chapter outline - western oregon university · chapter 7: stoichiometry - mass relations in...

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Chapter 7: Stoichiometry - Mass

Relations in Chemical Reactions

How do we balance chemical equations?

How can we used balanced chemical equations to

relate the quantities of substances consumed and

produced in chemical reactions?

How can we determine a compound’s elemental

composition and chemical formula?

→ →

Chapter Outline

7.1 Chemical reactions and the Conservation of Mass

7.2 Balancing Chemical Equations

7.3 Combustion Reactions

7.4 Stoichiometric Calculations and the Carbon Cyle

7.5 Percent Composition and Empirical Formulas

7.6 Empirical and Molecular Formulas Compared

7.7 Combustion Analysis

7.8 Limiting Reactants and Percent Yield

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Law of Conservation of Mass

The law of conservation of mass states that the sum

of the masses of the reactants of a chemical equation

is equal to the sum of the masses of the products.

C(s) + O2(g) → CO2(g)

12 g 32 g 44 g

1 mole 1 mole 1 mole

+ =

2 C(s) + O2(g) → 2 CO(g)

2 x 12 = 24 g

2 mole 1 mole 2 mole

32 g + 2 x 28 = 56 g =

Law of Conservation of Mass

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Chemical Equations

2 C(s) + O2(g) → 2 CO(g)

The “” symbol means “reaction proceeds in this direction” a “ “ symbol means the reaction is at equilibrium (s) = solid phase (l) = liquid phase (g) = gas phase (aq) = aqueous phase = heat

The 2’s are called “stoichiometric coefficients”

4 2

Stoichiometric Coefficients:

the ratios between reactants and/or products

1 mol O2

2 mol N2O5

2 mol N2O5

4 mol NO2

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Chapter Outline









Guidelines for Balancing

Chemical Equations 1. Write an expression using correct chemical formulas for the

reactants and products, separated by an arrow (). Include

symbols indicating physical states.

2. For each element, add up the numbers of atoms on each

side. Check whether the expression is already balanced. If

so, you’re done!

3. Otherwise - if present - choose an element that appears in

only one reactant and product to balance first. Insert the

appropriate coefficient(s) to balance this element.

4. Choose the element that appears in the next fewest total

reactants and products and balance it. Repeat the process

for additional elements if necessary.

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1. Write an expression using correct chemical formulas for the

reactants and products, separated by an arrow (). Include

symbols indicating physical states.

2. For each element, add up the numbers of atoms on each

side. Check whether the expression is already balanced. If

so, you’re done!

Example 1: balance the reaction that occurs between nitrogen

dioxide and water to form nitric acid and nitrogen monoxide

NO2(g) + H2O(l) → HNO3(aq) + NO(aq)

N = 1 N = 2

O = 3 O = 4

H = 2 H = 1

3. Otherwise - if present - choose an element that appears in

only one reactant and product to balance first. Insert the

appropriate coefficient(s) to balance this element.

= H, so try a 2 in front of HNO3

NO2(g) + H2O(l) → HNO3(aq) + NO(aq)

NO2(g) + H2O(l) → 2 HNO3(aq) + NO(aq)

N = 1 N = 3

O = 3 O = 7

H = 2 H = 2

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4. Choose the element that appears in the next fewest total

reactants and products and balance it. Repeat the process

for additional elements if necessary.

= N, try a 3 in front of NO2

NO2(g) + H2O(l) → 2 HNO3(aq) + NO(aq)

N = 3 N = 3

O = 7 O = 7

H = 2 H = 2

3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(aq)

Done!

Chapter Outline









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Combustion Reactions

The reaction of an organic compound with oxygen

to produce CO2 + H2O, for example, balance -

TIP FOR ALL COMBUSTION REACTIONS: Since oxygen

appears by itself, balance the other elements first, and then O2

CH4(g) + O2(g) → CO2(g) + H2O(g)

CH4(g) + O2(g) → CO2(g) + 2 H2O(g)

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

Done!

