Chapter 3. Stoichimetry
Calculations with chemical
formula and reactions are
called Stoichiometry
2-1
KEY CONCEPTS
Atomic, molecular and formula mass
Mole, Avogadro's number and Molar mass
Mass percent of elements
Mass % from analytical data
Mass % from formula
Empirical formula from Mass %
Chemical formula from composition
Chemical formula from mass %
Chemical Equations
Stoichiometric coefficients
Balancing chemical equations Stoichimetry
Stoichiometric coefficients and Limiting reactant
Yields of chemical reactions
Actual yield and Thoretical yield
Solutions
concentration and dilution of solutions
Solution stoichiometry
2-2
Question
You were given a chance to
pick 100 kg of
gold(I)periodate - AuIO4 or
gold(I)nitrate- AuNO3
Which one you would pick?
Rationalize your choice.
2-3
Gold
1 oz = $ 400
1 oz = 28.35 g
1 kg = 35.27 oz
1 kg Au = $ 14,109
2-4
Molecular mass vs. formula mass
Formula mass
Add the masses of all the atoms in formula
- for molecular and ionic compounds.
Molecular mass
Calculated the same as formula mass
- only valid for molecules.
Both have units of either u or grams/mole.
Examples of M.W.
H3PO4:
H = 1.01 g/mol,
P = 30.97 g/mol,
O = 16.00 g/mol
1 amu is equal to 1 g/mol
m.w. H3PO4 =
3 x 1.01+1 x 30.97+ 4 x 16.00
= 98.00 g/mol
2-6
Examples of F.W.
K2CO3:
K = 39.10 g/mol;
C = 12.01 g/mol;
O = 16.00 g/mol
f.w. K2CO3 =
2 x 39.10 + 1 x 12.01 + 3 x 16.00
= 138.2 g/mol
2-7
Mole Concept
We use masses weighed in grams in chemical
reaction. Need a conversion factor to
convert grams to atoms and molecules.
atomic weight or molecular weight taken in
grams contains
6.022 x 1023
atoms, molecules or particles.
The number 6.022 x 1023 is called Mole or
Avagadro’s Number
2-8
Can you guess this?
• How long is 1 mole of seconds?
19
• 1.9096 x 10 years
10
• 2 x 10 billion years
• How deep is the layer of
marbles, if 1mole of marbles
are spread over the surface of
earth?
• ~50 miles high
2-9
The Mole and Avogadro's number
Number of atoms in 12.000 grams of
1 mol
1g
1u
=
=
=
12C
6.022 x 1023 atoms
6.022 x 1023 u
1 g/mol
Mole is the converssion factor between M.W
and grams
1 mol =
grams / molecular weight
mole
= g/m.w. Or
g/f.w.
Conversion factors in
Stoichiometry?
6.022 x 1023 atoms = gram atomic weight
6.022 x 1023 molecules = gram molecular weight
6.022 x 1023 atoms C = 12.01 grams of carbon (C)
6.022 x 1023 molecules H2O = 18.02 g of H2O
6.022 x 1023 = 1 mol
1 g = 6.022 x 1023 amu (u)
1 amu = 1 g/mol
2 - 11
2 - 12
Atomic Masses
Atomic mass in grams ?
Mass of a copper atom in grams
63.55 g Cu
1 mole
1 mole
6.022 x 1023
= 1.055 x 10-22 g
Atomic mass in the periodic table divided by
Avagadro’s Number
2 - 13
An atom weighs 7.47 x 10-23 g.
What is the name of the element
this atom belongs to?
2 - 14
An atom weighs 7.47 x 10-23 g.
What is the name of the
element this atom belongs to?
Conversion factor: 1g = 6.022 x 1023 u
7.47 x 10-23 g x
6.022 x 1023 u
1g
= 44.98 u or g/mole
The element is Sc.
2 - 15
Example - (NH4)2SO4
How many atoms are in 20.0 grams of
ammonium sulfate?
Formula weight = 132.14 grams/mol
Atoms in formula
= 15 atoms / unit
1 mol
moles = 20.0 g x
= 0.151 mol
132.14 g
atoms
units
23
atoms = 0.151 mol x 15 unit x 6.02 x10
mol
atoms = 1.36 x1024
How do you convert grams to
mole and vice versa?
