Download Chapter 3: Stoichiometry

Chapter 3: Stoichiometry
●
“measuring elements”
●
Must account for ALL atoms in a chemical reaction
→
+
2 H2
+
O2
→
2 H 2O
Chapter 3: Stoichiometry
→
+
2 CO
+
+
2 CO2
→
+
CH4
→
O2
4 Cl2
→
+
CCl4
+
4 HCl
Chapter 3: Stoichiometry
C 2H 4
3 O2
→
2 HCl
→
2 AlCl3
+
3 H2
→
2
3
+
H2
+
2 Al
+
2CO2
2 H 2O
+
6
or:
2
3
Al
2 Al
2 HCl
+
+
6 HCl
→
AlCl3
2 AlCl3
+
3 H2
x3
Chapter 3: Stoichiometry
NH4NO3
→
N2
+
1
O
2 2
+
2 H2O
2 NH4NO3
→
2 N2
+
O2
+
4 H 2O
x2
Chapter 3: Stoichiometry
Three basic reaction types:
●
Combination Reactions
●
Decomposition Reactions
●
Combustions (in air)
Chapter 3: Stoichiometry
Combination Reactions
Two or more reactants combine to form a single product
C (s) + O2 (g)
→
CO2 (g)
2 Na (s) + Cl2 (g)
→
2 NaCl (s)
Decomposition Reactions
A single reactant breaks into two or more products
2 KClO3 (s)
→
2 KCl (s)
+
3 O2 (g)
Chapter 3: Stoichiometry
Combustions in Air = reactions with oxygen
Write the balanced reaction equation for the
combustion of magnesium to magnesium oxide:
metal + nonmetal = ionic compound: Mg2+ O2- => MgO
2 Mg (s)
+ O2 (g)
→
2 MgO (s)
Chapter 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 C2H4 gas
C2H4 (g)
+
3 O2 (g)
→
2 CO2 (g)
+
2 H2O (g)
Chapter 3: Stoichiometry
C2H4 (g)
+
3 O2 (g)
→
→
+
2 CO2 (g)
+
2 H2O (g)
+
How many C2H4 molecules are in the flask?
If you know the weight of one molecule of C2H4
and the total weight of gas in the flask, you can
calculate the number of molecules in the flask
●
Chapter 3: Stoichiometry
Molecular weight / Formula weight:
=> sum of all atomic weights in molecular formula
MW of C2H4 = 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: 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: 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: Stoichiometry
1 dozen eggs
=
12 individual eggs
How many sixpacks of eggs are in an egg carton that holds 12 eggs?
12individual eggs ×
1 sixpack
= 2 sixpacks
6 individual eggs
How many moles of eggs are in an egg carton that holds 12 eggs?
12individual eggs ×
1 mole of eggs
-23
2.0
x
10
=
moles of eggs
23
6.02× 10 individual eggs
Chapter 3: Stoichiometry
Ca(NO3)2
Type of compound:
ionic
Ions:
Ca2+
Total number of oxygen atoms:
Name:
Calcium nitrate
NO36
Chapter 3: Stoichiometry
Ca(NO3)2
How many moles of calcium nitrate are in 394g of Ca(NO3)2 ?
mol
MM
→
gram
Molar Mass of Ca(NO3)2 = 164.1 g/mol
394 g Ca(NO3)2 x
1 mol
164.1 g
= 2.4 moles Ca(NO3)2
Chapter 3: Stoichiometry
What is the mass in grams of 0.527 moles of
calcium sulfate dihydrate (gypsum), CaSO4· 2 H2O ?
mol
MM
→
gram
(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
172g CaSO 4  2H2O
0.527 moles CaSO4· 2 H2O 
1mol CaSO 4  2H2O
=
90.6 g CaSO4· 2 H2O
Chapter 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 ×
6 moles oxygen
1 mole Ca ( NO3 )2
= 14.4 moles oxygen
Chapter 3: Stoichiometry
Gretel(shoes)2
1 unit of Gretel(shoes)2 contains 2 shoes
1 mole ofGretel(shoes)2 contains 2 moles of shoes
How many moles of shoes are in 2.4 moles of Gretel(shoes)2 ?
2.4 moles of Gretel(shoes)2 x
2 moles of shoes
= 4.8 moles shoes
1 mol of Gretel(shoes)2
Chapter 3: Stoichiometry
Gretel(shoes)2
What is the percentage of shoes, by mass, in Gretel(shoes)2?
(1) total mass of Gretel(shoes)2
mass of Gretel(shoes)2
=
=
55.2 kg + 0.5kg x 2
56.2 kg
(2) mass of shoes in Gretel(shoes)2
2 x 0.5kg
=
1.0 kg
(3) percentage of shoe mass
mass % shoes
mass of shoes
total mass
100 %
1.0 kg
56.2 kg
100 %
1.8%
Chapter 3: Stoichiometry
Ca(NO3)2
What is the percentage of oxygen, by mass, in calcium nitrate?
(1) total mass of Ca(NO3 )2 in amu
FW of Ca(NO3)2 =
=
40.1 amu + (14.0 amu + 3 x 16.0 amu) x 2
164.1 amu
(2) mass of oxygen in compound, in amu
6 x 16.0 amu
=
96.0 amu
(3) percentage of oxygen, by mass
mass % O = mass of oxygen × 100% =
total mass
96amu
× 100% = 58.5%
164.1 amu
Chapter 3: Stoichiometry
Quantitative Information from Balanced Equations
How many grams of CO2 would be produced by the combustion
of 2 moles of CO?
2 CO
+
→
O2
2 CO2
→
+
2 moles CO
+ 1 mole O2
2 x 28 g
+
32 g
→
→
2 moles CO2
2 x 44 g
= 88 g
Chapter 3: Stoichiometry
Quantitative Information from Balanced Equations
2 CO
+
O2
→
2 CO2
You can write a series of stoichiometric factors for this reaction:
2 mol CO
1 mol O2
1 mol O2
2 mol CO
1 mol O2
2 mol CO2
Chapter 3: Stoichiometry
How many grams of H2O are formed from the complete combustion
of 2.0 g of C2H4?
need balanced reaction equation !!
18 g/mol
28 g/mol
1 C2H4 (g)
+
→
3 O2 (g)
2 CO2 (g)
+
2 H2O (g)
grams C2H4 → moles C2H4 → moles H2O → grams H2O
2 g C2H4