Example: C2H6 (ethane) is combusted in oxygen

Combustion Reactions

C2H6(g) + O2(g) → CO2(g) + H2O(g)

C2H6(g) + O2(g) → 2 CO2(g) + H2O(g)

C2H6(g) + O2(g) → 2 CO2(g) + 3 H2O(g)

C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g)

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)

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Chapter Outline




7.4 Stoichiometric Calculations and the Carbon Cycle





Stoichiometric Calculations and the Carbon Cycle

Photosynthesis: 6 CO2(g) + 6 H2O(g) C6H12O6(aq) + 6 O2(g)

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Mauna Loa Observatory CO2 Concentrations since 1960 -

CO2 emissions over the last 800,000 years -

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1. Write the balanced chemical equation

2. Convert the mass of the reactant into moles

3. Use coefficients in the balanced equation to calculate the

number of moles of product (stoichiometric ratio)

4. Convert moles of products into grams (or other desired

quantities)

Amounts of Reactants & Products

= STOICHIOMETRY

1/MM ratio MM

Stoichiometry Example, p. 282

If the combustion of fossil fuels adds 8.2 x 1012 kilograms of

carbon to the atmosphere each year as CO2, what is the mass

of added carbon dioxide?

Step 1: Write the balanced chemical equation

Step 2: Convert quantities of known substances (C) into moles

C(s) + O2(g) CO2(g)

8.2 x 1012 kg C

1000 g C

kg x

1 mol C

12.0 g C x 6.83 x 1014 mol C =

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Step 3: Use coefficients in balanced equation to calculate the

number of moles of CO2 (stoichiometric ratio)

Step 4: Convert moles of CO2 into grams

6.83 x 1014 mol C 1 mol CO2

1 mol C x = 6.83 x 1014 mol CO2

6.83 x 1014 mol CO2 44.0 g CO2

1 mol CO2

x = 3.00 x 1016 g CO2


Suppose we wish to prepare 1.00 kg of acetylsalicylic acid. How

many grams of salicylic acid and how many grams of acetic

anhydride are needed?

C9H8O4

MW =

180.16

C4H6O3

MW =

102.09

C7H6O3

MW =

138.12

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SA + AA ASA + Ac

1.00 kg ? g ? g


Step 2: Convert quantities of known substances (ASA) into moles

1.00 kg ASA

1 mol ASA

180.16 g ASA x

1000 g ASA

kg ASA x = 5.551 mol ASA



number of moles of SA and AA (stoichiometric ratio)

mol SA = 5.551 mol ASA

mol AA = 5.551 mol ASA

1 mol SA

1 mol ASA x

1 mol AA

1 mol ASA x

= 5.551 mol SA

= 5.551 mol AA

SA + AA ASA + Ac

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Step 4: Convert moles of SA and AA into grams


g SA = 5.551 mol SA

g AA = 5.551 mol AA

138.12 g SA

1 mol SA x

102.09 g AA

1 mol AA x

= 767 g SA

= 567 g AA

Sample Exercise 7.3 – Calculating the Mass of a

Product from the Mass of a Reactant.

Each year, power plants in the U.S. consume about 1.1 x 1011

kg of natural gas (CH4). How many kg of CO2 (MW = 44.01) are

released into atmosphere from these power plants. Given that

natural gas is mainly CH4 (MW = 16.04), base the calculation

on its combustion reaction –

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

1.1 x 1011 kg ? g


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Step 2: Convert quantities of known substances (CH4) into moles

Sample Exercise 7.3 – Calculating the Mass of a

Product from the Mass of a Reactant.

