Chapter 3.6-3.10 Student Notes Stoichiometry. Chapter 3 Table of Contents Copyright © Cengage Learning. All rights reserved 2 3.1 Counting by Weighin3.1

Chapter 3.6-3.10Student Notes

Stoichiometry

Chapter 3

Table of Contents

Copyright © Cengage Learning. All rights reserved 2

3.1 Counting by Weighing

3.2 Atomic Masses

3.3 The Mole

3.4 Molar Mass

3.5Percent Composition of Compounds

SEE PREVIOUS LESSONS ON OTHER SLIDE SHOW

3.6Determining the Formula of a Compound

3.7Chemical Equations

3.8Balancing Chemical Equations

3.9Stoichiometric Calculations: Amounts of Reactants and Products

3.10The Concept of Limiting Reagent

3

Chapter 3

Table of Contents

• Pick up calculator, notes packet, and pre-lab packet.• CW: Notes 3.1 to 3.5 • CW: Finish Test part 2 ch.1-2 if not done yet• CW/HW: Read lab and complete Pre-lab questions.

Note that the reading helps with the pre-lab questions due Thursday for lab. Try to be here by 7:45 for setup to give us time to discuss pre-lab and complete lab.

• HW: Section 3.1-3.5 pg. 117 #21, 26, 27, 29, 30, 31, 33, 35, 37, 39b,45b, 47b, 49b, 52, 57a-g, 61, 63 due Monday 9/23/13 to go over.

TUESDAY - B DAY - SEPT. 17, 2013

4

Section 3.6

Determining the Formula of a Compound

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Assignments Monday 9/23/13

• Pick up handouts for use later & have out ch. 3 notes and HW ch. 3

• CW: Notes 3.5-3.6 finished together• CW: Empirical and Molecular Formula Race Game• CW/HW: Notes 3.7-3.8 with computers.• HW: Formal Lab report due on Wednesday.• Be sure you completed homework 3.1-3.5 previously

assigned and turn in by Wednesday.• HW: Empirical Formula and Balancing Equations

w/sheets due on Friday.• HW: Be reading chapter 3. The next test is over ch.3-4

and is tentatively scheduled for October 16th. Ch. 4 is mostly new material not covered in Gen. Chemistry.

4

5

Section 3.5

Percent Composition of Compounds

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ASSIGNMENTS Ch. 3 Homework - Tuesday - 9/17/13

• HW: Section 3.1-3.5 pg. 117 #21, 26, 27, 29, 30, 31, 33, 35, 37, 39b,45b, 47b, 49b, 52, 57a-g, 61, 63 due Monday 9/23/13 to go over.

• You must show your work for most of these with calculations and units. No work for math problems will receive no points.

• CW: Finish test part 2 if you haven’t.• HW: Read and complete Pre-lab Determination of the

Molecular Weight of an Acid for lab on Thursday.

Section 3.6


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• Empirical formula = CH Simplest whole-number ratio

• Molecular formula = (empirical formula)n

[n = integer]

• Molecular formula = C6H6 = (CH)6

Actual formula of the compound

Formulas

FormulasFormulas

molecular formula = (empirical molecular formula = (empirical formula)formula)nn [n = integer] [n = integer]

molecular formula = Cmolecular formula = C66HH66 = (CH) = (CH)66

empirical formula = CHempirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.Molecular formula: the true number of atoms of each element in the formula of a compound.

FormulasFormulas (continued)(continued)

Formulas for Formulas for ionic compounds ionic compounds are are ALWAYSALWAYS empirical (lowest whole number ratio).empirical (lowest whole number ratio).

Examples:Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

FormulasFormulas (continued)(continued)

Formulas for Formulas for molecular compoundsmolecular compounds MIGHTMIGHT be empirical (lowest whole number ratio).be empirical (lowest whole number ratio).

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:Empirical:

H2O

CH2O C12H22O11

10

• We can “reverse” the process of finding percentage composition.

• First we use the percentage or mass of each element to find moles of each element.

• Then we can obtain the empirical formula by finding the smallest whole-number ratio of moles.– Find the whole-number ratio by dividing each number

of moles by the smallest number of moles.

Chemical Formulas from Mass Percent Composition

Empirical Formula DeterminationEmpirical Formula Determination

1.1. Base calculation on 100 grams of Base calculation on 100 grams of compound. (If given masses, go to compound. (If given masses, go to #2.)#2.)

