# Unit # 3 Stoichiometry: Calculations with Chemical Formulas and Equations - PowerPoint PPT Presentation

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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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CHM 1045: General Chemistry and Qualitative Analysis

## Unit # 3Stoichiometry:Calculations with Chemical Formulas and Equations

Dr. Jorge L. Alonso

Miami, FL

• Textbook Reference:

• Module #3 & 4

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?

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)

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 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

• 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

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}

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

 Na2O + CO2 + H2O

Ammonium carbonate powder is heated strongly

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*}

Bromine liquid is poured over aluminum metal

Hydrogen chloride and ammonia gas are mixed together.

Sulfur dioxide gas is bubbled into water.

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 *}

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*}

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

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.

Most Active Nonmetal

Most Active Metal

H+OH-

### 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*}

When oxygen is scarce….

*

## 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

*

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)

• 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

(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 %

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

Dermatological

Chemical

Biological

## The Mole Concept

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

pentillion

trillion

sextillion

billion

million

thousand

602 sextillion

### 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-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-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.

*

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

CxHy

Combustion furnace

### 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

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

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).

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 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 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

2006 A

?g C =

?g N =

2003 B

## Stoichiometry

(mass relationships within chemical equations)

### 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 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 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

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

• 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

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 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

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 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?

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

*

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 ?