Stoichiometry
Chapter 3
Chemical Stoichiometry
Stoichiometry - The study of quantities of
materials consumed and produced in
chemical reactions.
Detector
plate
03_34
Ion-accelerating
electric field
Accelerated
ion beam
Least massive ions
Positive ions
Sample
Most
massive
ions
Electron beam
Slits
Magnetic field
Heating device
to vaporize sample
Schematic diagram of a mass spectrometer.
Average Atomic Mass
Elements occur in nature as mixtures of
isotopes
Carbon =
98.89% 12C
1.11% 13C
<0.01% 14C
(0.9889)(12.0000 amu) = 11.87 amu
(0.0111)(13.0034 amu) = 0.144 amu
12.01 amu
The Mole
The number equal to the number of carbon
atoms in exactly 12 grams of pure 12C.
1 mole of anything = 6.022  1023 units of
that thing
Equal moles of
substances have equal
numbers of atoms,
molecules, ions,
formula units, etc.
Counting Atoms
Atoms are too small to be seen or counted
individually.
Atoms can only be counted by weighing them.
• all jelly beans are not identical.
• jelly beans have an average mass of 5 g.
• How could 1000 jelly beans be counted?
Jelly Beans & Mints
Mints have an average
mass of 15 g.
How would you count out
1000 mints?
Why do 1000 mints have
a mass greater than
1000 jelly beans?
Atomic Mass Unit
Atoms are so tiny that the gram is much too
large to be practical.
The mass of a single carbon atom is 1.99
x 10-23 g.
The atomic mass unit (amu) is used for atoms
and molecules.
AMU’s and Grams
1 amu = 1.661 x 10 -24 g
Conversion Factors
1.661 x 10-24g/amu
6.022 x 1023amu/g
The Mole
• One mole of rice grains is more than the
number of grains of all rice grown since the
beginning of time!
• A mole of marshmallows would cover the
U.S. to a depth of 600 miles!
• A mole of hockey pucks would be equal in
mass to the moon.
The Mole
Substance Average Atomic Mass
(g)
Na
22.99
Cu
63.55
S
32.06
Al
26.98
# Moles
1
1
1
1
# Atoms
6.022 x 1023
6.022 x 1023
6.022 x 1023
6.022 x 1023
Measurements
dozen = 12
gross = 12 dozen = 144
ream = 500
mole = 6.022 x 1023
Avogadro’s number
equals
23
6.022  10 units
Unit Cancellation
How many dozen eggs would 36 eggs be?
(36 eggs)(1 dozen eggs/12 eggs) = 3 dozen eggs
How many eggs in 5 dozen?
(5 dozen eggs)(12 eggs/1 dozen eggs) = 60 eggs
Calculating Mass Using
AMU’S
1 N atom = 14.01 amu
(23 N atoms)(14.01 amu/1N atom) = 322.2 amu
Calculating Number of Atoms
from Mass
1 O atom = 16.00 amu
(288 amu)(1 O atom/16.00 amu) = 18 atoms O
AMU’s & Grams
1 atom C = 12.011 amu = 1.99 x 10-23g
1 mol C = 12.011 g
Use TI-83 or TI-83 Plus to store 6.022 x 1023
to A.
Calculating Moles & Number of Atoms
1 mol Co = 58.93 g
(5.00 x 1020 atoms Co)(1mol/6.022 x 1023 atoms)
= 8.30 x 10-4 mol Co
(8.30 x 10-4 mol)(58.93g/1 mol) = 0.0489 g Co
Moles are the doorway
grams <---> moles <---> atoms
Molar Mass
A substance’s molar mass (molecular
weight) is the mass in grams of one mole of
the compound.
