# 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.

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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 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. By definition:1 atom 12 C “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 12 1 2

Atomic masses 62.93amu + 64.9278 amu A.A.M = (0.6909)(62.93amu)+(0.3091x64.9278amu) = 63.55amu  Ex:3.1 Calculate the average atomic mass of copper. 3 The average atomic mass is the weighted average of all of the naturally occurring isotopes of the element.

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) 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 amu O16.00 amu Cl35.5 amu Na22.99 amu C12.01 amu 5 ***For any element atomic mass (amu) = molar mass (grams)

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  1 amu = 6

7 x 6.022 x 10 23 atoms K 1 mol 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 10 23 atoms K 0.551 g K 1 mol K 39.10 g K x 8.49 x 10 21 atoms K

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

9 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 H = 6.022 x 10 23 atoms H 5.82 x 10 24 atoms H 1 mol C 3 H 8 O molecules = 8 mol H atoms 72.5 g C 3 H 8 O 1 mol C 3 H 8 O 60 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O x 6.022 x 10 23 H atoms 1 mol H atoms x =

10 Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. 1Na22.99 amu 1Cl + 35.45 amu NaCl 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 NaCl

11 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 P2 x 30.97 8 O + 8 x 16.00 310.18 amu

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 = (2x1.008 +32) = 34.016 g / mol %H = %O = 12

13 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 C2H6OC2H6O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14%H = 6 x (1.008 g) 46.07 g x 100% = 13.13%O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0%

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 0.500 : 1.50: 0.25 Smallest whole number ratios 14

15 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 = 0.6330 mol K 1 mol K 39.10 g K n Mn = 34.77 g Mn x = 0.6329 mol Mn 1 mol Mn 54.94 g Mn n O = 40.51 g O x = 2.532 mol O 1 mol O 16.00 g O

16 Percent Composition and Empirical Formulas K : ~ ~ 1.0 0.6330 0.6329 Mn : 0.6329 = 1.0 O : ~ ~ 4.0 2.532 0.6329 n K = 0.6330, n Mn = 0.6329, n O = 2.532 KMnO 4

Ex:3.11  COMPOUND Nitrogen 1.52g Oxygen 3.47g Mole= : 1 : 2  Empirical Formula NO 2  Empirical molar mass = 14.01+(16x2) = 46.01g Smallest whole number ratio 17

 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  Molecular Formula= (NO 2 ) 2 = N 2 O 4

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 19

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

21 In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. g Almol Almol Fe 2 O 3 neededg Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al neededg Al needed 124 g Al 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = 367 g Fe 2 O 3 Start with 124 g Alneed 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent

22 Use limiting reagent (Al) to calculate amount of product that can be formed. g Almol Almol Al 2 O 3 g Al 2 O 3 124 g Al 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 x = 234 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe At this point, all the Al is consumed and Fe 2 O 3 remains in excess.

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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