Chapter 3 part 1 ( 3.1 3.4 )

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Chapter 3 part 1 ( 3.1 3.4 )

  1. 1. Chapter 3 : Chemical Formulae and Equations 3.1 Relative Atomic Mass & Relative Molecular Mass 3.2 Relationship between the Number of Moles and the Number of Particles 3.3 Relationship between the Number of Moles of a Substance and its Mass 3.4 Relationship between the Number of Moles of a Gas and its Volume 3.5 Chemical Formulae 3.6 Chemical Equations 3.7 Scientific Attitudes and Values in Investigating Matter
  2. 2. 3.1 Relative Atomic Mass and Relative Molecular Mass  Relative Atomic Mass (Ar) – the number of times one atom of the element is heavier than one twelfth of the mass of a carbon-12 atom
  3. 3. Example: Sodium atom , Na is 23 time heavier than onetwelfth of the mass of one carbon-12 atom. Thus the relative atomic mass of Na is 23.  Mass of one Na atom (23) 1 × mass of one carbon - 12 atom 12 = 23
  4. 4.  Relative Molecular Mass (Mr) – the number of one molecule of the compound is heavier than one-twelfth of the mass of a carbon-12 atom
  5. 5. Example: A methane molecule , CH4 is 16 time heavier than one-twelfth of the mass of one carbon-12 atom. Thus the relative molecular mass of CH4 is 16.  Mass of one CH 4 molecule (16) = 16 1 × mass of one carbon - 12 atom 12
  6. 6. 3.2 Relationship between the Number of Moles and the Number of Particles  Mole – the amount of substance which contains the same number of particles (atoms/ions/molecules) as there are in 12 grams of carbon-12.  The number of atoms in 12grams of carbon12 is 6.02 x 1023 (Avogadro’s Number or Avogadro’s Constant (NA)
  7. 7.  Example: - 1 mol of gold contains 6.02 x 1023 of gold atoms - 1 mol of magnesium ions contains 6.02 x 1023 Mg2+ ions - 1 mol of magnesium chloride (MgCl2) contains 6.02 x 1023 Mg2+ ions and 2 x 6.02 x 1023 Cl- ions. - 1 mol of carbon dioxide contains 6.02 x 10 23 CO2 molecules CO2 is a covalent compounds; chemical bond that involves the sharing of electron pairs between atoms.
  8. 8.  Conversion of the number of moles to the numbers of particles and vice versa:- Number of Particles = Number of mole x NA Number of Moles = Number of particles ÷ NA  Example:- Calculate the number of particles in 0.75 mol of aluminium atoms,Al. Solution: 0.75 mol x 6.02 x 1023 Al atoms = 4.52 x 1023 Al atoms.
  9. 9. 3.3 Relationship between the Number of Moles of a Substance and Its Mass  Molar mass – mass of a substance that contains one mole of the substance  The molar mass of any substance contains 6.02 x 1023 particles  The mole atom = relative atomic mass of an atom but expressed in gram. Eg: 1 mole atom of Al = 27g  The mole molecule = relative molecular mass of a compund expressed in gram. Eg: 1 mole molecule of water ,H2O = 18g
  10. 10.  Conversion of the number of moles of a substance to its mass and vice versa:Number of mole-atom = mass in gram ÷ relative atomic mass Number of mole-molecule = mass in gram ÷ relative molecular mass Mass in gram = Number of mole x relative atomic mass or relative molecular mass
  11. 11.  Example: Calculate the number of moles of 23.5g of copper (II) nitrate , Cu(NO3)2. [ RAM: Cu = 64, N=14, O=16] Solution: 1 mol of Cu(NO3)2 = 64 + 2[14+3(16)] g = 188 g  RMM 23.5 g of Cu(NO3)2 = 23.5 × 1 mol 188 = 0.125 mol
  12. 12.  Example: Determine the mass for 0.08 mol of ascorbic acid , C6H8O6. [RAM: C=12,H=1,O=16] Solution: 1 mol of C6H8O6 = 6(12) + 8(1)+ 6(16) 0.08 mol of C6H8O6 = 176g = 0.08 x 176g =14.08g
  13. 13. 3.4 Relationship between the Number of Moles of a Gas and Its Volume  One mole of any gas at room temperature and 1 atm presure occupies a volume of 24dm3 ( 24000 cm3)  At standard temperature and pressure,s.t.p ( 0oC and 1 atm), one mole of gas occupies a volume of 22.4dm3 ( 22400 cm3).  Molar volume – volume occupied by one mole of any gas.
  14. 14.  Conversion of the number of moles of a gas to its volume and vice versa:Number of mole of a gas = volume of gas ÷ molar volume Volume of gas = Number of mole of a gas x molar volume
  15. 15.  Example: Calculate the number of moles of 4.8dm3 of chlorine gas at room temperature. [1 mol of gas occupies a volume of 24dm3 at room temperature] Solution: 3 4.8dm Number of moles= ×1 mol 3 24dm = 0.2 mol
  16. 16.  Example: Calculate the volume of 0.75 mol of nitrogen gas at s.t.p. [ 1 mol of gas occupies a volume of 22.4dm3 at s.t.p] Solution: Volume of nitrogen gas = 0.75 mol x 22.4dm3 = 16.8dm3
  17. 17. 3.5 Chemical Formulae  Used to represent a chemical compound  It shows:- the elements (denoted by symbols) - the relative numbers (indicated by subscript after the symbol)  Example:- H2 O
  18. 18. Chemical formulae of some covalent compounds Name of compound Chemical formula Number of each element in the compound Oxygen O2 2 oxygen atoms Water H2O 2 hydrogen atoms 1 oxygen atom
  19. 19. Chemical formulae of some ions (cations) Charge Cation Symbol +1 Sodium ion Na+ +2 Magnesium ion Mg 2+ +3 Iron(III) ion Fe3+
  20. 20. Chemical formulae of some ions (anions) Charge Anion Symbol -1 Fluoride ion F- -2 Oxide ion O2- -3 Nitride ion N3-
  21. 21.  To write the chemical formula of an ionic compounds:- write the formula of the ions involved in forming the compound - balance the positive and negative charge (use subscript) - Write the chemical formula of the ionic compound without the charges.
  22. 22. Example:-
  23. 23. Formulae of some ionic compounds Cation Anion Chemical formula Na+ Cl- NaCl Ca2+ Cl- CaCl2 Al3+ N3- AlN
  24. 24.  Empirical formulae of a compound:- shows the simplest ratio of the atoms of the elements that combine to form a compound - steps to determine the empirical formula of a compound:1) write the mass / percentage of each element in the compound 2) calculate the number of moles for each element 3) Divide each number by the smallest number to obtain simplest ratio 4) Write the empirical formula of the compound
  25. 25. Example:-
  26. 26.  Molecular formulae of the compound:- shows the actual numbers of the atoms of the elements that combine to form the compound Compound Molecular formula Simplest ratio of the elements Empirical formula Water H2O H:O = 2:1 H2O Ethene C2H4 C:H=1:2 CH2 Glucose C6H12O6 C:H:O=1:2:1 CH2O
  27. 27. 3.6 Chemical Equations  Chemical reaction can be represented by a chemical equation  Reactants – chemicals that are reacting. Written on LHS.  Products – chemicals formed in the reaction. Written on the RHS
  28. 28.  Writing a chemical equation:1) write the correct formulae of all reactants on the LHS of the equation 2) write the correct formulae of all products on the RHS of the equation 3) the equation is then balanced. 4) make sure the number of atoms before and after reaction are the same 5) Write the physical state of each reactants and products
  29. 29. Example:-

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