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Chapter 3 Mass Relationships in Chemical Reactions Semester 2/2011

1. Chapter 3 Mass Relationships in Chemical Reactions Semester 2/2011. 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

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Chapter 3 Mass Relationships in Chemical Reactions Semester 2/2011

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  1. 1 Chapter 3 Mass Relationships in Chemical Reactions Semester 2/2011 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. • By definition:1 atom 12C “weighs” 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  The average atomic mass is the weighted average of all of the naturally occurring isotopes of the element.

  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 Chemist Measure Atoms and molecules in a unit called “moles”( A unit to count numbers of particles) 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 ***For any element atomic mass (amu) = molar mass (grams) 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. 1 mol K 6.022 x 1023 atoms K x x = 1 mol K 39.10 g K Example:How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K 0.551 g K 8.49 x 1021 atoms K

  8. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 3.3 Molecular mass (molecular weight) Sum of atomic masses (in amu) in the molecule Ex: SO2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 8

  9. 8 mol H atoms 6.022 x 1023 H atoms 1 mol C3H8O x x x = 1 mol C3H8O 1 mol H atoms 60 g C3H8O How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 72.5 g C3H8O 5.82 x 1024 atoms H

  10. 1Na 22.99 amu NaCl 1Cl + 35.45 amu NaCl 58.44 amu Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl

  11. 3 x 40.08 3 Ca 2 P 2 x 30.97 8 O + 8 x 16.00 What is the formula mass of Ca3(PO4)2 ? 1 formula unit of Ca3(PO4)2 310.18 amu

  12. 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 = 12

  13. 2 x (12.01 g) 6 x (1.008 g) 1 x (16.00 g) n x molar mass of element %C = %H = %O = x 100% = 34.73% x 100% = 13.13% x 100% = 52.14% x 100% 46.07 g 46.07 g 46.07 g molar mass of compound C2H6O Percent composition of an element in a compound = n is the number of moles of the element in 1 mole of the compound 52.14% + 13.13% + 34.73% = 100.0%

  14. 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 14

  15. nK = 24.75 g K x = 0.6330 mol K 1 mol Mn nMn = 34.77 g Mn x = 0.6329 mol Mn 54.94 g Mn 1 mol O nO = 40.51 g O x = 2.532 mol O 16.00 g O 1 mol K 39.10 g K Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent.

  16. = 1.0 K : Mn : O : 1.0 4.0 0.6330 0.6329 2.532 ~ ~ ~ ~ 0.6329 0.6329 0.6329 Percent Composition and Empirical Formulas nK = 0.6330, nMn = 0.6329, nO = 2.532 KMnO4

  17. 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 17

  18.  Molecular Formula= (NO2)2 = N2O4 Molecular Mass(or) Molar Mass = 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 18 2 mol + 1 mol 2 mol

  19. 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 19

  20. 2NO + O2 2NO2 Limiting Reagent: Reactant used up first in the reaction. NO is the limiting reagent O2 is the excess reagent

  21. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Fe2O3 needed g Fe2O3 needed g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Fe2O3 160. g Fe2O3 1 mol Al = x x x 27.0 g Al 2 mol Al 1 mol Fe2O3 Start with 124 g Al need 367 g Fe2O3 In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. OR 367 g Fe2O3 124 g Al Have more Fe2O3 (601 g) so Al is limiting reagent

  22. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Al2O3 g Al2O3 1 mol Al x 27.0 g Al 1 mol Al2O3 102. g Al2O3 = x x 2 mol Al 1 mol Al2O3 Use limiting reagent (Al) to calculate amount of product that can be formed. 234 g Al2O3 124 g Al At this point, all the Al is consumed and Fe2O3 remains in excess.

  23. 3.10 Reaction Yield Theoretical yield is the amount of product that would result if all the limiting reagent reacted. [can be obtained from calculation based on balanced equation.] Actual yield is the amount of product actually obtained from a reaction. 23

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