CO2 = 44.01 grams per mole
Calculating Mass from Moles
CaCO3
1 Ca = 1 (40.08 g) = 40.08 g
1 C = 1 (12.01 g) = 12.01 g
3 O = 3 (16.00 g) = 48.00 g
100.09 g
(4.86 molCaCO3)(100.09 g/1 mol) = 486 g CaCO3
Calculating Moles from Mass
Juglone
10 C = 10(12.01g) = 120.1 g
6 H = 6(1.008 g) = 6.048 g
3 O = 3(16.00 g) = 48.00 g
174.1 g
(1.56 g juglone)(1 mol/174.1 g) = 0.00896
mol juglone
Percent Composition
Mass percent of an element:
mass of element in compound
mass % 
 100%
mass of compound
For iron in iron (III) oxide, (Fe2O3)
111.69
mass % Fe 
 100%  69.94%
159.69
% Composition
CuSO4. 5 H2O
1 Cu = 1 (63.55 g) = 63.55 g
1 S = 1 (32.06 g) = 32.06 g
4 O = 4 (16.00 g) = 64.00 g
5 H2O = 5 (18.02 g) = 90.10 g
249.71 g
% Composition
(Continued)
% Cu = 63.55 g/249.71g (100 %) = 25.45 % Cu
% S = 32.06 g/249.71 g (100 %) = 12.84 % S
% O = 64.00 g/249.71 g (100 %) = 25.63 % O
% H2O = 90.10 g/249.71 g (100 %) = 36.08 % H2O
Check: Total percentages. Should be equal to 100 %
plus or minus 0.01 %.
Formulas
Ionic compounds -- empirical formula
NaCl
CaCl2
Covalent compounds -- molecular formula C6H12O6
C2H6
Formula of a Compound Calculations
Example for Problems 75-78 on page 125.
A compound contains only carbon, hydrogen,
and nitrogen. When 0.1156 g of it is reacted
with oxygen, 0.1638 g of CO2 and 0.1676 g
of water are collected.
Molar mass of CO2 is 44.01 g
Molar mass of HOH is 18.02 g
Formula of a Compound Calculations
(0.1638g CO2)(12.01g C/44.01g CO2) =
0.04470g C
%C = (0.04470g/0.1156g)(100%) = 38.67% C
(0.1676 g HOH)(2.016 g H/18.02g HOH) =
0.01875g H
%H = (0.01875g/0.1156g)(100%) = 16.22% H
%H + %C + %N = 100%
%N = 100% -38.67% -16.22% = 45.11% N
Formula of a Compound Calculations
(38.67g C)(1 mol/12.01g) = 3.220 mol C = 1 C
3.219 mol
(16.22g H)(1 mol/1.008g) = 16.09 mol H = 5 H
3.219 mol
(45.11g N)(1 mol/14.01g) = 3.219 mol N = 1 N
3.219 mol
Empirical Formula is CH5N.
Formulas
molecular formula = (empirical formula)n
[n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH
Empirical Formula Determination
1. Base calculation on 100 grams of
compound.
2. Determine moles of each element in 100
grams of compound.
3. Divide each value of moles by the smallest
of the values.
4. Multiply each number by an integer to
obtain all whole numbers.
Calculating Empirical Formulas
4.151 g Al & 3.692 g O
(4.151 g Al)(1 mol/26.98 g) = 0.1539 mol Al/0.1539 mol = 1.000
(3.692 g O)(1 mol/16.00 g) = 0.2308 mol O/0.1539 mol = 1.500
1 Al (2) = 2 Al
1.5 O (2) = 3 O
Al2O3
Molecular Formulas
71.65 % Cl, 24.27 % C, & 4.07 % H
(71.65g Cl)(1 mol/35.45g) = 2.021 mol/2.021 mol = 1
(24.27 g C)(1 mol/12.01g) = 2.021 mol/2.021 mol = 1
(4.07 g H )(1 mol/1.01g) = 4.03 mol/2.021 mol = 2
(EM)x = (MM)
(49.46)x = (98.96)
x = 2
(EF)x
= (MF)
(ClCH2)2 = Cl2C2H4
Chemical Equations
Chemical change involves a
reorganization of the atoms in one or
more substances.
Chemical Equation
A representation of a chemical reaction:
C2H5OH + 3O2  2CO2 + 3H2O
reactants
products
Physical States
•
•
•
•
solid (s)
liquid (l)
gas (g)
aqueous (aq)
Important Equation Symbols
• yields
------>
• heat

------->
cat.
• catalyst ------->
• light
light
-------->
H2SO4
• catalyst ------>
elect.
• electricity ------>
Chemical Equations
Quantitative Significance
4 Al(s) + 3 O2(g)
---> 2 Al2O3(s)
This equation means
4 Al atoms + 3 O2 molecules
---give--->
2 molecules of Al2O3
4 moles of Al + 3 moles of O2
---give--->
2 moles of Al2O3
Balancing Equation Prerequisites
Student must have memorized:
• 44 chemical symbols
• Table 2.3 on page 61 in text
• Table 2.4 on page 62
• Table 2.5 on page 66
• Table 2.6 on page 67
• Type I, II, III, and acid nomenclature
• Count HOFBrINCl
Four Steps in Balancing Equations
1. Get the facts down.
2. Check for diatomic molecules (subscripts).
3. Balance charges on compounds containing a
metal, ammonium compounds, and acids
(subscripts).
