TOPIC IV: Chemical Stoichiometry LECTURE SLIDES Balancing Equations Simple Stoichiometry Limiting reagent % Yield Combustion Analysis Kotz &Treichel, Chapter.

TOPIC IV: Chemical Stoichiometry

LECTURE SLIDES• Balancing Equations• Simple Stoichiometry• Limiting reagent• % Yield• Combustion Analysis

Kotz &Treichel, Chapter 4

When a substance reacts with another substance or interacts with energy to form one or more different substances, a chemical reaction occurs .

The changes that have taken place are represented by the chemical equation, which keeps track of all the atoms involved in the reactants and the products.

The chemical equation must be carefully balanced to insure that it adequately describes the reaction without loosing or gaining any atoms in the process!

One cannot create or destroy matter in a chemical reaction, so all reactant atoms must show up somewhere in the product(s)...

To completely describe the reaction that has taken place, the equation should include not only the formula or symbol for each reactant and product, but also their physical state, and the energy process involved.

We will indicate physical state by subscripts (s) solid, (l)liquid, (g) gas, (aq) aqueous, in water solution.

We will not at this point add the energy component to our chemical equations but we must keep in mind that all reactions involve energy in some form:

Either:The reaction requires some form of heat and is termed “endothermic”

orThe reaction gives off heat and is described as “exothermic”

Typical chemical reactions include predictable types:

Combination: A + B --> C

Decomposition: AB ---> A + B

Double Replacement in Aqueous solution: AB + CD --> AD + CB

Combustion: CxHy + O2 --> CO2 + H2O

(We’ll consider another type, oxidation/reduction later in the unit!)

As we review balancing equations, let’s describe them by type:

Combination or “Synthesis”: A + B --> C

H2(g) + O2(g) --> H2O(g)

2 atoms O

1 atom O

UNBALANCED!

2 H2(g) + O2(g) --> 2 H2O(g)

We have now accounted for all atoms4 atoms H, left and right2 atoms O, left and rightEquation is balanced...

H2(g) + O2(g) --> 2 H2O(g)

Rebalance:

To balance O’s, place coefficient before formula:

Now 4 atoms H

NaCl(l) --> Na(l) + Cl2(g)

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

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

Balancing process, Decomposition Reactions: energy AB ----------> A + B

Al2(CO3)3(s) --> Al2O3(s) + CO2(g)

Al2(CO3)3(s) --> Al2O3(s) + 3CO2(g)

Balance Cl

Balance Na

Balance C,then O ok

Balancing Combustion Reactions: (C,H) + O2 --> CO2 + H2O + heat (C,H,O) + O2 --> CO2 + H2O + heat

C5H12 + O2 --> CO2 + H2O

Methodology: a) Balance C b) Balance H c) Balance O

Organic molecules:hydrocarbons, carbohydrates, alcohols: “fuels”

c) Balance O:

C5H12 + 8O2 --> 5 CO2 + 6 H2O

a) Balance C: C5H12 + O2 --> 5 CO2 + H2O

b) Balance H:

C5H12 + O2 --> 5 CO2 + 6 H2O

10 O 6 O

Now, a more interesting one: C8H18 + O2 --> CO2 + H2O

a) C8H18 + O2 --> 8CO2 + H2O

b) C8H18 + O2 --> 8CO2 + 9 H2O

c) C8H18 + 12.5 O2 --> 8 CO2 + 9 H2O

16 O + 9 O = 25 O

25 O 16 O 9 O

At this point we have:

C8H18 + 25/2 or 12.5 O2 --> 8CO2 + 9 H2O

Should have whole number coefficients, so MULTIPLY ALL SPECIES BY 2:

2 C8H18 + 25 O2 --> 16CO2 + 18 H2O

Checking for balance:

16 C 36 H 50 O -->; 16 C 36 H; 32 O +18 O

GROUP WORK 4.1, BALANCE:

C6H12O6 + O2 --> CO2 + H2O (sugar)

C5H11OH + O2 --> CO2 + H2O (alcohol)

a) Balance Cb) Balance Hc) Balance Od) Check

Balancing Double Replacement Reactions: AB + CD --> AD + CB

General directions: save H for next to last, O for last:

a) balance P:

2 H3PO4 + Ca(OH)2 --> Ca3(PO4)2 + H2O

b) balance Ca:

2H3PO4 + 3 Ca(OH)2 --> Ca3(PO4)2 + H2O

H3PO4 + Ca(OH)2 --> Ca3(PO4)2 + H2O

c) balance H

2 H3PO4 + 3 Ca(OH)2 --> Ca3(PO4)2 + H2O 6H 6H 2H

2 H3PO4 + 3 Ca(OH)2 --> Ca3(PO4)2 + 6 H2O

6H 6H 12H

d) checkout O’s:

8O + 6O ---> 8O + 6O

BALANCED!

Stoichiometry: Introduction

Let us balance one more equation, using techniquesjust introduced:

Fe2O3(s) + HCl(aq) ---> H2O + FeCl3(aq)

a) Fe2O3(s) + HCl(aq) ---> H2O + 2FeCl3(aq)

b) Fe2O3(s) + 6HCl(aq) ---> H2O + 2FeCl3(aq)

c) Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq)

Now let us go one step further and determine thequantitative relationships a BALANCED equation implies, utilizing the below:

Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq)

On the most basic level, the equation counts number of “molecules” or ionic “formula units”of each species are needed in order for this reaction to occur:

1 “formula unit” Fe2O3 + 6 molecules HCl--->

3 molecules H2O + 2 “formula units” FeCl3

Fe 3+

O2-

Fe 3+

O2-O2-

1 Fe2O3 + 6 HCl

H Cl

H Cl

H Cl

H Cl

H Cl

H Cl

3 H2O

O

HH

O

HHO

HH

+ 2 FeCl3

Fe 3+

Cl -

Cl -Cl -

Fe 3+

Cl -

Cl -Cl -

But, we know we can’t reach into a box and pull out molecules: we have to weigh out a specific mass of the molecules and then figure out how many we have because they are too tiny to count.

To count by weighing, we use the mole, knowing that

1 MOLE of molecules = 6.022 X 1023 molecules

and we can work with our equation in terms of moles...

The relationship:

1 “formula unit” Fe2O3 + 6 molecules HCl--->

3 molecules H2O + 2 “formula units” FeCl3

BECOMES:

1 mole Fe2O3 + 6 moles HCl--->

3 moles H2O + 2 moles FeCl3

and...

When we talk moles, we can talk mass in grams bycalculating the molar masses of the compounds involved:

Fe2O3 : 2 Fe= 2X 55.85 = 111.70 3O = 3X 16.00 = 48.00 159.70 g/mol Fe2O3

HCl: 1H = 1.01 1Cl = 35.45 36.46 g/mol HCl

FeCl3:

1 Fe = 1 X 55.847 = 55.85 3 Cl = 3 X 35.45 = 106.35 162.20 g/mol FeCl3

H2O:2H = 2X1.01 = 2.021O = 1X 16.00 = 16.00 18.02 g/mol H2O

1 mole Fe2O3 = 1 mol X 159.70 g/mol Fe2O3 = 159.70 g

6 moles HCl = 6 mol X 36.46 g/mol HCl = 218.76 g

3 moles H2O = 3 mol X 18.02 g/mol H2O = 54.06 g

2 moles FeCl3 = 2 mol X 162.20 g/mol FeCl3 = 324.40 g

Equivalent:1 mol Fe2O3 + 6 mol HCl---> 3 mol H2O + 2 mol FeCl3

159.70 g Fe2O3 + 218.76 g HCl ---> 54.06 g H2O + 324.40 g FeCl3

159.70 g Fe2O3 + 218.76 g HCl ---> 54.06 g H2O + 324.40 g FeCl3

Note how the Law Of Conservation of Matter has been upheld:

159.70 g + 218.76 g ---> 54.06 g + 324.40 g

378.46 g total mass reactants -----> 378.46 g total mass products

Let us now apply our new knowledge to problem solving situations:

In all cases, we will use the equation we have developed:


159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol

If we start with 25.00 g of Fe2O3, how many moles andhow many g of HCl are required for the reaction, andhow many g of FeCl3 and moles of water will we theoretically obtain?