1.1 x 1011 kg CH4 1 mol CH4

16.04 g CH4

x 1000 g CH4

kg CH4

x = 6.86 x 1012 mol

CH4


number of moles of CO2 (stoichiometric ratio)

= 6.86 x 1012 mol CH4

1 mol CO2

1 mol CH4

x = 6.86 x 1012 mol CO2

Step 4: Convert moles of CO2 into grams

6.86 x 1012 mol CO2

44.01 g CO2

1 mol CO2

x = 3.0 x 1014 g CO2

Chapter Outline









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Percent Composition: the composition of a compound in terms

of the percentage by mass of each element in the compound

n x molar mass of element molar mass of compound

x 100%

n is the number of moles of the element in 1 mole

of the compound

%H = 3 x (1.008 g)

97.99 g x 100% = 3.086%

%P = 1 x (30.97 g)

97.99 g x 100% = 31.60%

%O = 4 x (16.00 g)

97.99 g x 100% = 65.31%

3.086% + 31.60% + 65.31% = 99.96%

= 100%

H3PO4 MM = 97.99

g/mol

Sample Exercise 7.4 – Calculating Percent

Composition from a Chemical Formula

The mineral forsterite: Mg2SiO4,

MW = 140.71

%Mg = 2 x (24.31 g Mg)

140.71 g x 100% = 34.55%

%Si = 1 x (28.09 g Si)

140.71 g x 100% = 19.96%

%O = 4 x (16.00 g O)

140.71 g x 100% = 45.48%

34.55% + 19.96% + 45.48% = 99.99%

= 100%

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A formula showing the smallest whole number ratio of elements in a compound, e.g.

Benzene:

» Empirical = CH

» Molecular = C6H6

Glucose

» Empirical = CH2O

» Molecular = C6H12O6

Empirical Formula from % Composition

Empirical Formula from % Composition

1. Assume 100 g

2. Convert to moles

3. Divide by fewest number of moles

4. Convert the mole

ratio from step 3

into small whole

numbers if

necessary

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Sample Exercise 7.6

A sample of the carbonate mineral dolomite is 21.73% Ca,

13.18% Mg, 13.03% C, and the rest is oxygen. What is its

empirical formula?

Ca = 21.73 g

Mg = 13.18 g

C = 13.03 g

% O = 100 – 21.73 – 13.18 – 13.03 = 52.06 %

O = 52.06 g

1 mol Ca

40.078 g x = 0.54219 mol Ca

1 mol Mg

24.305 g x = 0.54228 mol Mg

1 mol C

12.011 g x = 1.0848 mol C

1 mol O

15.999 g x = 3.2540 mol O

1. Assume 100 g

2. Convert to moles

Sample Exercise 7.6

0.54219

0.54228

1.0848

3.2540


0.54219 Ca =

Mg =

C =

O =

0.54219

0.54219

0.54219

= 1.0

= 1.0

= 2.0

= 6.0

4. Convert the mole ratio from step 3 into small whole

numbers if necessary – NOT REQUIRED HERE

Empirical formula

= CaMgC2O6

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Example Illustrating Step 4

Vanillin is a common flavoring agent. It has a molar mass of 152 g/mol and is 63.15 %C and 5.30 %H; the rest is oxygen. What are the empirical and molecular formulas?

% O = 100 – 63.15 – 5.30 = 31.55 %

C = 63.15 g

H = 5.30 g

O = 31.55 g

1 mol C

12.011 g x = 5.258 mol C

1 mol H

1.008 g x = 5.258 mol H

1 mol O

15.999 g x = 1.972 mol O

1. Assume 100 g

2. Convert to moles

5.258

5.258

1.972


1.972 C =

H =

O =

1.972

1.972

= 2.67

= 2.67

= 1.0

4. Convert the mole ratio from step 3 into small whole numbers if necessary

Example Illustrating Step 4

Note that 2.67 = 2 2

3

So multiply by 3 which = 8

C =

H =

O =

2.67 x 3 = 8

1.0 x 3 = 3

2.67 x 3 = 8 Empirical formula =

C8H8O3

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Chapter Outline









Empirical and Molecular

Formulas Compared

The molecular formula can be determined from the empirical

formula if the molecular weight of the compound is known.

