Chapter 1 : MATTER• 1.1 Atoms and Molecules • 1.2 Mole Concept • 1.3 Stoichiometry 1 1.1 Atoms and molecules 2 Learning outcomes At the end of the lesson, you should be ABLE to: (a) Describe proton, electron and neutron in terms of mass and charge (b) Define proton no, nucleon no and isotope (c) Write isotopic notation (d) Define relative atomic mass and relative molecular mass (e) Calculate average atomic mass of an element 3 Introduction Matter: Anything that occupies space and has mass. Example: Air, water, animals, trees, atoms Matter consists of particles: atoms, molecules or ions. 4 States of Matter 1.1 Atoms and Molecules Atoms • An atom is the smallest unit of a chemical element/compound. • In an atom, there are three subatomic particles: - Proton (p) Packed in a small nucleus - Neutron (n) - Electron (e) Move rapidly around the nucleus of an atom 7 Atomic Structure (n) (e) (p) Subatomic Particles Mass Charge Relative Particle (gram) (coulomb) charge Electron 9.1 x 10-28 -1.6 x 10-19 -1 (e) Proton 1.67 x 10-24 +1.6 x 10-19 +1 (p) Neutron 1.67 x 10-24 0 0 (n) DEFINITION • Proton number (Z) : A number of protons in the atomic nucleus • Nucleon number (A) : The number of protons and neutrons in the atomic nucleus. Isotope Notation (symbol) An atom can be represented by an isotope notation (atomic symbol) : Nucleon number Element symbol (can be an Proton atom or an number ion) 11 Isotope Notation (symbol) Nucleon number of mercury, A = 202 Total charge on the ion The number of neutrons =A–Z = 202 – 80 = 122 proton number of mercury, Z = 80 * No of e = 80 -2 = 78 Isotopes • two or more atoms of the same element that have the same number of protons in their nucleus but different number of neutrons. • Examples: 235 2 H (D) 1 200 80 Hg 92 U 3 1 H (T) 200 80 Hg 238 92 U Ion Cation Neutral Anion (+ ion) Atom atom Atom (- ion) loses e accepts e Na Na + +e Cl + e Cl- Z = 11 Z = 11 e = 11 e = 10 Z = 17 Z = 17 e = 17 e = 18 Exercise 1 • Give the number of protons, neutrons, electrons and charge in each of the following species: Number of : Symbol Charge Proton Neutron Electron 200 80 Hg 63 29 Cu 17 8 O2 59 3 27 Co Exercise 1 • Give the number of protons, neutrons, electrons and charge in each of the following species: Symbol Number of : (notation) Charge Proton Neutron Electron 200 80 Hg 63 29 Cu 17 8 O2 59 3 27 Co Exercise 2 • Write the appropriate notation for each of the following nuclide : Number of : Notation Species for nuclide Proton Neutron Electron 4 A 2 2 2 2 A 3 B 1 2 0 1 B 2 C 1 1 1 1 C 14 3 D 7 7 10 7 C Learning outcomes….. √ (a) Describe proton, electron and neutron in terms of mass and charge (b) Define proton no, nucleon no √ and isotope √ (c) Write isotopic notation (d) Define relative atomic mass and NEXT…. relative molecular mass (e) Calculate average atomic mass of an element 19 Revise previously…. Atom The atom is a BASIC UNIT of matter that consists of a dense, central nucleus surrounded by a cloud of negatively charged electrons. Molecule • A molecule consists of a small number of atoms joined together by bonds. Relative Mass i. Relative Atomic Mass (Ar) Average mass of one atom of an element compared to 1/12 mass of one 12C atom with the mass 12.000g. Average mass of one atom of element (amu) Ar 1 mass of one atom of 12C (amu) 12 Ar is dimensionless, so no unit.. Example 1 Determine the relative atomic mass of an element Y if the ratio of the atomic mass of Y to carbon-12 atom is 0.45. ANSWER : Ar = Average mass of one atom of Y 1 Mass of one atom of C - 12 12 = 12 x 0.45 1 = 5.4 ii) Relative Molecular Mass, Mr • A mass of one molecule of a compound compared to 1/12 mass of one atom of 12C with the mass 