Stoichiometry
Dr. Ron Rusay
Chemical Stoichiometry

Stoichiometry is the study of mass in chemical reactions. It
deals with both reactants and products.

It quantitatively and empirically relates the behavior of
atoms and molecules in a balanced chemical equation to
observable chemical change and measurable mass effects.

It accounts for mass and the conservation of mass, just as the
conservation of atoms in a balanced chemical equation.
Chemical Reactions
Atoms, Mass & Balance: eg. Zn(s) + S(s)
Chemical Equation
Representation of a chemical reaction:
_ C2H5OH + _ O2  _ CO2 + _ H2O
Reactants
Products
• C=2; H =5+1=6; O=2+1
•
C=1; H=2; O=2+1
1 C2H5OH + 3 O2  2 CO2 + 3 H2O
Chemical Equation
C2H5OH + 3 O2  2 CO2 + 3 H2O
The balanced equation can be completely stated as:
1
mole of ethanol reacts with 3 moles of
oxygen to produce 2 moles of carbon
dioxide and 3 moles of water.
Chemical Equation
1
All Balanced Equations relate on a molar mass basis. For
the combustion of octane:
2 C8H18(l)+ 25 O2(g)
16 CO2(g) +18 H2O(l)
2 moles of octane react with 25 moles of oxygen to
produce 16 moles of carbon dioxide and 18 moles of
water.
QUESTION
The fuel in small portable lighters is butane (C4H10). After using a
lighter for a few minutes, 1.0 gram of fuel was used. How many
moles of carbon dioxide would it produce?
A.
B.
C.
D.
58 moles
0.017 moles
1.7  10–24 moles
0.068 moles
ANSWER
The fuel in small portable lighters is butane (C4H10). After using a
lighter for a few minutes, 1.0 gram of fuel was used. How many
moles of carbon dioxide would it produce?
A.
B.
C.
D.
58 moles
0.017 moles
1.7  10–24 moles
0.068 moles
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
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
The Chemical Equation:
Mole & Masses
C2H5OH + 3 O2  2 CO2 + 3 H2O
 46g
(1 mole) of ethanol reacts with 3 moles
of oxygen (96g) to produce 2 moles of
carbon dioxide and 3 moles of water.
 How
many grams of carbon dioxide and
water are respectively produced from 46g (1
mole) of ethanol ?
2 mol x 44 g/mol = 88g
3 mol x 18 g/mol = 54 g
The Chemical Equation:
Moles & Masses
• C2H5OH + 3 O2  2 CO2 + 3 H2O
 How
many grams of oxygen are needed to
react with 15.3g of ethanol in a 12oz. beer ?
15.3gethanol x molethanol /46.0gethanol = 0.333molethanol
0.333molethanol x 3moloxygen / 1molethanol = 0.999moloxygen
32.0goxygen /moloxygen x 0.999moloxygen = 32.0goxygen
NOTE: It takes approximately 1 hour for the biologically equivalent
amount of oxygen available from cytochrome p450 to consume the
alcohol in a human in 1 beer to a level below the legal limit of 0.08%.
Chemical Stoichiometry
Epsom salt (magnesium sulfate heptahydrate) is one
of five possible hydrates: mono-, di-, tri-, hexa-, or
hepta- hydrate.
 How can stoichiometry be used to determine, which
hydrate is present in a pure unknown sample, by
heating the sample in a kitchen oven at 400 o C for
45 minutes?