1mol C2 H 4
28 g C2 H 4

2mol H 2O
1mol C2 H 4
from balanced
equation

18 g H 2O
1mol H 2O
= 2.6 g H2O
Chapter 3: Stoichiometry
Summary
1) determine equation for the reaction
2) balance equation
3) formulate problem:
how much of A
=>
gets converted into how much of B
4) determine MW/FW of substances involved
5) determine stoichiometric factors from balanced equation
Chapter 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 soda
a bottle of Grenadine syrup, and 10 maraschino cherries?
a. 2
b. 4
c. 6
d.8
Chapter 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 soda
a bottle of Grenadine syrup, and 100 grams of maraschino cherries?
You need to know how many maraschino cherries are in 100 grams!
Chapter 3: Stoichiometry
Limiting Reactants
+
is the limiting reactant!
The amount of
limits the amount of product that can be formed
Chapter 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: Stoichiometry
Limiting Reactants: how much NH3 can be formed from 3 moles N2 and 6 moles H2?
1 N2
Available (given):
+
3 mol N2
3 H2
,
→ 2 NH3
6 mol H2
How much H2 would we need to completely react 3 mol N2:
3 mol N2 
3 H2
1 N2
= 9 mol H2
compare with what is available
We only have 6 mol H2 available – it is limiting !
How much NH3 can we form with the available reagents?
6 mol H2 
2 NH3
3 H2
continue with limiting reagent
= 4 mol NH3
Chapter 3: Stoichiometry
Limiting Reactants
1 N2
+
3 H2
Available (given): 3 mol N2
,
→ 2 NH3
6 mol H2
How much N2 is left over (in excess)?
H2 is limiting, there is plenty of N2
How much N2 is actually reacting?
6 mol H2

1 N2
3 H2
= 2 mol N2
continue with limiting reagent
this is the amount that reacts
Available – amount reacted = left over
3 mol – 2 mol = 1 mol N2 left over
Chapter 3: Stoichiometry
Limiting Reactants
2 Al + 3 Cl2 → 2 AlCl3
Available (given):
0.5 mol Al , 2.5 mol Cl2
How much Cl2 would we need to completely react 0.5 mol Al:
0.5 mol Al 
3 Cl2
2 Al
= 0.75 mol Cl2
compare with what is available
We have more than enough Cl2 available – it is in excess !
If Cl2 is in excess, Al must be limiting !
How much AlCl3 can we form with the available reagents?
0.5 mol Al 
2 AlCl3
2 Al
continue with limiting reagent
= 0.5 mol AlCl3
Chapter 3: Stoichiometry
Theoretical Yield
2 Al
Available (given):
+
3 Cl2
→ 2 AlCl3
0.5 mol Al , 2.5 mol Cl2
What mass do 0.5 mol AlCl3 correspond to?
0.5 mol AlCl3
133.5 g AlCl3

mol AlCl3
= 67 g AlCl3
The maximum mass of product that can be formed is the
theoretical yield
Chapter 3: Stoichiometry
Theoretical Yield
2 Al
Available (given):
+
3 Cl2
→ 2 AlCl3
0.5 mol Al , 2.5 mol Cl2
Fritz does the reaction with the available reagents he only ends up with
34g. What is the % yield of the reaction?
% yield =
actual yield
× 100 %
theoretical yield
% yield 
34 g
 100 % = 51 %
67 g
Chapter 3: Stoichiometry
Summary
Determine availabe
quantity of reactants
in moles
Determine % yield
of the reaction
Determine if one of
the reactants is a
limiting reactant
Compare with
actual
amount of product
recovered
(actual yield)
Determine
the maximum
# of moles of product
that can be formed
Convert into
grams of
product
(theoretical
yield)