The Mole,Stoichiometry, &
Formulas
Dr. Ron Rusay
Spring 2008
© Copyright 2008 R.J. Rusay
Chemical Stoichiometry
 Stoichiometry
is the study of chemicals and
their quantities that are consumed and
produced in chemical reactions.
 It
quantitatively relates the behavior of atoms
and molecules to observable chemical change
and measurable mass effects.
QUESTION
The fuel in small portable lighters is butane (C4H10). Suppose after
using such a lighter for a few minutes (perhaps to encourage your
favorite concert performer to play one more encore) you had used 1.0
gram of fuel. How many moles of butane would this be?
1.
2.
3.
4.
58 moles
0.077 moles
1.7  10–24 moles
0.017 moles
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
ANSWER
4) 0.017 mol is the answer for a proper grams to mole conversion.
You need to know the molar mass of butane: 4  12g/mol = 48 for
carbon + 10  1g/mol = 10.0 for hydrogen. Total = 48.0 + 10.0 =
58.0 g/mol. Next; 1.0 gram of butane  1 mol/58.0 g = 0.017 mol
Section 3.3: Molar Mass
Mass Calculations:
Products
Reactants
Chemically Relate:
Something (S)
Another Thing (AT)
Mass (S)
Mass (AT)
grams (S)
grams (AT)
© Copyright 1995-2002 R.J. Rusay
Mass Calculations:
Products
Reactants
grams (S)
grams (S)
1 mol (S)
grams (AT)
Avogadro's Number
Atoms
Molecules
Stoichiometry
grams (AT)
(Molecular
Weight)
?
grams (AT)
grams (S)
(Molecular
Weight)
© Copyright 1995-2008 R.J. Rusay
?
"Gatekeeper"
1 mol (AT)
QUESTION
The fuel in small portable lighters is butane (C4H10). Suppose after
using such a lighter for a few minutes (perhaps to encourage your
favorite concert performer to play one more encore) you had used 1.0
gram of fuel. . How many moles of butane would this be?
0.017 moles
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
How many grams of carbon dioxide would this produce?
1.) 750 mg
2.) 6.0 g
3) 1.5 g
4.) 3.0 g
ANSWER
Choice 4.) 3.0 g is the answer for the correct grams to moles to grams
conversion.
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
C4H10
1.0 g
grams (S)
1.0 g C4H10
?g CO2
grams (S)
grams (AT)
mol
C4H10
1 mol (S)
Avogadro's Number
Atoms
Molecules
44 g CO2
8Stoichiometry
mol
grams (AT)
CO2
(Molecular
Weight)
?
3.0 g CO2
?g CO2
grams (AT)
grams (S)
(Molecular
Weight)
58 g C4H10
2 mol
?
C4H10
"Gatekeeper"
1 mol (AT)
mol CO2
Combustion Analysis
CnHm + ( n + m ) O2 (g)
2
n CO(g) + m H2O(g)
2
Combustion Analysis Calculation
Ascorbic Acid ( Vitamin C )
• Combustion of a 6.49 mg sample in excess oxygen,
yielded 9.74 mg CO2 and 2.64 mg H2O
• Calculate it’s Empirical formula!
• C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
= 2.65 x 10-3 g C
• H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
= 2.92 x 10-4 g H
• Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg
= 3.54 mg O
Vitamin C: Calculation
(continued)
• C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) =
= 2.21 x 10-4 mol C
• H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) =
= 2.92 x 10-4 mol H
• O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) =
= 2.21 x 10-4 mol O
• Divide each by 2.21 x 10-4
• C = 1.00 Multiply each by 3 = 3.00 = 3.0
• H = 1.32
= 3.96 = 4.0
• O = 1.00
= 3.00 = 3.0
C3H4O3
Percent Composition
• Mass percent of an element:
mass of element in compound
mass % 
 100%
mass of compound
• For iron in (Fe2O3), iron (III) oxide = ?
11169
.
mass % Fe 
 100%  69.94%
159.69
QUESTION
Which of the following compounds has the same
percent composition by mass as styrene, C8H8?
1) Acetylene, C2H2
2) Benzene, C6H6
3) Cyclobutadiene, C4H4
4) -ethyl naphthalene, C12H12
5) All of these
ANSWER
5)
All of these
Section 3.5 Percent Composition of Compounds
(p. 89)
The ratio of C to H in C8H8 is 1:1. This is the
same ratio found for each of the compounds, so
all have the same percent composition by mass.
QUESTION
Morphine, derived from opium plants, has the potential for use and
abuse. It’s formula is C17H19NO3. What percent, by mass, is the
carbon in this compound?
1.
2.
3.
4.