2.2. Determine moles of each element in Determine moles of each element in 100 grams of compound. (or sample 100 grams of compound. (or sample given)given)

3.3. Divide each value of moles by the Divide each value of moles by the smallest of the values.smallest of the values.

4.4. Multiply each number by an integer Multiply each number by an integer to obtain all whole numbers.to obtain all whole numbers.

Empirical Formula DeterminationEmpirical Formula Determination

Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

In 100 g sample, we would have In 100 g sample, we would have

49.32 g Carbon 49.32 g Carbon

43.84 g Oxygen43.84 g Oxygen

6.85 g Hydrogen6.85 g Hydrogen

1.1. Base calculation on 100 grams of Base calculation on 100 grams of compound. compound.

Empirical Formula DeterminationEmpirical Formula Determination(part 2)(part 2)

2. Determine moles of each element 2. Determine moles of each element in 100 grams of compound.in 100 grams of compound.

Carbon:Carbon: 49.32 g 49.32 g / 12.01 g/mol = 4.107 mol / 12.01 g/mol = 4.107 mol

Hydrogen: Hydrogen: 6.85 g / 1.01 g/mol = 6.78 mol 6.85 g / 1.01 g/mol = 6.78 mol

Oxygen:Oxygen: 43.84 g 43.84 g / 16.00 g/mol = 2.740 mol / 16.00 g/mol = 2.740 mol


3. Divide each value of moles by the 3. Divide each value of moles by the smallest of the values.smallest of the values.

Carbon:Carbon: 4.107 mol /2.74 mol = 1.50 4.107 mol /2.74 mol = 1.50

Hydrogen: Hydrogen: 6.78 mol/2.74 mol = 2.47 6.78 mol/2.74 mol = 2.47

Oxygen:Oxygen: 2.74 mol /2.74 mol = 1.00 2.74 mol /2.74 mol = 1.00


4. Multiply each number by an 4. Multiply each number by an integer to obtain all whole numbers.integer to obtain all whole numbers.

Carbon:Carbon: 1.50 x 2 = 3.0 1.50 x 2 = 3.0

Hydrogen: Hydrogen: 2.47 = 2.5 x 2 = 5 2.47 = 2.5 x 2 = 5

Oxygen:Oxygen: 1.0 x 2 = 2.0 1.0 x 2 = 2.0

Empirical formula: C3H5O2

Finding the Molecular FormulaFinding the Molecular FormulaThe empirical formula for adipic acid is The empirical formula for adipic acid is CC33HH55OO22. The molecular mass of adipic acid . The molecular mass of adipic acid is 146 g/mol. What is the molecular is 146 g/mol. What is the molecular formula of adipic acid?formula of adipic acid?

1. Find the formula mass of C1. Find the formula mass of C33HH55OO22

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g


3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

2. Divide the molecular mass by the 2. Divide the molecular mass by the mass given by the emipirical formula.mass given by the emipirical formula.


3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

3. Multiply the empirical formula by this 3. Multiply the empirical formula by this number to get the molecular formula.number to get the molecular formula.

(C(C33HH55OO22) x 2 =) x 2 = CC66HH1010OO44

19

• A molecular formula is a simple integer multiple of the empirical formula.

• That is, an empirical formula of CH2 means that the molecular formula is CH2, or C2H4, or C3H6, or C4H8, etc.

• So: we find the molecular formula by:

= integer (nearly)

molecular formula massempirical formula mass

We then multiply each subscript in the empirical formula by the integer.

Relating Molecular Formulasto Empirical Formulas

Section 3.6


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• Device used to determine the mass percent of each element in a compound.

Analyzing for Carbon and Hydrogen

Combustion Analysis Method for determining molecular mass

21

Section 3.6


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Empirical & Molecular Formula Partner Team Race Activity• Working with a partner, you will solve one problem at a

time. Both of you will solve the problem on your paper and then decide if you agree. Consider the correct number of significant figures as well and race the answer to me when you think you have a consensus of the final answer. If you are correct, I will say “yes” and you will proceed to the next problem. If I say “no”, you will need to try again. If I say “no” Sig. Fig., your answer is right to the wrong significant figures. Use the AP periodic table values which are mostly to the hundredths for atomic mass.

