Stoichiometry: Calculations with
Chemical Formulas and Equations
Sponsored by
Lavoisier & Avogadro
Chapter 3
In this chapter
• The mole concept.
• Relationships between chemical formulas,
atomic masses and moles.
• Empirical and molecular formulas
• Writing chemical equations.
• Using moles to find the formulas of
compounds.
• Limiting reagent problems.
2
The Mole
• Atoms and molecules are extremely small
making it very difficult to measure their
masses individually.
– It is easier to weigh a large collection of these.
• A mole (abbreviated as mol) is defined as the
number of C atoms in exactly 12 grams of
pure C-12.
• This number is called Avogadro’s number.
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1
Fun facts about the Mole
• The American Chemical
Society celebrates the Mole
Day annually.
• If 10,000 people started to
count Avogadro’s # at the
rate of 100/minute every
day, it would take them 1
trillion years.
• 1 mole of dollars will provide
each person of the Earth’s
population (approx. 6 billion)
an income of $5000/s for
about 100 years.
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Mole of atoms: The Molar Mass
• The mass in grams of one mole of atoms of
any element is known as the molar mass of
the element. The unit used is g/mol (gmol-1).
• Molar mass of sodium (Na) = mass of exactly
one mole of Na atoms = 22.99 g/mol = mass
of 6.022 x 1023 atoms of Na.
• Molar mass of Uranium (U) = 238.03 g/mol.
• The atomic mass number of an element in a
periodic table = molar mass for the element
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6
2
Uses of the Mole
• Two important conversions that we must know:
• Grams to moles:
• The evil twin; moles to grams:
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Grams to moles
• Calculate the number of moles in 225.0 g of
Na.
• Need to look up the molar mass of Na from the
periodic table and then calculate the # of
moles.
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Moles to grams
• Calculate the number of grams of Na in 145.95
moles of Na.
9
3
Moles of Compounds
• 1 mole of any compound = 6.022 x 1023 units
of the compound.
• Molar masses of compounds are calculated by
adding the molar masses of the atoms present
in the compound.
• Calculate the molar mass of Ca3(PO4)2
• Need to look the molar masses of each of the
element present and multiply by the number of
the atoms present.
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Molar mass of Ca3(PO4)2
3 Ca
2P
8O
3 moles Ca x
40.078 gmol-1 =
2 moles P x
30.974 gmol-1 =
8 moles O x
15.999 gmol-1 =
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Working with Molar Masses
• Calculate the # of moles of Ca3(PO4)2 in 77.5 g
of Ca3(PO4)2.
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4
Another example
• Calculate the # of moles of O atoms in 77.5 g
of Ca3(PO4)2.
• The chemical formula gives us a clue (a
conversion factor):
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Percent composition of compounds
• Composition of compounds can be described
in 2 ways:
– By the number of its constituent atoms.
– By the mass (%) of each element present.
• The mass percents of elements are obtained
by comparing the mass of each element
present in 1 mole of the compound to the total
mass of the compound.
• Useful today with unknown compounds.
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Using % composition
• Calculate the % by weight of each element in
Ca3(PO4)2.
– Find the molar mass of Ca3(PO4)2 = (310.174 gmol-1).
– Find the mass of each element in 1 mole of Ca3(PO4)2,
divide by the molar mass of Ca3(PO4)2 and multiply by
100%
3 Ca
3 moles Ca x
40.078 gmol-1 =
120.234
(No rounding)
2P
2 moles P x
30.974 gmol-1 =
61.948
8O
8 moles O x
15.999 gmol-1 =
127.992
1 mole Ca3(PO4)2 =
310.174 gmol-1
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5
Using % composition
• Now divide the mass of each element by the
molar mass and multiply by 100.
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Empirical Formulas
• Molecular formulas tell us 2 things:
– the relative number of atoms (atom ratio) of each
element in a compound.
– the total number of atoms in a molecule
• Empirical formulas simplify the chemical
formula so that the molecular formula is
always a whole-number multiple of the
subscripts in the empirical formula.
