Chapter 3 Jan12

7/29/2019 Chapter 3 Jan12

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CHAPTER 3

STOICHIOMETRY:

CALCULATIONS OF CHEMICAL

FORMULAS AND EQUATIONS


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CONTENTS

3.1 Chemical Equations

3.2 Chemical Reactivity

3.3 Atomic and Molecular Weights

3.4 The Mole

3.5 Empirical Formulas from Analyses

3.6 Quantitative Information from Balanced

Equation

3.7 Limiting Reactants


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Learning Outcomes Able to calculate the molecular weight of any

given compound

Able to calculate the amount of reactants orproducts in mass or mole with informationgiven.

To be able to identify the limiting reactant in

a given reaction Able to determine the empirical & molecular

formula of a compound from an analysis


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Introduction

Stoichiometry study of quantitativerelationship between chemical formulas

and chemical equations. The law of conservation of mass is

observed: total mass of all substances

present before the reaction (reactants) isthe same as the total mass after the

reaction (products).


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3.1 Chemical Equations

Describe chemical reactions.

Eg: 2H2 + O22H2O

This chemical equation shows that hydrogen

react with oxygen to form water.

H2 and O2 (written to the left of the arrow) are

the reactants.

H2O (written to the right of the arrow) is the

product.


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Cont: 3.1 Chemical Equations

There are 2 types of number in a chemical

equation:

i. Stoichiometric coefficient - number infront of a chemical formula.

ii. Subscripts number of atoms of eachelement present in the chemical formula.


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Cont: 3.1 Chemical Equations

E.g:

2H2O:

2 water molecules present.

H2O - there are 2H atoms in one

molecule of water.


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3.1.1 Balancing ChemicalEquations

Formulas of the substances must be

correctly written.

The number of atoms of each type ofelement must be the same on both sides

of the arrow.

Only stoichiometric coefficients may beadjusted. Subscripts in chemical formulas

must not be changed.


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Cont: 3.1.1 Balancing Chemical

Equations

The sum of charges of ions on the left

hand side of the arrow must be the same

as on the right side.

Balancing chemical equation requires

some trial and error.


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Equations

Eg. methane reacting with oxygen.

CH4 + O2 CO2 + H2O : this is unbalanced.

o Firstly, balance the carbon and hydrogenatom without considering oxygen.

CH4 + O2 CO2 + 2H2O

o Secondly, balance the oxygen atom.CH4 + 2O2 CO2 + 2H2O


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Equations

Finally, the physical state of each chemical in

chemical equation should be written down.

Use the symbols: (g) for gas

(l) for liquid

(s) for solid (aq) for aqueous (water) solution

Eg: CH4(g) + O2(g) CO2(g) + 2H2O(l)


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Exercise 3.1

Balance the following equations:

a) HNO3(aq) + Na2CO3 (s) NaNO3 (aq) + H2O (l)+ CO2(g)

b) HCl (aq) + Al(OH)3(aq) H2O(l) + AlCl3 (aq)

c) ZnS(s) + O2(g) ZnO(s) + SO2(g)


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3.2 Chemical Reactivity

3.2.1 Using the Periodic Table

The pattern of chemical reactivity of elements

varies systematically, using the Periodic Table. Eg. All alkali metal react with water as follows:

2M(s) + 2H2O(l) 2MOH(s) + H2(g)

M= alkali metal

The reaction becomes more vigorous as wemove down the Periodic Table, i.e. from Li toCs.


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3.2.2 Combustion Reaction in Air

Combustion reactions produce a flame.

Most of the combustion reactions

involve oxygen as a reactant. Complete combustion of hydrocarbon

compounds and compounds containing C,

H and O atoms produces CO2 and H2O.Eg: C3H8(g) + 5O2 (g) 3CO2 (g) + H2O(l)


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Cont: 3.2.2 Combustion Reaction in Air

Incomplete combustion occurs when there is

insufficient amount of O2 present.

In this case, CO is produced instead of CO2.

Eg: C3H8 (g) + 7/2 O2 (g) 3CO (g) + 4H2O (l)


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3.2.3 Combination andDecomposition Reactions

i. Combination reaction

Combination reaction: reaction in which

two or more substances combine to formone product.

