MOLE/ STOICHIOMETRY
By: Tyler Lewis, Michael Stylc and Erich Johnson
DEFINITION OF STOICHIOMETRY

The relationship between the relative quantities
of substances taking part in a reaction or forming
a compound, typically a ratio of whole integers.
AVOGADRO’S NUMBER
6.02 x 10^23
WHAT IS A MOLE?

The mole is a unit of measurement used in
chemistry to express amounts of a chemical
substance.
MOLAR MASS
Definition:
The mass, in grams, of one mole of a substance
MASS TO MOLE CONVERSION
1 mol = g-formula-mass (periodic table)
Problem #1: How many moles in 28 grams of CO2 ?
Gram-formula-mass of CO2 1 C = 1 x 12.01 g = 12.01 g
2 O = 2 x 16.00 g =32.00 g
44.01 g/mol
28 g CO2 x 1 mole CO2 = 0.64 mole CO2
44.01 g CO2
Problem #2: How many moles in 480. grams of
Fe2O3 ?
2 Fe = 2 x 55.85 g = 111.7 g
3 O = 3 x 16.00 g = 48.0 g
159.7 g/mol
480. gram Fe2O3 x 1 moleFe2O3
= 3.01 mole
Fe2O3
159.7 gram Fe2O3
Problem #3: Find the number of moles of argon in 452 g
of argon.
Ar = 1 x 39.95 g = 39.95 g
39.95 g/mol
452 gram Ar x 1 mole Ar
= 11.3 mole Ar
39.95 gram Ar
PARTICLE TO MOLE CONVERSION
1 mol = 6.02 x 1023 particles
Problem #1: How many moles in 39.0 x 1028 particles of CO2 ?
39.0 x 1028 particles CO2 x 1 mole CO2
= 6.48 x 1028 mole CO2
6.02 x 1023 particles CO2
Problem #2:How many moles are 1.20 x 1025 particles of
phosphorous?
1.20 x 1025 particles P x 1 mole P
= 19.9 mole P
6.02 x 1023 particles P
Problem #3: How many moles in 48.0 x 1023 particles of
Cs ?
48.0 x 1023 particles Cs x 1 mole Cs
= 7.97 mole CS
6.02 x 1023 particles Cs
VOLUME TO MOLE CONVERSION
1 mol = 22.4 L for a gas at STP
Problem #1: How many moles in 22.4 Liters of CO3 ?
22.4 L CO3 x 1 mole CO3 = 1 mole CO3
22.4 L CO3
Problem #2: How many moles of argon atoms are
present in 11.2 L of argon gas at STP?
11.2 L Ar x 1 mole Ar = .500 mole CO3
22.4 L Ar
Problem #3: How many moles of argon atoms are
present in 403.2 L of Xenon gas at STP?
403.2 L Xe x 1 mole Xe = 18.00 mole Xe
22.4 L Xe
MOL-MOL CALCULATIONS
N2 + 3 H2 ---> 2 NH3
Problem #1: if we have 2 mol of N2 reacting with
sufficient H2, how many moles of NH3 will be produced?
NH3
 ratio to set up the proportion:
N2
2
 That means the ratio from the equation is:
1
Initial amount
Final Amount
2 mole N2 x 2 mole NH3 = 4mole NH3
1 mole N2
N2 + 3 H2 ---> 2 NH3
Problem #2: Suppose 6 mol of H2 reacted with sufficient
nitrogen. How many moles of ammonia would be
produced?
6 mole H2 x 2 mole NH3 = 4mole NH3
3 mole H2
N2 + 3 H2 ---> 2 NH3
Problem #2: We want to produce 2.75 mol of NH3. How
many moles of nitrogen would be required?
2.75 mole NH3 x 1 mole N2 = 1.38 mole NH3
2 mole NH3
MASS-MASS CALCULATION
2 AuCl3 ---> 2 Au + 3 Cl2
Problem #1: How many grams of chlorine can be liberated from the
decomposition of 64.0 g. of AuCl3?
