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Chapter 3 Mass Relationships in Chemical Reactions Semester 2/2012 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: Raymond Chang/Chemistry/Tenth Edition Prepared by A. Kyi Kyi Tin 1 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 1 amu = 12 times mass of one carbon –12 atom. By definition:1 atom 12C “weighs” 12 amu 1 amu = 1 x 12 amu 12 Ex: atomic mass of ‘H’ atom = 8.4% of carbon-12 Atom = 0.084 x 12.00 amu = 1.008 amu 2 Ex:3.1 Calculate the average atomic mass of copper. Cu (69.09%) Cu (30.91%) 63 29 65 29 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. 3 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) Atoms 1 mole = 6.02x1023 Molecule Ions Molar mass() mass [ in “g” (or) “Kg” ] of 1 mole of units 4 (atom (or) molecule (or) ion) From periodic Table Element Atomic Mass H 1.008 amu O 16.00 amu Cl 35.5 amu Na 22.99 amu C 12.01 amu Molar mass for “Atom”(g/mol) Molecule Molar mass for molecule (g/mol) ***For any element atomic mass (amu) = molar mass (grams) 5 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 = 12.00 g 1.993 x10 23 g 23 6.02 x10 1 atom of carbon-12 atom = 12amu 1.933x10 23 g 1.661x10 24 g 1 amu = 12 6 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” 1 mol K 0.551 g K x x 39.10 g K 6.022 x 1023 atoms K = 1 mol K 8.49 x 1021 atoms K 7 3.3 Molecular mass (molecular weight) Sum of atomic masses (in amu) in the molecule Ex: 1S 2O SO2 SO2 32.07 amu + 2 x 16.00 amu 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 8 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 8 mol H atoms 6.022 x 1023 H atoms 1 mol C3H8O 72.5 g C3H8O x x x = 1 mol C3H8O 1 mol H atoms 60 g C3H8O 5.82 x 1024 atoms H 9 Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. NaCl 1Na 22.99 amu 1Cl NaCl + 35.45 amu 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl 10 What is the formula mass of Ca3(PO4)2 ? 1 formula unit of Ca3(PO4)2 3 Ca 3 x 40.08 2P 2 x 30.97 8O + 8 x 16.00 310.18 amu 11 3.5 Percent Composition of the Compounds Ex: H2O2 2 mol of ‘H’ atom 1mol of H2O2 2 mol of ‘O’ atom Molar mass of H2O2 = (2x1.008 +32) = 34.016 g / mol %H = 2 x1.008g x100% 5.926% 34.016 g %O = 2 x16 g x100% 94.06% 34.016 g 12 Percent composition of an element in a compound = n x molar mass of element x 100% molar mass of compound n is the number of moles of the element in 1 mole of the compound C2H6O 2 x (12.01 g) %C = x 100% = 52.14% 46.07 g 6 x (1.008 g) %H = x 100% = 13.13% 46.07 g 1 x (16.00 g) %O = x 100% = 34.73% 46.07 g 52.14% + 13.13% + 34.73% = 100.0% 13 3.6 Empirical Formula Formula for a compound that contains the smallest whole number ratios for the elements in the compound. Ex Mole ratio Smallest whole number ratios C 0.500 : : 1.50 0.500mol : 0.25 2 H : : 0.25 0.25mol 0.25 1.50mol : 0.25 : O 6 : i.e C2H6O 1 14 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. nK = 24.75 g K x 1 mol K = 0.6330 mol K 39.10 g K nMn = 34.77 g Mn x 1 mol Mn= 0.6329 mol Mn 54.94 g Mn nO = 40.51 g O x 1 mol O = 2.532 mol O 16.00 g O 15 Percent Composition and Empirical Formulas nK = 0.6330, nMn = 0.6329, nO = 2.532 KMnO4 0.6330 ~ K: 1.0 ~ 0.6329 0.6329 = 1.0 Mn : 0.6329 2.532 ~ O: 4.0 ~ 0.6329 16 Ex:3.11 COMPOUND Nitrogen 1.52g Mole = Smallest whole number ratio 1.52 g 3.47 g : 0.108 : 0.217 14 g / mol 16 g / mol 0.108 0.108 1 Oxygen 3.47g : : 0.217 0.108 2 Empirical Formula NO2 Empirical molar mass = 14.01+(16x2) = 46.01g 17 molar _ mass 90 2 Empirical .Molar .Mass 46.01 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 molecules + 1 molecule 2 CO2(g) 2 molecules 18 2 mol + 1 mol 2 mol 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 INITIAL mole(given) 8 Balanced Equation 2mol + + O2 2NO2 7 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 Limiting Reagent: Reactant used up first in the reaction. 2NO + O2 2NO2 NO is the limiting reagent O2 is the excess reagent 20 In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe Calculate the mass of Al2O3 formed. g Al mol Al g Fe2O3 mol Fe2O3 needed OR mol Al needed mol Fe2O3 124 g Al x 1 mol Al 27.0 g Al x g Fe2O3 needed 1 mol Fe2O3 2 mol Al Start with 124 g Al g Al needed 160. g Fe2O3 = x 1 mol Fe2O3 367 g Fe2O3 need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent 21 Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al2O3 2Al + Fe2O3 124 g Al x 1 mol Al 27.0 g Al x g Al2O3 Al2O3 + 2Fe 1 mol Al2O3 2 mol Al 102. g Al2O3 = x 1 mol Al2O3 234 g Al2O3 At this point, all the Al is consumed and Fe2O3 remains in excess. 22 3.10 Reaction Yield actual.. yeild % yield x100% theoretical.. yeild 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