen = 12 one gross = 144 one mole = 6.02 x 1023

  30. Molar mass The molar mass of a substance is the mass of one mole of the substance For all substances, the molar mass in grams per mole is numericallyequal to the formula weight in atomic mass units. Molar mass = Ar or Mr (g mol1)

  31. Mole Concept • 1 mol = 6.02 x 1023particles (Avogadro’s constant = 6.02 x 1023 mol1 ) • 1 mol = Ar or Mr in unit gram • Ar of an element obtained from the Periodic Table • Mr of compound =  Ar of each constituent atom • 1 mol of any gas = 22.4 L at STP (0oC, 1 atm) • = 24 L at room temperature,RTP • (25oC, 1 atm)

  32. Case 1: mole mass a) Monatomic element 1 mol monatomic element = Ar g 1 mol copper, Cu = ____ g 6.36 0.1mol copper(II) ion, Cu2+ = 0.1 x ____ g = g 1 mol silver ion, Ag+ = ____ g 0.25mol silver, Ag = 0.25 x ____ g = g 27 Ar for Ag = 108 108 108 Cu = 63.5 63.5 63.5

  33. 2 x _____ g = Case 1: mole mass b) Polyatomic substance 1 mol Polyatomic compound = Mrg 2 x Ar (Cl) g 1 mol chlorine, Cl2 = = 71 g 1 mol carbonate ion, CO32 = [Ar (C) +3xAr (O)] g = [___+3x____] g = 60 g 60 = 0.4 x __ g 0.4mol CO32 = 24 g Ar for C = 12 12 Cl = 35.5 35.5 O = 16 16

  34. Points to remember • 1 mole 1 molecule • Units: the atomic mass = Aramu the molecular mass = Mramu the molar mass = Aror Mrg mol1 the mass of 1 mol of X = Ar or Mrg Ar, Mr, the weight of X --- dimensionless

  35. = Exercise: A sample of iron, Fe weighed 1.00 kg. What is the amount (mole) of Fe? Solution: Note: Mass, g Mole, n Molar mass, g mol1 Remember that massmust be expressed in grams here!

  36. Case 2: mole particles 1 mol substance = 6.02 x 1023 particles 1 mol sodium, Na = 6.02 x 1023atoms 1 mol fluorine, F2 = 6.02 x 1023molecules F2 = 2 x 6.02 x 1023atoms F (I molecule of fluorine is made up of 2 atoms) 1 mol ammonium ion, NH4+ = 6.02 x 1023 cations = 5 x 6.02 x 1023atoms (I particle of NH4+ consists of 5 atoms)

  37. x = Exercise: Calculate the number of atoms in 0.20 mol of magnesium. Solution: The number of Mg atoms atoms = 0.20 mol mol = Note: No. particles Mole, n Avogadro’s constant

  38. Case 3: mole volume gas 1 mol gas = 22.4 L at STP 1mol Ar = 22.4 L at STP 1mol N2 = 22.4 L at STP 1mol C2H2 = 22.4 L at STP = 13.4 L = 0.6 x 22.4 L 0.6mol C2H2

  39. x Exercise: How many moles are there in 6.5 L oxygen at STP ? Solution: mol Mole O2 = 6.5 L L = mol

  40. Mole X 2 = Mole Y 3 Case 4: mole X mole Y a) Utilise stoichiometric proportions 2X + 3Y M + 2Q a) 0.2 mole of X reacts with_______ mole of Y 0.3 b) 0.1 mole of Y needs _________ mole of X to react completely 0.067

  41. Exercise: An antacid tablet contained 450 mg Na2CO3. When swallowed, the Na2CO3 reacted with stomach acid (HCl), according to the reaction equation, Na2CO3 + 2HCl 2NaCl + CO2 + H2O How many grams of HCl were neutralized by the tablet?

  42. Solution: Strategy: Since the Na2CO3 is in units of mg, we need to convert to g by a metric conversion before changing to moles: Mole of Na2CO3 = = mol 1 mole Na2CO3 reacts with 2 moles of HCl If mol of Na2CO3 Ξ x 2/1 mol HCl = mol HCl

  43. Therefore, Mass of HCl = mol x 36.5 gmol-1 = g = g

  44. Case 4: mole X mole Y b) Ratio by atoms in a molecular formula 1 mole of C2H4 consists of ___ moles of C and ___ moles of H 4 2 6 3 moles of C2H4 give____ moles of C give____ moles of H 12 1 0.5 mole of C2H4 gives___ mole of C gives___ moles of H 2

  45. Exercise: Calculate the mass of (NH4)2CO3 that contains a) 0.300 mol NH4+b) 6.02 x 1023 H atoms Solution (a): 1 unit of mass of (NH4)2CO3 contains 2 units of NH4+ 2 mol NH4+ Ξ1 mol (NH4)2CO3 If 0.300 molNH4+ Ξ0.300 mol x ½mol (NH4)2CO3 = 0.15 mol (NH4)2CO3 1 mol of (NH4)2CO3 = 96 g If 0.15 mol of(NH4)2CO3 = 0.15 mol x 96 g mol-1 = 14.4 g

  46. Exercise: Calculate the mass of (NH4)2CO3 that contains a) 0.300 mol NH4+ b) 6.02 x 1023 H atoms Solution (b): 6.02 x 1023 H atoms = 1 mol of H atoms From (NH4)2CO3: 8 mol mol of H atoms = 1 mol of (NH4)2CO3 If 1 mol of H atom = 1 x 1/8 mol of (NH4)2CO3 = 0.125 mol of (NH4)2CO3 Then, 1 mol of (NH4)2CO3 = 96 g Therefore, 0.125 mol (NH4)2CO3 = 0.125 mol x 96g/1 mol = 12.0 g

  47. HOMEWORK • Sulphur is nonmetallic element. Its presence in coal gives rise to the acid rain phenomenon. How many atoms are in 16.3 g of sulphur. (3.06 x 1023 atoms) • How many hydrogen atoms are present in 25.6 g of urea [(NH2)2CO], which is used as a fertilizer, in animal feed, and in the manufacture of polymers? The molar mass of urea is 60.06 g. (1.03 x 1024 atoms)

  48. Empirical Formula: indicates which elements are present and the simplest whole-number ratio of their atoms in a molecule. Molecular Formula shows the exact number of atoms of each element in a molecule. n x empirical formula = molecular formula

  49. Compound EF MF n formaldehyde CH2O CH2O 1 acetic acid CH2O C2H4O2 2 glucose CH2O C6H12O6 6 All 3 compounds are 40.00 % C 6.714 % H 53.27 % O Compounds with different molecular formulaecan have the same empirical formula, and such substances will have the same percentage composition.

  50. Consider the following flow-diagram: Per cent composition Mass Composition Number of moles of each element Divide by smallest number of moles to find the molar ratios Multiply by appropriate number to get whole number subscripts

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