Chapter 3 Mass Relationships in Chemical Reactions Semester 1/2011

1. 1 Chapter 3 Mass Relationships in Chemical Reactions Semester 1/2011 3.1 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 Ref: Raymong Chang/Chemistry/Ninth Edition Prepared by A. Kyi Kyi Tin

2. 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 12.00 amu = 1.008 amu 2 1 12

3. 3 Ex:3.1 Calculate the average atomic mass of copper. Atomic masses 62.93amu + 64.9278 amu A.A.M = (0.6909)(62.93amu)+(0.3091x64.9278amu) = 63.55amu 

4. 3.2 Avogadro’s Number and the Molar Mass of an Element (Italian scientist..Amedeo Avogadro) Amedeo Avogadro’s number  NA Pair  2 items , Dozen = 12 items , Gross = 144 items Chemist Measure Atoms and molecules in a unit called “moles” 1 mole = 6.02x1023 Atoms Molecule Ions Molar mass()  mass [ in “g” (or) “Kg” ] of 1 mole of units (atom (or) molecule (or) ion) 4

5. From periodic Table 5

6. 1 mol of ‘H’ atom = 1.008 g = 6.02x1023 atoms of ‘H’ atom • 1 mol of ‘H2’ moleule = (1.008x2) g = 6.02 x1023 molecules of ‘H2’ molecule • 1 mol of ‘Na’ atom = 22.99 g = 6.02x1023 atoms of ‘Na’ atom • 1 mol of ‘O’ atom = 16.00 g = 6.02x1023 atoms of ‘O’ atom • 1 mol of ‘O2’ moleule = (16.00x2)g = 6.02x1023 molecule of ‘O2’ molecule • 1 mol of carbon-12 atom = 12g = 6.02x1023 atoms of carbon-12 atom • 6.02x1023 atoms of carbon-12 atom = 12 g 1 atom of carbon-12 atom = 1 atom of carbon-12 atom = 12amu  1 amu = 6

7. Ex: 3.2 Solve this problem in two ways: ‘He’ 1st method. 1 mol of ‘He’ atom = 4.003g = 6.02x1023 atoms of ‘He’ atom i.e 4.003g 1 mol of ‘He’ atom 6.46g  ? =1.61 mol of ‘He’ atom 2nd. Method Apply Conversion factor Of ‘He’ atom 7

8. 3.3 Molecular mass (molecular weight) Sum of atomic masses (in amu) in the molecule Ex: H2O 2(atomic mass of H)+1 atomic mass of O 2(1.008 amu) + 16.00amu = 18.02amu Note: For Ionic compounds like NaCl and MgO WE USE THE TERM “Formula Mass” Formula mass of NaCl = 22.99 amu + 35.45 amu = 58.44 amu Atomic mass + Atomic mass of “Na” of “Cl” 8

9. 3.5 Percent Composition of the Compounds Ex: H2O2 1mol of H2O2 2 mol of ‘H’ atom 2 mol of ‘O’ atom Molar mass of H2O2 = (2x1.008 +32) = 34.016 g / mol %H = %O = 9

10. 3.6 Empirical Formula Formula for a compound that contains the smallest whole number ratios for the elements in the compound. Ex C : H : O Mole ratio 0.500 : 1.50 : 0.25 Smallest whole number ratios 2 : 6 : 1 i.e C2H6O 10

11. Ex:3.11  COMPOUND Nitrogen 1.52g Oxygen 3.47g Mole = Smallest whole number ratio : 1 : 2  Empirical Formula NO2 Empirical molar mass = 14.01+(16x2) = 46.01g 11

12. (NO2)2 = N2O4 = 28.02+64 = 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) + O2 (g) 2 CO2(g) 2 molecules + 1 molecule 2 molecules 12 2 mol + 1 mol 2 mol

13. 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 + O2 2NO2 INITIAL mole(given) 8 7 Balanced Equation2mol + 1mol  2 mol 8 mol of “NO” yields…..8 mol of ”NO2” 7 mol of “O2”..yields …14 mol of “NO2” NO is Limiting O2 is Excess 13

14. 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