Unit 3 stoichiometry calculations with chemical formulas and equations
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CHM 1045 : General Chemistry and Qualitative Analysis. Unit # 3 Stoichiometry: Calculations with Chemical Formulas and Equations. Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL. Textbook Reference : Module #3 & 4. Chemical Reaction. H 2 O (g). CO 2 (g).

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Unit # 3 Stoichiometry: Calculations with Chemical Formulas and Equations

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Unit 3 stoichiometry calculations with chemical formulas and equations

CHM 1045: General Chemistry and Qualitative Analysis

Unit # 3Stoichiometry:Calculations with Chemical Formulas and Equations

Dr. Jorge L. Alonso

Miami-Dade College – Kendall Campus

Miami, FL

  • Textbook Reference:

  • Module #3 & 4


Unit 3 stoichiometry calculations with chemical formulas and equations

Chemical Reaction

H2O(g)

CO2 (g)

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

flame

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

O2 (g)

CH4 (g)

What is happening here?


Chemical equations

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

Chemical Equations

Reactants

Products

Symbolic representations of chemical reactions

States

Coefficients

Subscripts

Balanced 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)


Unit 3 stoichiometry calculations with chemical formulas and equations

Predicting Products: Types of Reactions

What happens when substances react?

AB  A + B

A + B AB

(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

(More Active Metal)

M

or N

MN +

(More Active Non-Metal)

  • : 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)


Chemical equations1

Chemical Equations

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

Sodium Chloride

  • Sodium + Chlorine 

  • Dihydrogen Monoxide 

  • Magnesium + Hydrochloric Acid 

  • Hydrochloric Acid + Calcium Hydroxide 

  • Combustion (burning with oxygen) of:

    Sucrose (C12H22O11)

    Octane

Hydrogen + oxygen

Magnesium Chloride + Hydrogen

Hydrogen hydroxide + Calcium chloride

+ Oxygen  Carbon dioxide + water

+ Oxygen  Carbon dioxide + water

Write balanced chemical equations for each.


Predicting products writing formulas and balancing equations

*

Predicting Products, writing Formulas and Balancing Equations

  • Alonso’s Rules for BE:

  • Easy element 1st hard elements last.

  • One element at a time.

  • Use fractions when necessary.

  • Sodium + Chlorine 

  • Dihydrogen Monoxide 

  • Magnesium + Hydrochloric Acid 

  • Hydrochloric Acid + Calcium Hydroxide 

  • Combustion (burning with oxygen) of:

  • Sucrose (C12H22O11)

  • Octane

Sodium Chloride

2

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

Hydrogen + oxygen

2

2

Magnesium Chloride + Hydrogen

2

Hydrogen hydroxide + Calcium chloride

2

2

Carbon dioxide + water

12

12

11

Carbon dioxide + water

16

18

2

12.5

8

9

25


Decomposition reactions

Decomposition Reactions

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

{AirBags Movie*}

sodium azide (N31-)

Important Exception:

Catalyst

2H2O2 2H2O + O2

{Peroxide Movie}


Decomposition reactions1

CaCO3 (s) CaO (s) + CO2 (g)

Decomposition Reactions

ComplexCompounds decompose into simpler compounds

Allcarbonatesbreak down to metal oxides and carbon dioxide

Chloratesbreak down to metal chlorides and oxygen

  • 2 KClO3 (s)  2 KCl (s) + 3O2 (g)

Acidsbreak down to nonmetal oxides and water

  • 2 H3PO4 (aq)  P2O5(g) + 3H2O (l)

  • 2HNO3 (aq)  N2O5(g) + H2O (l)

  • 2 NaOH (aq) Na2O (s) + H2O (l)

Basesbreak down to metal oxides and water


Unit 3 stoichiometry calculations with chemical formulas and equations

 Na2O + CO2 + H2O

Ammonium carbonate powder is heated strongly


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

Combination (Synthesis) Reactions

Now let’s balance equations

{Mg Movie}

{ZnS Movie*}

{H2O Movie*}

{AlBr3 Movie*}


Unit 3 stoichiometry calculations with chemical formulas and equations

Bromine liquid is poured over aluminum metal

Hydrogen chloride and ammonia gas are mixed together.

Sulfur dioxide gas is bubbled into water.


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

Metathesis (Double Displacement)

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

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


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)

Metathesis (Double Displacement): Acid-Base Neutralization Reaction

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*}


Single displacement reactions

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

Single Displacement Reactions

(Single Replacement Rx.)

