Jeffrey Mack
California State University,
Sacramento
Chapter 4
Stoichiometry:
Quantitative
Information about
Chemical Reactions
The study of mass relationships in chemical reactions is called
Stoichiometry.
Stoichiometry provides quantitative information about chemical
reactions.
Mass must be conserved in a chemical reaction.
Total mass of
reactants
Total mass of
products =
Chemical equations must therefore be balanced for mass.
Numbers of atoms
on the reactant side =
Numbers of atoms
on the product side
Stoichiometry
• The stoichiometric balancing coefficients are
the numbers in front of the chemical
formulas.
• They give the ratio of reactants and
products.
• The balancing coefficients allow us to convert
between numbers of reactants and products.
ratio Conversion
factors
Stoichiometry
The ratio of any two
species (reactants or products) in a balanced
chemical reaction.
Stoichiometric ratio:
The branch of chemistry that deals
with the mole proportions of chemical reactions.
Stoichiometry:
2A + 3B A2B3
2 A’s combine with 3B’s
CONVERSION FACTORS!!!
2A
3B
3B
2A
or
Stoichiometry
2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)
Given the following reaction
How many moles of water are produced when 3.0 mols of
oxygen react?
Stoichiometry
2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)
Given the following reaction
How many moles of water are produced when 3.0 mols of
oxygen react?
2 2.6 mol H O23.0 mol O 2
2
6 mol H O×
7 mol O
mol O2 mol H2O
Stoichiometry
2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)
Given the following reaction
How many moles of water are produced when 3.0 mols of
oxygen react?
2 2.6 mol H O23.0 mol O 2
2
6 mol H O×
7 mol O
mol O2 mol H2O
Stoichiometry
2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)
Given the following reaction
How many moles of water are produced when 3.0 mols of
oxygen react?
2 2.6 mol H O23.0 mol O 2
2
6 mol H O×
7 mol O
mol O2 mol H2O
2 sig. figs. 2 sig. figs. exact
conversion
factor
Stoichiometry
2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)
Given the following reaction
How many moles of C2H6 must react to produce 1.75 mols
of CO2?
Stoichiometry
2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)
Given the following reaction
How many moles of C2H6 must react to produce 1.75 mols
of CO2?
1.75 mol CO2 2 6
2
2 mol C H×
4 mol CO= 0.875 mol C2H6
Stoichiometry
2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)
Given the following reaction
How many moles of C2H6 must react to produce 1.75 mols
of CO2?
1.75 mol CO2 2 6
2
2 mol C H×
4 mol CO= 0.875 mol C2H6
Stoichiometry
• Mole/Mass relationships.
• It is not possible to relate masses in reactions
without going through moles.
Stoichiometry
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Strategy Map for the solution to the problem:
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Data Information:
Mass of reactant
Write the balanced
chemical equation Step 1:
Eq. gives mole ratios
(stoichiometry)
Convert mass to
moles Step 2:
amount of reactant in
moles
Convert moles reactant
to moles product Step 3:
amount of products
in moles
Convert moles of
products to mass Step 4:
mass of products in
grams
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 1: Write the balanced chemical equation
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many grams of
dinitrogen monoxide and water are formed?
Step 1: Write the balanced chemical equation
NH4NO3(s) + 2 H2O(l) N2O(g)
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 1: Write the balanced chemical equation
NH4NO3(s) + 2 H2O(l) N2O(g)
Step 2: Convert mass of reactant (454 g NH4NO3) to moles.
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 1: Write the balanced chemical equation
NH4NO3(s) + 2 H2O(l) N2O(g)
Step 2: Convert mass of reactant (454 g NH4NO3) to moles.
4 34 3
4 3
1mol NH NO454g NH NO ×
80.04g NH NO= 5.67 mol NH4NO3
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 3: Using the balanced chemical equation, convert moles
of reactant (NH4NO3) to moles of products.
NH4NO3(s) + 2 H2O(g) N2O(g)
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 3: Using the balanced chemical equation, convert moles
of reactant (NH4NO3) to moles of products.
NH4NO3(s) + 2 H2O(g) N2O(g)
5.67 mol NH4NO3 2
4 3
1 mol N O×
1 mol NH NO= 5.67 mol N2O
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 3: Using the balanced chemical equation, convert moles
of reactant (NH4NO3) to moles of products.
NH4NO3(s) + 2 H2O(g) N2O(g)
5.67 mol NH4NO3 2
4 3
1 mol N O×
1 mol NH NO= 5.67 mol N2O
5.67 mol NH4NO3 2
4 3
2 mol H O×
1 mol NH NO= 11.3 mol H2O
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 3: Using the balanced chemical equation, convert moles
of reactant (NH4NO3) to moles of products.