4. Balance the number of atoms (coefficients).
a. Balance most complicated molecule first.
b. Balance other elements.
c. Balance hydrogen next to last.
d. Balance oxygen last.
Balancing Equations Caution
The identities (formulas) of
the compounds must never be
changed in balancing a
chemical equation!
Only coefficients can be used
to balance the equationsubscripts will not change!
Chemical Equation
C2H5OH + 3O2  2CO2 + 3H2O
The equation is balanced.
1 mole of ethanol reacts with 3 moles of
oxygen
to produce
2 moles of carbon dioxide and 3 moles of
water
Balancing Equations
When solid ammonium dichromate
decomposes, it produce solid chromium(III)
oxide, nitrogen gas, and water vapor.

(NH4)2Cr2O7(s) ----> Cr2O3(s) + N2(g) + HOH(g)

(NH4)2Cr2O7(s) ----> Cr2O3(s) + N2(g) +
4HOH(g)
Calculating Masses of
Reactants and Products
1.
2.
3.
4.
Balance the equation.
Convert mass to moles.
Set up mole ratios.
Use mole ratios to calculate moles of
desired substituent.
5. Convert moles to grams, if necessary.
Gram to Mole & Gram to Gram
__Al(s) + __I2(s) ---> __AlI3(s)
2Al(s) + 3I2(s) ---> 2AlI3(s)
How many moles and how many grams of
aluminum iodide can be produce from 35.0
g of aluminum?
Gram to Mole & Gram to Gram
2Al(s) + 3I2(s) ---> 2AlI3(s)
(35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol
Al) = 1.30 mol AlI3
(35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol
Al)(407.68 g/1 mol) = 529 g AlI3
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the
amounts of products formed.
Solving a Stoichiometry Problem
1.
2.
3.
4.
Balance the equation.
Convert masses to moles.
Determine which reactant is limiting.
Use moles of limiting reactant and mole
ratios to find moles of desired product.
5. Convert from moles to grams.
Limiting Reactant Problem
If 56.0 g of Li reacts with 56.0 g of N2, how
many grams of Li3N can be produced?
__Li(s) + __N2(g) ---> __Li3N(s)
6 Li(s) + N2(g) ---> 2 Li3N(s)
(56.0 g Li) (1 mol/6.94g)(1 mol N2/6 mol Li)
(28.0 g/1 mol) = 37.7 g N2
Since there were 56.0 g of N2 and only 37.7 g
used, N2 is the excess and Li is the Limiting
Reactant.
Limiting Reactant Problem
6 Li(s) + N2(g) ---> 2 Li3N(s)
(56.0 g Li)(1 mol/6.94g)(2 mol LiN3/6 mol Li)
(34.8 g/1 mol) = 93.6 g Li3N
How many grams of nitrogen are left?
56.0g N2 given - 37.7 g used = 18.3 g excessN2
% Yield
Values calculated using stoichiometry are
always theoretical yields!
Values determined experimentally in the
laboratory are actual yields!
Limiting Reactant & % Yield
If 68.5 kg of CO(g) is reacted with 8.60 kg of
H2(g), what is the theoretical yield of methanol
that can be produced?
__H2(g) + __CO(g) ---> __CH3OH(l)
2 H2(g) + CO(g) ---> CH3OH(l)
(68.5 kg CO)(1 mol/28.0 g)(2 mol H2/1 mol CO)
(2.02 g/1mol) = 9.88 kg H2
Limiting Reactant & % Yield
2 H2(g) + CO(g) ---> CH3OH(l)
Since only 8.60 kg of H2 were provided, the H2 is
the limiting reactant, and the CO is in excess.
(8.60 kg H2)(1000 g/1 kg)(1 mol/2.02 g)(1 mol CH3OH/2
mol H2)(32.0 g/1 mol) = 6.85 x 104 g CH3OH
Limiting Reactant & % Yield
2 H2(g) + CO(g) ---> CH3OH(l)
If in the laboratory only 3.57 x 104 g of
CH3OH is produced, what is the % yield?
actual _ yield
100%
%Yield 
theoretical _ yield
3.57 x10 4 g
100%
%Yield 
4
6.85 x10 g
% Yield = 52.1 %