chapter 3: stoichiometry

Chapter 3: StoichiometryChapter 3: Stoichiometry

● “measuring elements”

● Must account for ALL atoms in a chemical reaction

+ →

H2

+ O2

→ H2O22

Chapter 3: StoichiometryChapter 3: Stoichiometry

CO + O2

→ CO2

22

+ →

CH4

+ Cl2

→ CCl4

+ HCl

→+ +

4 4

Chapter 3: StoichiometryChapter 3: Stoichiometry

C2H

4+ O

2→ CO

2+ H

2O2 23

Al + HCl → AlCl3

+ H2

2 2

6

2 3

Al + HCl → AlCl3

+ H2

2 23

23 x 3

2 Al + 6 HCl → 2 AlCl3

+ 3 H2

or:

Chapter 3: StoichiometryChapter 3: Stoichiometry

NH4NO

3→ N

2 + O

2 + H

2O21

2x 2

2 NH4NO

3→ 2 N

2 + O

2 + 4 H

2O

Chapter 3: StoichiometryChapter 3: Stoichiometry

Three basic reaction types:

● Combination Reactions

● Decomposition Reactions

● Combustions (in air)

Chapter 3: StoichiometryChapter 3: Stoichiometry

Combination Reactions

Two or more reactants combine to form a single product

C (s) + O2 (g) → CO

2 (g)

Decomposition Reactions

A single reactant breaks into two or more products

2 KClO3 (s) → 2 KCl (s) + 3 O

2 (g)

2 Na (s) + Cl2 (g) → 2 NaCl (s)

Chapter 3: StoichiometryChapter 3: Stoichiometry

Combustions in Air = reactions with oxygen

Write the balanced reaction equation for the combustion of magnesium to magnesium oxide:

Mg (s) + O2 (g) →

Mg2+ O2-metal + nonmetal = ionic compound: => MgO

MgO (s)22

Chapter 3: StoichiometryChapter 3: Stoichiometry

Combustions of Hydrocarbons in Air

= reactions with oxygen to form carbon dioxide and water(complete combustion)

Write the balanced reaction equation for the combustion of C2H

4 gas

C2H

4 (g) + O

2 (g) → CO

2 (g) + H

2O (g)2 23

Chapter 3: StoichiometryChapter 3: Stoichiometry

C2H

4 (g) + O

2 (g) → CO

2 (g) + H

2O (g)2 23

+ → +

How many C2H

4 molecules are in the flask?

● If you know the weight of one molecule of C2H

4

and the total weight of gas in the flask, you cancalculate the number of molecules in the flask

Chapter 3: StoichiometryChapter 3: Stoichiometry

Molecular weight / Formula weight:

=> sum of all atomic weights in molecular formula

MW of C2H

4 = 2 x 12.0 amu + 4 x 1.0 amu = 28.0 amu

FW of Mg(OH)2 = 1 x 24.3 amu + (16.0 amu + 1.0 amu) x 2

= 24.3 amu + 34.0 amu

= 58.3 amu

Chapter 3: StoichiometryChapter 3: Stoichiometry

Molar Mass

= mass of one mole of a substance in grams

FW or MW of substance in amu's = mass of 1mole of substance in grams

FW of Ca(NO3)

2 = 164.1 amu

Molar Mass of Ca(NO3)

2 = 164.1 g/mol

MW of O2 = 2 x 16.0 amu = 32 amu

Molar Mass of O2 = 32 g/mol

Chapter 3: StoichiometryChapter 3: Stoichiometry

Number of individual molecules are difficult to deal with

=> definition of a “package” of molecules or particles

1 dozen eggs = 12 individual eggs

1 sixpack of cans = 6 cans

1 mole of molecules = 6.02 x 1023 individual molecules

Avogadro's Number.

... .... ..

. ...... . ..

..

. . ...

...

.....

...

Chapter 3: StoichiometryChapter 3: Stoichiometry

1 dozen eggs = 12 individual eggs

How many sixpacks of eggs are in an egg carton that holds 12 eggs?

12 individual eggs ×1 mole of eggs

6.02×1023 individual eggs= moles of eggs

How many moles of eggs are in an egg carton that holds 12 eggs?

12 individual eggs × = 21 sixpack

6 individual eggssixpacks

2.0 x 10-23

Chapter 3: StoichiometryChapter 3: Stoichiometry

Ca(NO3)

2

Type of compound: ionic

Ions: Ca2+NO

3-

Total number of oxygen atoms: 6

Name: Calcium nitrate

Chapter 3: StoichiometryChapter 3: Stoichiometry

Ca(NO3)

2

How many moles of calcium nitrate are in 394g of Ca(NO3)

2 ?

