TOPIC 3 ATOMIC MASS AND THE CONCEPT OF

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

RELATIVE ATOMIC MASS (R. A. M) Relative mass of an atom = Mass of an atom of the element 1/12 of the mass of one atom carbon-12 Why carbon-12 is used as a standard? Ø 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 Eg. RAM of Mg = RAM of Pb = 24 1/12 x 12 207 1/12 x 12 = 24 = 207

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

EXAMPLE Molecule Molecular formula RMM Chlorine Cl 2 35. 5 x 2 = 72 Nitrogen N 2 14 x 2 = 28 Ammonia NH 3 14 + (1 x 3) = 17 Ethanol C 2 H 5 OH Carbon dioxide CO 2 (12 x 2) + (1 x 5) + 16 + 1 = 46 12 + (16 x 2) = 44

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. Cu. O ii. CH 4 iii. SO 2 iv. HCl v. Na 2 CO 3

ANSWER Find Relative molecular mass for: (RAM: Cu=64 ; O=16 ; C=12 ; H=1 ; S=32 ; Cl=35. 5 ; Na=23) i. Cu. O = 80 ii. CH 4 = 16 iii. SO 2 = 64 iv. HCl = 36. 5 v. Na 2 CO 3 = 106

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.

1 pair of shoes = 2 shoes 1 dozen of eggs = 12 eggs 1 mole of carbon = 6. 02 x 1023 atoms

1 mole of any element contains 6. 02 x 1023 atoms. The value of 6. 02 x 1023 atoms is called as Avogadro’s constant AVOGADRO’S CONSTANT = 6. 02 x 1023 Key: 1 mole of atom = 6. 02 x 1023 particles

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

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

Solution: 1. 2. 5 g of nitrogen, N 5 /14 =0. 357 moles 10 g of sodium, Na 10/23 = 0. 435 moles MASS 3. 80 g of carbon, C 80/12 = 6. 67 moles MOLE R. A. M

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 MOLE R. A. M

Solution: 1. 2 moles of nitrogen, N 2 x 14 = 28 g 2. 3. 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: N=14 ; Na=23 ; C=12) MASS MOLE R. A. M

Determined The Number Of Particles Number of particles = moles x 6. 02 x 1023 1. How many atoms are there in: 0. 5 moles of carbon 0. 5 x 6. 02 x 1023 = 3. 01 x 1023 atoms 2. 1. 2 moles of sodium 1. 2 x 6. 02 x 1023 = 7. 224 x 1023 atoms

Determined The Number Of Moles From The Number Of Atoms Number of moles of atoms = Number of atoms 6. 02 x 1023 1. How many moles are there in: 12. 04 x 1023 atoms of chlorine 12. 04 x 1023 / 6. 02 x 1023 = 2 moles 2. 1. 02 x 1046 atoms of sodium 1. 02 x 1046 / 6. 02 x 1023 = 1. 69 x 1023 moles

÷ RAM MASS × NA NUMBER OF PARTICLES MOLE × RAM ÷ NA Number Avogadro (NA) = 6. 02 x 1023 particle

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

m+ X n. Y Xn Ym = X n. Y m

Example: Chemical compound Cation Anion Chemical formula Iron (II) chloride Iron (III) chloride Fe 2+ Fe 3+ Cl. Cl- Fe. Cl 2 Fe. Cl 3 Copper (I) sulphate Copper (II) sulphate Cu+ Cu 2+ SO 2 -4 Cu 2 SO 4 Cu. SO 4 Manganese (II) oxide Manganese (IV) oxide Mn 2+ Mn 3+ Mn 4+ O 2 O 2 O 2 - Mn. O Mn 2 O 3 Mn. O 2

Empirical formula 1. 2. 3. 4. 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

Example Determine the empirical formula of this Element Ca Cl compound Mass 1. 6 g 4. 26 g RAM 40 35. 5 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 Element Solution: So, empirical formula = Ca. Cl 3

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 Ca. Cl 3 is 293 (Ca. Cl 3) n = 293 (40+ (3 x 35. 5)) n = 293 (146. 5) n = 293 n=2 So, molecular formula is (Ca. Cl 3) n = (Ca. Cl 3) 2

Let’s try 1. 2. 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) The emipirical formula of ethene is (CH 3)n. Its molecular formula mass is 30. Calculate the molecular formula of this compound.

Solution 1. Element Al Mass RAM Moles 0. 91 g 27 0. 91 / 27 = 0. 034 O 0. 79 g 16 0. 79 / 16 = 0. 05 Ratio x 2 0. 034 / 0. 034 = 1 2 0. 05 / 0. 034 = 1. 5 3 So, empirical formula is Al 2 O 3

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

THANK YOU…