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Chapter 3
Mass Relationships; Stoichiometry.
Atomic Weights


weighted average of the masses of the constituent
isotopes
lower number on periodic chart

How do we know what the values of these numbers are?
Atomic and Molecular Weights
The Atomic Mass Scale
1H
weighs 1.6735 x 10-24 g and 16O 2.6560 x 10-23 g.
We define: mass of 12C = exactly 12 amu.
Using atomic mass units:
1 amu = 1.66054 x 10-24 g
1 g = 6.02214 x 1023 amu
Atomic and Molecular Weights
Atomic Masses are determined with a mass
spectrometer
What is the mass of one atom of 50V in amu?
Experimentally Determined
Mass Spectrometer
Mass Spectrometer
Atomic and Molecular Weights
The Atomic Mass Scale
What is the mass of one atom of 50V in amu?
Experimentally Determined
49.9472 amu
•What is the mass of one atom of
50V
in grams?
Atomic and Molecular Weights
The Atomic Mass Scale
What is the mass of one atom of 50V in amu?
Experimentally Determined
49.9472 amu
What is the mass of one atom of 50V in grams?
1g
 23
(49.9472amu)(
)

8
.
29
x
10
g
23
6.02214 x10 amu
Atomic and Molecular Weights
Average Atomic Mass
Relative atomic mass: average masses of isotopes:
Naturally occurring C: 98.892 % 12C + 1.108 % 13C.
Average mass of C:
(0.98892)(12 amu) + (0.0108)(13.00335) = 12.011 amu.
Atomic weight (AW) is also known as average atomic
mass.
Atomic weights are listed on the periodic table.
Atomic and Molecular Weights
Determine the percent abundance of the two
isotopes of Br.
79Br 78.9183 amu
81Br 80.9163 amu

Atomic and Molecular Weights
Determine the percent abundance of the two isotopes
of Br.
79Br 78.9183 amu
81Br 80.9163 amu

Avg. Atomic Weight = (78.9183)(x) + (80.9163)(1-x)
79.904 amu = 78.9183x + 80.9163 – 80.9163x
1.9980x = 1.0123
79Br = x = 1.0123 / 1.9980 = 0.50666
50.67%
81Br = 1-x = 0.4933
49.33%
Atomic and Molecular Weights
Formula and Molecular Weights
Formula weights (FW): sum of AW for atoms in formula.
FW (H2SO4) = 2AW(H) + AW(S) + 4AW(O)
= 2(1.0 amu) + (32.0 amu) + 4(16.0)
= 98.0 amu
Molecular weight (MW) is the weight of the molecular
formula.
MW(C6H12O6) = 6(12.0 amu) + 12(1.0 amu) +
6(16.0 amu)
Atomic and Molecular Weights
What is the molecular weight of the
following molecules?
 CH4
 H2SO4
 Ca(NO3)2
 CuSO4 •5H2O

Atomic and Molecular Weights

What is the molecular weight of the
following molecules?


CH4
H2SO4
Ca(NO3)2

CuSO4 •5H2O

12.011 + 4(1.008) = 16.043 amu
Atomic and Molecular Weights

What is the molecular weight of the following
molecules?


CH4
12.011 + 4(1.008) = 16.043 amu
H2SO4
2(1.008) + 32.06 + 4(16.00) = 96.08 amu
Ca(NO3)2

CuSO4 •5H2O

Atomic and Molecular Weights

What is the molecular weight of the following
molecules?

CH4
12.011 + 4(1.008) = 16.043 amu
H2SO4
2(1.008) + 32.06 + 4(16.00) = 96.08 amu
Ca(NO3)2
40.08 + 2(14.007) + 6(16.00) = 164.09 amu
CuSO4 •5H2O



Atomic and Molecular Weights

What is the molecular weight of the following
molecules?

CH4
12.011 + 4(1.008) = 16.043 amu
H2SO4
2(1.008) + 32.06 + 4(16.00) = 96.08 amu
Ca(NO3)2
40.08 + 2(14.007) + 6(16.00) = 164.09 amu
CuSO4 •5H2O
63.546+32.06+9(16.00)+10(1.008) = 249.68 amu



Atomic and Molecular Weights
Percentage Composition from Formulas
Percent composition is the atomic weight for each element
divided by the formula weight of the compound multiplied
by 100:

Atoms of Element AW 
% Element 
 100
FW of Compound
Atomic and Molecular Weights

Determine the Mass Percent Composition of
nitrogen in NH4NO3?
Atomic and Molecular Weights

Determine the Mass Percent Composition of
nitrogen in NH4NO3?

