Stoichiometry

Unit # 3Stoichiometry:

Calculations with Chemical Formulas and Equations

CHM 1045: General Chemistry and Qualitative Analysis

Dr. Jorge L. AlonsoMiami-Dade College –

Kendall CampusMiami, FL

Textbook Reference: •Module #3 & 4

Stoichiometry

CO2 (g)H2O(g)

O2 (g)

CH4 (g)

Chemical Reaction

The actual phenomenon that occurs when chemical interact with each other.

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

flame

What is happening here?

Stoichiometry

Chemical EquationsSymbolic representations of chemical reactions

Reactants Products

CoefficientsSubscriptsStatesBalanced Chemical Equations represent events that occur at the atomic level which we cannot perceive; but they explain the mass ratio of substances involved in a chemical equation (stoichiometry)

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

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 EquationsWhat 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 2

2

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:

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) Na2O (s) + H2O (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 Br2 (l) 2 AlBr3 (s)

Simple:• Two or more elements react to form one compound

A + B AB

{Mg Movie}

{H2O Movie*}

{AlBr3 Movie*}

{ZnS Movie*}

Now let’s balance equations

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 two compounds exchange partners

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

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

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*}

Stoichiometry

Single Displacement Reactions

A more active element displacing a less active elements from a compound.

MN + M

MN + NActivity (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) + Br2 (aq)

No Reaction

Activity series can also be found in form of Reduction Potential table.

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*}

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 of a 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 be2 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 CompositionPercentage mass of a element (Na) in compound (NaCl):

100 x whole

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 Pb5Au(TeSb)4S5)x 100

Nagyagite Gold Ore:

% Au =(1) ( 197)

( 2,301)x 100

= 8.56 %

Oreg1000Auofg?

Oreg

Aug

100

56.8Aug 5.68

100x NaCl mass

Na mass Na %

Pb5Au(TeSb)4S5

g-MM = 2,301g/

40%100x 58g

23g

Stoichiometry

Using mass % to determine mass of one particular element in a sample

What is the mass of carbon in a 25g sample of carbon dioxide?

(1) What is the percentage mass of carbon in carbon dioxide?

(2) What is the mass of carbon in a 25 g sample of carbon dioxide?

There are two parts to this problem:

% C = 100

whole

part100

CO of mass

C of ass

2

m

100g 44

g 21 % 27

? g C = 25 g CO2

2CO g 001

C g 72g 6.75

x 27%

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 sextillionthousandmillionbilliontrillionquadrillionpentillionsextillion

Stoichiometry

Using Equivalences as Mole Ratios:

MM-g

mole 1

particles 10 x 6.022

mole 123

mole 1

MM-g

mole 1

particles 10 x 6.022 23

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

particles 10 x 6.022 23

g-MM

g-MM

particles 10 x 6.022 23

or

Stoichiometry

Mole Calculations: g-MM Moles # of Particles

mol 0.0548

g 187

NaCl mol 1

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

NaCl mol 1

or

f.u. 10 x 023.6

NaCl mol 123

NaCl mol 1

f.u. 10 x 023.6 23

or

g .44358

NaCl mol 1

NaCl mol 1

g .44358

f.u. 10 x 023.6

NaCl mol 123

NaCl mol 1

f.u. 10 x 023.6 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. 10 x 6.022

g 58.44323

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. 10 x 6.022

mol 123

4.1 x 1012 g

g 58.443

mole 13.3 x 1022 f.u.

4.1 x 1012 g

g 58.443

f.u 10 x 6.022 23

3.3 x 1022 f.u.

mol 1

g 58.443

mole 1

f.u. 10 x 6.022 23

Stoichiometry

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

atoms 15

? atoms = 0.50 mole Fe(CO)3(PH3)2 mole 1

f.u. 10 x 6.022 23

= 4.5 x 1024 atoms

*

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

OgHmolecules 229?

mole

molecules

1

10022.6 23

OgH

mole

218

1 molecules23102.2

OHofmoleculesatoms 223 102.2? atoms 10 x 6.6 23

OHmolecule

atoms

2 1

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 of

Compounds by Elemental Analysis

CxHy

Combustion furnace

Stoichiometry

Types of Formulas

• Empirical formulas give the lowest whole-number ratio of atoms of each element in a compound.

• Molecular formulas give the exact number of atoms of each element in a compound.

Why are empirical formulas needed?

• Structural formulas (skeletal or space-filling) show the order in which atoms are bonded and their three-dimensional shape.

Benzene, C6H6

HO CH H2O

Stoichiometry

Elemental Analyses

Compounds are broken down and 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?

H 4H 1

1

g

mole

O 4O 16

1

g

mole

H 14

O 14

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

Calculating Empirical FormulasAssuming 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 mol12.01 g

1 mol14.01 g

1 mol1.01 g

1 mol16.00 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 mol0.7288 mol5.09 mol

0.7288 mol0.7288 mol0.7288 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

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

CxHy + O2 CO2 + H2Omass

mass of C?

mass of H2O

2CO g 44.011

C g 12.011

OH g 18.0

H g 1.0

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)

Calculating Empirical Formulas

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?