(More Active Metal)

element

MN + M

MN + N

M

or N

compound

MN +

(More Active Non-Metal)

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


Single displacement reactions1

Examples:

Cu (s) + 2 AgNO3 (aq)

Cu (s) + Zn(NO3)2 (aq)

Cl2(g) + 2 NaBr (aq)

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

Single Displacement Reactions

{Movie:Cu+AgNO3}

  • 2 Ag +Cu(NO3)2 (aq)

  • No Reaction

  • 2 NaCl (aq) + Br2 (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

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


Unit 3 stoichiometry calculations with chemical formulas and equations

Most Active Nonmetal

Most Active Metal


Unit 3 stoichiometry calculations with chemical formulas and equations

H+OH-


Reactions with oxygen

Reactions with Oxygen

What is the difference?

Oxidation Rx.

(1) Oxidation Reactions: are combination

reactions involving oxygen.

{Movie: Mg, Fe, P, S +conc. O2

{Metal Oxides*}

(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)

Combustion Rx

{Movie: CH3OH + O2*}


Unit 3 stoichiometry calculations with chemical formulas and equations

When oxygen is scarce….

*


Gram molar mass g mm atomic weigh formula weigh or molecular weight

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

Mole = 6.022 x 1023 particles

Mass : Weight (in grams)


Gram molar mass g mm aw fw mw

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 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)


  • Formula weight fw

    2 x C: 2x(12.0 amu)

    6 x H: 6x(1.0 amu)

    30.0 amu

    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


    Mass percent composition

    (1) ( 197)

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

    x 100

    % element =

    % Au =

    x 100

    ( 2,301)

    (MW of Pb5Au(TeSb)4S5)

    (Mass) Percent Composition

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

    g-MM = 2,301g/

    Nagyagite Gold Ore:

    Pb5Au(TeSb)4S5

    Problem: (1) calculate mass % Au in Nagyagite. (2) If you buy 1 kg of the ore, how much gold does it have?

    = 8.56 %


    Unit 3 stoichiometry calculations with chemical formulas and equations

    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?

    There are two parts to this problem:

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

    % C =

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

    x 27%

    ? g C = 25 g CO2


    The mole concept

    Dermatological

    Chemical

    Biological

    The Mole Concept

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

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

    pentillion

    trillion

    sextillion

    quadrillion

    billion

    million

    thousand

    602 sextillion


    Using equivalences as mole ratios

    Using Equivalences as Mole Ratios:

    From Equivalences we obtain useful Ratios or Conversion factors:

    g-MM = 1 Mole () = 6.022 x 10 23

    particles

    NaCl = 58g/η(Atoms or molecules)

    or

    or

    or


    Mole calculations g mm moles of particles

    *

    Mole Calculations: g-MMMoles# of Particles

    Which ratios will you need?

    g-MM Moles:

    ? g = 3.20 mol of NaCl

    or

    ? mol = 3.20 g of NaCl

    = 1.9 x 10 24 f.u.

    Moles# of Particles:

    ? f.u. = 3.2 mol NaCl

    or

    = 5.3 x 10 28 mol

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


    Mole calculations g mm moles of particles1

    Mole Calculations: g-MMMoles# of Particles

    g-MM # of Particles:

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

    4.1 x 1012 g

    4.1 x 1012g

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

    3.3 x 1022 f.u.

    3.3 x 1022f.u.


    Unit 3 stoichiometry calculations with chemical formulas and equations

    *

    Mole Calculations: g-MMMoles# of Particles

    (atoms or molecules)

    How many molecules of H2O in 29g of water?

    How many atoms?

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

    = 3.0 X 1023 f.u.

    2+

    15

    1+

    3+

    3+

    6 =

    = 4.5 x 1024 atoms


    The determination of empirical formulas of compounds by elemental analysis

    The Determination of Empirical Formulas of Compounds by Elemental Analysis

    CxHy

    Combustion furnace


    Types of formulas

    Types of Formulas

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

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

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

    H2O

    HO

    CH

    Why are empirical formulas needed?

    Benzene, C6H6


    Elemental analyses

    Elemental Analyses

    How do we determine the formula of a compound?

    Compounds are broken down and the massesof their constituent elements are measured. From these masses the empirical formulas can be determined.

    HxOy

    moles

    Mole Ratio

    Expt. Data: (68g) 4g H

    64g O

    EmpF

    HO


    Calculating empirical formulas

    *

    Calculating Empirical Formulas

    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 (61.31g), hydrogen (5.14g),nitrogen (10.21g),oxygen (23.33g)

    carbon (61.31%), hydrogen (5.14%),nitrogen (10.21%),oxygen (23.33%).

    Percent means out of 100, so assume a 100g sample of the compound, then….

    Calculate the empirical formula (mole ratio) from the percent composition (% mass).