• Molecular formula = (Empirical formula) x a
whole number
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Empirical formulas
• Following are possible combinations between
Nitrogen and Hydrogen:
N2 H4
N3 H6
N4 H 8
All of these are whole number
multiples of the simplest possible ratio
between N and H which is NH2
The empirical formula for all of these
compounds = NH2
• Sometimes the empirical and molecular formula can
be the same; H2O, CO, CO2.
• To find the molecular formula of a compound the
molar mass must be known.
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6
Mass % to empirical formula
• Determine the empirical and molecular formula
of a compound that has the following mass %:
71.65% Cl, 24.27% C and 4.07% H.
The molar mass of this compound is known to
be = 98.96 g/mol.
– Assume we have 100.00 g of the compound.
– Convert the mass % to mass in grams.
Thus in 100.00 grams of this compound there are
71.65 g of Cl, 24.27 g of C and 4.07 g H.
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– Convert these grams to moles.
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– Divide the moles by the smallest # of moles to
obtain the empirical formula.
The smallest # of moles is 2.021 and dividing
throughout gives us the empirical formula as
ClCH2, (empirical formula mass = 49.48 g mol-1).
– Obtain the molecular formula by dividing the molar
mass by the empirical formula mass.
Molar mass
98.96 g mol −1
=
=2
Empirical formula mass 49.48 g mol −1
Molecular formula = (ClCH2) x 2 = Cl2C2H4. This
compound is called dichloroethane and it used to be
a gasoline additive.
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7
Another one
• Eugenol is the active ingredient of oil of
cloves. It has a molar mass of 164.2 g/mol
and is 73.14% C and 7.37% H, the remainder
is oxygen. Find the empirical and molecular
formulas for eugenol.
– Assume 100.00 g of the compound.
– Convert the mass % to mass in grams; in 100.00
g of this compound there are 73.14 g C and 7.37 g
H.
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• Since the total sample mass is 100.00 g, the mass of
the oxygen is
100.00 g = 73.14 g C + 7.37 g H + mass of O
Mass of O = 19.49 g
• Now find the number of moles and proceed as
before:
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• The empirical formula is C H O and the empirical formula
mass =
g mol-1.
• Divide the known molar mass by this number to get the
molecular formula.
• Thus, the molecular formula is C H O
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8
Combustion Analysis
• Especially useful when the compound is composed
of C, H and O only.
• The assumption here is all the C in the compound
will be changed to CO2 and the H to H2O. The mass
odf the O can be found by subtracting the masses of
the C and the H from the mass of the original
compound.
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An acid isolated from clover leaves is known to
contain only C, H and O. 0.513 g of this acid
produces 0.501g of CO2 and 0.103 g of H2O.
The molar mass of the acid has been
determined = 90.04 g mol-1. Find the empirical
and molecular formulas of the acid.
CxHyOz + some O2
0.513 g
x CO2 + y/2 H2O
0.501 g
0.103 g
– Convert the grams of CO2 and H2O to moles, using their
molar masses (44.010 g mol-1 for CO2 and 18.015 g mol-1
for H2O).
– Moles of CO2 = 0.0114 moles, moles of H2O = 0.00572.
– These moles need to be converted to moles and then to
grams of the C and H.
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• From their formulas we can tell that 1 mole of CO2
contains 1 mole of C and 1 mole of H2O contains 2
moles of H. Using these relationships:
⎛ 1 mole C ⎞ ⎛ 12.01 g C ⎞
0.0114 moles of CO 2 × ⎜⎜
⎟ = 0.137 g C
⎟⎟ × ⎜
⎝ 1 mole CO 2 ⎠ ⎝ 1 mole C ⎠
⎛ 2 mole H ⎞ ⎛ 1.0079 g H ⎞
⎟⎟ × ⎜
0.00572 mole H 2O × ⎜⎜
⎟ = 0.0115 g of H
⎝ 1 mole H 2 O ⎠ ⎝ 1 mole H ⎠
• These calculations show that 0.513g of the sample
contain 0.137 g C and 0.0115 g H. The mass of the
oxygen present in the original sample =
0.513 g – (0.137 g C + 0.0115 g H) = 0.365 g.
