The Mole
Stoichiometry: Cookbook Chemistry
The Mole
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A mole is a number
Avogadro’s number = 6.02x1023
Named after Amadeo Avogadro
Loschmidt determined the number
of particles in one cubic centimeter
of a gas at ordinary temperature
and pressure
Counting atoms by counting moles
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By counting moles, atoms or
molecules are counted
Counting atoms by using moles
eliminates waste in chemical
reactions
Coefficients in chemical equations
represent mole quantities
Counting atoms by counting moles
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2Na + Cl2  2NaCl
“Two moles sodium and one mole
chlorine gas react to give two moles
sodium chloride”
4 moles sodium require 2 moles Cl2
5.2 moles sodium require 2.6 moles
Cl2
3.1 moles Cl2 require 6.2 moles
sodium
2:1 is the sodium/chlorine mole ratio
Counting atoms by counting moles
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Counting atoms allows prediction of
product quantities
2Fe + 6HCl  2FeCl3 + 3H2
How many moles iron (III) chloride can
be made using 4.3 moles HCl?
Set up a proportion
Coeff. 2mol FeCl3 = x mol FeCl3
prob.
side
6mol HCl
4.3mol HCl
side
x=2(4.3)/6=1.43 mol FeCl3
Limiting reagents
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If mole quantities are not exact,
one of the reactants will run out
first – this reactant is the limiting
reagent
2H2 + O2  2H2O
If 3 moles H2 are reacted with 1
mole O2, what is the limiting
reagent?
Limiting reagents
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Divide each mole quantity by the
coefficient to find equivalents.
H2  3/2=1.5eq
O2  1/1=1eq  limiting reagent
The reactant with the fewest
equivalents (O2) is the limiting
reagent. The other (H2) is “in
excess”.
Limiting reagents
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2Fe + 6HCl  2FeCl3 + 3H2
0.0037 mol Fe is reacted with 0.017
mol HCl. What is the limiting
reagent?
Fe  0.0037/2 = 0.00185eq Fe
HCl  0.017/6 = 0.00283eq HCl
Using limiting reagents
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The quantity of product obtained is
limited by the amount of the
limiting reagent
2H2 + O2  2H2O
If 4.5 moles hydrogen gas and 1.9
moles oxygen are reacted, how
many moles water will be formed?
Using limiting reagents
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Solution: First determine the
limiting reagent.
H2: 4.5/2=2.25eq
O2: 1.9/1=1.9eq limiting reagent
Then set up a proportion between
the limiting reagent and the desired
product.
O2
1 = 1.9
x=3.8 moles
H2O 2
x
Molar Mass
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Molar mass is the mass of one mole of
particles
Atomic mass – found in the bottom of
each square of the periodic table – units
are grams/mole
Atomic mass is the weighted average of
the mass numbers of all the isotopes of
an element.
Molecular mass – the mass of one mole of
molecules
It is equal to the sum of the atomic
masses of all the atoms in the molecule.
Molar mass
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H2O – (H) 2x1 = 2
(O) 1x16 = 16 total = 18g/mol
NH3 – (N) 1x14 = 14
(H) 3x1 = 3 total = 17g/mol
glucose (C6H12O6)
(C) – 6x12 = 72
(H) – 12x1 = 12
(O) – 6x16 = 96 total = 180g/mol
Molar mass
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Formula mass is the sum of all the
atomic masses in a formula unit (for
salts)
NaCl – (Na) 23
(Cl) 35.5 total = 58.5g/mol
Mg(NO3)2
(Mg) 1x24.3 = 24.3
(N) 2x14 = 28
(O) 6x16 = 96
total = 148.3g/mol
Using molar mass
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Mass to moles conversions
mass/(molar mass) = moles
g  g/mol =
g x mol/g = moles
Example: How many moles are
represented by 2.5 grams of water?
Solution: 2.5g/(18g/mol) = 0.14mol
Using molar mass
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Moles to mass conversions
molesx(molar mass) = mass
mol x g/mol = g
Example: What is the mass of
0.094 moles sodium chloride?
