TOPIC 3 ATOMIC MASS AND THE CONCEPT OF MOLECULES

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TOPIC 3ATOMIC MASS AND THE CONCEPT OF MOLECULES

• By
• Maisara Shahrom binti Raja Shahrom
• Chemistry Lecturer
• School of Allied Health Sciences
• City University College of Science and Technology

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RELATIVE ATOMIC MASS (R.A.M)

• Relative mass of an atom
• Why carbon-12 is used as a standard?
• Its mass can be more easily measured with a mass
  spectrometer

Mass of an atom of the element 1/12 of the mass
of one atom carbon-12
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Mass of an atom of the element 1/12 of the mass
of one atom carbon-12

• Eg.
• RAM of Mg 24 24
• 1/12 x 12
• RAM of Pb 207 207
• 1/12 x 12

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RELATIVE MOLECULAR MASS (R.M.M)

• A molecule is made up of two or more atoms
• Calculation of RMM
• Determined the molecular formula
• Find the RAM of each element in the molecule
• Add up of all the RAM of the element

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EXAMPLE
Molecule Molecular formula RMM
Chlorine Cl2 35.5 x 2 72
Nitrogen N2 14 x 2 28
Ammonia NH3 14 (1 x 3) 17
Ethanol C2H5OH (12 x 2) (1 x 5) 16 1 46
Carbon dioxide CO2 12 (16 x 2) 44
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LETS TRY!!

• Find Relative molecular mass for
• (RAM Cu64 O16 C12 H1 S32 Cl35.5
  Na23)
• CuO
• CH4
• SO2
• HCl
• Na2CO3

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ANSWER

• Find Relative molecular mass for
• (RAM Cu64 O16 C12 H1 S32 Cl35.5
  Na23)
• CuO 80
• CH4 16
• SO2 64
• HCl 36.5
• Na2CO3 106

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THE MOLE AND THE NUMBER OF PARTICLES

• In our daily lives, we need to count the amount
  of objects
• Pairs and dozen are examples of units that we
  used
• In chemistry, we use the unit of mole to measure
  the amount of substances.

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1 pair of shoes 2 shoes
1 dozen of eggs 12 eggs
1 mole of carbon 6.02 x 1023 atoms
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• 1 mole of any element contains 6.02 x 1023 atoms.
• The value of 6.02 x 1023 atoms is called as
  Avogadros constant
• Key
• 1 mole of atom 6.02 x 1023 particles

AVOGADROS CONSTANT 6.02 x 1023
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THE MOLE AND THE MASS OF SUBSTANCE

• Number of mole atoms
• Mass
• RAM

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Problem solvingDetermined the moles

• How many moles of matter are there in
• 5 g of nitrogen, N
• 10 g of sodium, Na
• 80 g of carbon, C
• (RAM N14 Na23 C12)

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• Solution
• 5 g of nitrogen, N
• 5 /14 0.357 moles
• 10 g of sodium, Na
• 10/23 0.435 moles
• 80 g of carbon, C
• 80/12 6.67 moles

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Problem solvingDetermined the mass

• How many grams of matter are there in
• 2 moles of nitrogen, N
• 0.5 moles of sodium, Na
• 6.0 moles of carbon, C
• (RAM N14 Na23 C12)

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• Solution
• 2 moles of nitrogen, N
• 2 x 14 28 g
• 0.5 moles of sodium, Na
• 0.5 x 23 11.5 g
• 6.0 moles of carbon, C
• 6.0 x 12 72 g
• (RAM N14 Na23 C12)

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Determined The Number Of Particles
Number of particles moles x 6.02 x 1023

• How many atoms are there in
• 0.5 moles of carbon
• 0.5 x 6.02 x 1023 3.01 x 1023 atoms
• 1.2 moles of sodium
• 1.2 x 6.02 x 1023 7.224 x 1023 atoms

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Determined The Number Of Moles From The Number Of
Atoms

• How many moles are there in
• 12.04 x 1023 atoms of chlorine
• 12.04 x 1023 / 6.02 x 1023 2 moles
• 1.02 x 1046 atoms of sodium
• 1.02 x 1046 / 6.02 x 1023 1.69 x 1023 moles

Number of moles of atoms Number of
atoms 6.02 x 1023
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RAM
NA
MASS
MOLE
NUMBER OF PARTICLES
NA
RAM
Number Avogadro (NA) 6.02 x 1023 particles
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CHEMICAL FORMULA

• Cation ve charge
• eg Na, Mg2, Al3
• Anion -ve charge
• eg Cl-, O2-, N3-.
• If the cation Xm and the anion is Yn-, then the
  formula of the compound is XnYm
• If mn, then the formula XY

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Xm
Yn-
Xn
Ym
XnYm
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Example
Chemical compound Cation Anion Chemical formula
Iron (II) chloride Iron (III) chloride Fe2 Fe3 Cl- Cl- FeCl2 FeCl3
Copper (I) sulphate Copper (II) sulphate Cu Cu2 SO2-4 SO2-4 Cu2SO4 CuSO4
Manganese (II) oxide Manganese (III) oxide Manganese (IV) oxide Mn2 Mn3 Mn4 O2- O2- O2- MnO Mn2O3 MnO2
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Empirical formula

• Step
• Write the mass of element
• Calculate the moles
• Calculate the simplest mole ratio by dividing the
  each number with the smallest number
• Write the empirical formula

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Example

• Determine the empirical formula of this compound
• Solution
• So, empirical formula CaCl3

Element Ca Cl
Mass 1.6 g 4.26 g
RAM 40 35.5
Element Ca Cl
Mass 1.6 g 4.26 g
Moles 1.6 g / 40 0.04 4.26 / 35.5 0.12
Simplest ration 0.04 / 0.04 1 0.12 / 0.04 3
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Molecular formula

• Shows the actual number of atoms of elements that
  combine to form the compound.
• (Empirical formula)n Relative molecular mass
• Eg
• Relative molecular mass of CaCl3 is 293
• (CaCl3) n 293
• (40 (3x35.5)) n 293
• (146.5) n 293
• n2
• So, molecular formula is (CaCl3) n (CaCl3) 2
• Ca2Cl6

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Lets try

• 0.19 g of aluminium and 0.79 g atom oxygen are
  burnt to form aluminium oxide. What is the
  empirical formula of aluminium oxide RAM Al27
  O16)
• The emipirical formula of ethene is (CH3)n. Its
  molecular formula mass is 30. Calculate the
  molecular formula of this compound.

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Solution

• 1.
• So, empirical formula is Al2O3

Element Al O
Mass 0.91 g 0.79 g
RAM 27 16
Moles 0.91 / 27 0.034 0.79 / 16 0.05
Ratio 0.034 / 0.034 1 0.05 / 0.034 1.5
x2 2 3
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• The emipirical formula of ethene is (CH3)n. Its
  molecular formula mass is 30. Calculate the
  molecular formula of this compound.
• (CH3)n 30
• (12 3)n 30
• (15)n 30
• n 2
• So, molecular formula is C2H6

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THANK YOU