NH4NO3(s) + 2 H2O(g) N2O(g)
5.67 mol NH4NO3 2
4 3
1 mol N O×
1 mol NH NO= 5.67 mol N2O
5.67 mol NH4NO3 2
4 3
2 mol H O×
1 mol NH NO= 11.3 mol H2O
Recall that the coefficients represent the molar ratios of reactants to
products and vice versa…
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 4: Now convert the moles of products (N2O and H2O) to
grams using the molar mass of each.
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 4: Now convert the moles of products (N2O and H2O) to
grams using the molar mass of each.
5.67 mol N2O 2
2
44.01 g N O×
1 mol N O= 250. g N2O
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 4: Now convert the moles of products (N2O and H2O) to
grams using the molar mass of each.
5.67 mol N2O 2
2
44.01 g N O×
1 mol N O= 250. g N2O
11.3 mol H2O 218.02 g H O×
1 mol H O= 204 g H2O
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
Step 4: Now convert the moles of products (N2O and H2O) to
grams using the molar mass of each.
5.67 mol N2O 2
2
44.01 g N O×
1 mol N O= 250. g N2O
11.3 mol H2O 218.02 g H O×
1 mol H O= 204 g H2O
…or 454g NH4NO3 250. g N2O = 204 g H2O
Consider the following reaction:
If 454 g of solid ammonium nitrate decomposes to form
dinitrogen monoxide gas and liquid water, how many
grams of dinitrogen monoxide and water are formed?
0 mol + 5.67 mol +11.3 mol
0 g 250. g 204 g
454 g 0 g 0 g
Compound NH4NO3 N2O H2O
Initial (g)
Initial (mol)
Change (mol)
Final (mol)
Final (g)
5.67 mol 0 mol 0 mol
Look what happens!
5.67 mol + 5.67 mol +11.3 mol
Mass is conserved but Moles are not!
grams moles moles grams
Using
molar
mass
Using
mole
ratios
Using
molar
mass
• In order to solve stoichiometry problems, one must
go through moles using molar masses and mole
ratios as conversion factors.
• One cannot do this without writing a balanced
chemical equation first.
Stoichiometry Problems
• In some cases involving two or more
reactants, there is insufficient amount of one
reactant to consume the other reactants
completely.
• The reactant that is in short supply therefore
LIMITS the quantity of product that can be
formed.
• The theoretical yield of products is limited by
this “Limiting Reactant”.
Reactions Involving a Limiting
Reactant
32
Limiting Reactant Analogy
If we have 10
sandwiches, 18
cookies, and 12
oranges …
… how many
packaged meals
can we make?
• The theoretical yield is the maximum product yield that
can be expected based on the masses of the reactants
and the reaction stoichiometry.
• The actual yield is the experimentally measured amount of
products that results upon completion of the reaction.
• The percent yield is a measure of the extent of the reaction
in terms of the actual vs. the theoretical yield.
% Yield Actual Yield (in grams or moles)
Theoretical Yield (in grams or moles)100
Reaction Yields
Consider the reaction of aluminum and oxygen:
• Which is the limiting reactant if we start with 50.0 g
Al and 50.0 g O2?
• What is the Theoretical Yield for the reaction?
Consider the reaction of aluminum and oxygen:
• Which is the limiting reactant if we start with 50.0 g Al
and 50.0 g O2?
• What is the Theoretical Yield for the reaction?
Strategy Map: Data Information: Masses of two
reactants. Recognize as a limiting
reactant problem.
Write the balanced
chemical equation Step 1:
Balanced Eq. gives mole ratios
(stoichiometry).
Convert masses of each
reactant to moles Step 2:
Consider the reaction of aluminum and oxygen:
• Which is the limiting reactant if we start with 50.0 g Al
and 50.0 g O2?
• What is the Theoretical Yield for the reaction?
Strategy Map:
amount of each reactant in moles.
Decide which reactant limits by comparing
mole amounts to stoichiometric ratios. Step 3:
Limiting reactant is know.
Calculate the moles of product
based on limiting reactant. Step 4:
Consider the reaction of aluminum and oxygen:
• Which is the limiting reactant if we start with 50.0 g Al
and 50.0 g O2?
• What is the Theoretical Yield for the reaction?
Strategy Map: Amount of product based on the
L.R.
Calculate the mass of product
based on moles. Step 5:
This is the Theoretical Yield
Consider the reaction of aluminum and oxygen:
• Which is the limiting reactant if we start with 50.0 g Al
and 50.0 g O2?
• What is the Theoretical Yield for the reaction?
4Al(s) + 3O2(g) 2Al2O3(s)
Consider the reaction of aluminum and oxygen:
• Which is the limiting reactant if we start with 50.0 g Al
and 50.0 g O2?
• What is the Theoretical Yield for the reaction?
4Al(s) + 3O2(g) 2Al2O3(s)
g Al(initial) mols Al(initial)
g O2(initial) mols O2(initial)
1 mol Al50.0g Al 1.85 mols Al
26.98g Al
22 2
2
1 mol O50.0g O 1.56 mols O
32.00g O
Consider the reaction of aluminum and oxygen:
• Which is the limiting reactant if we start with 50.0 g Al
and 50.0 g O2?