Formula weight 80.043 amu
Atomic and Molecular Weights

Determine the Mass Percent Composition of
nitrogen in NH4NO3?

Formula weight 80.043 amu
2 x14.007amu
%N 
x100  34.999%
80.043amu
The Mole

strictly a convenience unit


amount that is large enough to see and handle in lab
mole = number of things
dozen = 12 things
 mole = 6.022 x 1023 things


Avogadro’s number = 6.022 x 1023 = NA
The Mole

how do you know when you have a mole?
count it out
 weigh it out


molar mass - mass in grams equal to the atomic
weight
H is 1.00794 g of H atoms = 6.022 x 1023 atoms
 Mg is 24.3050 g of Mg atoms = 6.022 x 1023 atoms

The Mole
Mole: convenient measure chemical quantities.
1 mole of something = 6.0221367 x 1023 of that thing.
Experimentally, 1 mole of 12C has a mass of 12 g.
Molar Mass
Molar mass: mass in grams of 1 mole of substance (units
g/mol, g.mol-1).
Mass of 1 mole of 12C = 12 g.
The Mole
Molar Mass
Molar mass: sum of the molar masses of the atoms:
molar mass of N2 = 2 x (molar mass of N).
Molar masses for elements are found on the periodic
table.
Formula weights are numerically equal to the molar
mass.
Molar Masses of Compounds
add atomic weights of each atom
 The molar mass of propane, C3H8, is:
3  C  3  12.01 amu  36.03

8  H  8  1.01 amu
Molar
mass

amu
8.08 amu
 44.11 amu
One mole of propane is 44.11 g of propane.
The Mole
Interconverting Masses, Moles, and Numbers of
Particles
The Mole

Calculate the mass of a single Mg atom, in
grams, to 3 significant figures.
The Mole

Calculate the mass of a single Mg atom, in
grams, to 3 significant figures.
 1 mol Mg atoms

 
? g Mg  1 Mg atom 
23
 6.022 10 Mg atoms 
 24.30gMg 

  4.04 10  23 g Mg
 1 mol Mg atoms 
The Mole

Calculate the number of atoms in one-millionth of
a gram of Mg to 3 significant figures.
The Mole

Calculate the number of atoms in one-millionth of
a gram of Mg to 3 significant figures.
 1 mol Mg 

? Mg atoms  1.00 10 g Mg
 24.30 g Mg 
6
 6.022 10 23 Mg atoms 

  2.48 1016 Mg atoms
 1 mol Mg atoms 
The Mole

How many atoms are contained in 1.67 moles of
Mg?
The Mole

How many atoms are contained in 1.67 moles of
Mg?
 6.022 10 23 Mg atoms 

? Mg atoms  1.67 mol Mg
1 mol Mg


24
 1.00 10 Mg atoms
The Mole

How many moles of Mg atoms are present in 73.4
g of Mg?
The Mole

How many moles of Mg atoms are present in 73.4
g of Mg?
 1molMg atoms 
  3.02molMg
?molMg  73.4g Mg 
 24.30g Mg 
Mole Problems with Compounds

Calculate the number of C3H8 molecules in 74.6 g
of propane.
Mole Problems with Compounds

Calculate the number of C3H8 molecules in 74.6 g
of propane.
? C3H8 molecules  74.6 g C3H8 
 6.022 10 23 C3H8 molecules

44.11 g C3H8


  1.02 10 24 molecules

Mole Problems with Compounds

Calculate the number of O atoms in 26.5 g of
Li2CO3.
Mole Problems with Compounds

Calculate the number of O atoms in 26.5 g of
Li2CO3.
1 mol Li 2CO3
? O atoms  26.5 g Li 2CO3 

73.8 g Li2CO3
6.022  1023 form.units Li2CO3
3 O atoms


1 mol Li2CO3
1 form.unit Li2CO3
6.49  10 O atoms
23
1 Mole of Some Common
Molecular Substances

1 Mole

Br2

159.81g
Contains



C4H10
6.022 x 1023 Br2 molecules
2(6.022 x 1023 ) Br atoms
1 Mole of Some Common
Molecular Substances

