Chapter 3 Mass Relationships in Chemical ReactionsSemester 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

1Ref: 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 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

12

1

12

1

2

%)91.30(%)09.69( 6529

6329 CuCu

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.

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The average atomic mass is the weightedaverage of all of the naturally occurringisotopes of the element.

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

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

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***For any element

atomic mass (amu) = molar mass (grams)

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 =

gxx

g 2323

10993.11002.6

00.12

1 atom of carbon-12 atom = 12amu

1 amu = gxgx 24

23

10661.112

10933.1

6

7

x 6.022 x 1023 atoms K1 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 1023 atoms K

0.551 g K 1 mol K

39.10 g Kx

8.49 x 1021 atoms K

3.3 Molecular mass (molecular weight)

Sum of atomic masses (in amu) in the molecule

Ex:

8

For any molecule

molecular mass (amu) = molar mass (grams)

SO2

1S 32.07 amu

2O + 2 x 16.00 amu

SO2 64.07 amu

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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 H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =

10

Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound.

1Na 22.99 amu

1Cl + 35.45 amuNaCl 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

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What is the formula mass of Ca3(PO4)2 ?

1 formula unit of Ca3(PO4)2

3 Ca 3 x 40.08

2 P 2 x 30.97

8 O + 8 x 16.00

310.18 amu

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 =

%926.5%100016.34

008.12x

g

gx

%06.94%100016.34

162x

g

gx12

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Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%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%

3.6 Empirical Formula

Formula for a compound that contains the smallest whole number ratios for the elements in the compound.

Ex C : H : O

:25.0

500.0 mol:

25.0

50.1 mol25.0

25.0 mol

2 : 6 : 1

i.e C2H6O

Mole ratio 0.500 : 1.50 : 0.25

Smallest whole number ratios

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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 = 0.6330 mol K1 mol K39.10 g K

nMn = 34.77 g Mn x = 0.6329 mol Mn1 mol Mn54.94 g Mn

nO = 40.51 g O x = 2.532 mol O1 mol O16.00 g O

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Percent Composition and Empirical Formulas

K : ~~ 1.00.63300.6329

Mn : 0.63290.6329

= 1.0

O : ~~ 4.02.5320.6329

nK = 0.6330, nMn = 0.6329, nO = 2.532

KMnO4

Ex:3.11 COMPOUND

Nitrogen 1.52g

Oxygen 3.47g

Mole = :/14

52.1

molg

g217.0:108.0

/16

47.3

molg

g

108.0

108.0

108.0

217.0:

1 : 2

Empirical Formula NO2

Empirical molar mass = 14.01+(16x2) = 46.01g

Smallest whole number ratio

17

201.46

90

..

_

MassMolarEmpirical

massmolar

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

2 mol + 1 mol 2 mol18

Molecular Formula= (NO2)2 = N2O4

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 2NO2INITIAL mole(given) 8 7

Balanced Equation 2mol + 1mol 2 mol

8 mol of “NO” yields…..8 mol of ”NO2” 7 mol of “O2”..yields …14 mol of “NO2”

O2 is ExcessNO is Limiting19

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Limiting Reagent:

2NO + O2 2NO2

NO is the limiting reagent

O2 is the excess reagent

Reactant used up first in the reaction.

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In one process, 124 g of Al are reacted with 601 g of Fe2O32Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

ORg Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al

27.0 g Alx

1 mol Fe2O3

2 mol Alx

160. g Fe2O3

1 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent

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Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102. g Al2O3

1 mol Al2O3

x = 234 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

At this point, all the Al is consumed and Fe2O3 remains in

excess.

3.10 Reaction Yield

%100..

..% x

yeildltheoretica

yeildactualyield

Theoretical yield is the amount of product that wouldresult if all the limiting reagent reacted.

[can be obtained from calculation based on balanced equation.]Actual yield is the amount of product actually obtainedfrom a reaction.

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