• What is the Theoretical Yield for the reaction?
4Al(s) + 3O2(g) 2Al2O3(s)
Compare moles of Al to moles of O2
= =2 2 2 2
1.85 mols Al 1.19 mols Al 4 mols Al 1.33 mols Al vs.
1.56 mols O 1.00 mols O 3 mols O 1.00 mols O
Since 1.19 < 1.33, Aluminum is the Limiting Reactant
Consider the reaction of aluminum and oxygen:
• Which is the limiting reactant if we start with 50.0 g Al
and 50.0 g O2?
• What is the Theoretical Yield for the reaction?
4Al(s) + 3O2(g) 2Al2O3(s)
Calculate the moles and grams of product based on the
limiting reactant, Al.
2 3 2 32 3
2 3
2 mols Al O 101.96 g Al O1.85 moles Al 94.3 g Al O
4 mols Al 1 mol Al O
Consider the reaction of Al and O2 previously
discussed: From the calculations it was determined that 94.3g of Al2O3 could be
formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the
theoretical yield.
Consider the reaction of Al and O2 previously
discussed: From the calculations it was determined that 94.3g of Al2O3 could be
formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the
theoretical yield.
If 48.5 g of Al2O3 was collected at the completion of the
reaction (actual yield), what is the % yield for the reaction?
Consider the reaction of Al and O2 previously
discussed: From the calculations it was determined that 94.3g of Al2O3 could be
formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the
theoretical yield.
If 48.5 g of Al2O3 was collected at the completion of the
reaction (actual yield), what is the % yield for the reaction?
=% Yield Actual Yield
Theoretical Yield ´ =100
2 3
2 3
48.5 g Al O
94.3 g Al O´ 100
= 51.4 %
Consider the reaction of Al and O2 previously
discussed:
From the calculations it was determined that 94.3g of Al2O3 could be
formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the
theoretical yield.
How many grams of oxygen remain?
Consider the reaction of Al and O2 previously
discussed:
From the calculations it was determined that 94.3g of Al2O3 could be
formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the
theoretical yield.
How many grams of oxygen remain?
By conservation of mass:
mass of reactants = mass of products
mass Al + O2 (initial) = mass of Al2O3 + O2 (left over)
Consider the reaction of Al and O2 previously
discussed:
From the calculations it was determined that 94.3g of Al2O3 could be
formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the
theoretical yield.
How many grams of oxygen remain?
2
2 3
50.0 g Al
50.0 g O
94.3 g Al O
+
-
5.7g O2 remains
48
Here's a nice limiting reagent problem we will practice
Consider the reaction:
2 Al + 3 I2 ------> 2 AlI3
Determine the limiting reagent if one starts with:
a) 1.20 mol Al and 2.40 mol iodine.
b) 1.20 g Al and 2.40 g iodine
c) How many grams of Al are left over in part b?
ANSWES a) Al b) Iodine c) 1.03g Al
Now Try This
Chemical Analysis
• A 0.123 g sample of the mineral thenardite
contains sodium sulfate along with several inert
compounds.
• The sodium sulfate in the sample is converted to
insoluble barium sulfate by adding aqueous
barium chloride in excess to the dissolved
thenardite sample.
• The mass of recovered barium sulfate is 0.177 g
• What is the mass percent of sodium sulfate in
the mineral?
Chemical Analysis
• Recognize that all of the sulfate in sodium sulfate
ends up as barium sulfate.
Chemical Analysis
Chemical Analysis
• Recognize that all of the sulfate in sodium sulfate
ends up as barium sulfate.
Na2SO4(aq) + BaCl2(aq) BaSO4(aq) + 2NaCl(aq)
• Recognize that all of the sulfate in sodium sulfate
ends up as barium sulfate.
Na2SO4(aq) + BaCl2(aq) BaSO4(aq) + 2NaCl(aq)
Chemical Analysis
Chemical Analysis
• Recognize that all of the sulfate in sodium sulfate
ends up as barium sulfate.
Na2SO4(aq) + BaCl2(aq) BaSO4(aq) + 2NaCl(aq)
• The moles and mass of sodium sulfate can
therefore can be found from the mass of barium
sulfate
• Recognize that all of the sulfate in sodium sulfate
ends up as barium sulfate.
Na2SO4(aq) + BaCl2(aq) BaSO4(aq) + 2NaCl(aq)
• The moles and mass of sodium sulfate can
therefore can be found from the mass of barium
sulfate
g BaSO4 mols BaSO4 mols Na2SO4 g Na2SO4 % Na2SO4
Chemical Analysis
Chemical Analysis
• Recognize that all of the sulfate in sodium sulfate
ends up as barium sulfate.