Represented by: A + B CEg. 2 Mg(s) + O2(g) 2 MgO (s)


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Cont: 3.2.3 Combination and

Decomposition Reactions

ii. Decomposition reaction

Breakdown of a compound into two or

more components.

May be represented by: C A + B

Eg. CaCO3(s) CaO (s) + CO2(g)


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3.3 Atomic Weight and MolecularWeight

3.3.1 Relative Masses and Atomic Mass Unit

Atoms are too small to be weighed. Atomic mass unit (amu) is used in dealing with

these extremely small masses:

1 amu = 1.66054 10-24 g

1 g = 6.02214 1023 amu The atomic mass of any isotope is then

determined relative to 12C nuclide.


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3.3.2 Average Atomic Masses

Most elements occur in nature as a

mixture of isotopes.

For example, naturally occurring chlorineis 75.53 % 35Cl (atomic mass of 34.969

amu) and 24.47 % 37Cl (atomic mass

36.966 amu). The average of the masses of isotopes will

give the relative atomic masses.


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Cont: 3.3.2 Average Atomic

Masses

Average Atomic Mass ( atomic weight)

= [(isotope mass) (fractional isotope abundance)]

for all isotopes of the element. Eg:

Average atomic mass of chlorine:

= (34.969 amu)(0.7553) + (36.966amu)(0.2447)= 26.41 amu + 9.05 amu

= 35.46 amu


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3.3.3 Formula and MolecularWeight

Formula weights (FW) are the sum of atomic

weights for the atoms in the formula.

Eg Sulfuric Acid H2SO4

FW (H2SO4)

= 2AW (H) + AW (S) + 4AW (O)= [2(1.0)+ (32.0) + 4(16.0)] amu

= 98.0 amu


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Cont: 3.3.3 Formula and Molecular

Weight

Molecular weight (MW) is used in place of formula

weight if the chemical formula of a substance is its

molecular formula. E.g. MW of glucose (C6H12O6):

= 6AW (12.0 amu) + 12AW (1.0 amu) + 6AW (16.0

amu)= 72.0 amu + 12.0 amu + 96.0 amu

= 180.0 amu


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Cont: 3.3.3 Formula and Molecular

Weight

For ionic substances like NaCl which

exists as 3D array of ions, it is not a

molecule. Therefore we cannot write it asmolecular formula or molecular weight

(MW).

The formula weight of NaCl:

FW (NaCl) = 23.0 amu + 35.5 amu

= 58.5 amu


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3.3.4 Percentage Compositionfrom Formulas

The percent composition of a substance isthe percent by mass of each element in thesubstance.

The sum of all percentage of elements in asubstance must be 100 %.

% mass of an element:= mass of an element in substance 100

formula weight of substance


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Example 1

Calculate the percentage composition ofC12H22O11 (FW 342).

The percentage of a given element in a

compound:

= (atom of element) (AW)

100FW of compound


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Cont: Example 1

% C = 12 (12.0 amu) 100 = 42.1 %

342 amu

% H = 22 (1.0 amu) 100 = 6.4 %

342 amu

% O = 11 (16.0 amu) 100 = 51.5 %

342 amu

Total % = (42.1 + 6.4 + 51.5)% = 100 %


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3.4 The Mole

The unit for dealing with atoms, ions, and molecules is themole (mol).

A mole is defined as the amount of matter that contains the

same amount of particles (molecules, atoms, ions) as thenumber of atoms in exactly 12 g of12C.

In 12 g of12C, there are 6.022 1023 atoms which is calledthe Avogadros number.

1 mole = 6.022

10

23

atoms Thus a mole of water contains 6.022 1023 molecules of

water.

A mole of NaCl contains 6.022 1023 sodium ions (Na+)

and 6.022

1023

chloride ions (Cl-

).


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Example 2

How many oxygen atoms in 0.25 mol of Ca(NO3)2?

