Unit.03.Stoichiometry1.Formulas&Equations.lecture

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Stoichiometry

Unit # 3Stoichiometry:

Calculations with ChemicalFormulas and Equations

CHM 1045: General Chemistry and

Qualitative Analysis

Dr. Jorge L. Alonso

Miami-Dade College – Kendall Campus

Miami, FL

Textbook Reference:

•Module #3 & 4



Stoichiometry

CO2 (g)H2O(g)

O2 (g)

CH4 (g)

Chemical

Reaction

The actual

phenomenon that

occurs when chemicalinteract with each other.

Methane gas is mixed with air and thenit is light-up by a spark.

flame

What is happening here?





Stoichiometry

(1) Decomposition:

(2) Combination (Synthesis):

(3) Double Displacement (Replacement) or Metathesis,

Exchange

(4) Single Displacement (Replacement)

(5) Combustion

AB + CD AD + CB where A & C are Metals, B & D Nonmetals

MN + M

MN + N

M

or

N

MN +

AB A + B

A + B AB

: reactions of oxygen with an organic compounds

(hydrocarbons, alcohols) that produce CO2 + H2O and a flame.

C3H8 (g ) + 5 O2 (g ) 3 CO2 (g ) + 4 H2O (g )

Predicting Products: Types of ReactionsWhat happens when substances react?



Stoichiometry

• Sodium + Chlorine

• Dihydrogen Monoxide

• Magnesium + Hydrochloric Acid

• Hydrochloric Acid + Calcium Hydroxide

• Combustion (burning with oxygen) of:

Sucrose (C12H22O11)

Octane

Chemical Equations

What happens when you mix (cause a reaction of) the following?

Write balanced chemical equations for each.

Sodium Chloride

Hydrogen + oxygen

Magnesium Chloride + Hydrogen

Hydrogen hydroxide + Calcium chloride

+ Oxygen Carbon dioxide + water

+ Oxygen Carbon dioxide + water



Stoichiometry

Predicting Products, writing Formulas

and Balancing Equations

Na + Cl2

H2O Mg + HCl

HCl + Ca(OH)2

C12H22O11 + O2

C8H18 + O2

NaCl

H2 + O2 MgCl2 + H2

HOH + CaCl2

CO2 + H2O

CO2 + H2O

2

2 22

2 2

12 11

8 912.52 25 16 18

2

12

*

• Sodium + Chlorine

• Dihydrogen Monoxide

• Magnesium + Hydrochloric Acid

• Hydrochloric Acid + Calcium Hydroxide

Combustion (burning with oxygen) of:Sucrose (C12H22O11)

Octane

Sodium Chloride

Hydrogen + oxygen

Magnesium Chloride + Hydrogen

Hydrogen hydroxide + Calcium chloride

Carbon dioxide + water

Carbon dioxide + water

Alonso’s Rules for BE: (1) Easy element 1st hard

elements last.

(2) One element at a time.

(3) Use fractions when

necessary.



Stoichiometry

Decomposition Reactions

{AirBags Movie*}

sodium azide (N31-) ∆

Simple: Binary compounds break down into their

constituent elements

2H2O 2H2 + O2

2NaCl(l) 2Na (l) + Cl2(g)

2NaN3(s) 2Na(s) + 3N2(g)

electrolysis

electrolysis

heat

2H2O2 2H2O + O2

Catalyst

{Peroxide Movie}

Important Exception:

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/Airbags.MOV




http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/04peroxide%20Decoposition%20with%20Catalyst.MOV

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/04peroxide%20Decoposition%20with%20Catalyst.MOV




Stoichiometry

Decomposition Reactions

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

∆ 2 KClO3 (s) 2 KCl (s) + 3O2 (g )

Chlorates break down to metal chlorides and oxygen

Complex Compounds decompose into simpler compounds

2 NaOH (aq)

Na2

O (s)

+ H2

O (l )

2 H3PO4 (aq) P2O5(g ) + 3H2O (l )

Acids break down to nonmetal oxides and water

Bases break down to metal oxides and water

2HNO3 (aq) N2O5(g ) + H2O (l )

All carbonates break down to metal oxides and carbon dioxide



Stoichiometry

Ammonium carbonate powder is heated strongly

Na2O + CO2 + H2O



Stoichiometry



Stoichiometry

Combination (Synthesis) Reactions

2 Mg (s) + O2 (g ) 2 MgO (s)

Zn (s)

+ S (s)

ZnS (s)

2 H2 (g ) + O2 (g ) 2 H2O (l )

2 Al (s) + 3 Br 2 (l ) 2 AlBr 3 (s)

Simple:

• Two or more elements react to form one compound

A + B AB

{Mg Movie}

{H2O Movie*}

{AlBr 3 Movie*}

{ZnS Movie*}

Now let’s balance equations

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/02.Dalton.Rx.%20Mg%20+%20O%20=%20MgO.MOV

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/FormationofWater.MOV




http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/04AlBr3Formation.MOV



http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/03Zn%20+%20S%20=%20ZnS.MOV

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/03Zn%20+%20S%20=%20ZnS.MOV



http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/02.Dalton.Rx.%20Mg%20+%20O%20=%20MgO.MOV



Stoichiometry

Bromine liquid is poured over aluminum metal

Hydrogen chloride and ammonia gas are mixed together.

Sulfur dioxide gas is bubbled into water.



Stoichiometry



Stoichiometry

Metathesis (Double Displacement)

Example: what quantity of Baking Soda will react with 100mL of vinegar?

NaHCO3 (s) + HC2H3O2 (l ) NaC2H3O2 (aq) + HHCO3 (aq)

H2CO3 (aq) H2O (l ) + CO2 (g ) (2nd Rx decomposition)

• Involve two Compounds

• Elements (or polyatiomic

groups) in the twocompounds exchange

partners

{Movie: Bicarb + Vineg with Stoichio&LimitReag *}

AB + CD AD + CB where A & C are Metals, B & D Nonmetals

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/05Stoichi%20of%20BakingSoda%20+%20Vinegar.MOV






Stoichiometry

Metathesis (Double Displacement):

Acid-Base Neutralization Reaction

Examples:

HCl (aq ) + NaOH (aq ) NaCl (aq ) + HOH (l )

2 HCl (aq ) + Ca(OH)2(aq ) CaCl2 (aq ) + 2 HOH (l )

HCl (aq ) + NH4OH (aq ) NH4Cl (aq ) + 2 HOH (l )

Acid: compound containing hydrogen and a non metal (HN)

Bases: a metal hydroxide (MOH)

HN + MOH MN + HOH Acid + Base Salt + Water

{Movie: A-B RxNo Ind} {Movie: A-B Rx Ind+pH meter*}

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/Copy%20of%2004AcidBaseRx%20NoIdicator.MOV


http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/Copy%20of%2004AcidBaseRx%20withIndicator%20and%20pH%20meter.MOV






Stoichiometry

Single Displacement Reactions

A more active element displacing a less active elementsfrom a compound.

MN + M

MN + N

Activity (Electromotive) Series:

Metals: Li > K > Ba > Sr > Ca > Na > Mg > Al > Zn > Cr > Fe >

Cd > Co > Ni > Sn > Pb > (H) > Sb > Bi > Cu > Hg > Ag > Pd >

Pt > Au

Halogens: F >Cl > Br > I

M

or

N

(Single Replacement Rx.)

MN +

compound

element



Stoichiometry

Single Displacement Reactions

• Examples:

Cu (s) + 2 AgNO3 (aq) Cu (s) + Zn(NO3)2 (aq)

Cl2 (g) + 2 NaBr (aq)

Activity (Electromotive) Series:

Metals: Li > K > Ba > Sr > Ca > Na > Mg > Al > Zn > Cr > Fe >

Cd > Co > Ni > Sn > Pb > (H) > Sb > Bi > Cu > Hg > Ag > Pd >

Pt > Au

Halogens: F >Cl > Br > I

{Movie:Cu+AgNO3}

A more active element displacing a less active

elements from a compound.

2 Ag + Cu(NO3)2 (aq)

2 NaCl (aq) + Br 2 (aq)

No Reaction

Activity series can also be found in

form of Reduction Potential table.

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/04Cu%20+%20AgNO3%20=%20Reaction.MOV






Stoichiometry

Most

Active

Nonmetal

Most

Active

Metal



Stoichiometry

H+OH-



Stoichiometry

Reactions with Oxygen

(2) Combustion Reaction: Rapid reactions of

oxygen with an organic compounds

(hydrocarbons, alcohols) that produce CO2 + H2O

and a flame.

Examples:

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

C3H8 (g ) + 5 O2 (g ) 3 CO2 (g ) + 4 H2O (g )

Oxidation Rx.

Combustion Rx

(1) Oxidation Reactions: are combination

reactions involving oxygen.

{Movie: Mg, Fe, P, S + conc. O2

{Metal Oxides*}

What is the difference?

{Movie: CH3OH + O2*}

Wh i

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/ReactionswithOxygen.MOV


http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/05Methanol%20+%20O2%20=%20Limiting%20Reactant.MOV
















Stoichiometry

*When oxygen is scarce….