Na2SO4(aq) + BaCl2(aq) BaSO4(aq) + 2NaCl(aq)
• The moles and mass of sodium sulfate can
therefore can be found from the mass of barium
sulfate
• This yields the % of sodium sulfate in the sample.
g BaSO4 mols BaSO4 mols Na2SO4 g Na2SO4 % Na2SO4
Na2SO4(aq) + BaCl2(aq) BaSO4(s) + 2 NaCl(aq)
= ´2 42 4
mass Na SOMass % Na SO 100
mass Sample
4 2 4 2 44 2 4
4 4 2 4
1 mol BaSO 1 mol Na SO 142.0 g Na SO0.177 g BaSO 0.108 g Na SO
233.4 g BaSO 1 mol BaSO 1 mol Na SO´ ´ ´ =
molar
mass
BaSO4
mole
ratio
molar
mass
Na2SO4
´ = 2 4
0.108g100 61.0% Na SO
0.177g
Chemical Analysis
• Combustion involves the addition oxygen to another
element.
• When hydrocarbon molecules burn completely, the
products are always carbon dioxide gas and water.
Determining the Formula of a Hydrocarbon by Combustion
• A streaming flow of pure O2 is passed over a weighed mass
of the unknown sample in an apparatus like the one above.
• The heat from the furnace converts the carbon in the
compound to CO2 and the hydrogen into H2O.
• The CO2 and H2O are collected in pre-weighed traps.
A hydrocarbon is a
compound that contains
predominately carbon
and hydrogen and
possibly smaller amount
of elements such as O, N
and S.
Combustion Analysis
When combustion is completed, the traps are weighed and the
masses of CO2 and H2O are determined.
mass trap + H2O
mass trap
= mass H2O
mass trap + CO2
mass trap
= mass CO2
Combustion Analysis
• All of the carbon in the sample ends up in the CO2.
• All of the hydrogen in the sample ends up in the water.
• Using stoichiometry, one can relate the carbon in CO2 and
the hydrogen in H2O back to the carbon and hydrogen in the
original sample.
• Any oxygen in the sample can be deduced via conservation
of mass.
Combustion Analysis
Combustion Analysis Problem:
• 1.516 g of a compound containing carbon, hydrogen and oxygen
(CXHYOZ) is subjected to combustion analysis.
• The results show that 2.082 g of CO2 and 1.705 g of H2O were
produced.
• What is the empirical formula for the compound?
• If the molecular weight of the compound is 160.2 g/mol,
what is the molecular formula of the compound?
Combustion Analysis Problem:
• 1.516 g of a compound containing carbon, hydrogen and oxygen
(CXHYOZ) is subjected to combustion analysis.
• The results show that 2.082 g of CO2 and 1.705 g of H2O were
produced.
Data Information: Mass of
CO2 & H2O, MM of the
compound.
calculate moles of CO2 and
H2O using molar masses Step 1:
Amounts of CO2 & H2O in moles
Convert mols CO2 & H2O into mols
C and H from the sample. Step 2:
Strategy Map
Combustion Analysis Problem:
• 1.516 g of a compound containing carbon, hydrogen and oxygen
(CXHYOZ) is subjected to combustion analysis.
• The results show that 2.082 g of CO2 and 1.705 g of H2O were
produced.
Strategy Map Moles of Carbon and hydrogen in
the sample.
Convert the moles of C
and H to grams Step 3:
Grams of Carbon and hydrogen in
the sample.
Subtract the mass of C &
H from the sample mass Step 4:
Combustion Analysis Problem:
• 1.516 g of a compound containing carbon, hydrogen and oxygen
(CXHYOZ) is subjected to combustion analysis.
• The results show that 2.082 g of CO2 and 1.705 g of H2O were
produced.
Strategy Map This gives the mass of Oxygen in the
sample. If the mass is zero then there is
no oxygen in the sample.
Convert grams of oxygen
to moles of oxygen Step 5:
moles of O, C and H are now known. This
becomes and empirical formula problem.
Divide each mole quantity by the smallest mole
value to get the elemental mole ratios. Step 6:
Combustion Analysis Problem:
• 1.516 g of a compound containing carbon, hydrogen and oxygen
(CXHYOZ) is subjected to combustion analysis.
• The results show that 2.082 g of CO2 and 1.705 g of H2O were
produced.
Strategy Map This gives x, y and z in
CXHYOZ, the empirical
formula.
Determine the ratio of the experimental
molar mass to the empirical molar mass. Step 6:
This yields the molecular formula..
Solution to the problem:
22.082 g CO 2
2
1 mol CO 44.01g CO 2
1 mol C 1 mol CO
g CO2 mol CO2 mol C
0.04731 mols C
Solution to the problem:
22.082 g CO 2
2
1 mol CO 44.01g CO 2
1 mol C 1 mol CO
0.04731 mols C
g CO2 mol CO2 mol C
21.705 g H O 2
2
1 mol H 0 18.02 g H 0 2
2 mol H 1 mol H O
0.1892 mol H
g H2O mol H2O mol H
Solution to the problem:
22.082 g CO 2
2
1 mol CO 44.01g CO 2
1 mol C 1 mol CO
0.04731 mols C
g CO2 mol CO2 mol C
21.705 g H O 2
2
1 mol H 0 18.02 g H 0 2
2 mol H 1 mol H O
0.1892 mol H
g H2O mol H2O mol H
Don’t forget the mole ratio…
But what about the oxygen?