There are 6 O atoms in 1 molecule of Ca(NO3)2. 0.25 mol Ca(NO3)2 6.022 1023 molecule Ca(NO3)2

1 molCa(NO3)26 Oxygen atom

1 molecule Ca(NO3)2

= 9.0 1023 Oxygen atoms in 0.25 mole of Ca(NO3)2


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3.4.1 Molar Mass

Molar mass (MM) of a substances is themass (in grams) of 1 mole of that substance.

Unit g/mol (g mol-1

). The mass of a single atom of an element (in

amu) is numerically equal to the mass (ingrams) of 1 mol of atoms of that element.

E.g: One 12C atom weighs 12.0 amu

One mol 12C weighs 12.0 g


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Cont: 3.4.1 Molar Mass

E.g:

One 24Mg atom weighs 24.0 amu

One mol 24Mg weighs 24.0 g

The molar mass (in grams) of any substance

is always numerically equal to its formulaweight, FW (in amu).


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Cont: 3.4.1 Molar Mass

One H2O molecule weighs 18.0 amu

One H2O mol weighs 18.0 g

One NO3-

ion weighs 62.0 amu One mol NO3

- ion weighs 62.0 g

One Cl2 molecule weighs 70.90 amu (35.45 amu 2)

One mol Cl2 weighs 70.90 g

One BaCl2 formula weighs 208 amu

One mol BaCl2 weighs 208 g


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Formula Formula

weight

(amu)

Mass of 1

mol of

formulaunits (g)

Number and kind of

particles in 1 mol

N 14.0 14.0 6.022 1023

atoms

N2 28.0 28.0 6.022 1023

molecules

2 (6.022 1023

)atoms

BaCl2 208.2 208.2 6.022 1023

BaCl2 unit

6.022

10

23

Ba

+

ions2 (6.022 10

23) Cl

-ions


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3.4.2 Interconverting Masses,Moles, and Numbers of Particles

Use dimensional analysis

Units involve are:

mass : g

moles : mol

molar mass : g/mol

number of particles: 6.022 1023 mol-1

(Avogadros number)


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Cont: 3.4.2 Interconverting Masses,

Moles, and Numbers of Particles

a) To convert between gramsand moles,use molar mass.

mol g: mol x g / mol = gg mol: g g/mol = mol

b) To convert between molesandmolecules, use Avogadros number.

moles Avogadros number = molecules


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Cont: 3.4.2 Interconverting Masses,

Moles, and Numbers of Particles

use use Ions/Grams Moles Molecules/

Molar Avogadros AtomsMass Number


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Example 3

1. How many atoms in 1 mole N2 and O3

2. How many molecules in 1 mole N2 and O3

Answers

1. Atoms : 2(6.022 1023 atoms) N.

: 3(6.022 1023 atoms) O.2. Molecules: 6.022 1023 molecules N2.

: 6.022 1023 molecules O3


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Example 4

Ethanol C2H6O (C:12.01, H:1.01, O:16.00)

a) What is its molecular weight?

b) What is the mass of 1 mol of ethanol molecules?

c) Calculate the number of moles of ethanol in 1.00 g.

d) Calculate the number of molecules in 1.00 g of

ethanol.e) Calculate the percentage of carbon in one molecule

of ethanol.


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Example 4 (Answer)

a) Molecular Weight (MW) of ethanol, C2H6O:

b) Therefore 1 mol of C2H6O = 46.08 g

Hatom

amuHatomsCatom

amuCatoms

1

01.1

61

01.12

2

Oatom

amuOatom

1

00.161

amu08.46


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Example 4 (Answer)

c) You need a conversion factor that will change

1.00g of ethanol to moles.