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9
• Now to find the empirical formula of the compound
calculate the moles of each element in the sample:
⎛ 1 mole C ⎞
0.137 g C × ⎜⎜
⎟⎟ = 0.0114 mole C
⎝ 12.01g C ⎠
⎛ 1 mol H ⎞
⎟⎟ = 0.0114 mole H
0.0115 g H × ⎜⎜
⎝ 1.0079 g H ⎠
• Find the mole ratio of the
elements by dividing
throughout by the smallest
# of moles (C or H).
⎛ 1 mole O ⎞
⎟⎟ = 0.0228 mole O
0.365 g O × ⎜⎜
⎝ 15.999 g O ⎠
0.0114 mol H 1.00 mol H
=
0.0114 mol C 1.00 mol C
0.0228 mol O 2.00 mol O
=
0.0114 mol C 1.00 mol C
• The empirical formula of the acid is CHO2 and the
molecular formula is C2H2O4.
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Problematic Mole Ratios
• Sodium dichromate is a bright orange
compound with the following mass
percentages: 17.5% Na, 39.7% Cr and 42.8%
O. What is the empirical formula of this
compound?
– Assume 100.0 g of the compound
– In 100.0 g of the compound; 17.5 g = Na, 39.7 g =
Cr and 42.8 g = O.
– Convert all grams to moles.
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17.5g Na ×
1 mole Na
= 0.761 mole Na
22.990g Na
1 mole Cr
39.7g Cr ×
= 0.764 mole Cr
51.996g Cr
1 mole O
42.8g O ×
= 2.68 mole O
15.999g O
For Na :
• Now divide by the
smallest number of
moles (Na moles)
to get:
0.761 mol
= 1.00
0.761 mol
For Cr :
0.764 mol
= 1.00
0.761 mol
For O :
2.68 mol
= 3.53
0.761 mol
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10
Hydrated Compounds
• Ionic compounds that have water molecules
associated with the ions of the compound.
• Common in occurrence.
• Plaster board is made of Gypsum CaSO4·2H2O
(calcium sulfate dihydrate) and CaSO4 (Anhydrous
calcium sulfate).
• On heating Gypsum produces CaSO4·½H2O which is
also known as Plaster of Paris.
• There is no easy way to tell how much water a
hydrated compound has, so it has to be determined
experimentally.
• Heating the compound and finding the amount of
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water released by the compound is one way.
• To find the value of “y” in the blue hydrated
CuSO4·yH2O, 1.023 g of the hydrated salt was
heated. After heating the solid, 0.654 g of the
anhydrous white solid CuSO4 remained.
CuSO4·yH2O + heat
1.023 g
CuSO4 + yH2O
0.654g
• The mass of the water = (Mass of the hydrated
compound – mass of the anhydrous compound)
= 1.023 g – 0.654 g = 0.369 g.
• Need to convert the grams of water and the
anhydrous salt to moles.
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• Determine the ratio between the H2O and the CuSO4
moles.
• The water:CuSO4 ratio is 5:1 and the formula of the
hydrated compound is CuSO4·5H2O.
• The name of this compound is copper (II) sulfate
pentahydrate.
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11
The Chemical Reaction
• Reorganization of atoms in one or more substances
resulting in the production of one or more different
substances.
• Lavoisier proposed the law of conservation of matter
which states that matter can neither be created nor
destroyed.
• All chemical reactions are denoted by a chemical
equation.
Reactants
Products
• Some reactions have special names: combination,
decomposition and combustion (reaction of an
organic substance with oxygen).
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Chemical Equations
• The chemical equation for a reaction shows:
– The formulas of the reactants and the products.
– The relative number of each.
• The physical states of the reactants & the
products are also sometimes included:
– Solid (s), liquid (l), gas (g) and (aq) = aqueous
solution indicating dissolved in water.
• Chemical equations must be balanced:
– same number of atoms on both sides.
– same number of charges on both sides.
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What does an equation tell us?