1 Mole

Br2

159.81g
Contains



C4H10
58.04 g



6.022 x 1023 Br2 molecules
2(6.022 x 1023 ) Br atoms
6.022 x 1023 C4H10 molecules
4 (6.022 x 1023 ) C atoms
10 (6.022 x 1023 ) H atoms
Formulas from Elemental
Composition

empirical formula - simplest molecular formula,
shows ratios of elements but not actual numbers of
elements

molecular formula - actual numbers of atoms of
each element in the compound

determine empirical & molecular formulas of a
compound from percent composition

percent composition is determined experimentally
Empirical Formulas
Simplest whole number ratio of atoms or
ions present in a compound
Empirical Formula
 H2O
 C6H12O6
 H2O2

Empirical Formulas
Simplest whole number ratio of atoms or
ions present in a compound
Empirical Formula
 H2O
H2O
 C6H12O6
 H2O2

Empirical Formulas
Simplest whole number ratio of atoms or
ions present in a compound
Empirical Formula
 H2O
H2O
 C6H12O6
CH2O
 H2O2

Empirical Formulas
Simplest whole number ratio of atoms or
ions present in a compound
Empirical Formula
 H2O
H2O
 C6H12O6
CH2O
 H2O2
HO

Empirical Formulas and Molecular
Formula
Once we know the empirical formula, we need the MW to
find the molecular formula.
Subscripts in the molecular formula are always wholenumber multiples of subscripts in the empirical formula.
Empirical Formulas from Analyses
Combustion Analysis
Percent Composition
mass of that element divided by the mass of
the compound x 100%
 percent composition of C in C3H8

mass C
% C 
mass C 3 H
3  12.01 g

 100%
44.11 g
 100%
8
 81.68%
Percent Composition

percent composition of H in C3H8
Percent Composition

percent composition of H in C3H8
mass H
8 H
%H
 100% 
 100%
mass C 3 H 8
C3H 8
8  1.01 g

 100%  18.32%  or 
44.11 g
100%  81.68%  18.32%
Percent Composition

Calculate the percent composition of Fe2(SO4)3 to
3 sig. fig.
Percent Composition

Calculate the percent composition of Fe2(SO4)3 to
3 sig. fig.
2  Fe
2  55.8g
%Fe 
100% 
100%  27.9%Fe
Fe 2 (SO 4 ) 3
399.9g
3 S
3  32.1g
%S
100% 
100%  24.1%S
Fe 2 (SO 4 ) 3
399.9g
12  O
12 16.0g
%O
100% 
100%  48.0%O
Fe 2 (SO 4 ) 3
399.9g
Total  100%
Empirical Formulas from Analyses
Start with mass % of elements (i.e. empirical data) and
calculate a formula, or
Start with the formula and calculate the mass % elements.
Empirical Formulas from Analysis

Determine the empirical formula (CrxOy)
for a compound with the percent
composition: 68.42% Cr, 31.58% O.
1.) Assume 100g
 2.) Determine mol of each element
 3.) Divide through by smallest subscript
 4.) Convert all subscripts to integers

Empirical Formulas from Analysis

Determine the empirical formula (CrxOy)
for a compound with the percent
composition: 68.42% Cr, 31.58% O.

1.) Assume 100g
68.42g Cr
and
31.58 g O
Empirical Formulas from Analysis

Determine the empirical formula (CrxOy)
for a compound with the percent
composition: 68.42% Cr, 31.58% O.
1.) Assume 100g
68.42g Cr
and 31.58 g O
2.) Determine mol of each element

1molCr
molCr  68.42 gCr
 1.316molCr
51.996 gCr
1molO
molO  31.58 gO
 1.974molO
16.00 gO
Empirical Formulas from Analysis

Divide through by smallest whole number
1.316/1.316 = 1 Cr
1.974/1.316 = 1.50 O
Convert to whole numbers
Cr1O1.5 x 2 = Cr2O3
Empirical Formulas from Analysis


A compound contains 24.74% K, 34.76% Mn, and
40.50% O by mass. What is its empirical formula?
Make the simplifying assumption that you have 100.0 g
of compound.

in 100.0 g of compound there is



24.74 g of K
34.76 g of Mn
40.50 g of O
Empirical Formulas from
Analysis
? mol K
1 mol K
 24.74 g K 
 0.6327 mol K
39.10 g K
1 mol Mn
? mol Mn  34.76 g Mn 
 0.6327 mol Mn
54.94 g Mn
1molO
? mol O  40.50 g O 
 2.531 mol O
16.00gO
obtain smallest whole number ratio
0.6327
0.6327
for K 
1K
for Mn 
 1Mn
0.6327
0.6327
2.531
for O 
 4 O  KMnO 4
0.6327
Empirical Formulas from
Analysis