glycolaldehyde

empirical = CH2O

formula weight (FW) = 30 g/mol

molecular = C2H4O2

Molecular weight (MW) = 60 g/mol

Note that molecular formula = empirical x 2

C2H4O2 = (CH2O) x 2 = C2H4O2

Or more generally –

molecular = (empirical) x R

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Empirical and Molecular

Formulas Compared

molecular = (empirical) x R it follows that -

MW = (FW) x R and therefore - R = MW

FW

e.g. glucose – assume we know the

MW = 180.0

Empirical = CH2O FW = 30.0

So molecular = (CH2O) x 6 = C6H12O6

R = 180.0

30.0 = 6

Molecular Mass and Mass

Spectrometry

Acetylene

C2H2

Benzene

C6H6

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Sample Exercise 7.7 – Using Percent Composition

and Molecular weight to Derive a Molecular Fomula

Pheromones are chemical substances secreted by members of a species

to stimulate a response in other individuals of the same species. The

percent composition of eicosene, a compound similar to the Japanese

beetle mating pheromone, is 85.63% C and 14.37% H. Its molecular mass,

as determined by mass spectrometry, is 280 amu. What is the molecular

formula of eicosene?

C = 85.63 g

H = 14.37 g

1 mol C

12.011 g x = 7.129 mol C

1 mol H

1.008 g x = 14.26 mol H

1. Assume 100 g

2. Convert to moles

7.129

14.26


7.129 C =

H = 7.129

= 1.00

= 2

4. Convert the mole ratio from step 3 into small whole numbers if necessary

– NOT NEEDED

Sample Exercise 7.7 – Using Percent Composition

and Molecular weight to Derive a Molecular Fomula

R = MW

FW

280

14 = 20 =

So molecular = (CH2) x 20 = C20H40

Empirical formula = CH2

FW = 14.0

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Chapter Outline









Experimental Determination of Empirical

Formulas: Combustion Analysis

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Experimental Determination of Empirical

Formulas: Combustion Analysis

Calculation Outline:

CxHyOz + O2(g) x CO2(g) + y/2 H2O(g) excess mass

sample

Some of the oxygen comes

from the sample, some from

the excess O2

g CO2 mol CO2 mol C g C

g H2O mol H2O mol H g H

g of O = g of sample – g of C - g of H

Sample Exercise 7.8: Deriving an Empirical

Formula from Combustion Analysis Data

Combustion of 1.000 grams of an organic compound known to

contain only C, H and O produces 2.360 g CO2 and 0.640 g

H2O. What is the empirical formula of the compound?

mol C = 2.360 g CO2

12.011 g C

1 mol C x

1 mol CO2

44.0 g CO2 x

g C =

= 0.05364 mol C

0.05364 mol C

1 mol C

mol CO2 x

= 0.6442 g C

mol H = 0.640 g H2O

1.008 g H

1 mol H x

1 mol H2O

18.0 g H2O x

g H =

= 0.07111 mol H

= 0.07111 mol H

2 mol H

mol H2O x

= 0.07168 g H

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g of O = g of sample – g of C - g of H

g of O = 1.000 g – 0.6442 g C - 0.07168 g H

g of O = 0.2841 g

and so the moles of O = 0.2841 g 1 mol O

15.999 g O x

= 0.01776 mol O



0.05364 mol C

0.07111 mol H

0.01776 mol O

0.01776 C =

H =

O =

0.01776

0.01776

= 3

= 4

= 1

Empirical formula =

C3H4O

Now divide by the smallest number of moles -

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Chapter Outline









Limiting Reagents

Limiting Reagents - a reactant that is consumed completely

in a chemical reaction before the other reactant(s) run out.

The amount of product formed depends on the amount of the

limiting reagent available.

