CHAPTER 11 Stoichiometry 11.1 Analyzing a Chemical Reaction

CHAPTER 11

Stoichiometry

11.1 Analyzing a Chemical Reaction

2 11.1 Analyzing a Chemical Reaction

Chemical equations tell stories… 2CO(g) + O2(g) → 2CO2(g)


Chemical equations tell stories…

… and stories can be put into different categories

Nonfiction

Science fiction

Adventure

Romance

History

Psychology

Children’s literature

…

2CO(g) + O2(g) → 2CO2(g)




Nonfiction

Science fiction

Adventure

Romance

History

Psychology

Children’s literature

…

Synthesis / Decomposition

Single / Double replacement

Precipitate reaction

Polymerization reaction

2CO(g) + O2(g) → 2CO2(g)




Synthesis / Decomposition

Single / Double replacement

Precipitate reaction

Polymerization reaction

2CO(g) + O2(g) → 2CO2(g)



But what exactly do they tell us?

2CO(g) + O2(g) 2CO2(g)




2CO(g) + O2(g) 2CO2(g)

They tell us what compounds we start with:Carbon monoxide (CO) gas

Oxygen (O2) gas




2CO(g) + O2(g) 2CO2(g)

They tell us what compounds we start with:Carbon monoxide (CO) gas

Oxygen (O2) gas

what compounds are formed:Carbon dioxide (CO2) gas



What else do they tell us?

2CO(g) + O2(g) 2CO2(g)




2CO(g) + O2(g) 2CO2(g)

2 CO molecules 2 CO2 molecules1 O2 molecules

They tell us how much of each compound is involved




2CO(g) + O2(g) 2CO2(g)

stoichiometry: the study of the amounts of substances involved in a chemical reaction.

2 CO molecules 2 CO2 molecules1 O2 molecules

They tell us how much of each compound is involved


2CO(g) + O2(g) 2CO2(g)

2 CO molecules2 dozen CO molecules

1 O2 molecules1 dozen O2 molecules

2 CO2 molecules2 dozen CO2 molecules


2CO(g) + O2(g) 2CO2(g)

2 CO molecules2 dozen CO molecules2 moles CO molecules


1 mole O2 molecules

2 CO2 molecules2 dozen CO2 molecules2 moles CO2 molecules


2CO(g) + O2(g) 2CO2(g)

2 CO molecules2 dozen CO molecules2 moles CO molecules

2 x (6.022 x 1023) CO molecules


1 mole O2 molecules(1 x) 6.022 x 1023 O2 molecules

2 CO2 molecules2 dozen CO2 molecules2 moles CO2 molecules

2 x (6.022 x 1023) CO2 molecules


2CO(g) + O2(g) 2CO2(g)

2 moles

CO molecules

1 mole

O2 molecules

2 moles

CO2 molecules

Number of moles is not conserved

Is that okay?

≠+


2CO(g) + O2(g) 2CO2(g)

2 moles

CO molecules

1 mole

O2 molecules

2 moles

CO2 molecules

Number of moles is not conserved

Is that okay?

Yes, as long as the chemical equation is balanced!

≠+


2CO(g) + O2(g) 2CO2(g)

2 C atoms

2 O atoms2 O atoms

2 C atoms

4 O atoms

Number of atoms is

conserved

This chemical equation is balanced

=

=+


2CO(g) + O2(g) 2CO2(g)

2 moles

CO molecules

1 mole

O2 molecules

2 moles

CO2 molecules


The coefficients are correct

Coefficients


2CO(g) + O2(g) 2CO2(g)

2 moles

CO molecules

1 mole

O2 molecules

2 moles

CO2 molecules


The coefficients are correct

CoefficientsThese are important!


Coefficients are important

1 bagcake mix

+ 3 eggs + ¼ cup oil + 1 cup water 1 batch cupcakes



+ 3 eggs + ¼ cup oil + 1 cup water 1 batch cupcakes1 bagcake mix

I need ¼ cup of oil to make 1 batch of cupcakes




1 14114

cup oil batch cupcakesor

batch cupcakes cup oilWrite as a ratio:




I need 1 cup of oil to make 4 batches of cupcakes


Write as a ratio:



1 4

4 1

cup oil batches cupcakesor

batches cupcakes cup oil



Fermentation of sugar (glucose) into alcohol:

C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)





1 moleglucose

2 molesethanol

2 molescarbon dioxide





1 moleglucose

2 molesethanol


1 moles

3 molesglucose will yield

2 moles

6 moles

ethanol/ carbon dioxide

x 3





1 moleglucose

2 molesethanol


1 moles

3 moles

7.5 moles

glucose will yield

2 moles

6 moles

15 moles

ethanol/ carbon dioxide

x 7.5x 3





1 moleglucose

2 molesethanol


7.5C6H12O6(aq) 15C2H5OH(aq) + 15CO2(g)

7.5 molesglucose

15 molesethanol


You MUST multiply all the coefficients by the same number




1 moleglucose

2 molesethanol


Write as a ratio: 1 2

2 1

mole glucose moles ethanol

moles ethanol mole glucose

These are stoichiometric equivalents




1 moleglucose

2 molesethanol


Write as a ratio: 1 2

2 1



mole ratio: a ratio comparison between substances in a balanced equation. It is obtained from the coefficients in the balanced equation.