A sample of a compound contains 6.541g of Co
and 2.368g of O. What is empirical formula for
this compound?
Empirical Formulas from
Analysis
A sample of a compound contains 6.541g of Co
and 2.368g of O. What is empirical formula for
this compound?
1 m ol C o
? m ol C o  6 .541 g C o 
 0 .1110 m ol C o
58 .93 g C o

1 m ol O
? m ol O  2 .368 g O 
 0 .1480 m ol O
16 .00 g O
find sm allest w hole num ber ratio
Empirical Formulas from
Analysis

A sample of a compound contains 6.541g of Co
and 2.368g of O. What is empirical formula for
this compound?
for C o 
0.1110
 1 Co
0.1110
for O 
0.1480
 1.333 O
0.1110
Empirical Formulas from
Analysis

A sample of a compound contains 6.541g of Co
and 2.368g of O. What is empirical formula for
this compound?
01110
.
01480
.
for Co 
 1 Co
for O 
 1333
.
O
01110
.
01110
.
multipy both by 3 to turn fraction to whole number
1 Co  3  3 Co
1333
.
O3 4O
thus our formula is:
Co3 O 4
Chemical Equations



Lavoisier: mass is conserved in a chemical
reaction. (Law of Conservation of Mass)
Chemical equations: symbolic representation of a
chemical reaction.
Two parts to an equation: reactants and products:
2H2 + O2  2H2O



reactants on left side of reaction
products on right side of equation
relative amounts of each with stoichiometric
coefficients
Chemical Equations

attempt to show on paper what is happening
at the molecular level
Chemical Equations

look at what an equation tells us

Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2
reactants
yields
1 form. unit 3 mol.
1 mole
3 moles
159.7g
84g
products
2 atoms 3 mol.
2 moles 3 moles
111.7g 132g
Chemical Equations

Stoichiometric coefficients: numbers in front of the
chemical formulas; give ratio of reactants and products.
Chemical Equations
Law of Conservation of Matter
no detectable change in quantity of matter in an
ordinary chemical reaction
 discovered by Lavoisier
 balance chemical reactions to obey this law
 propane,C3H8, burns in oxygen to give carbon
dioxide and water

C3H8 + 5 O2  3 CO2 + 4 H2O
note that there are equal numbers of atoms of each
element on both sides of equation
Law of Conservation of Matter

NH3 burns in oxygen to form NO & water

2 NH 3 + O 2 
 2 NO + 3 H 2 O
5
2
or correctly

4 NH 3 + 5 O 2 
 4 NO + 6 H 2 O
Law of Conservation of Matter

C5H12 burns in oxygen to form carbon
dioxide & water

C 5 H 12 + 8 O 2  5 CO 2 + 6 H 2 O

balancing equations is a skill acquired only
with lots of practice

work many problems
Balance the following Equations
CH4(g) + O2(g)  CO2(g) + H2O(g)
 CaCO3(s) CaO(s) + CO2(g)
 AgNO3(aq) + NaI(aq)  NaNO3(aq) + AgI(s)
 H2SO4(aq) + KOH  K2SO4(aq) + H2O(l)

Balance the following Equations

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Balance the following Equations
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
 CaCO3(s) CaO(s) + CO2(g)

Balance the following Equations
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
 CaCO3(s) CaO(s) + CO2(g)
 AgNO3(aq) + NaI(aq)  NaNO3(aq) + AgI(s)

Balance the following Equations
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
 CaCO3(s) CaO(s) + CO2(g)
 AgNO3(aq) + NaI(aq)  NaNO3(aq) + AgI(s)
 H2SO4(aq) + 2KOH  K2SO4(aq) + 2H2O(l)

Quantitative Information from Balanced
Equations
Balanced chemical equation gives number of molecules
that react to form products.
Interpretation: ratio of number of moles of reactant
required to give the ratio of number of moles of product.
These ratios are called stoichiometric ratios.
NB: Stoichiometric ratios are ideal proportions
Real ratios of reactants and products in the laboratory
need to be measured (in grams and converted to moles).
Quantitative Information from
Balanced Equations
The ratio
of grams of
reactant
cannot be
directly
related to
the grams
of product.
Calculations Based on Chemical
Equations
can do this in moles, formula units, etc.
 most often work in grams (or kg or pounds or tons)

Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2
How many CO molecules are required to react with 25
formula units of Fe2O3?