Conversion factor is 1 mol ethanol

46.08 g ethanol

Moles ethanol = 1.00 g ethanol 1 mol ethanol

46.08 g ethanol

= 0.0217 mol ethanol


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Example 4 (Answer)

or use simple direct proportionality

46.08 g ethanol equals to 1 mol ethanol1.00 g ethanol equals to 1.00 g 1 mol ethanol

46.08 g

= 0.0217 mol ethanol


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Example 4 (Answer)

d) The conversion from:

grams moles molecules

46.08 g ethanol = 1 mol ethanol= 6.022 1023 molecules of ethanol

Molecule of ethanol in 1.00 g= 1.00 g ethanol 6.022 1023 molecules ethanol

46.08 g ethanol

= 1.31 1022 molecules ethanol


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Example 4 (Answer)

e) The percentage of element carbon, % C :

= (no of C atoms per molecule) (AW of C) 100MW of one C2H6O molecule

= (2 atom C/1 molecule C2

H6

O)(12.01 amu/1 atom C) 100

46.08 amu/1 molecule

= 52.13 %


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3.5 Empirical Formulas FromAnalyses

Empirical formula is the relative numberof atom in the molecule. Empirical formula

gives the ratioof atoms of each type ofcompound.

E.g: the empirical formula of C2H4 is CH2.

An empirical formula is determined fromexperimental percent composition data.


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Cont: 3.5 Empirical Formulas

From Analyses

a) If we have the mass % of the elements, it iseasier if we assume that we start with 100 g

of sample.

b) Translate the mass % as the no. of grams of

each element in 100 g of sample.

c) From these masses, the number of moles

can be calculated.

The subscripts in an empirical formula are

calculated as follows:


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Cont: 3.5 Empirical Formulas

From Analyses

d) Determine the simplest whole-number ratio of

atoms by dividing the number of moles of each

element by the smallest number of moles.

e) If the ratios are not whole numbers, multiply the

ratios by an integer that clears the denominators

of the fractions

e.g: 0.5 and 1.75 are expressed as fractions: 1/2and 7/4

Multiply by 4 1/2 4 and 7/4 4, the ratios are converted to whole

numbers 2 and 7, respectively.


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Example 5

Ascorbic acid contains 40.92 % C, 4.58 % H

and 54.50% O by mass. What is the empirical

formula of the ascorbic acid?


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Example 5 (Answer)

i. Assume we have 100 g of ascorbic acid. Wehave 40.92 g of C, 4.58 g H and 54.50 g of O.

ii. Calculate the number of moles:Moles C= (40.92 g C) 1 mol C = 3.407 mol

12.01 g C

Moles H= (4.58 g H) 1 mol H = 4.544 mol

1.008 g H


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Example 5 (Answer)

Moles O

= (54.50 g O) 1 mol O = 3.406 mol

16.0 g Oiii. Determine the simplest whole-number.

The smallest number of moles = 3.406

C = 3.407 =1

H = 4.544 =1.33

3.406 3.406

O = 3.406 = 1

3.406


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Example 5 (Answer)

The ratio of H is 1.33 which is equal to 4/3.

Multiply the simplest whole number by 3.

C : H : O = 3(1 : 1.33 : 1)

= 3 : 4 : 3

The whole-number ratio gives us the subscripts

for the empirical formula.Therefore, the empirical of ascorbic acid

is C3H

4O

3.


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3.5.1 Molecular Formula fromEmpirical Formula

Molecular formula is the actual ratio ofelements in the molecule.

The empirical formula may not be the molecular

formula. Eg The empirical formula of ascorbic acid is

C3H4O3 and its molecular formula is C6H8O6. To obtain the molecular formula from the

empirical formula, we need to know themolecular weight, MW.

The ratio of the molecular weight (MW) to

formula weight (FW) must be a whole number.


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Example 6

A compound contains only aluminium

and oxygen elements. Its elemental

composition is determined to be 53.0 %

aluminium and 47.0 % oxygen. The mass

of one mole of the compound is 102 g.

What is the molecular formula?


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Example 6 (Answer)

First determine the empirical formula.

Assume 100 g of compound:

Al = 53.0 g ; O = 47.0 g

Moles Al = (53.0 g Al) 1 mol Al = 1.96 mol Al

27.0 g Al

Moles O = (47.0 g O) 1 mol O = 2.94 mol O

16.0 g O


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Example 6 (Answer)

Divide by 1.96:

Al = 1.96 = 1.00O = 2.94 = 1.50

1.96 1.96

Al 1.00 O 1.50

Multiply by 2 to convert 1.50 into an integer.

The empirical formula is Al2O3


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Example 6 (Answer)

Calculate the mass of Al2O3 to be 102 g.

