Download 2 - C7Chemistry

Basic
Stoichiometry
Part One
Pisgah High School
M. Jones
Revision history: 5/16/03, 02/04/12, 04/27/12
The word stoichiometry
comes from the Greek
words stoicheion which
means “element” and
metron which means
“measure”.
Stoichiometry deals
with the amounts of
reactants and products
in a chemical reaction.
Stoichiometry deals
with moles.
Recall that …
1 mole = 6.022
23
x 10
atoms or
molecules
1 mole = the molar mass
1 mole = 22.4 L of any
gas at STP
The word mole is one that
represents a very large
number.
Much like “dozen” means 12,
23
… “mole” means 6.022 x 10
The key to doing
stoichiometry is the
balanced chemical
equation.
2 H2 + O2  2 H2O
The coefficients give the
relative number of atoms or
molecules of each reactant or
product … as well as the number
of moles.
2 H2 + O2  2 H2O
2 H2 + O2  2 H2O
2 molecules 1 molecule
of hydrogen of oxygen
2 molecules
of water
Two molecules of hydrogen
combine with one molecule of
oxygen to make two molecules
of water.
2 H2 + O2  2 H2O
2 molecules 1 molecule
of hydrogen of oxygen
2 molecules
of water
The balanced chemical
equation also gives the smallest
integer number of moles.
2 H2 + O2  2 H2O
2 moles
1 mole
of hydrogen of oxygen
2 moles
of water
The balanced chemical
equation also gives the smallest
integer number of moles.
2 H2 + O2  2 H2O
2 moles
1 mole
of hydrogen of oxygen
2 moles
of water
Two moles of hydrogen
combine with
one mole of oxygen
to make two moles of water.
How can we show that
this is really true?
Consider the combustion
of hydrogen in oxygen …
2 H2 + O2  2 H2O
2 moles
1 mole
of hydrogen of oxygen
2 moles
of water
What do each of the reactants
and product weigh?
2 H2 + O2  2 H2O
2 moles
1 mole
of hydrogen of oxygen
2 moles
of water
2 x 2.0 g 1 x 32.0 g 2 x 18.0 g
4.0 g
+ 32.0 g = 36.0 g
2 H2 + O2  2 H2O
2 moles
1 mole
of hydrogen of oxygen
2 moles
of water
The Law of Conservation of Matter
4.0 g
+ 32.0 g = 36.0 g
2 H2 + O2  2 H2O
2 moles
1 mole
of hydrogen of oxygen
2 moles
of water
Matter can neither be created nor
destroyed, only changed in form.
4.0 g
+ 32.0 g = 36.0 g
The oxidation of iron
Consider the oxidation of iron:
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
4 moles
3 moles
2 moles
The coefficients give
the ratio of moles.
Consider the oxidation of iron:
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
If these
react … then we have the following:
4 moles
8 moles
2 moles
0.50 mol
3 moles
6 moles
1.5 moles
0.375 mol
2 moles
4 moles
1 mole
0.25 mol
Consider the oxidation of iron:
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
The 0.375 moles was not as easy to predict.
3 mole O 2
0.5 mole Fe 
 0.375 mole O2
4 mole Fe
Consider the oxidation of iron:
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
The 0.375 moles was not as easy to predict.
3 mole O 2
0.5 mole Fe 
 0.375 mole O2
4 mole Fe
Use a conversion factor to determine the
number of moles of an unknown.
Consider the oxidation of iron:
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
3 mole O 2
0.5 mole Fe 
 0.375 mole O2
4 mole Fe
The conversion factor comes from the
coefficients in the balanced equation.
Back to the oxidation of iron:
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
4 moles
3 moles
2 moles
Calculate the masses of
these moles.
Back to the oxidation of iron:
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
4 moles 3 moles
2 moles
4 x 55.8 g 3 x 32.0 g 2 x 159.6 g
223.2 g + 96.0 g = 319.2 g
Mass is conserved.
The decomposition of
ammonium carbonate
Now consider the decomposition of
solid ammonium carbonate.
(NH4)2CO3  2 NH3 + CO2 + H2O
Suppose 96.0 grams of ammonium
carbonate decomposes. How
many grams of each of the gases
will be produced?
