chapter 3 chemical formulae

10
CHEMICAL FORMULAE AND EQUATIONS

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Page 1: Chapter 3 chemical formulae

CHEMICAL FORMULAE AND EQUATIONS

Page 2: Chapter 3 chemical formulae

An atom size is very tiny and descrete

How to determine the mass of atom?

Page 3: Chapter 3 chemical formulae

Relative Atomic Mass,Ar

• = the mass of one atom of the element

1/12 x the mass of carbon-12 atom

Page 4: Chapter 3 chemical formulae

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

Page 5: Chapter 3 chemical formulae

• 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?

Page 6: Chapter 3 chemical formulae

Relative Atomic Mass,Ar

• = the mass of one atom of the element

1/12 x the mass of carbon-12 atom

Example: Ar for Magnesium = 24 .

1/12 x 12

= 24

Page 7: Chapter 3 chemical formulae

Relative Atomic Mass,Ar

= Nucleon number (p+n)

Page 8: Chapter 3 chemical formulae

• Adding up the Ar of all atoms in a molecule

• Example: MgCO3 ;

Ar Mg = 24 , C=12 , O=16

Mr of MgCO3 = (1x24)+(1x12)+(3x16)

=24+12+48

=84

Relative Molecular Mass,Mr

Page 9: Chapter 3 chemical formulae

Example: HPO4: ;

Ar H = 1 , P=31 , O=16

Mr of HPO4 =(1x1)+(1x31)+(4x16)

=1+31+64

=96

Page 10: Chapter 3 chemical formulae

Let’s try!