Thus the empirical and molecular formulaare identical, Al2O3.

MW = 1

FW


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Example 7

Given the empirical formula is C3H4. Fromexperiment, the molecular weight is 121 amu.

What is the molecular formula?Solution:

The formula weight of C3H4 = 40.0 amu

Thus MW = 121 = 3

FW 40.0

The molecular formula is C9H

12.


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3.5.2 Combustion Analysis

i. Empirical formulas are routinelydetermined by combustion analysis.

ii. Hydrocarbon compounds and compoundscontaining C, H and O are combusted inexcess oxygen to produce CO2 and H2O.

All the carbon is converted to CO2All the hydrogen is converted to H2O


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Cont: 3.5.2 Combustion Analysis

O2 sample CuO H2O CO2

furnace absorber absorber

(Mg(ClO4

)2

) NaOH

iii. The amount of CO2 produced can bemeasured by determining the mass

increase in the CO2 absorber. This givesthe amount of C originally present in thesample.(1 mol CO2 contains 1 mol C).


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Cont: 3.5.2 Combustion Analysis

iv. The amount of H2O produced can bemeasured by determining the mass

increase in the H2O absorber. This givesthe amount of H originally present in thesample. (1mol H2O contains 2 mol H).

v. The amount of O originally present in the

sample is given by the difference in theamount of sampleand the amount of Cand Haccounted for.


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Example 8

Isopropyl alcohol contains C, H and O.Combustion of 0.255 g of isopropyl alcohol

produces 0.561 g CO2 and 0.306 g H2O.

1. Calculate the quantities of C, H and O in the

sample.

2. The empirical formula of isopropyl alcohol.


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Example 8 (Answer)

a) No of grams of C and H in 0.561 g of CO2 and0.306 g of H2O.

CgCmol

Cg

COmol

Cmol

COg

COmolCOg 153.0

1

0.12

1

1

0.44

1561.0

22

2

2

HgHmol

Hg

OHmol

Hmol

OHg

OHmolOHg 034.0

1

0.1

1

2

0.18

1306.0

22

2

2


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Example 8 (Answer)

oxygen amount = 0.255 g - (0.153 g + 0.034 g

= 0.068 g O

Empirical Formula:

Calculate the moles of each element first:

moles C = 0.153 g C 1 mol C = 0.01275 moles C

12.0 g C


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Example 8 (Answer)

moles H = 0.034 g H 1 mol H = 0.034 moles H

1.0 g H

moles O = 0.068 g O 1 mol O = 0.00425 moles O

16.0 g O

Divide each no by the smallest no. ie 0.00425

The ratio of H : O : C

= 8 : 1 : 3

Therefore the empirical formula is C3H

8O

3 6 Q tit ti I f ti f


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3.6 Quantitative Information fromBalanced Equations

Objective: To calculate the mass of aparticular substance produced or used in a

chemical reaction.

A balance chemical equation gives the

number of molecules which react to formproduct(s).

C t 3 6 Q tit ti I f ti


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Cont: 3.6 Quantitative Information

from Balanced Equations

This can be interpreted as the ratio of number

of moles of reactant required to give the ratio

of number of moles of product.

e.g. consider the following balanced equation:

CH4(g) + 2O2(g) CO2(g) + 2H2O (l)



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On the atomic-molecular level, the equation

states:

1 molecule CH4 + 2 molecules O2 give 1molecule CO2 + 2 molecules H2O.

The numerical coefficients in a balanced

chemical equation can be interpreted both as

the relative numbers of molecules and as therelative numbers of moles.



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Thus:

1 mol CH4 + 2 mol O2 1 mol CO2 + 2 mol H2O

From the equation, we know that 2 mol of O2 is

required to react with 1 mol of CH4.

This can be represented as

1 mol CH4 2 mol O2

means stoichiometrically equivalent to.