To find the moles of oxygen in
the sample you could use the
grams of CO2 or H2O…
Right?
Survey says….
NO!
To find the moles of oxygen in
the sample you could use the
grams of CO2 or H2O…
You can’t determine the oxygen in the sample from
either CO2 or H2O…
Think about what is happening to the sample:
CxHyOz + O2(g) CO2(g) + H2O (l)
The oxygen in the CO2 and H2O came from both the sample
and the O2 in the furnace!
Since there is no way to know how much of that oxygen
came from the sample and the O2, you cannot determine the
oxygen in the sample directly!
Combustion Analysis Problem
Mass of O in the sample:
1.516 g 0.5682 g 0.1907 g 0.757 g O
mols of O: 0.757 g O1 mol O
16.00 g O
0.0473 mols O
original sample mass
mass of C and H in sample
Combustion Analysis Problem
0.04731 mols C
0.1892 mol H
0.0473 mols O
Collecting the results:
Both carbon and oxygen have the smallest numbers of
mols: Divide each set of mols by 0.04731.
CXHYOZ
0.0473 1
0.0X
473
0.1892 4
0.0473Y
0.0473 1
0.0Z
473
yields
The empirical formula is: CH4O
Combustion Analysis Problem
Determining the Molecular Formula
For some compounds, the molecular formula is a multiple of the
empirical formula:
empirical formula molecular formula
Since the molecular formula is a multiple is scaled by a factor “n”,
the molecular and empirical molar masses must also scale by
the same ratio.
n x CXHYOZ = CnXHnYOnZ n = 2,3,4…
gmolar formula mass mol
gempirical formula mass mol
n
Combustion Analysis Problem
From the combustion analysis it was found that the empirical
formula for the unknown compound was CH4O.
This yields an empirical molar mass or empirical weight of:
[12.01 + 4x(1.0079) 16.00] g/mol =
The ratio of the molar mass to the empirical mass is:
g160.2 mol
g32.05.
4mol
000
4Therefore, the molecular formula is 5 CH O
5 20 5C H O
32.04 g/mol
Combustion Analysis Problem
78
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
ANSWER: Empirical formula C2H6O
Now Try This for Combustion Analysis
• Most chemical studies require quantitative
measurements.
• Experiments involving aqueous solutions are
measured in volumes of solutions rather than
masses of solids, liquids, or gases.
• Solution concentration, expressed as
molarity, relates the volume of solution in
liters to the amount of substance in moles.
Measuring Concentrations of
Compounds in a Solution
Solution = solute + solvent
That which is
dissolved (lesser
amount)
That which is
dissolves
(greater amount)
Terminology
per liter of solution. Molarity: Moles of solute
Molarity of X (cX) = moles of solute
L of Solution {units: mol/L}
Molarity is a conversion factor that transforms units of volume
to mole and vise–versa.
cNaCl = [NaCl] mol NaCl
= 1.00 1L
= 1.00 M
Solutions and Concentration
Molarity
moles
Liters
Molarity relates
mols and
volume (L)
knowing 2
corners, you can
find the 3rd
If you know moles & L, you know Molarity (M) and so on.
Moles, Liters, & Molarity
Molarity Volume = moles
mols
L
moles = Volume
mol L
mol
L
If you know Molarity and volume, you know moles!
If you know mols and molarity, you know volume!
L = moles
Moles, Liters, & Molarity
Preparing Solution of Known
Concentration
PROBLEM:
Dissolve 5.00 g of NiCl2 • 6 H2O in enough
water to make 250. mL of solution. Calculate
molarity.
Step 1: Calculate moles of NiCl2 • 6H2O
2 2 2 2
1 mol5.00g NiCl 6H O 0.0210 NiCl 6H O
237.7 g
Step 2: Calculate molarity
3
2 2
1 10 mL0.0210 NiCl 6 H O 0.0841 M
250. mL 1 L
[CuCl2] = 0.30 M = [Cu2+] = 0.30 M
[Cl-] = 2 0.30 M = 0.60 M
CuCl2(aq)
Cu2+(aq) + 2Cl (aq)
For every one copper
(II) ion in solution,
there are two chloride
ions.
Ion Concentrations in a Solution
Concentration Problem:
Determine the nitrate ion concentration (in mols/L) of a
solution that is made from 50.1451 g of iron (III) nitrate
diluted to a final volume of 250.0 mL.