Grams ---> mole
grams /molecular (formula) weight
Moles ----> grams
moles x molecular (formula) weight
2 - 17
Calculations
Calculate how many moles are in 100 g of sulfur-S8
Calculate the grams of sugar in 3 moles of -glucose-
C6H12O6
2 - 18
% Element Composition in
Compounds
n x Atomic weight
% mass = --------------------- x 100
molecular weight
n = subscript of the element in the formula
2 - 19
Calculation
Al2(SO4)3
calculate % Al, % S and % O
2 - 20
Examples
a) Al2(SO4)3:
f.w. = 342.14 g/mol,
Atomic weight of oxygen = 16.00 g/mol,
n = 12
12 x 16.00
% O mass = ------------- x 100 = 56.12% oxygen.
342.14
c) CuSO4: f.w. = 159.61 g/mol
Atomic weight of oxygen = 16.00 g/mol n = 4
4 x 16.00
% O mass =-------------- = 40.01% oxygen.
159.61
2 - 21
Calculate % composition
from analysis
A 12.5 g sample of a compound that
contains only phosphorus and sulfur
was analyzed and found to contain
7.04 g of phosphorus and 5.46 g of
sulfur.
Calculate the percent composition of
elements.
2 - 22
Examples
Glucose has a molecular formula of
C6H12O6 (M.W. 180.16 g/mol).
a) How many grams of C, H and O are
available in 1 mole of glucose?
b) Calculate mass percents of
elements C, H and O in glucose.
2 - 23
How many grams of C, H and O are
available in 1 mole of glucose?
1 mol C6H12O6= 6 mol C = 6 x 12.01 =72.06g C
1 mol C6H12O6=12 mol H= 12 x 1.01
= 12.12g H
1 mol C6H12O6=6 mol O =6 x 16.00
= 96.00g O
2 - 24
Mass percent of element. C6H12O6 =
180.16 g/mol
%C =
%H =
%O =
6 x 12
--------- x 100 =
40.00% C
180.16
12 x 1.01
------------- x 100 =
6.73% H
180.16
6 x 16.00
------------ x 100 = 53.29% O
180.16
----------100.02%
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What is Empirical Formula
Simple whole number ratio of
each atom expressed in the
subscript of the formula.
Molecular Formula = C6H12O6 of glucose
Empirical Formula = CH2O
Emiprical formula is calculated from % composition
2 - 26
How do you get Empirical
Formula from % composition
and vice versa?
2 - 27
% composition to Empirical
formula
• Divide percent composition by
atomic weight.
• Divide answer by lowest number to
get simple ratio of moles or atoms.
• Multiply by a factor to get whole
number ratio.
• Write a formula with whole number
ratio as subscripts to get empirical
formula.
2 - 28
Examples
• A 12.5 g sample of a compound that
contains only phosphorus and sulfur was
analyzed and found to contain 7.04 g of
phosphorus and 5.46 g of sulfur.
• a) Calculate the percent composition of
elements.
• b) The empirical formula of the compound
2 - 29
Calculation
mass of element
% Element = ------------------- x 100
mass of sample
7.04
%P =
--------- x 100 = 56.3% P
12.5
5.46
%S =
------- x 100
= 43.7% S
12.5
100.0
2 - 30
Calculation
% of
P
and
56.3
Moles of P and S:
56.3g
------- = 1.82 mol P
30.97
Atom ratio:
1.82 atoms P
1.33 atoms P
S:
43.7
43.7g
------ = 1.36 mol S
32.06
1.36 atoms S
1.00 atoms S
2 - 31
Calculation..continued
1.82 atoms P
1.36 atoms S
Divide by lowest number:
1.33 atoms P
1.00 atoms S
simple atom ratio:
1.33 atom P
1 atom S
to get a simple whole number ratio. Multiply
both numbers by 2, 3, and 4 etc. until 1.33
becomes close to a whole number.
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Calculation..continued
To get simple whole number ratio:
P
S
1.33 atom
1 atom
2 x 1.33 = 2.66
2 x 1
2.66
2
3 x 1.33= 3.99
3 x 1
4.00
3.0
Therefore, Formula should contain 4 P
atoms and 3 S atoms.
Therefore, the empirical Formula of the
compound is: P4S3
2 - 33
More Example
Glucose contains the elements C, H, and O
in 39.99% C, 6.71% H and 53.28% O,
respectively, by mass.
a) Calculate the empirical formula of
glucose.
b) Molecular weight determination in
solution for glucose has shown a
molecular weight close to 180
g/mol. What is the molecular formula of
glucose?