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CH4(g) + 2O2(g) CO2(g) + 2H2O (l)

Therefore, the stoichiometrically equivalentratio for the products are as follows:

1 mol CH4 2 mol H2O

1 mol CH4 1 mol CO2

2 mol O2 1 mol CO2

1 mol CH4 2 mol H2O

2 mol O2 2 mol H2O



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CH4(g) + 2O2(g) CO2(g) + 2H2O (l)

These stoichiometric relations can be

converted to mass equivalences by

converting the mole of a substance to its

mass using molar mass.

HOW?


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Example 9

Combustion of butane C4H10 is given bythe equation:

2C4H10(l) + 13O2(g) 8CO2(g) + 10H2O (l)

Calculate the mass of CO2 that is producedin burning 1.00 g of C4H10.


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Example 9 (Answer)

Ensure that the equation is balanced!!!

2C4H10(l) + 13O2(g) 8CO2(g) + 10H2O (l)

Write the stoichiometrically equivalent ratio

between C4H10 and CO2.

2 mol C4H10 8 mol CO2


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Example 9 (Answer)

104

104

104104

0.58

100.1

HgC

HCmolHCgHCmoles

104

21072.1 HCmol

104

2

104

2

2 2

81072.1

HCmol

COmol

HCmolCOmoles

2

21088.6 COmol


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Example 9 (Answer)

Mass of CO2 in grams:

Thus the conversion sequence is:gram moles moles gramreactant reactant product product

producedCOg

COmol

COgCOmol

2

2

2

2

2

03.3

1

0.44

1088.6


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Exercise 3.2

Determine how many grams of HI are requiredto form 1.20 moles of H2 when HI reacts

according to the equation:HI(g) H2(g) + I2(g)

(1 mol HI = 127.91 g HI)

g984.306HImol1

HIg91.127x

Hmol1

HImol2x)Hmol20.1(

22

Answer: Balance the equation first

= 307 g


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3.7 Limiting Reactants

Limiting Reactant - The reactant in a chemicalreaction that limits the amount of product(s)that can be formed. The reaction will stop

when all of the limiting reactant is consumed. Excess Reactant - The reactant in a chemical

reaction that remains when a reaction stopswhen the limiting reactant is completelyconsumed. The excess reactant remainsbecause there is nothing with which it canreact.


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Cont: 3.7 Limiting Reactants

No matter how many tires there are, if there areonly 8 car bodies, then only 8 cars can be made.

Likewise with chemistry, if there is only a certainamount of one reactant available for a reaction,the reaction must stop when that reactant isconsumed whether or not the other reactant hasbeen used up.


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Example 10

A 2.00 g sample of ammonia is mixed with

4.00 g of oxygen. Which is the limitingreactant and how much excess reactant

remains after the reaction has stopped?


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Example 10 (Answer)

First, we need to create a balanced equation for

the reaction:

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) Next we can use stoichiometry to calculate how

much product is produced by each reactant.

NOTE: It does not matter which product is

chosen, but the same product must be used forboth reactants so that the amounts can be

compared.


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Example 10 (Answer)

The reactant that produces the lesseramount of product: in this case theoxygen.


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Example 10 (Answer)

Next, to find the amount of excessreactant, we must calculate how much of

the non-limiting reactant (ammonia)actually did react with the limitingreactant (oxygen).


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Example 10 (Answer)

We're not finished yet though. 1.70 g is theamount of ammonia that reacted, not what isleft over. To find the amount of excess

reactant remaining, subtract the amount thatreacted from the amount in the originalsample.

2.00g NH3 (original sample) - 1.70g (reacted)

= 0.30g NH3 remaining


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3.7.1 Theoretical Yield

The quantity of product that is calculated to

form when all the limiting reactant reacts is

called the theoretical yield. The amount of product actually obtained in a

reaction is called the actual yield.

The percent yield of a reaction:Percent yield = actual yield 100

theoretical yield


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Exercise 3.3

Calculate the % yield of Al2(SO4) from the reaction:

2Al(OH)3(s) + 3H2SO4(aq) Al2(SO4)3(s) + 6H2O(l)

given that 205 g of Al(OH)3 react with 751 g of

H2SO4 to yield 252 g Al2(SO4)3.

Answer: % yield Al2(SO4) =56.0%


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END of CHAPTER 3


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END of CHAPTER 3

Chapter 3 Jan12

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