Now consider the decomposition of
solid ammonium carbonate.
(NH4)2CO3  2 NH3 + CO2 + H2O
96.0 grams
?g
?g
?g
The coefficients tell moles, not grams.
Convert 96.0 g (NH4)2CO3 to moles.
Now consider the decomposition of
solid ammonium carbonate.
(NH4)2CO3  2 NH3 + CO2 + H2O
96.0 grams
?g
?g
?g
Find the molar mass of ammonium
carbonate.
2x14 + 8 +12 + 3x16 = 96.0 g/mol
Now consider the decomposition of
solid ammonium carbonate.
(NH4)2CO3  2 NH3 + CO2 + H2O
96.0 grams
?g
?g
?g
Isn’t that convenient!
We have one mole of ammonium
carbonate.
Now consider the decomposition of
solid ammonium carbonate.
(NH4)2CO3  2 NH3 + CO2 + H2O
96.0 grams
?g
?g
?g
1 mol
2 mol 1 mol 1 mol
2 moles of ammonia are produced,
along with 1 mole of carbon dioxide
and 1 mole of water vapor.
Now consider the decomposition of
solid ammonium carbonate.
(NH4)2CO3  2 NH3 + CO2 + H2O
96.0 grams
?g
?g
?g
1 mol
2 x 17.0g
2 mol
1 mol 1 mol
44.0 g
18.0 g
How many grams of each product are formed?
Now consider the decomposition of
solid ammonium carbonate.
(NH4)2CO3  2 NH3 + CO2 + H2O
96.0 grams
?g
?g
?g
1 mol
2 x 17.0g
2 mol
1 mol 1 mol
44.0 g
18.0 g
34.0 g + 44.0 g + 18.0 g = 96.0 g
The reaction between
dinitrogen pentoxide and
water
Consider the reaction between
dinitrogen pentoxide and water.
N2O5 + H2O  2 HNO3
There is only one product.
What kind of reaction is it?
It must be a synthesis reaction.
Consider the reaction between
dinitrogen pentoxide and water.
N2O5 + H2O  2 HNO3
There are several kinds of synthesis reactions.
Check the reference tables.
1.
2.
3.
4.
Hydrogen + nonmetal = binary acid
Metal + nonmetal = salt
Metal + water = base
Nonmetal oxide + water = ternary acid
Nitric acid is a ternary acid.
N2O5 + H2O  2 HNO3
Suppose we needed to make
100.0 grams of nitric acid.
How many grams of dinitrogen
pentoxide would we need to
react with excess water?
N2O5 + H2O  2 HNO3
??? g
0.794 mole
1.59 mol/2
100.0 g
1.59 mole
mole HNO 3
100.0 g HNO 3 
= 1.59 mole HNO3
63.0 g HNO 3
Divide 1.59 mole HNO3 by 2 to get moles of N2O5
108.0 g N 2 O 5
0.794 mole N 2 O 5 
= 85.7 g N2O5
1 mole N 2 O5
N2O5 + H2O  2 HNO3
85.7 g
0.794 mole
1.59 mol/2
100.0 g
1.59 mole
But we don’t have 85.7 grams of N2O5.
There are only 60.0 grams available.
How many grams of nitric
acid could we actually make?
N2O5 + H2O  2 HNO3
60.0 g
0.556 mol x 2
0.556 mole
??? g
1.11 mole
1 mol N 2 O5
60.0 g N 2 O5 
= 0.556 mol N2O5
108 g N 2 O5
Multiply 0.556 mol by 2 to get moles of HNO3.
63.0 g HNO3
1.11 mol HNO3 
= 70.0 g HNO3
1.00 mol HNO3
Description of
stoichiometry problems
Stoichiometry problems will usually
take one of the following forms:
1. Mole-mole problem where you
might be given moles and asked to
find moles of another substance.
2. Mole-mass problem where you
might be given moles and asked
find the mass of another substance.
3. Mass-mass problem where you
might be given a mass and asked to
find the mass of another substance.
4. Mass-volume problem where you
might be given a mass and asked to
find the volume of a gas.
5. Volume-volume problem where you
might be given a volume and asked
to find another volume.