NaHCO3 (s) + HCl (aq)
Reactants
NaCl (aq) + CO2 (g) + H2O (l)
Products
1 molecule of NaHCO3 +
1 molecule of HCl
1 molecule of NaCl + 1
molecule of CO2 + 1 molecule
of H2O
1 mole of NaHCO3 + 1
mole of HCl
1 mole of NaCl + 1 mole of CO2
+ 1 mole of H2O
6.022 x 1023 molecules of
NaHCO3 + 6.022 x 1023
molecules of HCl
6.022 x 1023 molecules of NaCl
+ 6.022 x 1023 molecules of
CO2 + 6.022 x 1023 molecules
of H2O
84.0059 g NaHCO3 +
36.4609 g HCl =
120.4668g reactant
58.443 g NaCl + 44.009g CO2 +
18.0148 g H2O = 120.4668 g
products
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12
Balancing an equation
• Do not change the formulas of the reactants or
products.
• Ensure that you have the same number of
atoms of each element by placing
stoichiometric coefficients in the front of the
reactant or product.
– Non-zero whole numbers, try to avoid fractions.
– These numbers indicate the number of moles of
the particular reactant or product.
• The relationship between the amounts of
reactants and products is called stoichiometry
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Balancing equations
• Systematic trial and error method.
• Balance the most complicated compounds first.
• You may also follow:
– M: metals
– I: ions, look for polyatomic ions that cross over from
reactants to products unchanged. Balance them as a
group.
– N: non-metals
– O & H: often involves water on one side or the other.
• Write a balanced equation for the combustion of
propane (C3H8).
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C3H8 (g) + O2 (g)
CO2 (g) + H2O (l)
• Balance the C atoms first, by placing a __ in front of CO2
C3H8 (g) + O2 (g)
__CO2 (g) + H2O (l)
• Now, balance the H atoms, by placing a __ in front of H2O
C3H8 (g) + O2 (g)
__CO2 (g) + __H2O (l)
• On the right side of the equation we now have (__ x __) + (__
x __) = __ O atoms. Balance these on the left by placing a __
in front of the O2.
C3H8 (g) + __O2 (g)
__ C atoms
__CO2 (g) + __H2O (l)
__ C atoms
__ H atoms
__ H atoms
__ O atoms
__ O atoms
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13
• Write a balanced equation for the combustion of C4H10
C4H10 (g) + O2 (g)
CO2 (g) + H2O (l)
• Following the procedure as before we can write:
C4H10 (g) + O2 (g)
__CO2 (g) + __H2O (l)
• Here there are an ____ number of O atoms on the products
side (_ x _) + (_ x _) = __ O atoms. Need __ moles of O2 on
the reactant side.
C4H10 (g) + ___ O2 (g)
___CO2 (g) + ___H2O (l)
• Coefficients can only be whole numbers and hence to avoid
this fraction we multiply the equation by ___ to get:
__C4H10 (g) + __O2 (g)
__CO2 (g) + __H2O (l)
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Stoichiometry
• Using balanced chemical equations we can
predict how much of the products will be
produced from the reactants.
• In real life when the reactants are mixed they
don’t always change completely to the
products.
• Must work in moles.
• Convert all grams to moles and then use the
stoichiometric coefficients.
• Limiting reactant problems.
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Typical Steps in a Stoichiometry
Problem
1. Convert grams of known substance to its
moles.
2. Multiply the moles from step 1 by the ratio of
the moles of the unknown to the known
(from the balanced equation).
3. Multiply the result from step 2 by the molar
mass of the unknown to get the grams of the
unknown.
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14
X moles reactant A
Y moles product B
X
Grams of reactant A
Grams of product B
⎛ 1 mole A ⎞
⎟⎟
× ⎜⎜
⎝ gA ⎠
Moles of reactant A
⎛ 1 mole B ⎞
⎟⎟
× ⎜⎜
⎝ gB ⎠
⎛ Y moles of B ⎞
×⎜
⎟
⎝ X moles of A ⎠
Moles of product B
Y and X are the stoichiometric coefficients from the balanced equation
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Calculating product amounts
• How many grams of CO2 will be produced by the combustion
of 50.00 g of C3H8?
– Start by writing a balanced equation.
– Convert the grams of C3H8 to moles of C3H8.
C3H8 (g) + __O2 (g)
__CO2 (g) + __H2O (l)
• From the stoichiometric coefficients in the balanced equation
obtain a stoichiometric factor and use it to convert the moles
of C3H8 to moles of CO2.