To solve, copy balanced equation, and insert under each reactant and product the molar mass and any question or information given in the problem:



25.00g ?g, mol ?moles ?g

We can now proceed to solve each question:



25.00g ?g, mol

Question #1: 25.00 g Fe2O3 = ? mol HCl

Relationships:

159.70 g Fe2O3 = 1 mol Fe2O3

1 mol Fe2O3 = 6 mol HCl

25.00 g Fe2O3

159.70 g Fe2O3

1 mol Fe2O3

1 mol Fe2O3

6 mol HCl= .939261 mol HCl

= .9393 mol HCl

PATHWAY: g Fe2O3 ---> mol Fe2O3 ---> mol HCl

Conversionfactor: M

Conversion factor: balancedequation

25.00 g Fe2O3

159.70 g Fe2O3

1 mol Fe2O3

1 mol Fe2O3

6 mol HCl

1 mol HCl

36.46 g HCl= 34.2455 g

= 34.25 g HCl

Conversionfactor: M

Conversionfactor: M

Conversion factor: balancedequation

Question #2: 25.00 g Fe2O3 = ? g HCl




answer: .9393 mol, 34.25 g

To finish:

Question #3: 25.00 g Fe2O3 = ? mol H2O

Question #4: 25.00 g Fe2O3 = ? g FeCl3

25.00 g Fe2O3

159.70 g Fe2O3

1 mol Fe2O3

1 mol Fe2O3

3 mol H2O= .4696 mol H2O

Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq

25.00 g Fe2O3

159.70 g Fe2O3

1 mol Fe2O3

1 mol Fe2O3

2 mol FeCl3

1 mol FeCl3

162.20 g FeCl3= 50.7827 g

= 50.78 g FeCl3

Question #3: 25.00 g Fe2O3 = ? mol H2O

Question #4: 25.00 g Fe2O3 = ? g FeCl3

To prove we did everything in a correct fashion, let’s seeif it all adds up:




answer: .9393 mol, .4696 mol 34.25 g ---> 8.462 g 50.78 g

25.00 + 34.25 = 59.25 g reactants8.462 + 50.78 = 59.245 g products

GROUP WORK 4.2:

How many g of Fe2O3 are required to make 75.00 g ofFeCl3?



? g 75.00 g

PATHWAY: g FeCl3 --->mol FeCl3 ---> mol Fe2O3 ---> g Fe2O3

Next, let’s consider problems we’ll call “LIMITING REAGENT PROBLEMS”

In this type of problem, amounts OF BOTH reactants will be given to you, and you’ll be asked to calculate how much of one or more products can be formed.

In most cases, one given value will be in excess, to insure that all of the other reagent is used up.

We can call the the reactant present in excessive amount “the excess reagent”

and we can call the reactant we expect to get used up“the limiting reagent” because it limits the amount ofproduct which we can form.....

Never try to guess from amounts given which reactantis which; you should always calculate to determinethe nature of both given reactants.

There are several ways this can be accomplished; I willpresent the method I am most comfortable with:

Let’s use the same equation we have been considering today:



35.00 g 100.0 g ? g

Now, suppose you were given 100.0 g HCl and 35.00 g Fe2O3.How many g of FeCl3 will you theoretically obtain?

Amounts of both:Limiting Reagent problem

My method of choice: Calculate g of product from each reactant.

The reactant which gives you the larger amount of product is the excess reagent. The amount of product calculated from this reagent answer is incorrect, too much...

The reactant that gives you the smaller amount of product is the limiting reagent. (The calculation from the limiting reagent gives is the correctanswer!).....



35.00 g 100.0 g ? g

The question becomes:

• 35.00 g Fe2O3 = ? g FeCl3

• 100.0 g HCl = ? g FeCl3

• which answer is correct?

GROUP WORK 4.3: CALCULATE BOTH, DECIDE!

