Chapter 3 Mass Relationships in Chemical Reactions Semester 1/ Atomic Mass 3.2 Avogadro’s Number and the Molar Mass of an element 3.3 Molecular Mass 3.5 Percent Composition of Compounds 3.6 Experimental Determination of Empirical Formulas 3.7 Chemical Reactions and Chemical Equations 3.8 Amounts of Reactants and Products 3.9 Limiting Reagents 3.10 Reaction Yield 1 Ref: Raymong Chang/Chemistry/Ninth Edition Prepared by A. Kyi Kyi Tin

3.1 Atomic Mass(Atomic weight) Mass of the atom in atomic mass units (amu), which is based on the carbon-12 isotope scale. amu = atomic mass unit Define: 1amu  1 amu = times mass of one carbon –12 atom. Mass of one carbon-12 atom = 12 amu  1 amu = x 12 amu Ex:atomic mass of ‘H’ atom=8.4% of carbon-12 Atom =0.084 x amu =1.008 amu

Atomic masses 62.93amu amu A.A.M = (0.6909)(62.93amu)+(0.3091x amu) = 63.55amu  Ex:3.1 Calculate the average atomic mass of copper. 3

3.2 Avogadro’s Number and the Molar Mass of an Element (Italian scientist..Amedeo Avogadro) Amedeo Avogadro’s number  N A Pair  2 items, Dozen = 12 items, Gross = 144 items Chemist  Measure Atoms and molecules in a unit called “moles” 1 mole = 6.02x10 23 Atoms Molecule Ions Molar mass(  )  mass [ in “g” (or) “Kg” ] of 1 mole of units (atom (or) molecule (or) ion) 4

From periodic Table ElementAtomic MassMolar mass for “Atom”MoleculeMolar mass for molecule H1.008 amu1.008 g / molH2H x2 = 2.016g/mol O16.00 amu16.00 g / molO2O x2 = 32.00g/mol Cl35.5 amu35.5 g / molCl 2 HCl 35.5x2 = g / mol ( ) = g / mol Na22.99 amu22.99 g / molNa22.99 g / mol C12.01 amu12.01 g / molCO( ) = g / mol 5

1 mol of ‘H’ atom = g = 6.02x10 23 atoms of ‘H’ atom 1 mol of ‘H 2 ’ moleule = (1.008x2) g = 6.02 x10 23 molecules of ‘H 2 ’ molecule 1 mol of ‘Na’ atom = g = 6.02x10 23 atoms of ‘Na’ atom 1 mol of ‘O’ atom = g = 6.02x10 23 atoms of ‘O’ atom 1 mol of ‘O 2 ’ moleule = (16.00x2)g = 6.02x10 23 molecule of ‘O 2 ’ molecule 1 mol of carbon-12 atom = 12g = 6.02x10 23 atoms of carbon-12 atom  6.02x10 23 atoms of carbon-12 atom = 12 g 1 atom of carbon-12 atom = 1 atom of carbon-12 atom = 12amu  1 amu = 6

Ex: 3.2 Solve this problem in two ways: ‘He’ 1 st method. 1 mol of ‘He’ atom = 4.003g = 6.02x10 23 atoms of ‘He’ atom i.e 4.003g  1 mol of ‘He’ atom 6.46g  ? =1.61 mol of ‘He’ atom Of ‘He’ atom 2 nd. Method Apply Conversion factor 7

3.3Molecular mass (molecular weight) Sum of atomic masses (in amu) in the molecule Ex:H 2 O2(atomic mass of H)+1 atomic mass of O 2(1.008 amu) amu = 18.02amu  Note: For Ionic compounds like NaCl and MgO WE USE THE TERM “Formula Mass” Formula mass of NaCl = amu amu = amu Atomic mass + Atomic mass of “Na” of “Cl” 8

3.5 Percent Composition of the Compounds Ex:H 2 O 2 1mol of H 2 O 2 2 mol of ‘H’ atom 2 mol of ‘O’ atom Molar mass of H 2 O 2 = (2x ) = g / mol %H = %O = 9

3.6Empirical Formula Formula for a compound that contains the smallest whole number ratios for the elements in the compound. ExC:H:O 2:6:1 i.e C 2 H 6 O Mole ratio : 1.50: 0.25 Smallest whole number ratios 10

Ex:3.11  COMPOUND Nitrogen 1.52g Oxygen 3.47g Mole= : 1 : 2  Empirical Formula NO 2  Empirical molar mass = (16x2) = 46.01g Smallest whole number ratio 11

 (NO 2 ) 2 = N 2 O 4 = = 92.02g/mol 3.8 Amounts of Reactants and Products Stoichiometry is the quantitative study of reactants and products in a balanced chemical reaction. 2 CO (g) + O 2 (g)2 CO 2 (g) 2 molecules + 1 molecule 2 molecules 2 mol+ 1 mol 2 mol 12

3.9 Limiting Reagents (L.R) Limiting Reagent….. The reactant used up first in a reaction. Excess Reagent.. The reactants present in quantities greater than necessary to react with the quantity of the limiting reagent. Ex: 2NO +O 2  2NO 2 INITIAL mole(given) 8 7 Balanced Equation 2mol + 1mol  2 mol 8 mol of “NO” yields…..8 mol of ”NO 2 ” 7 mol of “O 2 ”..yields …14 mol of “NO 2 ” O 2 is ExcessNO is Limiting 13

3.10 Reaction Yield Theoretical yield can be obtained from calculation based on balanced equation. Actual yield can be obtained from the given problem. 14