3 CO molecules
? CO molecules = 25 FU Fe 2 O 3 

1 FU Fe 2 O 3
= 75 CO molecules
Calculations Based on Chemical
Equations

How many iron atoms can be produced by the
reaction of 2.50 x 105 formula units of iron (III)
oxide with excess carbon monoxide?
Calculations Based on Chemical
Equations

How many iron atoms can be produced by the
reaction of 2.50 x 105 formula units of iron (III)
oxide with excess carbon monoxide?
2 Fe atoms
? Fe atoms = 2.50  10 FU Fe 2 O 3 

1 FU Fe 2 O 3
5
= 5.00  105 Fe atoms
Calculations Based on Chemical
Equations

Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2
What mass of CO is required to react with 146 g of
iron (III) oxide?
Calculations Based on Chemical
Equations

Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2
What mass of CO is required to react with 146 g of
iron (III) oxide?
1 mol Fe 2 O 3
3 mol CO
? g CO = 146 g Fe 2 O 3 

159.6 g Fe 2 O 3 1 mol Fe 2 O 3
28.0 g CO

 76.8 g CO
1 mol CO
Calculations Based on Chemical
Equations

What mass of carbon dioxide can be produced by
the reaction of 0.540 mole of iron (III) oxide with
excess carbon monoxide?
Calculations Based on Chemical
Equations

What mass of carbon dioxide can be produced by
the reaction of 0.540 mole of iron (III) oxide with
excess carbon monoxide?

Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2
Calculations Based on Chemical
Equations

What mass of carbon dioxide can be produced by
the reaction of 0.540 mole of iron (III) oxide with
excess carbon monoxide?

Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2
3 mol CO 2
44.0 g CO 2
? g CO 2  0.540 mol Fe 2 O 3 

1 mol Fe 2 O 3 1 mol CO 2
= 71.3 g CO 2
Calculations Based on Chemical
Equations

What mass of iron (III) oxide reacted with excess
carbon monoxide if the carbon dioxide produced by
the reaction had a mass of 8.65 grams?
Calculations Based on Chemical
Equations
1 molCO2 1mol Fe2O3
? g Fe2O3  8.65 g CO2 


44.0 g CO2 3 mol CO2
1596
. g Fe2O3
 105
. g Fe2O3
1 mol Fe2O3
Calculations Based on Chemical
Equations

How many pounds of carbon monoxide would
react with 125 pounds of iron (III) oxide?
Calculations Based on Chemical
Equations

Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2
159.6 g
84.0 g
111.6 g
132.0 g
or
159.6 lb
84.0 lb
111.6 lb
132.0 lb
84.0 lb CO
? lb Co = 125 lb Fe2O3 
 658
. lb CO
159.6 lb Fe2O3
Other Interpretations of Chemical
Formulas

What mass of ammonium phosphate, (NH4)3PO4,
would contain 15.0 g of N?
molar mass of (NH 4 )3 PO 4  149.0 g/mol
1 mol N
? mol N  15.0 g of N 
 1.07 mol N
14.0 g N
Other Interpretations of Chemical
Formulas

What mass of ammonium phosphate, (NH4)3PO4,
would contain 15.0 g of N?
molar mass of (NH 4 )3 PO 4  149.0 g/mol
1 mol N
? mol N  15.0 g of N 
 1.07 mol N
14.0 g N
1 mol (NH 4 )3 PO 4
1.07 mol N 
 0.357 mol (NH 4 )3 PO 4
3 mol N
Other Interpretations of Chemical
Formulas
What mass of ammonium phosphate, (NH4)3PO4,
would contain 15.0 g of N?
molar mass of (NH 4 )3 PO 4  149.0 g/mol
1 mol N
? mol N  15.0 g of N 
 1.07 mol N
14.0 g N
1 mol (NH 4 )3 PO 4
1.07 mol N 
 0.357 mol (NH 4 )3 PO 4
3 mol N
149.0 g (NH 4 )3 PO 4
0.357 mol (NH 4 )3 PO 4 
 53.2 g (NH 4 )3 PO 4
1 mol (NH 4 )3 PO 4

Patterns of Chemical Reactivity
Using the Periodic Table

Properties of compounds vary systematically
because of good ordering in the periodic table.
2K(s) + 2H2O(l)  2KOH(aq) + H2(g)
2M(s) + 2H2O(l)  2MOH(aq) + H2(g)
Combustion in Air