42.5%
27.9%
71.6%
This cannot be solved until the mass of the sample is given.
ANSWER
Choice 3 is correct. First determine the molar mass of the
compound, then divide that into the total mass of carbon present,and
finally, multiply that by 10.
(17  12) / ((17  12) + (19  1) + (1  14) + 3  16)) = 0.716
0.716  100 = 71.6 %
Section 3.4: Percent Composition of Compounds
QUESTION
How many grams of potassium are in 12.5 g of
K2CrO7?
1) 2.02 g
2) 8.80 g
3) 4.04 g
4) 78.2 g
5) 25.0 g
ANSWER
3)
4.04 g
Section 3.5 Percent Composition of Compounds
(p. 89)
The molar mass of K2CrO7 is 2  39.10 + 52.00 +
7  16.00 = 242.2. The mass fraction of potassium
is (2  39.10)/242.2 = 0.3229.
0.3229  12.5 g = 4.04 g.
Formulas: Dalton’s Law
• Dalton’s law of multiple proportions:
When two elements form different compounds,
the mass ratio of the elements in one
compound is related to the mass ratio in the
other by a small whole number.
Formulas: Multiple Proportions
QuickTime™ and a
Sorenson Video decompressor
are needed to see this picture.
Formulas & Multiple Proportions
Components of acid rain, SO2(g) and SO3(g)
• Compound A contains:
1.000 g Sulfur & 1.500 g Oxygen
• Compound B contains:
1.000 g Sulfur & 1.000 g Oxygen
• Mass ratio A: 2 to 3; Mass ratio B: 1 to 1
• Adjusting for atomic mass differences: AW
sulfur is 2x the AW oxygen; the atom ratios
therefore are S1O3 and S1O2 respectively
Formulas &
Molecular Representations
 molecular
formula = C6H6 Benzene
 empirical formula = CH = C6/6H6/6
 molecular formula = (empirical formula)n
[n = integer] (CH)6
• Other representations: Lewis Dot formulas,
structural formulas, 2-D, 3-D
Formulas &
Molecular Representations
QuickTime™ and a
Sorenson Video decompressor
are needed to see this picture.
Empirical Formulas from Analyses
Empirical Formula Determination
• 1. Use percent analysis.
Let 100 % = 100 grams of compound.
• 2. Determine the moles of each element.
(Element % = grams of element.)
• 3. Divide each value of moles by the
smallest of the mole values.
• 4. Multiply each number by an integer to
obtain all whole numbers.
QUESTION
The dye indigo is a compound with tremendous economic
importance (blue jeans wouldn’t be blue without it.) Indigo’s percent
composition is: 73.27% C; 3.84% H; 10.68%N and 12.21% O. What
is the empirical formula of indigo?
1.
2.
3.
4.
C6H4NO
C8H3NO
C8H5NO
I know this should be whole numbers for each atom, but I do not
know how to accomplish that.
ANSWER
Choice 3 is the smallest whole number ratio of the atoms that make
up a molecule of indigo. The percentage must be converted to a
mass, then the mass is converted to moles of the atoms and finally,
the smallest is divided into the others to obtain the proper ratio.
Section 3.5: Determining the Formula of a Compound
Empirical & Molecular Formula
Determination
Quinine:
C 74.05%, H 7.46%, N 8.63%, O 9.86%
• 74.05/12.01, 7.46/1.008, 8.63/14.01, 9.86/16.00
C6.166 H7.40 N0.616 O0.616
• Empirical Formula: C10 H12 N1 O1
Empirical Formula Weight = ?
• Molecular Weight = 324.42
Molecular Formula = 2x empirical formula
• Molecular Formula = C20 H24 N2 O2
A Mass Spectrometer
Records a mass spectrum
A mass spectrum records only positively charged
fragments
m/z = mass to charge ratio of the fragment
http://chemconnections.llnl.gov/general/chem120/quinine.html
From the structures, determine the
molecular formula of quinine.
A Carbon atom is at each angle.
Each C has 4 bonds (lines + Hs).
Hs are not always drawn in &
must be added.
H
H 2C
N
HO
H
H3CO
N
C20H24N2O2
C = 20
H = 24
N=2
O=2
QUESTION
The empirical formula of styrene is CH; its molar
mass is 104.1. What is the molecular formula of
styrene?
1) C2H4
2) C8H8
3) C10H12
4) C6H6
5) None of these
ANSWER
2)
C8H8
Section 3.6 Determining the Formula of a
Compound (p. 91)
The mass of CH is 13.018. 13.018 divides into
104.1 about 8 times. Therefore there are 8 CH
groups in this compound. C8H8 is the molecular
formula.
More Chemical Equations /
Stoichiometric Calculations