Stoichiometry

gram-Molar Mass (g-MM)= Atomic Weigh, Formula

Weigh or Molecular Weight

Mass : Weight (in grams)

Mole = 6.022 x 1023 particles



Stoichiometry

gram-Molar Mass (g-MM): AW, FW, MW

the mass in grams of 1 mole ofa substance (units= g/mol)

For an element we find it on the

periodic table.

For compounds the same as the formula & molecular weight

(but in g/mol)

Example: the g-MM of Al2(SO4)3, would be

2 Al: 2x(26.98 amu) = 53.96

+ 3 S: 3x(32.06 amu) = 96.18

+3x4 O: 12x(16.00 amu) =192.00

342.14 amu (g/mol)

*



Stoichiometry

Formula Weight (FW)

• Sum of the atomic weights for the atoms in a chemical formula unit

(ionic compound)• So, the formula weight of calcium chloride, CaCl2, would be

Ca: 1x(40.1 amu)

2 x Cl: 2x(35.5 amu)

111.1 amu

Molecular Weight (MW)

• Sum of the atomic weights of the atoms in a molecule (covalent

compound)

• For the molecule ethane, C2H6, the molecular weight would be

2 x C: 2x(12.0 amu)

6 x H: 6x(1.0 amu)

30.0 amu



Stoichiometry

(Mass) Percent Composition

Percentage mass of a element (Na) in compound (NaCl):

100 xwhole

part %

Problem: (1) calculate mass % Au in Nagyagite.

(2) If you buy 1 kg of the ore, how much gold

does it have?

% element =

(# atoms of element) (atomic weight of Au)

(MW of Pb5 Au(TeSb)4S5) x 100

Nagyagite Gold Ore:

% Au =(1) ( 197)

( 2,301)

x 100 = 8.56 %

Oreg1000Auof g?

Ore g

Au g

100

56.8 Au g 5.68

100x NaClmass

Namass Na%

Pb5 Au(TeSb)4S5

g-MM = 2,301g/

40%100x58g

23g

http://en.wikipedia.org/wiki/File:Nagyagit_-_Nagyag,_Rum%C3%A4nien.jpg





Stoichiometry

The Mole ConceptDermatological BiologicalChemical

Avogadro's Number:6.022,141,410,704,090,840,990,72 x 1023

602,214,141,070,409,084,099,072 .

602 sextillion

thousandmillionbilliontrillionquadrillionpentillionsextillion

http://www.photostogo.com/store/Chubby.asp?ImageNumber=1027539

http://en.wikipedia.org/wiki/Image:Mole_(body_part).jpg



Stoichiometry

Using Equivalences as Mole Ratios:

MM-g

mole1

particles10x6.022

mole123

mole1

MM-g

mole1

particles10x6.02223

or

g-MM = 1 Mole ( ) = 6.022 x 10 23 particles

NaCl = 58g/η (Atoms or molecules)

From Equivalences we obtain useful Ratios or Conversion factors:

or

particles10 x6.02223

g-MM

g-MM

particles10 x6.02223

or

*



Stoichiometry

Mole Calculations:

g-MM Moles # of Particles

mol0.0548

g187

NaClmol1 g.44358

g-MM Moles:

Moles # of Particles:

? g = 3.20 mol of NaCl

? mol = 3.20 g of NaCl

? f.u. = 3.2 mol NaCl

? mol = 3.2 x 10 52 f.u. NaCl

g.44358

NaClmol1

or

f.u.10x023.6

NaClmol1

23

NaClmol1

f.u.10x023.6 23

or

g.44358

NaClmol1

NaClmol1 g.44358

f.u.10x023.6

NaClmol1

23

NaClmol1

f.u.10x023.6 23

*

Which ratios will you need?

= 5.3 x 10 28 mol

= 1.9 x 10 24 f.u.



Stoichiometry

Mole Calculations:

g-MM Moles # of Particles

f.u.10x6.022

g58.44323

g-MM # of Particles:

? g = 4.2 x 10 34 f. u. of NaCl

? f. u. of NaCl = 3.2 g of NaCl

f.u.10x6.022mol1

23

4.1 x 1012 g

g58.443

mole13.3 x 1022 f.u.

4.1 x 1012 g

g58.443

f.u10x6.02223

3.3 x 1022 f.u.

mol1g58.443

mole1

f.u.10x6.02223

*M l C l l i



Stoichiometry

233 )(PHFe(CO)of f.u.1

atoms15

? atoms = 0.50 mole Fe(CO)3(PH3)2

mole1

f.u.10x6.02223

= 4.5 x 1024 atoms

*

How many molecules of H2O in 29g of water? How many atoms?