= moles Ca(NO3)

2 2.4

Molar Mass of Ca(NO3)

2 = 164.1 g/mol

394 g Ca(NO3)

2 x

164.1 g

1 mol

mol → gramMM

0.527 moles CaSO4· 2 H2O O2HCaSO1molO2HCaSO172g

24

24

Chapter 3: StoichiometryChapter 3: Stoichiometry

What is the mass in grams of 0.527 moles of

calcium sulfate dihydrate (gypsum), CaSO4· 2 H2O ?

(1) determine molar mass (MM) of CaSO4· 2 H2O

MW = (40+32+4*16)+2*(2*1+16) amu = 172 amu

MM = 172 g/mol

(2) use MM to convert moles into grams

= g CaSO4· 2 H2O90.6

mol → gramMM

Chapter 3: StoichiometryChapter 3: Stoichiometry

Ca(NO3)

2

How many moles of oxygen are in 2.4 moles of Ca(NO3)

2 ?

1 formula unit of Ca(NO3)

2 contains 6 oxygen atoms

1 mole of Ca(NO3)

2 contains 6 moles of oxygen atoms

2.4 moles Ca ( NO3 )2 × 6moles oxygen1moleCa ( NO3 )2

= moles oxygen14.4

Chapter 3: StoichiometryChapter 3: Stoichiometry

How many moles of shoes are in 2.4 moles of Gretel(shoes)2 ?

1 unit of Gretel(shoes)2 contains 2 shoes

1 mole ofGretel(shoes)2 contains 2 moles of shoes

= moles shoes4.8

Gretel(shoes)2

2.4 moles of Gretel(shoes)2 1 mol of Gretel(shoes)2

2 moles of shoesx

Chapter 3: StoichiometryChapter 3: Stoichiometry

What is the percentage of shoes, by mass, in Gretel(shoes)2?

mass of Gretel(shoes)2 = 55.2 kg + 0.5kg x 2

= 56.2 kg

(1) total mass of Gretel(shoes)2

(2) mass of shoes in Gretel(shoes)2

2 x 0.5kg = 1.0 kg

(3) percentage of shoe mass

Gretel(shoes)2

mass % shoesmass of shoes

total mass100%

1.0 kg56.2kg

100% 1.8%

Chapter 3: StoichiometryChapter 3: Stoichiometry

Ca(NO3)

2

What is the percentage of oxygen, by mass, in calcium nitrate?

FW of Ca(NO3)

2 = 40.1 amu + (14.0 amu + 3 x 16.0 amu) x 2

= 164.1 amu

(1) total mass of Ca(NO3 )

2 in amu

(2) mass of oxygen in compound, in amu

6 x 16.0 amu = 96.0 amu

(3) percentage of oxygen, by mass

% O = mass of oxygentotal mass

× 100 % = 96amu164.1 amu

× 100% = 58.5%mass

Chapter 3: StoichiometryChapter 3: Stoichiometry

Quantitative Information from Balanced Equations

CO + O2

→ CO2

22

How many grams of CO2 would be produced by the combustionof 2 moles of CO?

+ →

2 moles CO + 1 mole O2 → 2 moles CO2

2 x 28 g + 32 g → 2 x 44 g

= 88 g

Chapter 3: StoichiometryChapter 3: Stoichiometry

Quantitative Information from Balanced Equations

CO + O2

→ CO2

22

You can write a series of stoichiometric factors for this reaction:

2 mol CO

1 mol O2 2 mol CO

1 mol O2 1 mol O2

2 mol CO2

Chapter 3: StoichiometryChapter 3: Stoichiometry

C2H

4 (g) + O

2 (g) → CO

2 (g) + H

2O (g)2 23

How many grams of H2O are formed from the complete combustionof 2.0 g of C2H4?

28 g/mol 18 g/mol

2 g C2H442

42

28

1

HCg

HCmol42

2

12

HCmolOHmol

OHmolOHg

2

2

118 = 2.6 g H2O

from balanced equation

grams C2H

4→ moles C

2H

4→ moles H

2O → grams H

2O

1

need balanced reaction equation !!