MgSO4 . x H2O(s)
MgSO4 (s) + x H2O (g)
Mass Calculations
1
All Balanced Equations relate on a molar and mass basis.
For the combustion of octane:
2 C8H18(l)+ 25 O2(g)
16 CO2(g) +18 H2O(l)
228 g of octane (2 moles)* will react with 800 g of
oxygen (25 moles) to produce (16 moles) 704 g of
carbon dioxide and (18 moles) 324 g of water.
*(2 moles octane x 114 g/mol = 228 g )
Mass Calculations:
Reactants
Products
Chemically Relate:
Something (S)
Another Thing (AT)
Mass (S)
Mass (AT)
grams (S)
grams (AT)
Mass Calculations:
Reactants
Products
1.
2.
3.
4.
5.
Balance the chemical equation.
Convert mass of reactant or product
to moles.
Identify mole ratios in balanced equation:
They serve as the “Gatekeeper”.
Calculate moles of desired product or
reactant.
Convert moles to grams.
Mass Calculations:
Reactants
Products
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)
?
"Gatekeeper"
1 mol (AT)
Mass Calculations:
Reactants
Products
• How many grams of salicylic acid are needed to produce
1.80 kg of aspirin?
• Balanced Equation:
Salicylic Acid
Aspirin
COOH
H
COOH
OH
H
+
H
OCCH3
+
CH3COCl
H
H
H
C7H6O3
MW = 138.12
O
HCl
H
H
C2H3OCl
MW = 78.49
C9H8O4
MW = 180.15
HCl
MW= 36.45
Mass Calculations:
Reactants
Products
grams (Aspirin)
grams (Salicylic Acid)
Avogadro's Number
Atoms
Molecules
Stoichiometry
grams (SA)
1800 grams (A)
1 mol (A)
1 mol SA
(Molecular
Weight SA)
? (SA)
grams (A)
1 mol A
(Molecular
Weight A)
"Gatekeeper"
1 mol (SA)
Mass Calculations:
How many grams of
salicylic acid are needed to produce 1.80 kg of aspirin?
• Balanced Equation:
Salicylic Acid
Aspirin
COOH
H
COOH
OH
H
+
H
H
C7H6O3
MW = 138.12
OCCH3
+
CH3COCl
H
H
1
O
HCl
H
H
C2H3OCl
MW = 78.49
1 C9H8O4
MW = 180.15
HCl
MW= 36.45
?g C7 = 1.80 x 103 gC9 x [molC9 /180.15gC9 ] x [1 molC7/ 1 molC9] x 138.12 gC7/molC7
?g C7 = 1380 grams
Mass Calculations:
Reactants
Products
QUESTION
The fuel in small portable lighters is butane (C4H10). After using a
lighter for a few minutes, 1.0 gram (0.017 moles) of fuel was used.
How many grams of carbon dioxide would it produce?
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
How many grams of carbon dioxide would this produce?
A.) 750 mg
B.) 6.0 g
C) 1.5 g
D.) 3.0 g
ANSWER
D.) 3.0 g
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
• CxHy + (x + y/4) O2  x CO2 + y/2 H2O
Molecules with oxygen in their formula are more difficult to
solve for Oz knowing only the respective masses of CxHyOz
sample, CO2 and H2O.
• CxHyOz + (x + y/4 - z) O2  x CO2 + y/2 H2O
SEE: COMPARISON to wt % CALCULATIONS
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
QUESTION
Erythrose contains carbon, hydrogen and oxygen
(MM = 120.0 g/mol). It is an important sugar that is used in many
chemical syntheses.
Combustion analysis of a 700.0 mg sample yielded 1.027 g CO2 and
0.4194 g H2O. Mass Spectrometry produced a molecular ion @ 120
mass units (m/z). What is the molecular formula of erythrose?
A)
B)
C)
D)
CH2O
C6H12O6
C3H6O3
C4H8O4
ANSWER
D) C4H8O4
Alternative Solution Method
Mass ratio of C in CO2 =
mol C x M of C
=
mass of 1 mol CO2
= 1 mol C x 12.01 g C/ 1 mol C = 0.2729 g C / 1 g CO2
44.01 g CO2
mol H x M of H =
mass of 1 mol H2O
2 mol H x 1.008 g H / 1 mol H
=
= 0.1119 g H / 1 g H2O
18.02 g H2O
Mass ratio of H in H2O =
Calculating masses of C and H:
Mass of Element = mass of compound x mass ratio of element
Alternative Method
Mass (g) of C = 1.027 g CO2 x 0.2729 g C
1 g CO2
= 0.2803 g C
0.1119 g H
Mass (g) of H = 0.4194 g H2O x
= 0.04693 g H
1 g H2O
Calculating the mass of O:
Mass (g) of O = Sample mass -( mass of C + mass of H )
= 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O
Calculating moles of each element:
C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C
H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H
O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O
C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula
120 g /mol / 30 g / formula = 4 formula units / cmpd = C4H8O4