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Stoichiometric factor
• The numerical relation between the moles of any reactant(s)
and product(s) in a balanced equation.
• The last step is to convert the moles of CO2 to grams of CO2.
• Try to put all of these steps together:
• Calculate the grams of water produced in this reaction.
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15
Another example
• Hematite (Fe2O3) is a common and important
iron ore. Iron metal is obtained from this ore
by the reacting it with CO in a blast furnace.
How many grams of CO will be needed for the
reaction of 1.00 kg of Fe2O3?
• 1 kg = 1000 g, start by writing a balanced
equation:
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Limiting reactants: theoretical and
experimental yields
• Some times reactants are added in amounts
different from as required by the chemical
equation.
• Only one of the reactants may be completely
consumed while others may be left over.
• The reactant that is completely consumed at
the end of a reaction is known as the limiting
reactant.
• All reactants that are not completely used up
are said to be in excess.
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More about limiting reactants
• Once the limiting reactant is used up the
reaction stops.
• The moles of the product formed are always
decided by the moles of the limiting reactant.
• Amounts of all reactants will be provided.
• Just because the amount of one reactant is
smaller than the other does not make it a
limiting reactant.
• Must find the limiting reactant first and then
calculate the amount of the product produced.
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16
Limiting reactant example
• Zinc metal reacts with hydrochloric acid to produce
zinc (II) chloride and hydrogen gas. If 0.30 moles of
Zn is mixed with 0.52 moles of HCl, how much
hydrogen will be produced?
Zn (s) + ___HCl (aq)
ZnCl2 (aq) + H2 (g)
• Calculate the moles of H2 produced from each
reactant individually to determine the limiting
reactant.
• Since the ______ produces the smaller amount of
H2, the ______ is the limiting reactant.
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Another example
• If 5.00 g of Fe2O3 is reacted with 1.00 g of H2,
how many grams of Fe can be obtained?
• Need to convert to moles and then work with
the moles. Can’t decide from the number of
grams.
• Need to find the limiting reactant.
• Write a balanced equation for this reaction:
Fe2O3 + __H2
__Fe + __H2O
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• Convert all grams to moles and find the grams
of Fe produced, starting with each reactant.
• _____is the limiting reactant, as it produces
the smaller number of grams of Fe.
• This reaction will produce ______ g of Fe.
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17
Another one
• In an industrial process for producing acetic
acid (HC2H3O2), oxygen gas is bubbled in
acetaldehyde (CH3CHO). In a lab test for this
process 20.0 g of CH3CHO and 10.0 g of O2
were placed in a flask and allowed to react.
How many grams of acetic acid can be
produced? How many grams of the excess
reactant will be left over?
• Start with writing a balanced equation:
__ CH3CHO + O2
__ HC2H3O2
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• Convert all grams to moles and proceed to find the
limiting reactant, which is the reactant that produces
the least amount of product.
• The ________is the _________ reactant and the
____ is the reactant ________.
• This reaction will produce only ______ g of HC2H3O2.
• To find the amount of the left over O2 we need to work
with the amount of HC2H3O2 produced in this reaction
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(______):
• Convert the grams of HC2H3O2 to moles and then
calculate the moles of O2 required to produce these
grams.
• Convert the moles of O2 to grams of O2
• Find the difference between the starting grams and
the required grams of O2.
• The grams of O2 left over can now be found by:
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18
Theoretical yields and % yields
• Theoretical yield: The maximum amount of a product
that can be obtained from the reaction of the given
amounts of reactants.
– Typically the amount of product produced from the limiting
reactant.
– Has to be calculated.
• Actual yield: The amount produced by actually
performing the reaction.
– Incomplete reactions, product loss, competing reactions.
• The % yield is calculated by the formula:
⎛ Actual yield ⎞
% yield= ⎜⎜
⎟⎟ ×100%
⎝ Theoretical yield⎠
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% yield calculation
• If in the reaction between 20.0 g of CH3CHO
and 10.0 g of O2 only 20.5 g of HC2H3O2 is
produce, calculate the % yield of this reaction.
• Here, the actual yield = 20.5 g, theoretical
yield = 27.3 g (calculated form the limiting
reactant) and the % yield =
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