2 - 34
Calculation
Empirical formula
from % elemental composition.
% composition: 39.99% C, 6.71% H, 53.28% O
ii) mole ratio:
C
H
O
39.99
6.71
53.28
------- = 3.33;
-------- = 6.64; -------- = 3.33
12.01
1.01
16.00
3.33 mol C:
6.64 mol H:
3.33 mol O
2 - 35
Calculation..continued
Atom ratio
3.33 mol C:
6.64 mol H:
Simple atom:
3.33
6.64
------------3.33
3.33
1
1.99
Simple whole
number
1
2
3.33 mol O
3.33
------3.33
1
1
number ratio of atom
The empirical formula of the compound is
CH2O
2 - 36
Calculation..continued
Molecular Formula = n x empirical Formula
Molecular weight
180
n = -------------------------------------- = ------ = 6
Empirical Formula Weight
30
Molecular Formula = (CH2O)n = (CH2O)6
Molecular Formula = C6H12O6 of glucose
2 - 37
Question
A molecular compound
contains 92.3% carbon and
7.7% hydrogen by weight.
What is its empirical
formula?
2 - 38
Question
A molecular compound has
empirical formula CH. If
0.050 mol of the compound
weighs 3.90 g, what is its
molecular formula?
2 - 39
Chemical Equation
P4O10 (s) + 6H2O (l)
=
4 H3PO4(l)
reactants enter into a reaction.
products are formed by the reaction.
Parantheses represent physical state
stoichiometric coefficients are numbers
in front of chemical formula formula
gives the amounts (moles) of each
substance used and each substance
produced.
Equation Must be balanced!
2 - 40
Chemical
Reaction
Could be described in words
Chemical equation:
Reactants?
Products?
reaction conditions?
=, ---> , <==> or
?
stoichiometric coefficients?
Number in front of substances
representing moles, atoms, molecules
2 - 41
Steps in Stoichiometric
Calculations
• Check whether chemical equation is
balanced
• get the moles from grams of materials
• find the limiting reactant
• calculate moles of products from the
limiting reactant
• convert moles of the products to grams
• find the actual yield of the reaction
• calculate % yield of the reaction
2 - 42
Question
How much hydrogen
gas is produced when
1 kg of sodium reacts
with water?
2 - 43
Examples
Calculate the following using the chemical
equation given below:
4 NH3(g) + 5 O2(g) ----> 4 NO(g) + 6 H2O(g)
a) moles of NO(g) from 2 moles of NH3(g)
and excess O2(g).
b) moles of H2O(g) from 3 moles of O2(g)
and excess NH3(g).
2 - 44
Examples
How many moles of H2O will be
produced by 0.80 mole of O2 with
excess H2 according to the
equation?
2H2(g) + O2(g) = 2 H2O(l)
2 - 45
Question
2Al(s) + 6HCl(aq)--> 2AlCl3(aq) +3H2(g)
According to the equation above, how
many grams of aluminum are needed
to react with 0.582 mol of hydrochloric
acid?
2 - 46
5.23 g Al
2 - 47
What is the limiting reagent?
Limiting reagent is the reactant,
which is used up first.
To find the limiting reactant you
have to compare the amounts of
reactants in moles.
2 - 48
Examples
A 300.0 g sample of phosphorus
( M.W. 123.88 g/mol) burns in 500.0
g of oxygen (M.W. 32.00 g/mol)
according to following equation:
P4(s) + 5O2(g) = P4O10(s)
What is the limiting reagent?
2 - 49
Theoretical yield
Theoretical yield is the amount (grams)
of products formed according to
chemical equation.
Use the limiting reagent to calculate the
moles of the product and then convert
moles to grams.
2 - 50
Actual Yield
Actual yield is the grams of the
product obtained by an experiment.
Actual yield should be less than the
theoretical yield if the experiment was
carried out meticulously. If the
products are contaminated with
impurities or the formula of product
was wrong the actual yield could be
higher.
2 - 51
% Yield
actual yield
% yield = ------------------------ x 100
theoretical yield
If the products are contaminated with
impurities or the formula of product
was wrong the % yield could be higher
than 100%.