Concentration Problem:
Determine the nitrate ion concentration (in mols/L) of a
solution that is made from 50.1451 g of iron (III) nitrate
diluted to a final volume of 250.0 mL.
Solution:
3 3g Fe(NO ) 3 3mol Fe(NO ) 3mole NO 1
Vol. (L)
molar mass
mole ratio
3mole NO
1 L
Concentration Problem:
Determine the nitrate ion concentration (in mols/L) of a
solution that is made from 50.1451 g of iron (III) nitrate
diluted to a final volume of 250.0 mL.
Solution:
3 3g Fe(NO ) 3 3mol Fe(NO ) 3mole NO 1
Vol. (L)
molar mass
mole ratio
3mole NO
1 L
Fe(NO3)3 (aq) Fe3+(aq) + 3NO3– (aq)
Concentration Problem:
Determine the nitrate ion concentration (in mols/L) of a
solution that is made from 50.1451 g of iron (III) nitrate
diluted to a final volume of 250.0 mL.
Solution:
3 3g Fe(NO ) 3 3mol Fe(NO ) 3mole NO 1
Vol. (L)
molar mass
mole ratio
3mole NO
1 L
Fe(NO3)3 (aq) Fe3+(aq) + 3NO3– (aq)
3 350.1451g Fe(NO ) 3 3
3 3
1 mol Fe(NO ) 241.88 g Fe(NO )
3
3 3
3 mol NO 1 mol Fe(NO )
1 250.0 ml
1000 ml×
L= 2.488 M NO3
–
Concentration Problem:
Determine the nitrate ion concentration (in mols/L) of a
solution that is made from 50.1451 g of iron (III) nitrate
diluted to a final volume of 250.0 mL.
Solution:
3 3g Fe(NO ) 3 3mol Fe(NO ) 3mole NO 1
Vol. (L)
molar mass
mole ratio
3mole NO
1 L
Fe(NO3)3 (aq) Fe3+(aq) + 3NO3– (aq)
3 350.1451g Fe(NO ) 3 3
3 3
1 mol Fe(NO ) 241.88 g Fe(NO )
3
3 3
3 mol NO 1 mol Fe(NO )
1 250.0 ml
1000 ml×
L= 2.488 M NO3
–
from the balanced equation!
33NO
C = [NO ] = 2.488 M
Concentration Problem:
How many grams of sodium phosphate are in
35.0 mL of a 1.51 M Na3PO4 solution?
mL solution L mols Na3PO4 g Na3PO4
use M as a
conversion
factor
Solution:
molar mass
Concentration Problem:
How many grams of sodium phosphate are in
35.0 mL of a 1.51 M Na3PO4 solution?
mL solution L mols Na3PO4 g Na3PO4
use M as a
conversion
factor
Solution:
molar mass
35.0 mL3
L×
10 mL3 41.51 mol Na PO
× L
3 4
163.94 g×
1 mol Na PO
3 4= 8.66 g Na PO
• Weigh out the solid solute
and carefully transfer it to the
volumetric flask.
• Dissolve in a volume of
solvent less than the flask
volume.
• Fill the volumetric flask to the
calibration mark with solvent.
• Invert the flask many times to
ensure homogeneous mixing.
Preparing Solutions
Preparing a Solution by Dilution
When more solvent is added to a solution, the concentration
of solute decreases:
Why? molcon
es centra
Solutetion (
L SoluM) =
tion
as solvent is added, the
denominator increases in
magnitude…
since the moles of
solute remains
constant…
The magnitude of the concentration must decrease!!!
Dilutions: Volume by Volume
• A Student adds a 50.0 mL aliquot of a 0.515 M HCl
solution to a 500.0 mL volumetric flask.
• She then filled the volumetric flask to the calibration mark
with water.
• What is the new molarity of the HCl solution?
• A Student adds a 50.0 mL aliquot of a 0.515 M HCl
solution to a 500.0 mL volumetric flask.
• She then filled the volumetric flask to the calibration mark
with water.
• What is the new molarity of the HCl solution?
Solution
50.0 mL HCl3
L 10 ml
0.515 mol HCl
1 L
• A Student adds a 50.0 mL aliquot of a 0.515 M HCl
solution to a 500.0 mL volumetric flask.
• She then filled the volumetric flask to the calibration mark
with water.
• What is the new molarity of the HCl solution?
Solution
50.0 mL HCl3
L 10 ml
0.515 mol HCl
1 L
at this point I
have mols HCl
• A Student adds a 50.0 mL aliquot of a 0.515 M HCl solution
to a 500.0 mL volumetric flask.
• She then filled the volumetric flask to the calibration mark
with water.
• What is the new molarity of the HCl solution?
Solution
50.0 mL HCl3
L 10 ml
0.515 mol HCl
1 L
at this point I
have mols HCl
1 500.0 mL
• A Student adds a 50.0 mL aliquot of a 0.515 M HCl
solution to a 500.0 mL volumetric flask.