Solution: 0.094mol x 58.5g/mol =
5.5g
Mass-mass stoichiometry
How many grams hydrogen peroxide
(H2O2) are needed to make 2.43
grams iron (III) nitrate (Fe(NO3)3)
according to the reaction below?
2HNO3 + H2O2 + 2Fe(NO3)2  2Fe(NO3)3 + 2H2O
moles H2O2
x by
molar
mass
moles iron (III) nitrate
mole ratio
(2:1)
x g H2O2
 by
molar
mass
2.43 g iron (III) nitrate
Per cent yield
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Mass obtained from calculations is
“theoretical yield” – never obtained
in practice
Per cent yield =
actual yield x 100%
theoretical yield
Per cent yield
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Jorma makes drugs for a hobby
(aspirin, that is) and expects to
obtain 2.13g aspirin from his
synthesis reaction. In reality he
only gets 1.89g. What is his %
yield?
(1.89/2.13)x100% = 88.7%
Per cent composition by mass
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% composition by mass is a tool for
compound identification
To calculate: divide the molar mass
contribution of each element by the
total molar mass and multiply by
100%
Per cent composition by mass
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Example: H2SO4 (sulfuric acid)
total molar mass
H: 1x2=2
S: 32x1=32
O: 16x4=64
sum=98g/mol
%H=2(100%)/98=2.04%
%S=32(100%)/98=32.65%
%O=remainder=65.31%=64(100%)/98
Determining formulas from %
composition
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Formulas are a mole ratio of elements
Empirical formula: simplest mole ratio of
elements, like NaCl or Ca(NO3)2
Applies to any type of compound
Molecular formula: mole ratio of elements
in an actual molecule (all nonmetals), like
H2O or NH3
Often the molecular formula and the
empirical formula are the same, but not
always
Determining formulas from %
composition
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Hydrazine, a rocket fuel
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Hydrogen peroxide
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molecular formula – N2H4
empirical formula – NH2
molecular formula – H2O2
empirical formula – HO
Glucose, a sugar
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molecular formula – C6H12O6
empirical formula – CH2O
Determining formulas from %
composition
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% composition is a mass ratio – so
by converting mass to moles, the
empirical formula can be
determined.
Example: Laboratory analysis finds
a compound to consist of 28.05%
Na, 29.27% C, 3.67% H, and
39.02% O. What is the empirical
formula?
Determining formulas from %
composition
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Treat the % like grams
Convert grams to moles
Na: 28.05g/(23g/mol) = 1.22 mol
C: 29.27g/(12g/mol) = 2.44 mol
H: 3.67g/(1g/mol) = 3.67mol
O: 39.02g/(16g/mol) = 2.44 mol
Determining formulas from %
composition
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Convert to simplest whole number ratio –
divide all mol quantities by the smallest
one. These results become the subscripts
in the formula.
Na: 1.22/1.22 = 1
C: 2.44/1.22 = 2
H: 3.67/1.22 = 3
O: 2.44/1.22 = 2
So the empirical formula is NaC2H3O2
(sodium acetate).
Ideal gas law
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Boyle’s Law: PV = C (P1V1 = P2V2)
Factors that affect pressure/volume:
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Temperature (T)
Amount (moles) of gas (Avogadro’s
Principle) (n)
Ideal Gas Law: PV  nT
Constant of proportionality = R (gas
constant)
Ideal Gas Law
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Ideal gas Law: PV = nRT
V must be liters, T is Kelvins, n is
moles
Values for gas constant (depends on
pressure units)
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P in atm: R = 0.08206Latm/molK
P in kPa: R = 8.314LkPa/molK
P in torr: R = 62.4Ltorr/molK
Ideal Gas Law
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Example: Find the moles of oxygen
in a balloon of 2.3L volume and
1.3atm pressure if the temperature
is 45ºC.
Solution: PV = nRT
1.3(2.3) = n(0.08206)(45+273)
n = 1.3(2.3)/(0.08206)(45+273)
n = 0.115 mol
Ideal Gas Law
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Example 2: Find the molar volume
of a gas at STP.
Solution: STP = standard
temperature and pressure (273K
and 1 atm)
PV = nRT
1V = 1(0.08206)(273) = 22.4L/mol