2 - 52
Question
Sulfur trioxide, SO3 , is made from
the oxidation of SO2 and the
reaction is represented by the
equation
2SO2 + O2
-----> 2SO3
A 16.0-g sample of SO2 gives
18.0 g of SO3. The percent
yield of SO3 is
2 - 53
Examples
• A 300.0 g sample of phosphorus ( M.W. 123.88
g/mol) burns in 500.0 g of oxygen (M.W. 32.00
g/mol) according to following equation:
•
P4(s) + 5O2(g) = P4O10(s)
• a) What is the limiting reagent?
• b) How many moles of P4O10 are produced
theoretically?
• c) If 612 g of P4O10 is actually produced in this
reaction, calculate the percent yield.
2 - 54
How do you calculate moles
of substances in solutions
• Use concentration of solution to
convert L or mL of solution in to moles
• What is concentration of a solution?
• The relative amounts of solute and
solvent
• There are so many ways to show
amount: g, mole, equivalents,volume
2 - 55
Concentration Units
•
•
•
•
•
•
•
•
a) Molarity (M)
b) Normality (N)
c) Molality (m)
f) Mole fraction (Ca)
g) Mass percent (% weight)
h) Volume percent (% volume)
i) "Proof"
j) ppm and ppb
2 - 56
Molarity
M =
moles solute
=
liters of solution
mol
L
• [ ] - special symbol which means molar
( mol/L )
2 - 57
Molarity
What’s the molarity of a solution that has
18.23 g HCl in 2.0 liters?
First, you need the FM of HCl.
FMHCl
= 1.008 x 1 H + 35.45 x 1 Cl
= 36.46 g/mol
Next, find the number of moles.
molesHCl = 18.23 gHCl / 36.46 g/mol
= 0.50 mol
Finally, divide by the volume.
MHCl
= 0.50 mol / 2.0 L
2 - 58
Examples
Calculate the molarity of a solution prepared
by dissolving 200.0 g of K2SO4 in enough
water to make 500.0 mL solution.
moles of solute
Molarity(M) = ---------------------Liters of solution
2 - 59
solute = K2SO4; F.W. = 174.27 g/mol; mass= 200g
moles of K2SO4 = ?
200 g /174.27 g K2SO4
= 1.148 mol K2SO4
500.0 mL = ?
Liters of solution = 0.5 L
Molarity?
1.148 mol K2SO4
Molarity of K2SO4 sol. = -----------------------0.5 Liters of solution
= 2.30 mole/Liter = 2.30 M (M = moles/liters)
2 - 60
Normality (N)
Equivalents of solute
Normality(N) = ---------------------------Liters of solution
moles
Equivalents = ----------------------n
n is # of H, OH in or e- in acid or base
N(normality) = Molarity x n (# of H, OH or e-)
2 - 61
Examples
How many grams of KNO3 are contained in
500 mL of a 0.500 M solution of potassium
nitrate?
How many mL of 2.00 M solution of HNO 3 are
required with water to make a 250 mL of
1.50 M nitric acid solution?
2 - 62
Mole fraction (Ca)
Ca
moles of solute (substance)
= ------------------------------------moles of solute + solvent
2 - 63
Calculate the mole fraction of
benzene in a benzene(C6H6)chloroform(CHCl3) solution which
contains 60 g of benzene and 30 g
of chloroform.
M.W. = 78.12 (C6H6)
M.W. = 119.37 (CHCl3)
2 - 64
Weight/Weight %
Weight/Weight %
= Mass Solute x 100
Total Mass
Use the same units for both
If a ham contained 5 grams of fat in 200 g
of ham, what is the % wt/wt?
5 g / 200g * 100
=
2.5 wt/wt%
On the label, it would say 97.5 % fat free.
2 - 65
Volume/Volume %
Volume/Volume %
= Volume Solute x 100
Total Volume
Use the same units for both
If 10 ml of alcohol is dissolved in water to make
200 ml of solution, what is the concentration?
10 ml / 200 ml * 100
=
5 V/V%
Alcohol in wine is measured as a V/V%.
2 - 66
Weight/Volume %
Weight/Volume %
=
Mass solute x 100
Total Volume
use g and ml
If 5 grams of NaCl is dissolved in water to make
200 ml of solution, what is the concentration?
5 g / 200 ml * 100
=
2.5 wt/v%
Saline is a 0.9 wt/v% solution of NaCl in water.