2CO g 44.011

C g 12.011? 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

g 12.011

C mole 1

g 1.011

H mole 1

= 0.332 mol C

= 1.01 mol H

/ 0.332 = 1.00

/ 0.332 = 3.04

Empirical Formula

CH3

5.00 g 14.6 g

(12g:32g=44g)

Stoichiometry

2006 A

?g C =

?g N =

Stoichiometry

Stoichiometry

2003 B

Stoichiometry

Stoichiometry

Stoichiometry

(mass relationships within chemical equations)

Stoichiometry OH mol 2

O mol 1

2

2

O mol 1

H mol 2

2

2

OH mol 2

H mol 2

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

Stoichiometry

Stoichiometric Calculations

mole ratio from balanced equation

MgO g 2.35 O g ? 2

MgO g 2.35 O g ? 2

MgO g 40.304

MgO mol 1

MgO mol 2

O mol 1 2

2

2

O mol 1

O g 32.000 2O g 0.933

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?

Stoichiometry

Stoichiometric Calculations

(1) Starting with 1.00 g of C6H12O6…

we calculate the moles of C6H12O6…

(2) use the coefficients to find the moles of H2O…

(3) and then turn the moles of water to grams

C6H12O6 + 6 O2 6 CO2 + 6 H2O? g H2O

Grams C6H12O6

Moles C6H12O6

Grams H2O

Moles H2O

Balanced Eq. uses MOLE language

(1)

(2)

(3)

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 of one of the ingredients

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

Stoichiometry

Limiting & Excess Reactants• The limiting reactant is the reactant present in the

smallest stoichiometric amount– In other words, it’s the reactant you’ll run out of first

Limiting H2; Excess O2Which is Limiting which is Excess?

Stoichiometry

Limiting & Excess Reactants 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 left

unreacted ?(3) How much MgO will be produced ?

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

Stoichiometry

Limiting & Excess Reactants Problem

? g O2 = 5.0 g Mg

? g Mg = 5.0 g O2

2

2

O mol 1

g 32.0

Mg mol 2

O mol 1

g 24.3

Mg mol 1

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

?g MgO = 5.0 g Mg

Mg mol 1

g 24.3

O mol 1

Mg mol 2

g 32.0

O mol 1

2

2

MgO mol 1

g 40.3

Mg mol 2

MgO mol 2

g 24.3

Mg mol 1MgO g 8.29

= 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

Stoichiometry

Leveling Bulb: to maintain pressure of H2 same as that of atmosphere

Zn+HCl ZnCl2

H2

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

H2O

Experiments:

(1) 0.0025 η + 0.0050 η 61.0 mL

(2) 0.0012 η + 0.0050 η 30.5 mL

(3) 0.0031 η + 0.0050 η 61.0 mL

Limiting Reactant Experiments

{Movie:LimitReac}

Which are the limiting and excess reactants in each expt.?

Stoichiometry

Experiments:

0.0025 g + 0.0050 g

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

Limiting Reactant Experiments

gZngHCl 0025.0?

gHClgZn 0050.0?

Zn35.453g

1 Znmole

Znmole

Hmole

1

Cl 2

Cl 1

Cl 40.66

Hmole

Hg

gHCl0093.0

HCl g 66.40

Cl 1 Hmole

Cl 2

n 1

Hmole

Zmole

n 1

Zn453.35

Zmole

g

gZn0013.0

HCl is Limiting Reactant

Zn is Excess Reactant

How many grams of ZnCl2 are produced in this reaction?

gHClgZnCl 0050.0? 2

HCl g 66.40

Cl 1 Hmole

Cl 2

nCl 1 2

H

Z

2

2

nCl 1

ZnCl233.166

Zmole

g20063.0 gZnCl

Stoichiometry

Theoretical Yieldthe amount of product that can be made as

calculated by stoichiometry.

the amount reaction actually produces (less)

Actual Yield

Percent YieldActual Yield

Theoretical YieldPercent Yield = x 100

gHClgZnCl 0050.0? 2

HCl g 66.40

Cl 1 Hmole

Cl 2

nCl 1 2

H

Z

2

2

nCl 1

ZnCl233.166

Zmole

g20063.0 gZnCl

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

20045.0Balance Using gZnCl

%4.711000063.

0045.

How many grams of ZnCl2 are produced in this reaction?

Stoichiometry

HGeF g 160. 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 HGeF

g 130.58

GeH 1

HGeF 4

g 76.622

GeH

34

34

1

1

HGeF of g 152 0.95 H)GeF g (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

gYield %81

What would be the Theoretical yield if 120 g was a 75% yield ?

100

120%75

whole

g

100

%75

120gwhole g160