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# Chapter 3 chemical formulae

## by leonling on Jul 15, 2011

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## Chapter 3 chemical formulaePresentation Transcript

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