1. TOPIC 3 ATOMIC 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

2. 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

3.  Eg. RAM of Mg = 24= 24 1/12 x 12 RAM of Pb = 207= 207 1/12 x 12 Mass of an atom of the element 1/12 of the mass of one atom carbon-12

4. RELATIVE MOLECULAR MASS (R.M.M)  A molecule is made up of two or more atoms  Calculation of RMM 1. Determined the molecular formula 2. Find the RAM of each element in the molecule 3. Add up of all the RAM of the element

5. MoleculeMolecular formulaRMM ChlorineCl 2 35.5 x 2 = 72 NitrogenN2N2 14 x 2 = 28 AmmoniaNH 3 14 + (1 x 3) = 17 EthanolC 2 H 5 OH(12 x 2) + (1 x 5) + 16 + 1 = 46 Carbon dioxide CO 2 12 + (16 x 2) = 44

6. LET’S TRY!! Find Relative molecular mass for: (RAM: Cu=64 ; O=16 ; C=12 ; H=1 ; S=32 ; Cl=35.5 ; Na=23) i. CuO ii. CH 4 iii. SO 2 iv. HCl v. Na 2 CO 3

7. ANSWER Find Relative molecular mass for: (RAM: Cu=64 ; O=16 ; C=12 ; H=1 ; S=32 ; Cl=35.5 ; Na=23) i. CuO = 80 ii. CH 4 = 16 iii. SO 2 = 64 iv. HCl = 36.5 v. Na 2 CO 3 = 106

8. 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.

9. 1 pair of shoes = 2 shoes 1 dozen of eggs = 12 eggs 1 mole of carbon = 6.02 x 10 23 atoms

10.  1 mole of any element contains 6.02 x 10 23 atoms.  The value of 6.02 x 10 23 atoms is called as Avogadro’s constant  Key: 1 mole of atom = 6.02 x 10 23 particles AVOGADRO’S CONSTANT = 6.02 x 10 23

11. THE MOLE AND THE MASS OF SUBSTANCE  Number of mole atoms = Mass RAM MASS MOLER.A.M

12. Problem solving Determined the moles How many moles of matter are there in: 1. 5 g of nitrogen, N 2. 10 g of sodium, Na 3. 80 g of carbon, C (RAM: N=14 ; Na=23 ; C=12) MASS MOLE R.A.M

13. Solution: 1. 5 g of nitrogen, N 5 /14 =0.357 moles 2. 10 g of sodium, Na 10/23 = 0.435 moles 3. 80 g of carbon, C 80/12 = 6.67 moles MASS MOLER.A.M

14. Problem solving Determined the mass How many grams of matter are there in: 1. 2 moles of nitrogen, N 2. 0.5 moles of sodium, Na 3. 6.0 moles of carbon, C (RAM: N=14 ; Na=23 ; C=12) MASS MOLER.A.M

15. Solution: 1. 2 moles of nitrogen, N 2 x 14 = 28 g 2. 0.5 moles of sodium, Na 0.5 x 23 = 11.5 g 3. 6.0 moles of carbon, C 6.0 x 12 =72 g (RAM: N=14 ; Na=23 ; C=12) MASS MOLE R.A.M

16. Determined The Number Of Particles  How many atoms are there in: 1. 0.5 moles of carbon 0.5 x 6.02 x 10 23 = 3.01 x 10 23 atoms 2. 1.2 moles of sodium 1.2 x 6.02 x 10 23 = 7.224 x 10 23 atoms Number of particles = moles x 6.02 x 10 23

17. Determined The Number Of Moles From The Number Of Atoms  How many moles are there in: 1. 12.04 x 10 23 atoms of chlorine 12.04 x 10 23 / 6.02 x 10 23 = 2 moles 2. 1.02 x 10 46 atoms of sodium 1.02 x 10 46 / 6.02 x 10 23 = 1.69 x 10 23 moles Number of moles of atoms = Number of atoms 6.02 x 10 23

18. MASSMOLENUMBER OF PARTICLES ÷ RAM ÷ N A × N A × RAM Number Avogadro (N A ) = 6.02 x 10 23 particles

19. CHEMICAL FORMULA  Cation : +ve charge eg: Na +, Mg 2+, Al 3+ …  Anion : -ve charge eg: Cl -, O 2-, N 3- ….  If the cation X m+ and the anion is Y n-, then the formula of the compound is X n Y m  If m=n, then the formula XY

20. X m+ Y n- XnXn YmYm = X n Y m

21. Example: Chemical compoundCationAnionChemical formula Iron (II) chloride Iron (III) chloride Fe 2+ Fe 3+ Cl - FeCl 2 FeCl 3 Copper (I) sulphate Copper (II) sulphate Cu + Cu 2+ SO 2- 4 Cu 2 SO 4 CuSO 4 Manganese (II) oxide Manganese (III) oxide Manganese (IV) oxide Mn 2+ Mn 3+ Mn 4+ O 2- MnO Mn 2 O 3 MnO 2

22. Empirical formula  Step 1. Write the mass of element 2. Calculate the moles 3. Calculate the simplest mole ratio by dividing the each number with the smallest number 4. Write the empirical formula

23. Example  Determine the empirical formula of this compound  Solution:  So, empirical formula = CaCl 3 ElementCaCl Mass1.6 g4.26 g RAM4035.5 ElementCaCl Mass1.6 g4.26 g Moles1.6 g / 40 = 0.044.26 / 35.5 = 0.12 Simplest ration0.04 / 0.04 = 10.12 / 0.04 = 3

24. 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 CaCl 3 is 293 (CaCl 3 ) n = 293 (40+ (3x35.5)) n = 293 (146.5) n = 293 n=2 So, molecular formula is (CaCl 3 ) n = (CaCl 3 ) 2 = Ca 2 Cl 6

25. Let’s try 1. 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: Al=27 ; O=16) 2. The emipirical formula of ethene is (CH 3 ) n. Its molecular formula mass is 30. Calculate the molecular formula of this compound.

26. Solution 1. So, empirical formula is Al 2 O 3 ElementAlO Mass0.91 g0.79 g RAM2716 Moles0.91 / 27 = 0.0340.79 / 16 = 0.05 Ratio0.034 / 0.034 = 10.05 / 0.034 = 1.5 x223

27. 2. The emipirical formula of ethene is (CH 3 ) n. Its molecular formula mass is 30. Calculate the molecular formula of this compound. (CH 3 ) n = 30 (12 + 3) n = 30 (15) n = 30 n = 2 So, molecular formula is C 2 H 6

28. THANK YOU…