Chapter 3: StoichiometryChapter 3: Stoichiometry

Summary

1) determine equation for the reaction

2) balance equation

4) determine MW/FW of substances involved

5) determine stoichiometric factors from balanced equation

3) formulate problem:how much of A => gets converted into how much of B

Chapter 3: StoichiometryChapter 3: Stoichiometry

Limiting Reactants

½ cup Ginger ale

+ ½ cup lime soda

+ 1 Tbsp. Grenadine syrup

+ 1 Maraschino cherry

Shirley Temple Cocktail

How many ST’s can you make from 4 cups Ginger ale, 2 cups of lime sodaa bottle of Grenadine syrup, and 10 maraschino cherries?

a. 2 b. 4 c. 6 d.8

Chapter 3: StoichiometryChapter 3: Stoichiometry

Limiting Reactants

½ cup Ginger ale

+ ½ cup lime soda

+ 1 Tbsp. Grenadine syrup

+ 1 Maraschino cherry

Shirley Temple Cocktail

How many ST’s can you make from 4 cups Ginger ale, 4 cups of lime sodaa bottle of Grenadine syrup, and 100 grams of maraschino cherries?

You need to know how many maraschino cherries are in 100 grams!

Chapter 3: StoichiometryChapter 3: Stoichiometry

Limiting Reactants

+

is the limiting reactant!

The amount of limits the amount of product that can be formed

Chapter 3: StoichiometryChapter 3: Stoichiometry

Limiting Reactants

Limiting Reactant

- limits the amount of product that can be formed

- reacts completely (disappears during the reaction)

- other reactants will be left over, i.e. in excess

Chapter 3: StoichiometryChapter 3: Stoichiometry

Limiting Reactants: how much NH3 can be formed from 3 moles N2 and 6 moles H2?

N2 + 3 H

2 → 2 NH

3

3 mol N2

3 mol N2 , 6 mol H

2Available (given):

How much H2 would we need to completely react 3 mol N2:

2

2

N 1H 3

1

= 9 mol H2

We only have 6 mol H2 available – it is limiting !

How much NH3 can we form with the available reagents?

continue with limiting reagent

6 mol H2

2

3

H 3NH 2

= 4 mol NH3

compare with what is available

Chapter 3: StoichiometryChapter 3: Stoichiometry

Limiting Reactants

N2 + 3 H

2 → 2 NH

3

3 mol N2 , 6 mol H

2Available (given):

1

How much N2 is left over (in excess)?

continue with limiting reagent

6 mol H2

2

2

H 3N 1

= 2 mol N2

this is the amount that reacts

H2 is limiting, there is plenty of N2

How much N2 is actually reacting?

Available – amount reacted = left over

3 mol – 2 mol = 1 mol N2 left over

Chapter 3: StoichiometryChapter 3: Stoichiometry

Limiting Reactants

0.5 mol Al

Available (given):

Al 2Cl 3

2 = 0.75 mol Cl2

We have more than enough Cl2 available – it is in excess !

How much AlCl3 can we form with the available reagents?

continue with limiting reagent

0.5 mol AlAl 2

AlCl 2 3 = 0.5 mol AlCl3

compare with what is available

2 Al + 3 Cl2 → 2 AlCl

3

0.5 mol Al , 2.5 mol Cl2

How much Cl2 would we need to completely react 0.5 mol Al:

If Cl2 is in excess, Al must be limiting !

Chapter 3: StoichiometryChapter 3: Stoichiometry

Theoretical Yield

What mass do 0.5 mol AlCl3 correspond to?

0.5 mol AlCl33

3

AlCl molAlCl g 133.5

= 67 g AlCl3

2 Al + 3 Cl2 → 2 AlCl

3

The maximum mass of product that can be formed is thetheoretical yield

Available (given): 0.5 mol Al , 2.5 mol Cl2

Chapter 3: StoichiometryChapter 3: Stoichiometry

Theoretical Yield

Fritz does the reaction with the available reagents he only ends up with 34g. What is the % yield of the reaction?

2 Al + 3 Cl2 → 2 AlCl

3

% yield = actual yieldtheoretical yield

× 100 %

%1006734

% gg

yield = 51 %

Available (given): 0.5 mol Al , 2.5 mol Cl2

Chapter 3: StoichiometryChapter 3: Stoichiometry

Summary

Determine availabequantity of reactants

in moles

Determine if one ofthe reactants is a limiting reactant

Determine the maximum

# of moles of productthat can be formed

Convert into grams ofproduct

(theoreticalyield)

Compare with actual

amount of productrecovered

(actual yield)

Determine % yieldof the reaction