2 - 67
Very low concentrations
Pollutants in air and water are typically found
at very low concentrations. Two common
units are used to express these trace
amounts.
Parts per million - ppm
Parts per billion - ppb
Both are modifications of the % system
which could be viewed as parts per hundred
- pph.
Both mass and volume % systems are used.
2 - 68
Low concentrations in air
Trace amounts in are are expressed as
volume/volume ratios.
ppm
ppb
=
=
volume solute
volume solution
volume solute
volume solution
x 106
x 109
Example. One cm3 of SO2 in one m3 of air
would be expressed as 1 ppm or 1000 ppb.
2 - 69
Low concentrations in water
Mass percentages are used for water
pollutants.
ppm =
ppb =
mass solute
mass solution
mass solute
mass solution
x 106
x 109
Example. One ppm of a toxin in water is the
same as 1 mg / liter since one liter of water
has a mass of approximately 106 mg.
2 - 70
Solution preparation
Solutions are typically prepared by:


Dissolving the proper amount of solute and
diluting to volume.
Dilution of a concentrated solution.
Lets look at an example of the calculations
required to prepare known molar solutions
using both approaches.
2 - 71
Dilution Problems
Why we dilute solutions?
Preparing solutions by adding water
to concentrated solutions
moles before = moles after
MiVi = MfVf
Mi
Vi
Mf
Vf
=
=
=
=
initial molarity
initial volume
final molarity
final molarity
2 - 72
Examples
How many mL of 2.00 M solution of HNO 3 are
required with water to make a 250 mL of 1.50 M
nitric acid solution?
MiVi = MfVf
Mi = 2.00
Vi = ?
Mf = 1.50
Vf = 250 mL
MfVf
Vi = --------------- =
Mi
1.50 x 250
---------------- = 187.5 mL
2.00
2 - 73
Stoichiometric calculations
of solutions reactions
• Check whether chemical equation is
balanced
• get the moles from volume of solutions
• find the limiting reactant
• calculate moles of products from the
limiting reactant
• convert moles of the products to
grams
• find the actual yield of the reaction
• calculate % yield of the reaction
2 - 74
Solution stoichiometry example
Determine the volume of 0.100 M HCl that
must be added to completely react with 250
ml of 2.50 M NaOH
Balanced chemical equation
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O (l)
The first step is to determine how many
moles of NaOH we have.
2 - 75
Solution stoichiometry example
We have 250 ml of a 2.50 M solution.
molNaOH
= 0.250 L x 2.50 mol/L
= 0.625 molNaOH
From the balanced chemical equation, we
know that we need one mole of HCl for each
mole of NaOH.
That means we need 0.625 molHCl.
2 - 76
Solution stoichiometry example
Now we can determine what volume of our
0.100 M HCl solution is required.
L
= molHCl / MHCl
=
(
0.625 mol
1L
0.100 mol
)
= 6.26 L
2 - 77
Examples
How many mLs of 0.100 M BaCl2 are required
to react completely with 25 mL of 0.200 M
Fe2(SO4)3?
3 BaCl2(aq) + Fe2(SO4)3(aq)---> 3 BaSO4(s) + 2 Fe Cl3(aq)
3 BaCl2 = 1 Fe2(SO4)3
2 - 78
Calculate the moles of Fe2(SO4)3 :
moles = Molarity x
Liters of solution
0.200 M x 0.025 L = 0.005 mole Fe2(SO4)3
Then convert Fe2(SO4)3 to BaCl2 mole, BaCl2 moles
to liters and liters to mL.
0.005 mole Fe2(SO4)3 ----> BaCl2 moles
0.005 mol Fe2(SO4)3 x 3 = 0.015 BaCl2 moles
moles = Molarity x
Liters of solution
0.015 = 0.100 x Liters
Liters BaCl2 = 0.15L
= 150 mL of BaCl2
2 - 79
Examples
How many mLs of 0.300 M NaOH are
required to react with 500 mL of 0.170 M
H2SO4 completely?
Acid Base Reactions
.
Na xVa = Nb x Vb .
Na = Normality of acid,
Nb = Normality of base
V= volume
N(normality) = Molarity x n (# of H or OH)
Na (0.170 x 2) xVa (500) = Nb (0.300x 1)x Vb(?) .
Vb= 0.170 x 2 x 500/ 0.300 = 566.7 mL NaOH
2 - 80