32 11.1 Analyzing a Chemical Reaction11.1 Analyzing a Chemical Reaction



1 moleglucose

2 molesethanol


1 2

2 1



2

2

21

2 1

moles COmole glucose

moles CO mole glucose

2

2

22

2 2

moles COmoles ethanol

moles CO moles ethanol

mole ratios for this

chemical equation

Mole ratios


CO(g) + 2H2(g) CH3OH(l)

Mole ratios

Consider the following equation:

Compare the reactant CO to the product CH3OH.

carbon monoxide

hydrogen methanol



Mole ratios


carbon monoxide

hydrogen methanol

3

3

11

1 1

molemole

m

CH

ol

OHCO

e mCH OH Oole C

Compare the reactant CO to the product CH3OH.



Mole ratios


carbon monoxide

hydrogen methanol

If the reaction produces 5 moles of CH3OH, how many moles of H2 are consumed?



Mole ratios


carbon monoxide

hydrogen methanol


Asked: moles of H2

Given: moles of CH3OH



Mole ratios


carbon monoxide

hydrogen methanol


Asked: moles of H2


Write a mole ratio between H2 and CH3OH

Write moles of H2 (asked) “on top”


2

33

25

1

moles H

mole CH Omoles CH H

HO


Mole ratios


carbon monoxide

hydrogen methanol


“on top”Asked: moles of H2Write a mole ratio between H2 and CH3OH

Write moles of H2 (asked) “on top”



Mole ratios


carbon monoxide

hydrogen methanol


2

33

25

1

moles H

mole CH Omoles CH H

HO




Mole ratios


carbon monoxide

hydrogen methanol


2

323

25

110

moles Hmoles CH OH

molmo

eCH OH

Hles

Asked: moles of H2


A mixture of aluminum metal and chlorine gas reacts to form aluminum chloride (AlCl3): 2Al(s) + 3Cl2(g) → 2AlCl3(s). How many moles of aluminum chloride will form when 5 moles of chlorine gas react with excess aluminum metal?



Asked: moles AlCl3

Given: moles Cl2



Asked: moles AlCl3

Given: moles Cl2

Write a mole ratio between AlCl3 and Cl2

Write moles of AlCl3 (asked) “on top”



23

233.3

25

3moles C

moles AlCl

moles Cll moles AlCl

Asked: moles AlCl3

Given: moles Cl2“on top”

Write a mole ratio between AlCl3 and Cl2

Write moles of AlCl3 (asked) “on top”



Asked: moles AlCl3

Given: moles Cl2

23

233.3

25

3moles C

moles AlCl

moles Cmoll

les AlCl



Asked: moles AlCl3

Given: moles Cl2

23

233.3

25

3moles C

moles AlCl

moles Cmoll

les AlCl

units cancel out



Asked: moles AlCl3

Given: moles Cl2

32

23

2

33.35

moles AlClmoles Cl

molmoles A

es Cl

llC

units cancel out



Asked: moles AlCl3

Given: moles Cl2

32

23

2

33.35

moles AlClmoles Cl

molmoles A

es Cl

llC

Answer: 3.3 moles of AlCl3 are produced.


There is no scale that measures in moles!

How do you convert from moles to grams?

The mass of 1 mole of Al is not the same as the mass of 1 mole of Cl2.

How do you convert from grams of Al to grams of Cl2?


There is no scale that measures in moles!

How do you convert from moles to grams?

The mass of 1 mole of Al is not the same as the mass of 1 mole of Cl2.

How do you convert from grams of Al to grams of Cl2?

By using the molar mass (g/mole)

By using the molar mass (g/mole) and mole ratios


Process for calculating grams from grams given


If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are produced?

Calcium carbonate (CaCO3) is present in rocks, eggshells and most seashells.

CaCO3 decomposes when it is heated.



Asked: grams of CO2 Given: grams of CaCO3

Relationships: mole ratios molar mass of CaCO3 = 40.078 + 12.011 + (15.999 x 3) = 100.0 g/molemolar mass of CO2 = 12.011 + (15.999 x 2) = 44.01 g/mole

Strategy:




Relationships: mole ratios, molar mass of CaCO3 = 100.0 g/molemolar mass of CO2 = 44.01 g/mole

45.0 g CaCO3





33 3

3

145 0.45

100.09

mole CaCOg CaCO moles CaCO

g CaCO

45.0 g CaCO3 0.45 moles CaCO3





0.45 moles CaCO3





23 2

3

10.45 0.45

1

mole COmoles CaCO moles CO

mole CaCO

0.45 moles CaCO3 0.45 moles CO2





0.45 moles CO2





22

22

44.010.45

119.8

g COmoles CO

mole COg CO

0.45 moles CO2 19.8 g CO2





0.45 moles CO2 19.8 g CO245 g CaCO3 0.45 moles CaCO3

Answer: 19.8 g of CO2 are produced.



1 moleglucose

2 molesethanol


1 2

2 1



These are stoichiometric equivalents

Mole ratios:

Documents

CHAPTER 11 Stoichiometry 11.1 Analyzing a Chemical Reaction