Combustion is the burning of a substance in
oxygen from air:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
Patterns of Chemical Reactivity
Patterns of Chemical Reactivity
Combination and Decomposition Reactions
2Mg(s) + O2(g)  2MgO(s)
There are fewer products than reactants; the Mg has
combined with O2 to form MgO.
2NaN3(s)  2Na(s) + 3N2(g)
(the reaction that occurs in an air bag)
There are more products than reactants; the sodium azide
has decomposed into Na and nitrogen gas.
Patterns of Chemical Reactivity
Combination and Decomposition Reactions
Combination reactions: fewer reactants than products.
Decomposition reactions: more products than reactants.
Limiting Reactant Concept
If the reactants are not present in stoichiometric
amounts, at end of reaction some reactants are still
present (in excess).
Limiting Reactant: one reactant that is consumed.
Limiting Reactant Concept

What is the maximum mass of sulfur dioxide
that can be produced by the reaction of 95.6 g of
carbon disulfide with 110 g of oxygen?
CS 2  3 O 2  CO 2  2 SO 2
1 mol
3 mol
1 mol
76.2 g
3(32.0 g) 44.0 g
2 mol
2(64.1 g)
Limiting Reactant Concept
CS2  3 O2  CO2  2 SO2
? mol SO2  956
. g CS2 
1 mol CS2 2 mol SO2 641
. g SO2


 161 g SO2
76.2 g
1 mol CS2 1 mol SO2
What do we do next?
Limiting Reactant Concept
CS2  3 O2  CO2  2 SO2
1 mol CS2 2 mol SO2 641
. g SO2
? mol SO2  956
. g CS2 


 161 g SO2
76.2 g
1 mol CS2 1 mol SO2
1 mol O2 2 mol SO2 641
. g SO2
? mol SO2  110 g O2 


 147 g SO2
32.0 g O2 3 mol O2 1 mol SO2

What can we conclude from our data now?
Which is limiting reactant?
What is maximum mass of sulfur dioxide?


Limiting Reactant Concept
CS2  3 O2  CO2  2 SO2
1 mol CS2 2 mol SO2 641
. g SO2
? mol SO2  956
. g CS2 


 161 g SO2
76.2 g
1 mol CS2 1 mol SO2
1 mol O2 2 mol SO2 641
. g SO2
? mol SO2  110 g O2 


 147 g SO2
32.0 g O2 3 mol O2 1 mol SO2


limiting reactant is O2
maximum amount of SO2 is 147 g
Theoretical yield
The amount of product predicted from stoichiometry
taking into account limiting reagents is called the
theoretical yield.
The percent yield relates the actual yield (amount of
material recovered in the laboratory) to the theoretical
yield:
Actual yield
% Yield 
 100
Theoretical yield
Percent Yields from Reactions
actual yield
% yield =
 100%
theoretical yield


theoretical yield is what we have been calculating
actual yield is what you have in your flask
A 10.0 g sample of ethanol, C2H5OH, was boiled with excess
acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate,
CH3COOC2H5. What is the percent yield?
Percent Yields from Reactions
CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5  H 2 O
Calculate the theoretical yield
Percent Yields from Reactions
CH 3COOH + C 2 H 5OH  CH 3COOC 2 H 5  H 2 O
Calculate the theoretical yield
88.0 g CH 3COOC 2 H 5
? g CH 3COOC 2 H 5 = 10.0 g C 2 H 5OH 
46.0 g C 2 H 5OH
. g CH 3COOC 2 H 5
 191
14.8 g CH 3COOC 2 H 5
 100%  77.5%
% yield =
19.1 g CH 3COOC 2 H 5
Synthesis Problem

In 1986, Bednorz and Muller succeeded in
making the first of a series of chemical
compounds that were superconducting at
relatively high temperatures. This first compound
was La2CuO4 which superconducts at 35K. In
their initial experiments, Bednorz and Muller
made only a few mg of this material. How many
La atoms are present in 3.56 mg of La2CuO4?
Synthesis Problem
molar mass of La 2 CuO 4 = 405.3 g / mol
 1g 
3.56 mg La 2 CuO 4 

 1000 mg 
 1 mol La 2 CuO 4 
6

8
.
78

10
mol La 2 CuO 4


 405.3 g La 2 CuO 4 
(8.78  10  6
 6.022  10 23 molecules La 2 CuO 4 
mol La 2 CuO 4 )

1 mol La 2 CuO 4




2 La atoms
.  1019 La atoms

  106
 molecule La 2 CuO 4 
Group Activity

Within a year after Bednorz and Muller’s initial
discovery of high temperature superconductors,
Wu and Chu had discovered a new compound,
YBa2Cu3O7, that began to superconduct at 100 K.
If we wished to make 1.00 pound of YBa2Cu3O7,
how many grams of yttrium must we buy?