O gH molecules 229?

mole

molecules

1

10022.623

O gH

mole

218

1molecules

23102.2

O H of moleculesatoms 223 102.2? atoms10x6.6

23

O H moleculeatoms

21 3

Mole Calculations:

g-MM Moles # of Particles(atoms or molecules)

1+ 3+ 3+ 2+ 6 =

= 3.0 X 1023 f.u.

15



Stoichiometry

The Determination of

Empirical Formulas ofCompounds by

Elemental Analysis

CxHy

Combustion

furnace

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/05Combustion%20&%20Empirical%20Formula.MOV









Stoichiometry

Types of Formulas

• Empirical formulas give the lowestwhole-number ratio of atoms of eachelement in a compound.

• Molecular formulas give the exactnumber of atoms of each element in acompound.

Why are empirical formulas needed?

• Structural formulas (skeletal or space-

filling) show the order in which atoms arebonded and their three-dimensional shape.

Benzene, C6H6

HO CH H2O



Stoichiometry

Elemental Analyses

Compounds are broken downand the masses of their constituent

elements are measured. From

these masses the empirical

formulas can be determined.

Expt. Data: (68g) 4g H

64g O

EmpF

HO

How do we determine the formula of a compound?

H4H1

1

g

mole

O4O16

1

g

mole

H14

O14

Mole Ratio

HxOy

moles

*



Stoichiometry

Calculating Empirical Formulas

Calculate the empirical formula (mole ratio) from

the percent composition (% mass).

Problem:The compound para-aminobenzoic acid (you may have seen it

listed as PABA on your bottle of sunscreen) is composed of

carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and

oxygen (23.33%). Find the empirical formula of PABA.

*

carbon (61.31%),hydrogen (5.14%),

nitrogen (10.21%),

oxygen (23.33%).

carbon (61.31g),hydrogen (5.14g),

nitrogen (10.21g),

oxygen (23.33g)

Percent means out of 100, so assume a

100g sample of the compound, then….



Stoichiometry


Assuming 100.00 g of para-aminobenzoic acid,

? mol C = 61.31 g x = 5.105 mol C

? mol H = 5.14 g x = 5.09 mol H

? mol N = 10.21 g x = 0.7288 mol N

? mol O = 23.33 g x = 1.456 mol O

1 mol

12.01 g

1 mol

14.01 g

1 mol

1.01 g

1 mol

16.00 g

What is the smallest mole ratio of the elements in this compound?

= 7.005 7

= 6.984 7

= 1.000

= 2.001 2

5.105 mol

0.7288 mol

5.09 mol

0.7288 mol0.7288 mol

0.7288 mol1.458 mol

0.7288 mol

Calculate the mole ratio by dividing by the smallest number of moles.

These are the subscripts for the empirical formula: C7H7NO2

61% C

5% H

10% N

23% O

Combustion Analysis: i th d f



Stoichiometry

Combustion Analysis: is a method of

experimentally determining empirical formulas

• Compounds containing C, H and O are routinely analyzed through combustion in a

chamber like this

– C is determined from the mass of CO2 produced

– H is determined from the mass of H2O produced

– O is determined by difference after the C and H have been determined

{Movie}CxHy

Combustion

furnace

Magnesium

perchlorate

Sodium

hydroxide

Cx

Hy

+ O2

CO2

+ H2

O

mass

mass of C?

mass of H2O

2COg44.011

Cg12.011

OHg18.0

Hg1.0

2

mass of CO2

mass of H?

mass of H2Omass of CO2

How do you calculate the mass of C

in CO2 and that of H in H2O?














Stoichiometry

CxHy (g ) + O2 (g ) CO2 (g ) + H2O (g )


A 5.00 g sample of an unknown hydrocarbon was burned and produced

14.6 g of CO2. What is the empirical formula of the unknown

compound?

2COg44.011

Cg12.011? g C = 14.6 g CO2 = 3.98 g C

g H = 5.00 CxHy – 3.98 g C = 1.02 g H

? mol C = 3.98 g C

? mol H = 1.02 g H

g12.011

Cmole1

g1.011

Hmole1

= 0.332 mol C

= 1.01 mol H

/ 0.332 = 1.00

/ 0.332 = 3.04

EmpiricalFormula

CH3

5.00 g 14.6 g

(12g:32g=44g)

2006 A



Stoichiometry

2006 A

?g C =

?g N =



Stoichiometry

2003 B



Stoichiometry

2003 B



Stoichiometry



Stoichiometric Calculations



Stoichiometry

OHmol2

Omol1

2

2

Omol1

Hmol2

2

2

OHmol2

Hmol2

2

2

Mole Ratios from Balanced Equation:

The coefficients in the balanced equation can also be

interpreted as mole ratios of reactants and products

Stoichiometric Calculations2 2

Stoichiometric Calculations *



Stoichiometry

Stoichiometric Calculations

mole ratio from balanced equation

MgOg2.35 O g? 2

MgOg2.35 O g? 2

MgOg40.304

MgOmol1

MgOmol2

Omol1 2

2

2

Omol1

Og32.000 2Og0.933

2 Mg (s ) + O2 (g ) 2 MgO (s )

2 Mg (s ) + O2 (g ) 2 MgO (s )

gramsNo direct

calculation

grams

Change:1. grams of MgO mol MgO

2. mol of MgO mol O2

3. mol of O2 grams of O2

How many grams of O2 are

required to form 2.35 g of MgO?



St i hi t Li iti R t t



Stoichiometry

In this example the sugar would be the limiting reactant,

because it will limit the amount of cookies you can make.

{ MovieLimitingReactants: Zn + 2 HCl ZnCl2 + H2 }

Stoichiometry: Limiting Reactants

Make cookies until you run out ofone of the ingredients

(or, too much of one reactant and not enough of the other)

http://localhost/var/www/apps/conversion/tmp/Unit%203%20Media/05Zn%20+%20HCl%20=%20LimitReagent.MOV












Limiting & Excess Reactants



Stoichiometry


Problem

If 5.0g of both Mg and O2 are used:(1) Which is the limiting and the excess

reactants?

(2) How much of the excess will be leftunreacted ?

(3) How much MgO will be produced ?

2 Mg (s ) + O2 (g ) 2 MgO (s )




Stoichiometry


Problem

? g O2 = 5.0 g Mg

? g Mg = 5.0 g O2

2

2

Omol1

g32.0

Mgmol2

Omol1

g24.3

Mgmol1

2 Mg (s ) + O2 (g ) 2 MgO (s )

?g MgO = 5.0 g Mg

Mgmol1

g24.3

Omol1

Mgmol2

g32.0

Omol1

2

2

MgOmol1

g40.3

Mgmol2

MgOmol2

g24.3

Mgmol1

MgOg8.29

= 3.29 g O2

= 7.59 g Mg

Mg is limiting reactant and O2 is excess reactant

Which one do I use to determine the MgO produced by rx?

5 g 5 g



Limiting Reactant Experiments



Stoichiometry

Experiments:

0.0025 g + 0.0050 g

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

Limiting Reactant Experiments

gZn gHCl 0025.0?

gHCl gZn 0050.0?

Zn35.453g

1 Znmole

Znmole

H mole

1

Cl2

Cl1

Cl40.66

H mole

H g

gHCl 0093.0

HClg66.40

Cl1 H mole

Cl2

n1

H mole

Z mole

n1

Zn453.35

Z mole

g

gZn0013.0

HCl is Limiting Reactant

Zn is Excess Reactant

How many grams of ZnCl2 are produced in this reaction?

gHCl gZnCl 0050.0? 2

HClg66.40

Cl1 H mole

Cl2

nCl1 2

H

Z

2

2

nCl1

ZnCl233.166

Z mole

g 20063.0 gZnCl

How many grams of ZnCl2 are produced in this reaction?



Stoichiometry

Theoretical Yield

the amount of product that can be made as

calculated by stoichiometry.

the amount reaction actually produces (less)

Actual Yield

Percent Yield

Actual Yield

Theoretical YieldPercent Yield = x 100

gHCl gZnCl 0050.0? 2

HClg66.40

Cl1 H mole

Cl2

nCl1 2

H

Z

2

2

nCl1

ZnCl233.166

Z mole

g 20063.0 gZnCl

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

20045.0BalanceUsing gZnCl

%4.711000063.

0045.

*



Stoichiometry

HGeFg160. 3

The following reaction has a 95% yield:

GeH4 + 3GeF4

4GeF3Hg-MM: 76.622 148.5756 130.58

Problem:

How many grams of the product are formed, when 23.4 g of GeH4 are

reacted with excess GeF4?

?g GeF3H = 23.4 g GeH4 H GeF

g 130.58

GeH 1

H GeF 4

g 76.622

GeH

34

34

1

1

HGeFof g152 0.95H)GeFg(160. 33

Since % yield is only 95%, then actual yield is:

What would the yield be if reaction Actually produced 130g only?

100160

130%

g

g Yield %81

100

120%75

whole

g

100

%75

120 g whole g 160

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Unit.03.Stoichiometry1.Formulas&Equations.lecture

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