Each sandwich consists of 2 slices bread, 1 slice of cheese, and 1 slice of salami

8 slices

bread

4 slices

cheese 3 slices

salami

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2 Br + 1 Ch + 1 Sal 1 sandwich

Limiting Reagents

8 4 3

Br in excess, so need

more salami! = LR

Stoichiometric ratio: 2 Br = 2.0

1 Sal

Given ratio: 8 Br = 2.7 tip: use a ratio > 1

3 Sal

Here’s the logic – compare the given number of moles to

the stoichiometric ratio

Given:

Strategy for determining which

reactant is the Limiting Reagent:

aA + bB C

then A is in excess and B is the limiting reagent

𝑚𝑜𝑙𝑒𝑠 𝐴

𝑚𝑜𝑙𝑒𝑠 𝐵𝑔𝑖𝑣𝑒𝑛

> 𝑎

𝑏 𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐 if

1.Convert grams of each into moles

2.Calculate the stoichiometric ratio that’s the

largest, e.g. a/b or b/a

3.Calculate the given mole ratio in the same way

4.Compare to identify the LR

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Sample Exercise 7.9: Identifying the

Limiting Reagent in a Reaction Mixture

The flame in an acetylene torch reaches temperatures as high as 3500 oC as a result of the combustion of a mixture of acetylene (C2H2) and pure oxygen. If these two gases flow from high-pressure tanks at the rates of 52.0 g C2H2 and 188 g O2 per minute, which reactant is the limiting reagent, or is the mixture stoichiometric?

2 C2H2 + 5 O2 4 CO2 + 2 H2O

52.0 g 188 g

mol C2H2

Given ratio:

= 52.0 g C2H2

1 mol C2H2

26.0 g C2H2

x = 2.00 mol C2H2

mol O2 = 188 g O2

1 mol O2

32.0 g O2

x = 5.88 mol O2

2.00 mol C2H2

5.88 mol O2

= 2.94

tip: use a ratio > 1

Stoichiometric ratio:

Sample Exercise 7.9: Identifying the

Limiting Reagent in a Reaction Mixture

5 mol O2

2 mol C2H2 = 2.5

Given ratio:

= 2.94 2.00 mol C2H2

5.88 mol O2

therefore O2 is in excess and so C2H2 is the LR

𝑖𝑠 𝑚𝑜𝑙𝑒𝑠 𝑂2

𝑚𝑜𝑙𝑒𝑠 𝐶2𝐻2 𝑔𝑖𝑣𝑒𝑛

> 5 𝑚𝑜𝑙 𝑂2

2 𝑚𝑜𝑙 𝐶2𝐻2 𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐

? ? ?

2 C2H2 + 5 O2 4 CO2 + 2 H2O

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Theoretical Yield is the maximum amount of product

formed from given quantities of reactants (check if a

limiting reagent is present).

Actual Yield is the amount of product actually obtained

from a reaction.

% Yield = Actual Yield

Theoretical Yield x 100

Percent Yield

Sample Exercise 7.10

The industrial process for making the ammonia used in fertilizer, explosives, and many other products is based on the reaction between nitrogen and hydrogen at high temperature and pressure.

If 18.2 kg of NH3 (MW = 17.03) is produced by a reaction mixture that initially contains 6.00 kg of H2 (MW = 2.016) and an excess of N2, what is the percent yield of the reaction?

N2(g) + 3 H2(g) → 2 NH3(g)

Actual yield =

18.2 kg

6.00 kg Excess

(no LR)

Calculate the theoretical yield of NH3 based on H2, and then calculate the %

yield -



g NH3 = 6.00 kg H2 kg

1000 g x

3 mol H2

2 mol NH3 x mol NH3

17.03 g NH3 x

= 3.379 x 104 g NH3

mol H2

2.016 g H2

x

= 33.79 kg NH3 = 33.8 kg NH3

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Sample Exercise 7.10

Actual Yield = 18.2 kg

Theoretical Yield = 33.79 kg

% Yield = 18.2 kg

33.79 kg x 100

= 53.8 %

chapter outline - western oregon university · chapter 7: stoichiometry - mass relations in...

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