Volume-volume stoichiometry
problems are easiest when you
use Gay-Lussac’s Law.
“The ratio between the volumes of
the reactant gases and the
products can be expressed in
simple whole numbers.”
Simply put, Gay-Lussac’s Law
says this:
The volumes of the gases are in the
same ratio as the coefficients in the
balanced equation.
2 H2 + O2  2 H2O
2 moles
2L
1 mole
1L
2 moles
2L
Applications of
Gay-Lussac’s Law
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
1 mol
1L
5 mols
5L
3 mols
3L
4 mols
4L
Suppose 3.5 L of propane gas at STP is
burned in oxygen, how many L at STP of
oxygen will be required?
5 L O2
3.5 L C3H 8 
 17.5 L O2
1 L C3H 8
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
1 mol
1L
5 mols
5L
3 mols
3L
4 mols
4L
Suppose 3.5 L of propane gas at STP is
burned in oxygen, how many L at STP of
oxygen will be required?
17.5 L O
2
How many liters of carbon dioxide gas and
water vapor at STP would be produced?
10.5 L CO2 and 14.0 L H2O
CH4(g) + 4 Cl2(g)  CCl4(g) + 4 HCl(g)
When methane gas is allowed to react
with an excess of chlorine gas,
tetrachloromethane and hydrogen
chloride gas will be produced.
How many L of methane will react
with 0.800 L of chlorine gas at STP?
1 L CH 4
0.800 L Cl2 
 0.200 L Cl2
4 L Cl2
Stoichiometry problems
involving gases
Find the mass of HOCl that can be
produced when 2.80 L of chlorine gas
at STP reacts with excess hydrogen
peroxide.
Cl2(g) + H2O2 (l)  2 HOCl (l)
2.80 L
??? g
Convert 2.80 L of Cl2 gas at STP
to moles
Cl2(g) + H2O2 (l)  2 HOCl (l)
2.80 L
.125 x 2
0.125 mol
??? g
0.250 mol
1 mol Cl2
2.80 L Cl2 
 0.125 mol Cl2
22.4 L Cl2
52.5 g HOCl 13.1g
0.250 mol HOCl 

1 mol HOCl HOCl
The reaction between
copper and nitric acid
Will copper dissolve in acids?
Consider hydrochloric acid
Cu + 2HCl  No
CuCl
Reaction
2 + 2H2 (g)
Most metals react with HCl to produce
a metal chloride solution and H2 gas.
Not copper
Will copper dissolve in acids?
Consider hydrochloric acid
Cu + 2HCl  No
CuCl
Reaction
2 + 2H2 (g)
Copper is below hydrogen in the activity
series. Copper metal will only replace
elements that are below it in the activity
series.
What about other acids?
Cu + HBr  NR
Cu + HI  NR
Cu + HF  NR
Cu + H2SO4  NR
Cu + HC2H3O2  NR
The same is true for all acids
except nitric acid
Nitric acid is the only acid that will
dissolve copper.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
A beaker contains a
penny and some nitric
acid is added.
Nitric acid is the only acid that will
dissolve copper.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
The penny begins to
disappear and the solution
turns blue-green and a
brown gas is given off.
Nitric acid is the only acid that will
dissolve copper.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
The gas produced in the
reaction is NO, which is
colorless. Reddish brown
NO2 forms when NO reacts
with the oxygen in the air.
Nitric acid is the only acid that will
dissolve copper.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
The penny is gone and
the solution turns a dark
blue. The brown NO2
gas escapes from the
open beaker.
Nitric acid is the only acid that will
dissolve copper.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
Calculate the volume
of NO gas at STP
when 20.0 grams of
copper dissolves.
Nitric acid is the only acid that will
dissolve copper.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
20.0 g
0.315 mol
??? L
x 0.667
0.210 mol
1 mol Cu
20.0 g Cu 
 0.315 mol Cu
63.5 g Cu
Nitric acid is the only acid that will
dissolve copper.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
20.0 g
0.315 mol
??? L
x 0.667
0.210 mol
22.4 L NO
0.210 mol NO 
 4.70 L NO
1 mol NO
Part Two of
Basic Stoichiometry
will include the gas laws.