• She then filled the volumetric flask to the calibration mark
with water.
• What is the new molarity of the HCl solution?
Solution
50.0 mL HCl3
L 10 ml
0.515 mol HCl
1 L
at this point I
have mols HCl
1 500.0 mL
dividing by the “new
volume” results in a new
M
• A Student adds a 50.0 mL aliquot of a 0.515 M HCl
solution to a 500.0 mL volumetric flask.
• She then filled the volumetric flask to the calibration mark
with water.
• What is the new molarity of the HCl solution?
Solution
50.0 mL HCl3
L 10 ml
0.515 mol HCl
1 L
at this point I
have mols HCl
1 500.0 mL
dividing by the “new
volume” results in a new
M
310 ml
L
converting
to L gives:
1/10th the
original M =HClC 0.0515 M
What we call pH is actually a mathematical function, "p"
(the negative logarithm,
based 10 )
p is a shorthand notation for " log10"
Quick Review of logs… (see Appendix A, p. A2 of your text)
log x = n where x = 10n
log 1000 = log(103) = 3
log 10 = log(101) = 1
log 0.001 = log(10 3) = 3
The pH Scale
Example: A student is given a solution that is labeled pH = 4.72,
what is the molarity of H+ in this solution?
+ -= pH[H ] 10plugging in 10 4.72 into you calculators yields:
1.90546 10 5
but wait… how many sig. figs. are allowed?
- - -= = ´4.72 (0.28 5) 0.28 510 10 10 10
100.28 = 1.9 2 sig. figs.!
therefore the concentration
should be reported as: - +´ 51.9 10 M [H ]
The pH of a solution provides a way to express the acidity, or
the concentration of H+ in solution:
low pH = high [H+]
acidic solution
high pH = low [H+]
basic solution
a pH of 7 indicates that the solution is neutral
The pH Scale: 0 to 14
Prior to now, we have discussed reactions in solution
from a qualitative aspect.
Reactants Products
With the addition of molarity to our tools of chemistry,
we now can perform quantitative calculations for
reactions in aqueous solutions.
volume moles moles grams
grams moles moles volume
Solution Stoichiometry
Example:
How many grams of calcium carbonate can be
consumed by 35.5 mL of 0.125 M H2SO4 (aq) ?
Example:
How many grams of calcium carbonate can be
consumed by 35.5 mL of 0.125 M H2SO4 (aq) ?
H2SO4 (aq) + CaCO3 (s) H2O (l) + CO2(g) + CaSO4(s)
Solution: We know that acids react with carbonate salts to produce CO2(g)
Example:
How many grams of calcium carbonate can be
consumed by 35.5 mL of 0.125 M H2SO4 (aq) ?
H2SO4 (aq) + CaCO3 (s) H2O (l) + CO2(g) + CaSO4(s)
Solution: We know that acids react with carbonate salts to produce CO2(g)
mL H2SO4 mol H2SO4 mol CaCO3 g CaCO3
convert volume to grams
Example:
How many grams of calcium carbonate can be
consumed by 35.5 mL of 0.125 M H2SO4 (aq) ?
H2SO4 (aq) + CaCO3 (s) H2O (l) + CO2(g) + CaSO4(s)
Solution: We know that acids react with carbonate salts to produce CO2(g)
mL H2SO4 mol H2SO4 mol CaCO3 g CaCO3
convert volume to grams
35.5 mL 3
L
10 mL2 40.125 mol H SO
L3
2 4
1 mol CaCO
1 mol H SO3
3
100.1 g CaCO
1 mol CaCO
= 30.444g CaCO
• Some solutions cannot be accurately made by
weight and dilution methods if the solute is
impure or unstable.
• When this is the case, one can make up a
solution of approximate concentration then
“Standardize” the solution against a “Standard”
compound that reacts with the solute in solution.
• A standard compound is one that is very stable
with a known molar mass that yields a
necessary number of sig. figs. (4 or more)
Standard Solutions
• To standardize a solution, one performs a
“Titration” where a measured volume of the
solution is added to a known amount of the
standard.
• An “Indicator” is added to signal the point
where the moles of standard = moles of
solute reacted (Endpoint), knowing moles
and volume, one can compute the
concentration of the solution.
Standard Solutions
Setup for Titrating an Acid with a
Base
1. Add solution from the buret.
2. Reagent (base) reacts with
compound (acid) in solution in
the flask.
3. Indicator shows when exact
stoichiometric reaction has
occurred.
4. Net ionic equation
H3O+(aq) + OH-(aq)
2 H2O (l)
5. At equivalence point
moles H3O+ = moles OH-
Titrations
A common standard used in base (OH–) standardizations in
the mono–protic acid, KHP.
