Chapter 3 chemical formulae

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Chapter 3 chemical formulae

  1. 1. CHEMICAL FORMULAE AND EQUATIONS
  2. 2. An atom size is very tiny and descrete How to determine the mass of atom?
  3. 3. Relative Atomic Mass,A r <ul><li>= the mass of one atom of the element </li></ul><ul><li>1/12 x the mass of carbon-12 atom </li></ul>
  4. 4. Why pick carbon-12 as standard? <ul><li>Hydrogen-1 </li></ul><ul><li>Exist as gas at room temperature </li></ul><ul><li>Hard to handle </li></ul><ul><li>Oxygen-16 </li></ul><ul><li>Exist in three isotopes O-16, O-17,O-18 </li></ul>
  5. 5. <ul><li>Carbon-12 </li></ul><ul><li>Exist as solid at room temperature </li></ul><ul><li>Easily to handle </li></ul><ul><li>Also found three isotopes:C-12, C-13, C-14 but the C-12 is the most abundant, about 98.89% </li></ul>Why pick carbon-12 as standard?
  6. 6. Relative Atomic Mass,A r <ul><li>= the mass of one atom of the element </li></ul><ul><li>1/12 x the mass of carbon-12 atom </li></ul><ul><li>Example: A r for Magnesium = 24 . </li></ul><ul><li>1/12 x 12 </li></ul><ul><li>= 24 </li></ul>
  7. 7. Relative Atomic Mass,A r = Nucleon number (p+n)
  8. 8. <ul><li>Adding up the Ar of all atoms in a molecule </li></ul><ul><li>Example: MgCO 3 ; </li></ul><ul><li> Ar Mg = 24 , C=12 , O=16 </li></ul><ul><li>Mr of MgCO 3 = (1x24)+(1x12)+(3x16) </li></ul><ul><li>=24+12+48 </li></ul><ul><li>=84 </li></ul>Relative Molecular Mass,Mr
  9. 9. Example: HPO 4 : ; Ar H = 1 , P=31 , O=16 Mr of HPO 4 =(1x1)+(1x31)+(4x16) =1+31+64 =96
  10. 10. Let’s try!

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