22

Section 3.6


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Assignments Monday

• CW: Notes 3.5-3.6 finished together• CW: Empirical and Molecular Formula Race Game• CW/HW: Notes 3.7-3.8 with computers.• HW: Formal Lab report due on Wednesday.• Be sure you completed homework 3.1-3.5 previously

assigned and turn in by Wednesday.• HW: Empirical Formula and Balancing Equations

w/sheets due on Friday.

Combustion Analysis Method for determining molecular mass

• Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is the empirical formula for menthol?

Combustion Analysis Method for determining molecular mass (cont)

Combustion Analysis Method for determining molecular mass (cont.)

More complicated example:

You do not have to record but do look over it.

Nasty example - you don’t have to record but look

over.

• A chemical equation is a shorthand description of a chemical reaction, using symbols and formulas to represent the elements and compounds involved.

Writing Chemical Equations

Notes 3.7

Section 3.7

Chemical Equations

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• A representation of a chemical reaction:

C2H5OH + 3O2 → 2CO2 + 3H2O

reactants products

• Reactants are only placed on the left side of the arrow, products are only placed on the right side of the arrow.

Section 3.7

Chemical Equations

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C2H5OH + 3O2 → 2CO2 + 3H2O

• The equation is balanced.• All atoms present in the reactants are

accounted for in the products.• 1 mole of ethanol reacts with 3 moles of

oxygen to produce 2 moles of carbon dioxide and 3 moles of water.

Section 3.7

Chemical Equations

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• The balanced equation represents an overall ratio of reactants and products, not what actually “happens” during a reaction.

• Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed.

32

• Sometimes additional information about the reaction is conveyed in the equation.

Writing Chemical Equations

Copyright © by Holt, Rinehart and Winston. All rights reserved.

ResourcesChapter menu

Symbols Used in Chemical Equations

Chapter 8Section 1 Describing Chemical Reactions

Section 3.8

Balancing Chemical Equations

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1. Determine what reaction is occurring. What are the reactants, the products, and the physical states involved?

2. Write the unbalanced equation that summarizes the reaction described in step 1.

3. Balance the equation by inspection, starting with the most complicated molecule(s). The same number of each type of atom needs to appear on both reactant and product sides.

Writing and Balancing the Equation for a Chemical Reaction

Copyright © Houghton Mifflin Company. All rights reserved. 3–


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36

• If an element is present in just one compound on each side of the equation, try balancing that element first.

• Balance any reactants or products that exist as the free element last.

• In some reactions, certain groupings of atoms (such as polyatomic ions) remain unchanged. In such cases, treat these groupings as a unit.

• At times, an equation can be balanced by first using a fractional coefficient(s). The fraction is then cleared by multiplying each coefficient by a common factor.

Guidelines for BalancingChemical Equations

Section 3.8


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Exercise

Which of the following correctly balances the chemical equation given below? There may be more than one correct balanced equation. If a balanced equation is incorrect, explain what is incorrect about it.

CaO + C → CaC2 + CO2

I. CaO2 + 3C → CaC2 + CO2

II. 2CaO + 5C → 2CaC2 + CO2

III. CaO + (2.5)C → CaC2 + (0.5)CO2

IV. 4CaO + 10C → 4CaC2 + 2CO2

Section 3.8


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Concept Check

Which of the following are true concerning balanced chemical equations? There may be more than one true statement.I. The number of molecules is conserved.II. The coefficients tell you how much of each

substance you have.III. Atoms are neither created nor destroyed.IV. The coefficients indicate the mass ratios of the

substances used.V.The sum of the coefficients on the reactant side

equals the sum of the coefficients on the product side.

Section 3.8


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• The number of atoms of each type of element must be the same on both sides of a balanced equation.

• Subscripts must not be changed to balance an equation.

• A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction.

• Coefficients can be fractions, although they are usually given as lowest integer multiples.

Notice


Table 3.2 Information Conveyed by the Balanced Equation for the Combustion

of Methane

Section 3.9

Stoichiometric Calculations: Amounts of Reactants and Products

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• Chemical equations can be used to relate the masses of reacting chemicals.

Stoichiometric Calculations

42

• A stoichiometric factor or mole ratio is a conversion factor obtained from the stoichiometric coefficients in a chemical equation.