C
C
C
C
C
C
C
O
OH
C
OK
O
H
H
H
H
KHP is an acronym for:
potassium hydrogen phthalate (not potassium hydrogen phosphorous)
The H-atom in the upper right of the
compound dissociates to yield H+(aq)
1 mol of KHP yields 1 mol of H+ when dissolved in solution
KHP (aq) H+(aq) + KP–(aq)
The molar mass of KHP is 204.22 g/mol (C8H5O4K)
KPH
Problem: A student prepares a solution of approximately 0.5 M NaOH(aq).
He performs a standardization of the solution using KHP as the
standard to determine the exact concentration.
Determine the [OH–] from the data collected:
Problem: A student prepares a solution of approximately 0.5 M NaOH(aq).
He performs a standardization of the solution using KHP as the
standard to determine the exact concentration.
Determine the [OH–] from the data collected:
mass KHP + flask:
mass empty flask:
95.3641 g
95.0422 g
final buret reading:
initial buret reading:
30.12 mL
1.56 mL
Problem: A student prepares a solution of approximately 0.5 M NaOH(aq).
He performs a standardization of the solution using KHP as the
standard to determine the exact concentration.
Determine the [OH–] from the data collected:
mass KHP + flask:
mass empty flask:
95.3641 g
95.0422 g
final buret reading:
initial buret reading:
30.12 mL
1.56 mL
mass KHP: 0.3219 g vol. NaOH: 28.56 mL
Problem: A student prepares a solution of approximately 0.5 M NaOH(aq).
He performs a standardization of the solution using KHP as the
standard to determine the exact concentration.
Determine the [OH–] from the data collected:
mass KHP + flask:
mass empty flask:
95.3641 g
95.0422 g
final buret reading:
initial buret reading:
30.12 mL
1.56 mL
mass KHP: 0.3219 g vol. NaOH: 28.56 mL
- =[OH ]-mols OH
L of solution=
mol KHP
L titrated
Remember, [OH–] = [NaOH] Strong Electrolyte!!!
Problem: A student prepares a solution of approximately 0.5 M NaOH(aq).
He performs a standardization of the solution using KHP as the
standard to determine the exact concentration.
Determine the [OH–] from the data collected:
0.3219 g KHP1 mol KHP
204.22 g
1 mol OH 1 mol KHP
1 28.56 mL
310 mL
L
0.05519 M OH
[NaOH] = 0.05519 M
from the stoichiometry, one mole of
OH reacts with 1 mol of KHP
Problem: • Apples contain a compound know as malic acid, C4H6O5.
• Malic acid is a “diprotic” acid, meaning that one mole of the acid
reacts with two moles of base, OH (aq).
• If 76.80 g of apple requires 34.56 mL of 0.664 M NaOH for
titration, what is weight % of malic acid in the apples?
Problem: • Apples contain a compound know as malic acid, C4H6O5.
• Malic acid is a “diprotic” acid, meaning that one mole of the acid
reacts with two moles of base, OH (aq).
• If 76.80 g of apple requires 34.56 mL of 0.664 M NaOH for
titration, what is weight % of malic acid in the apples?
Data Information:
Apple contains acid
Write the balanced
chemical equation Step 1:
Eq. gives mole ratios
(stoichiometry)
Convert volume to
moles Step 2:
Moles of OH used
to titrate
Convert moles reactant
to moles product Step 3:
amount of malic acid
in moles
Convert moles of
products to mass Step 4:
mass of malic acid in
grams / wt. % in apples
Problem: • Apples contain a compound know as malic acid, C4H6O5.
• Malic acid is a “diprotic” acid, meaning that one mole of the
acid reacts with two moles of base, OH (aq).
• If 76.80 g of apple requires 34.56 mL of 0.664 M NaOH for
titration, what is weight % of malic acid in the apples?
The equation for the reaction is:
C4H6O5(aq) + 2NaOH(aq) Na2C4H4O5(aq) + 2H2O(liq)
Mass % of Malic Acid:
4 6 54 6 53
4 6 5
1 mol C H O1 L 0.664 mol OH 134.09 g34.56 mL 154 g C H O
10 1 L 2 mol OH 1 mol C H O
4 6 54 6 5
154g C H O100 2.01 C H O
76.80g Sample´ =
Spectrophotometry is the interaction between light
and matter that provides quantitative about solution
concentration.
Spectrophotometry
Light
Source
Wavelength
Selection
Sample
Container Spectrum
An Absorption Spectrophotometer
• Amount of light absorbed by a sample depends on
path length and solute concentration.
Different concentration of Cu2+ Same concentration, different
pathlengths
Spectrophotometry
BEER-LAMBERT LAW relates amount of light
absorbed and the path length and solute
concentration.
A = absorbance l = path length
= molar absorptivity c = concentration
• There is a linear relation between A and c for a
given path length and compound.
• Once the instrument is calibrated, knowing
absorbance one can determine concentration.
Spectrophotometry
To use the Beer-Lambert law you must first calibrate the
instrument.
The calibration plot can be
used to find the unknown
concentration of a solution
from a measured
Absorbance.
Spectrophotometry