    Calculating empirical formulas1

    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

    1.01 g

    1 mol

    14.01 g

    1 mol

    16.00 g

    Calculating Empirical Formulas

    5.105 mol

    0.7288 mol

    61% C

    = 7.005  7

    = 6.984  7

    = 1.000

    = 2.001  2

    5.09 mol

    0.7288 mol

    5% H

    0.7288 mol

    0.7288 mol

    10% N

    1.458 mol

    0.7288 mol

    23% O

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

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

    These are the subscripts for the empirical formula: C7H7NO2


    Combustion analysis is a method of experimentally determining empirical formulas

    Combustion Analysis: is amethod of experimentally determining empirical formulas

    CxHy + O2 CO2 + H2O

    mass of CO2

    mass of H2O

    mass

    mass of C?

    mass of H?

    How do you calculate the mass of C in CO2 and that of H in H2O?

    mass of H2O

    mass of CO2

    Magnesium perchlorate

    Sodium hydroxide

    {Movie}

    CxHy

    • 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

    Combustion furnace


    Calculating empirical formulas2

    Calculating Empirical Formulas

    • CxHy (g) + O2 (g) CO2 (g) + H2O (g)

    (12g:32g=44g)

    14.6 g

    5.00 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?

    Empirical Formula

    CH3

    ? 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

    = 0.332 mol C

    / 0.332 = 1.00

    ? mol H = 1.02 g H

    = 1.01 mol H

    / 0.332 = 3.04


    Unit 3 stoichiometry calculations with chemical formulas and equations

    2006 A

    ?g C =

    ?g N =


    Unit 3 stoichiometry calculations with chemical formulas and equations

    2003 B


    Stoichiometry

    Stoichiometry

    (mass relationships within chemical equations)


    Stoichiometric calculations

    Stoichiometric Calculations

    2

    2

    The coefficients in the balanced equation can also be interpreted as mole ratios of reactants and products

    Mole Ratios from Balanced Equation:


    Stoichiometric calculations1

    *

    Stoichiometric Calculations

    How many grams of O2 are required to form 2.35 g of MgO?

    2 Mg (s) + O2(g) 2 MgO (s)

    No direct calculation

    grams

    grams

    Change:

    • grams of MgO  mol MgO

    • mol of MgO  mol O2

    • mol of O2  grams of O2

    2 Mg (s) + O2(g)  2 MgO (s)

    mole ratio from balanced equation


    Stoichiometric calculations2

    Stoichiometric Calculations

    Balanced Eq. uses MOLE language

    C6H12O6 + 6 O2 6 CO2 + 6 H2O

    ? g H2O

    Grams H2O

    Grams C6H12O6

    (3)

    (1)

    (2)

    Moles C6H12O6

    Moles H2O

    (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


    Unit 3 stoichiometry calculations with chemical formulas and equations

    Stoichiometry: Limiting Reactants

    (or, too much of one reactant andnot enough of the other)

    Make cookies until you run out of one of the ingredients

    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 }


    Limiting excess reactants

    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

    • Which is Limiting which is Excess?

    • Limiting H2; ExcessO2


    Limiting excess reactants problem

    Limiting & Excess Reactants Problem

    2 Mg (s) + O2(g)  2 MgO (s)

    If 5.0g of both Mg and O2 are used:

    • Which is the limiting and the excess reactants?

    • How much of the excess will be left unreacted ?

    • How much MgO will be produced ?


    Limiting excess reactants problem1

    Limiting & Excess Reactants Problem

    5 g

    5 g

    2 Mg (s) + O2(g)  2 MgO (s)

    ? g O2 = 5.0 g Mg

    ? g Mg = 5.0 g O2

    = 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?

    ?g MgO = 5.0 g Mg


    Limiting reactant experiments

    Limiting Reactant Experiments

    H2

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

    Zn+HCl  ZnCl2

    H2O

    {Movie:LimitReac}

    Experiments:

    (1)0.0025η + 0.0050η 61.0 mL

    (2)0.0012 η+ 0.0050η 30.5 mL

    (3)0.0031 η+ 0.0050η 61.0 mL

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

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


    Limiting reactant experiments1

    Limiting Reactant Experiments

    Experiments:

    0.0025g + 0.0050g

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

    HCl is Limiting Reactant 

    Zn is Excess Reactant 

    How many grams of ZnCl2 are produced in this reaction?


    Theoretical yield

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

    How many grams of ZnCl2 are produced in this reaction?

    Actual Yield

    Theoretical Yield

    Percent Yield = x 100

    Theoretical Yield

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

    the amount of product that can be made as calculated by stoichiometry.

    Actual Yield

    the amount reaction actually produces (less)

    Percent Yield


    Unit 3 stoichiometry calculations with chemical formulas and equations

    *

    g-MM: 76.622 148.5756 130.58

    The following reaction has a 95% yield:

    GeH4 + 3GeF4 4GeF3H

    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

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

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

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


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