12.000g. Mass of one molecule of a substance (amu) Mr 1 Mass of one atom of 12C (amu) 12 The relative molecular mass of a compound is the summation of the relative atomic masses of all atoms in a molecular formula. EXAMPLE 1: NaOH Ar of Na = 23.0 Ar of O = 16.0 Ar of H = 1.0 Mr of NaOH= 23.0 + 16.0 + 1.0 = 40 25 Example 2 : Calculate the relative molecular mass (Mr) of C5H5N. Answer : Mr = 5(Ar of C) + 5(Ar of H) + Ar of N = 5(12.01) + 5(1.01) + 14.01 = 60.05 + 5.05 + 14.01 = 79.11 26 Learning outcomes….. √(a) Describe proton, electron and neutron in terms of mass and charge (b) Define proton no, nucleon no √ and isotope √ (c) Write isotopic notation √ (d) Define relative atomic mass and relative molecular mass (e) Calculate average atomic mass of NEXT…. an element 27 Device used to measure Ar : Mass Spectrometer Device used to measure Ar : Mass Spectrometer Mass Spectrum matter Device used to measure Ar : Mass Spectrometer Ionisation Chamber Acceleration Magnetic Vaporisation + Chamber Chamber Chamber - - Ion Beam Heated Vacuum Ion Detector Filament Pump Mass Spectrum AMPLIFIER Recorder Relative abundance (Q) Mass Spectrum of Magnesium • The mass spectrum of Mg shows that Mg consists of 63 3 isotopes : 24Mg, 25Mg and 26Mg. 9.1 • The height of each line is 8.1 proportional to the 24 25 26 abundance of each m/e isotope. (mass, amu) % • 24Mg is the most abundant Ratio m/e = of the three isotopes. Fraction mass/charge How to calculate the average atomic mass from mass spectrum? Average ∑ Qi mi amu atomic = mass ∑ Qi Q= the relative / fractional / ratio / percentage abundance of isotopes of the element in the mixture m= the isotopic mass of the element in unit amu or u. The unit of average atomic mass is amu or u Example 1 Copper, Cu consists of two isotopes which is 69% 63Cu and 31% 65Cu. The isotopic mass of 63Cu and 65Cu are 62.9 a.m.u and 64.9 a.m.u respectively. Calculate the average atomic mass of copper. 69 abundance Relative (%) 31 62.9 64.9 m/e (amu) SOLUTION: Average ∑ Qi mi atomic mass = ∑ Qi = (69 x 62.9 a.m.u) + (31 x 64.9 a.m.u) 69 + 31 = 63.52 a.m.u Example 2 Given below is a mass spectrum of rubidium element, Rb. abundance 18 Relative 7 m/e 85 87 (amu) a. What isotopes are present in Rb? b. What is the percentage abundance of each isotope? c. Calculate the relative atomic mass (Ar) of Rb. SOLUTION abundance Relative 18 7 m/e 85 87 (amu) (a) 85Rb and 87Rb (b) % abundance 85Rb = 18 X 100 = 72% (18+7) % abundance 85Rb = 7 X 100 = 28% (18+7) 36 CONTINUE SOLUTION Qi M i C. Average mass of Rb Qi (18x85) (7 x87) 25 85.56 amu 85.56 amu A r of Rb 1 x12.00 amu 12 85.56 Example 3 The ratio of relative abundance of naturally occurring of chlorine isotopes is as follow : 35 Cl 37 3.127 Cl Based on the carbon-12 scale, the relative isotopic mass of 35Cl = 34.9689 and 37Cl = 36.9659. Calculate the Ar of chlorine. SOLUTION 35Cl = 3.127 37Cl 1 3.127 Abundance, Q Relative 35Cl 1 37Cl 34.9689 36.9659 m/e (amu) CONTINUE SOLUTION Average ∑ Qi mi atomic = mass ∑ Qi = (3.127 x 34.9689 amu) + (1 x 36.9659 amu) 3.127 + 1 = 35.4528 amu Relative atomic = 35.4528 amu = 35.4528 mass 1 x 12.00 amu 12 Example 4 6 7 Li The relative atomic mass of 3 Li and 3 are 6.01 and 7.02 respectively. What is the percentage abundance of each isotope if the relative atomic mass of Li is 6.94? SOLUTION Abundance , Q x% (100-x) % Relative 6 Li (%) 3 7 Li 3 6.01 7.02 m/e (amu) CONTINUE SOLUTION Assume that, % abundance of 6Li = x % % abundance of 7Li = (100-x) % 6.94 = x (6.01amu) + (100 – x) 7.02amu (x) + (100 – x) 6.94 = 6.01 x + 702 – 7.02 x 100 694 - 702 = -1.01 x +8 = +1.01 x x = 7.92 So, % abundance of 6Li = 7.92 % And % abundance of 7Li = 100-x = 92.08 %