Theoretical, Actual and Percent Yield

So far we have considered how much product we couldtheoretically obtain from a given reaction by calculatingfrom a balanced equation. We have been working with the following equation, so let’s finish up the topic with it:



“theoretical yield” is the amount of any product you calculate from your balanced equation, using the limiting reagent when amounts of all reactants are given.

“actual yield” is the amount of product you actually obtain from a given experiment using the reactiondescribed in the balanced equation; this amount is rarely as large as the theoretical yield.

Percent yield = actual yield X 100% theoretical yield

Let’s recall the last problem we did:

If 35.00 g Fe2O3 were reacted with 100.0 g HCl, how manyg of FeCl3 would theoretically be obtained?

We calculated the amount of product from both reactants, and decided that 71.10 g of FeCl3 would bethe “expected” amount of product, based on the limiting reagent and balanced equation.

(See next slide)



? g, theoretical35.00 g 100.0 g ans: 71.10 g

159.70 g Fe2O3 1 mol Fe2O3

1 mol Fe2O3 2 mol FeCl3

1 mol FeCl3

162.20 g FeCl3=

35.00 g Fe2O3 71.096 g

= 71.10 g FeCl3

36.46 g HCl 6 mol HCl

1 mol HCl 2 mol FeCl3

1 mol FeCl3

162.20 g FeCl3=

100.0g HCl148.29 g

= 148.3 g FeCl3

Too much

Now suppose you ran the reaction in the laboratory anddiscovered that you could only isolate 65.88 g of FeCl3.

The next question asked would be, “what is your percent yield for this experiment?”



71.10 g theoretical35.00 g 100.0 g 65.88 g actual % Yield = ?

Percent yield = actual yield X 100 theoretical yield

% Yield = 65.88 g obtained or “actual” X 100 71.10 g calculated or “theoretical”

= 92.66 %

Combustion Analysis

To round off Chapter 4, let’s consider the analysis of organic hydrocarbons (CxHy) and compounds CxHyOz: This type of analysis is done by the complete combustion of these compounds with O2 and an analysis of the CO2 and H2O formed.

All the carbon in the original compound will be contained in the CO2, all the hydrogen in the water, and O can be determined by difference. Let’s see how that works:

CxHy or CxHyOz + O2 ---> CO2 + H2O

CxHy:

Since all the C in the original compound is in the CO2, we can get moles of C from grams of CO2 produced:

g, CO2 mol, CO2 mol, C

Since all the H in the original compound is in the H2O produced, we can get moles of H from moles of H2O, g, H2O mol H2O mol H

To get formula, obtain simplest mole ratio, C, H

A 0.523 g sample of unknown compound CxHy was burned in air to give 1.612 g CO2 and 0.7425 g H2O. Its molar mass was determined to be 114 g/mol. What is its empirical and molecular formula?

1.612 g CO2 = ? Moles C CO2, 44.010 g/mol

.7425 g H2O = ? Moles H H2O, 18.015 g/mol

1.612 g CO2

44.010 g CO2

1 mol CO2

1 mol CO2

1 mol C= .03663 mol C

.7425 g H2O

18.015 g H2O

1 mol H2O

1 mol H2O

2 mol H= .08243 mol H

.03663 mol C

.03663 mol C= 1

.08243 mol H

.03663 mol C

= 2.25

X 4 = 4

X 4 = 9

C4H9Empirical Formula

4C= 4 X 12.01 = 48.049H= 9 X 1.008= 9.072

57.11 g/mol Empirical Formula Weight

114

57.11= 1.996= 2.00 (C4H9)n = (C4H9)2 = C8H18

g CO2, H2O ---> mol CO2, H2O ---> mol C,H ---> g C,H

g original compound - g C,H = g O in original compound

g C,H,O ---> mol C,H,O ---> simplest mole ratio

If there is oxygen in the original compound, the problem is more complicated: the amount of oxygen must be detected by difference:

CHECKOUT EXAMPLE 4.7 IN TEXT!

TOPIC IV: Chemical Stoichiometry LECTURE SLIDES Balancing Equations Simple Stoichiometry Limiting reagent % Yield Combustion Analysis Kotz &Treichel, Chapter.

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