chapter 3: stoichiometry
DESCRIPTION
→. +. Chapter 3: Stoichiometry. “ measuring elements” Must account for ALL atoms in a chemical reaction. 2. 2. H 2 +O 2 →H 2 O. →. +. +. +. →. Chapter 3: Stoichiometry. 2. CO+O 2 →CO 2. 2. CH 4 +Cl 2 →CCl 4 +HCl. 4. 4. 2 3. 2 3. x 3. - PowerPoint PPT PresentationTRANSCRIPT
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