Chapter 3 Mass Relationships in Chemical Reactions Semester 1/2012 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: Raymong Chang/Chemistry/Ninth 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 amu = times mass of one carbon 12 atom. By definition:1 atom 12 C weighs 12 amu 1 amu = 1 12 1 12 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
Ex:3.1 Calculate the average atomic mass of copper. 63 Cu(69.09%) Cu(30.91%) 29 29 Atomic masses 62.93amu + 64.9278 amu A.A.M = (0.6909)(62.93amu)+(0.3091x64.9278amu) 65 = 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 N A 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 = 6.02x10 23 Molecule Ions Molar mass( ) mass [ in g (or) Kg ] of 1 mole of units (atom (or) molecule (or) ion) 4
From periodic Table Element Atomic Mass Molar mass for Atom Molecule Molar mass for molecule H 1.008 amu O 16.00 amu Cl 35.5 amu Na 22.99 amu C 12.01 amu ***For any element atomic mass (amu) = molar mass (grams) 5
1 mol of H atom = 1.008 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 = 22.99 g = 6.02x10 23 atoms of Na atom 1 mol of O atom = 16.00 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 12.00 1.993x10 23 6.02x10 g 23 g 1 amu = 23 1.933 x10 24 12 g 1.661 x10 g 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 10 23 atoms K 0.551 g K x 1 mol K 39.10 g K x 6.022 x 10 23 atoms K 1 mol K = 8.49 x 10 21 atoms K 7
3.3 Molecular mass (molecular weight) Sum of atomic masses (in amu) in the molecule Ex: SO 2 1S 2O SO 2 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 C 3 H 8 O? 1 mol C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 g C 3 H 8 O 1 mol C 3 H 8 O molecules = 8 mol H atoms 1 mol H = 6.022 x 10 23 atoms H 72.5 g C 3 H 8 O 1 mol C 3H 8 O x 60 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O 6.022 x 10 23 H atoms x = 1 mol H atoms 5.82 x 10 24 atoms H 9
Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. NaCl 1Na 1Cl NaCl 22.99 amu + 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 Ca 3 (PO 4 ) 2? 1 formula unit of Ca 3 (PO 4 ) 2 3 Ca 3 x 40.08 2 P 2 x 30.97 8 O + 8 x 16.00 310.18 amu 11
3.5 Percent Composition of the Compounds Ex: H 2 O 2 1mol of H2O2 2 mol of H atom 2 mol of O atom Molar mass of H 2 O 2 = (2x1.008 +32) = 34.016 g / mol %H = 2x1.008 g 34.016 g x100% 5.926% %O = 2x16 g 34.016 g x100 % 94.06% 12
Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C 2 H 6 O %C = 2 x (12.01 g) x 100% = 52.14% 46.07 g %H = 6 x (1.008 g) x 100% = 13.13% 46.07 g %O = 1 x (16.00 g) 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 C : H : O Mole ratio 0.500 : 1.50 : 0.25 Smallest whole number ratios 0.500mol 0.25 : 1.50mol 0.25 : 0.25mol 0.25 2 : 6 : 1 i.e C 2 H 6 O 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. n K = 24.75 g K x 1 mol K = 0.6330 mol K 39.10 g K n Mn = 34.77 g Mn x 1 mol Mn = 0.6329 mol Mn 54.94 g Mn n O = 40.51 g O x 1 mol O = 2.532 mol O 16.00 g O 15
Percent Composition and Empirical Formulas n K = 0.6330, n Mn = 0.6329, n O = 2.532 0.6330 K : 0.6329 ~1.0 Mn : 0.6329 0.6329 = 1.0 2.532 O : ~ 4.0 0.6329 KMnO 4 16
Ex:3.11 COMPOUND Nitrogen 1.52g Oxygen 3.47g 1.52 g 14 g / mol Mole = : 3.47 g 16 g / mol 0.108 : 0.217 Smallest whole number ratio 0.108 0.108 : 0.217 0.108 1 : 2 Empirical Formula NO 2 Empirical molar mass = 14.01+(16x2) = 46.01g 17
molar _ mass Empirical. Molar. Mass 90 46.01 2 Molecular Formula= (NO 2 ) 2 = N 2 O 4 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) + O 2 (g) 2 CO 2 (g) 2 molecules + 1 molecule 2 molecules 2 mol + 1 mol 2 mol 18
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 NO is Limiting O 2 is Excess 19
Limiting Reagent: Reactant used up first in the reaction. 2NO + O 2 2NO 2 NO is the limiting reagent O 2 is the excess reagent 20
In one process, 124 g of Al are reacted with 601 g of Fe 2 O 2Al + Fe 3 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. g Al mol Al mol Fe 2 O 3 needed g Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al needed g Al needed 124 g Al x 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al 160. g Fe 2 O x 3 = 367 g Fe 2 O 3 1 mol Fe 2 O 3 Start with 124 g Al need 367 g Fe 2 O 3 Have more Fe 2 O 3 (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 Al 2 O 3 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 124 g Al x 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al 102. g Al 2 O x 3 = 234 g Al 2 O 3 1 mol Al 2 O 3 At this point, all the Al is consumed and Fe 2 O 3 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