• In the equation: CO(g) + 2 H2(g) → CH3OH(l)

– 1 mol CO is chemically equivalent to 2 mol H2

– 1 mol CO is chemically equivalent to 1 mol CH3OH

– 2 mol H2 is chemically equivalent to 1 mol CH3OH

Stoichiometric Equivalenceand Reaction Stoichiometry

1 mol CO––––––––– 2 mol H2

1 mol CO––––––––––––– 1 mol CH3OH

2 mol H2––––––––––––– 1 mol CH3OH

43

One car may be equivalentto either 25 feet or 10 feet,depending on the method of parking.

One mole of CO may be equivalent to one mole of CH3OH, or to one mole of CO2, or to two moles of CH3OH, depending on the reaction(s).

Concept of Stoichiometric Equivalence

44

Outline of Simple Reaction Stoichiometry

Note: preliminary and/or follow-up calculations may be needed.

Section 3.9


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1. Balance the equation for the reaction.

2. Convert the known mass of the reactant or product to moles of that substance.

3. Use the balanced equation to set up the appropriate mole ratios.

4. Use the appropriate mole ratios to calculate the number of moles of desired reactant or product.

5. Convert from moles back to grams if required by the problem.

Calculating Masses of Reactants and Products in Reactions

46Outline of Stoichiometry Involving

Mass

To our simple stoichiometry

scheme …

… we’ve added a conversion from mass at

the beginning …

… and a conversion to mass at the

end.

Substances A and B may be two

reactants, two products, or reactant and

product.Think: If we are given moles of substance A

initially, do we need to convert A

to grams?

Section 3.9


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Calculating Masses of Reactants and Products in Reactions

48

Section 3.9


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WEDNESDAY - SEPT 25, 2013• CW: Notes 3.9• Turn in Formal Lab Report from last Thursday’s titration

lab AND HW 3.1 to 3.5 - go over any questions• CW: Limiting Reactants problem w/sheet handout - HW

if not finished in class due FRI• HW: Balancing Equations practice quick w/sheet due

FRI• HW: Complete the empirical vs. molecular formula and

% composition w/sheet from Monday - due Friday• CW/HW: pre-lab moles vs. molar mass graphical with

limiting reactants - using graphing to determine unknown.

StoichiometryStoichiometry

Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet

“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”

ReviewReview: Chemical Equations: Chemical EquationsChemical change involves a reorganization Chemical change involves a reorganization

of of

the atoms in one or more substances.the atoms in one or more substances.CC22HH55OH + 3OOH + 3O22 →→ 2CO 2CO22 + 3H + 3H22OO

reactantsreactants productsproducts

11 mole of ethanol mole of ethanol reacts with reacts with 33 moles of moles of oxygen oxygen to produce to produce 22 moles of carbon moles of carbon dioxide dioxide andand 33 moles of water moles of water

When the equation is balanced it has When the equation is balanced it has quantitative significance:quantitative significance:

Calculating Masses of Reactants Calculating Masses of Reactants and Productsand Products

1.1. Balance the equation.Balance the equation.

2.2. Convert mass or volume to moles, Convert mass or volume to moles, if necessary.if necessary.

3.3. Set up mole ratios.Set up mole ratios.

4.4. Use mole ratios to calculate moles Use mole ratios to calculate moles of desired substituent.of desired substituent.

5.5. Convert moles to mass or volume, Convert moles to mass or volume, if necessary.if necessary.

1.1. Balance the equation.Balance the equation.

2.2. Convert mass or volume to moles, Convert mass or volume to moles, if necessary.if necessary.

3.3. Set up mole ratios.Set up mole ratios.

4.4. Use mole ratios to calculate moles Use mole ratios to calculate moles of desired substituent.of desired substituent.

5.5. Convert moles to mass or volume, Convert moles to mass or volume, if necessary.if necessary.

Working a Stoichiometry ProblemWorking a Stoichiometry Problem6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.

1. Identify reactants and products and write the balanced equation.

AlAl ++ O2 Al2O3

b. What are the reactants?

a. Every reaction needs a yield sign!

c. What are the products?

d. What are the balanced coefficients?

4 3 2

Working a Stoichiometry ProblemWorking a Stoichiometry Problem6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?

4 Al + 3 O2 2Al2O3

=6.50 g Al

? g Al2O3

1 mol Al

26.98 g Al 4 mol Al

2 mol Al2O3

1 mol Al2O3

101.96 g Al2O3

6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =12.3 g Al2O3

Working a Stoichiometry ProblemWorking a Stoichiometry ProblemQuestion- How many grams of water can be produced from 10. grams of H2 with excess O2?

You try.

=

Working a Stoichiometry ProblemWorking a Stoichiometry ProblemQuestion- How many grams of water can be produced from 10. grams of H2 with excess O2?

Answer: 2H2 + O2 --> 2H2O

=1 mol H2

2.02 g H2

10 g H2 2 mol H2O

2 mol H2

18.02 g H2O

1 mol H2O89.2=89 g

Limiting ReactantLimiting Reactant The limiting reactant is the reactant that is consumed first, limiting the

amounts of products formed.

The reactant left over is in EXCESS. (remember nuts and bolts mini-lab)

57

• Many reactions are carried out with a limited amount of one reactant and a plentiful amount of the other(s).

• The reactant that is completely consumed in the reaction limits the amounts of products and is called the limiting reactant, or limiting reagent.

• The limiting reactant is not necessarily the one present in smallest amount.

Limiting Reactants

58

Limiting Reactant Analogy

If we have 10 sandwiches, 18 cookies,

and 12 oranges …

… how many packaged

meals can we make?

59

Limiting Reactant Analogy

If we have 10 sandwiches, 18 cookies,

and 12 oranges …

… how many packaged

meals can we make?

There would only be 9 because you need 20 cookies to use all of the sandwiches 1:2 ratio and you only need 10 oranges with 10 sandwiches so oranges and sandwiches are in excess.

60

When 28 g (1.0 mol) ethylene reacts with …

… 128 g (0.80 mol) bromine, we

get …

… 150 g of 1,2-dibromoethane,

and leftover ethylene!

1.Why is ethylene left over, when we started with more bromine than ethylene? (Hint: count the molecules.)

2.What mass of ethylene is left over after reaction is complete? (Hint: it’s an easy calculation; why?)

Molecular View of the Limiting Reactant Concept

61

• We recognize limiting reactant problems by the fact that amounts of two (or more) reactants are given.

• One way to solve them is to perform a normal stoichiometric calculation of the amount of product obtained, starting with each reactant.

• The reactant that produces the smallest amount of product is the limiting reactant.

Recognizing and Solving Limiting Reactant Problems

Section 3.9


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Exercise

Consider the following reaction:

P4 (s) + 5 O2 (g) 2P2O5 (s)

If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with?

Section 3.9


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Exercise


P4 (s) + 5 O2 (g) 2P2O5 (s)

If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with?

6.25 g / 124 g/mol = .0504 mol 1:5 ratio so

.0504... x 5 =0.252 mol x 32 g/mol = 8.06 g O2

8.07 g if you do everything and then round at the end

Section 3.10

The Concept of Limiting Reagent

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Concept Check

Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation:

2H2 + O2 → 2H2O

Identify limiting reactants & excess reactanta) 2 moles of H2 and 2 moles of O2

• 2 moles of H2 and 3 moles of O2

• 2 moles of H2 and 1 mole of O2

d) 3 moles of H2 and 1 mole of O2

e) Each produce the same amount of product.

Section 3.10


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• We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation.

Notice

Section 3.10


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Concept Check

• You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B. What information do you need to

know in order to determine the mass of product that will be produced?

Section 3.10


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• Where are we going? To determine the mass of product that will be

produced when you react 10.0 g of A with 10.0 g of B.

• How do we get there? We need to know:

• The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation.

• The molar masses of A, B, and the product they form.

Let’s Think About It

Section 3.10


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Exercise

You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is

20.0 g/mol, and C is 25.0 g/mol? They react according to the equation:

A + 3B → 2C

Section 3.10


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• An important indicator of the efficiency of a particular laboratory or industrial reaction.

Percent Yield

Section 3.10


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ExerciseConsider the following reaction:

P4(s) + 6F2(g) → 4PF3(g)

What mass of P4 is needed to produce 85.0 g of PF3 if the reaction has a 64.9% yield?

% yield = actual / theoretical

0.649 = 85.0 g / t t = 130.97 = 131 g

131 g PF3 / 88 g/mol = 1.49 mol (1.5 mol)

ratio 4 mol PF3 to 1 mol P4 1.49/4=.3725 mol P4

.3725 mol x 124 g/mol = 46.19 = 46.2 g P4

85.0 g?? g

Section 3.10


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Exercise


P4(s) + 6F2(g) → 4PF3(g)

What mass of P4 is needed to produce 85.0 g of PF3 if the reaction has a 64.9% yield?

46.1 g P4

• The theoretical yield of a chemical reaction is the calculated quantity of product in the reaction.

• The actual yield is the amount you actually get when you carry out the reaction.

• Actual yield will be less than the theoretical yield, for many reasons … can you name some?

Yields of Chemical Reactions

actual yieldPercent yield = ––––––––––––– ×

100 theoretical yield

Example - % Yield

• 2CO(g) + O2(g) --> 2CO2(g)

Calculate the % yield if 69.1g of CO combines with excess O2 to form an experimental yield of 48.3L of CO2 @STP

Example - % Yield

• 2CO(g) + O2(g) --> 2CO2(g)


69.1 g / 28 g mol⋆ -1 = 2.47 mol CO thus creates same amount of moles of CO2. For gases only, 22.4 L = 1 mol of any gas, so 2.47 mol x 22.4 L / 1 mol = 55.3 L (theoretically from stoichiometry)

% yield = 48.3 / 55.3 (x 100) = 87.3%

• 2CO(g) + O2(g) --> 2CO2(g)


8 87.3% yield

444

4

2.47 mol

76

Actual Yield of ZnS Is Less than the Theoretical Yield

77

• Solute: the substance being dissolved.• Solvent: the substance doing the dissolving.• Concentration of a solution: the quantity of a

solute in a given quantity of solution (or solvent).– A concentrated solution contains a relatively large

amount of solute vs. the solvent (or solution).

– A dilute solution contains a relatively small concentration of solute vs. the solvent (or solution).

– “Concentrated” and “dilute” aren’t very quantitative …

Solutions andSolution Stoichiometry

78

Molarity (M), or molar concentration, is the amount of solute, in moles, per liter of solution:

• A solution that is 0.35 M sucrose contains 0.35 moles of sucrose in each liter of solution.

• Keep in mind that molarity signifies moles of solute per liter of solution, not liters of solvent.

Molar Concentration

moles of soluteMolarity = –––––––––––––– liters of solution

79

Preparing 0.01000 M KMnO4

Weigh 0.01000 mol

(1.580 g) KMnO4.

Dissolve in water. How much water? Doesn’t matter, as long as we don’t go over a

liter.

Add more water to reach the 1.000 liter

mark.

80

Example 3.23What is the molarity of a solution in which 333 g potassium hydrogen carbonate is dissolved in enough water to make 10.0 L of solution?

Example 3.24We want to prepare a 6.68 molar solution of NaOH (6.68 M NaOH).(a) How many moles of NaOH are required to prepare 0.500 L of 6.68 M NaOH?(b) How many liters of 6.68 M NaOH can we prepare with 2.35 kg NaOH?

Example 3.25The label of a stock bottle of aqueous ammonia indicates that the solution is 28.0% NH3 by mass and has a density of 0.898 g/mL. Calculate the molarity of the solution.

81

• Dilution is the process of preparing a more dilute solution by adding solvent to a more concentrated one.

• Addition of solvent does not change the amount of solute in a solution but does change the solution concentration.

• It is very common to prepare a concentrated stock solution of a solute, then dilute it to other concentrations as needed.

Dilution of Solutions

82

Dilution Calculations …

• … couldn’t be easier.• Moles of solute does not change on dilution.• Moles of solute = M × V• Therefore …

Mconc × Vconc = Mdil × Vdil

83Visualizing the Dilution of a Solution

We start and end with the same amount

of solute.

Addition of solvent has decreased

the concentration

.

84Example 3.26

How many milliliters of a 2.00 M CuSO4 stock solution are needed to prepare 0.250 L of 0.400 M CuSO4?

85

• Molarity provides an additional tool in stoichiometric calculations based on chemical equations.

• Molarity provides factors for converting between moles of solute (either reactant or product) and liters of solution.

Solutions in Chemical Reactions

86

If substance A is a solution of known concentration …

If substance B is in solution, then …

… we can start with molarity of A

times volume (liters) of the

solution of A to get here. … we can go from

moles of substance B to either volume of B or molarity of

B. How?

Adding to the previous stoichiometry scheme …


Limiting Reactants

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Chapter 3.6-3.10 Student Notes Stoichiometry. Chapter 3 Table of Contents Copyright © Cengage Learning. All rights reserved 2 3.1 Counting by Weighin3.1