The amount of moles of
The molar ratio of
chemical
initial chemical
64.0 gram AuCl3 x 1 mole AuCl3
= .211 mole AuCl3
303.35 grams AuCl3
The mole ratio of chemicals
.211 mole AuCl3 x 3 mole Cl2
= .3165 mole Cl2
2 mole AuCl3
.3165 mole Cl x 70.9 gram Cl2
1 mole Cl2
= 22.4 gram Cl2
3 AgNO3 + AlCl3 --> 3 AgCl + Al(NO3)3
Problem #2: Calculate the mass of AgCl that can be
prepared from 200. g of AlCl3 and sufficient AgNO3?
200. gram AlCl3 x 1 mole AlCl3
= 1.50 mole AlCl3
133.33 grams AlCl3
1.50 mole AlCl3 x 3 mole AgCl
= 4.5 mole AgCl
1 mole AlCl3
4.5 mole AgCl x 143.35 gram AgCl
1 mole AgCl
= 645. gram Cl2
2 Au + 3 Cl2 ---> 2 AuCl3
Problem #4: How many grams of AuCl3 can be made
from 100.0 grams of chlorine?
100. gram Cl2x 1 mole Cl2
70.9 grams Cl2
= 1.41mole Cl2
1.41 mole Cl2 x 2 mole AuCl3 = .94mole AuCl3
3 mole Cl2
.94 mole AuCl3 x 303.35 grams AuCl3 = 285 gram AuCl3
1 mole AuCl3
PARTICLE-PARTICLE CALCULATION
2 AuCl3 ---> 2 Au + 3 Cl2
Problem #1: How many particles of chlorine can be liberated from the
decomposition of 6.02x1023 particles of AuCl3? The amount of moles of
initial chemical
The partial ratio of
chemical
6.02x1023 particles AuCl3 x 1 mole AuCl3
= 1 mole AuCl3
6.02x1023 particles AuCl3
The mole ratio of chemicals
1 mole AuCl3 x 3 mole Cl2
= 1.5mole Cl2
2 mole AuCl3
1.5 mole Cl2 x 6.02x1023 particles Cl2
1 mole Cl2
= 9.03x1023 particles Cl2
2 AuCl3 ---> 2 Au + 3 Cl2
Problem #2: How many particles of chlorine can be liberated from
the decomposition of 150. particles of AuCl3?
150. particles AuCl3 x 1 mole AuCl3
= 2.492x10-22 mole AuCl3
6.02x1023 particles AuCl3
2.492x10-22 mole AuCl3 x 3 mole Cl2
= 3.738x10-22 mole Cl2
2 mole AuCl3
3.738x10-22 mole Cl2 x 6.02x1023 particles Cl2
1 mole Cl2
= 225 particles Cl2
3 AgNO3 + AlCl3 --> 3 AgCl + Al(NO3)3
Problem #3: Calculate the particles of AgCl that can be
prepared from 132particals of AlCl3 and sufficient
AgNO3?
132. article AlCl3 x 1 mole AlCl3
= 2.19x1024 mole AlCl3
6.02x1023 particle AlCl3
2.19x1024 mole AlCl3 x 3 mole AgCl
1 mole AlCl3
= 6.57x1024 mole AgCl
6.57x1024mole AgCl x 6.02x1023 particle AgCl =396. particle Cl2
1 mole AgCl
VOLUME-VOLUME CALCULATION
2 AuCl3 ---> 2 Au + 3 Cl2
Problem #1: How many particles of chlorine can be liberated from
the decomposition of 22.4 liters of AuCl3? The amount of moles of
initial chemical
The volume ratio
22.4 liters AuCl3 x 1 mole AuCl3
22.4 liters AuCl3
= 1 mole AuCl3
The mole ratio of chemicals
1 mole AuCl3 x 3 mole Cl2
= 1.5mole Cl2
2 mole AuCl3
1.5 mole Cl2 x 22.4 liters Cl2
1 mole Cl2
= 33.6 liters Cl2
N2 + 3 H2 ---> 2 NH3
Problem #2: Suppose 5.6 liters of H2 reacted with
sufficient nitrogen. How many liters of ammonia would
be produced?
5.6 liters H2 x 1 mole H2
22.4 liters H2
.25 mole H2 x 2 mole NH3
3 mole H2
.167 mole NH3x 22.4 liters NH3
1 mole NH3
= .25 moles H2
= .167mole NH3
= 3.7 liters NH3
2 Na + Cl2 ---> 2 NaCl
Problem #3: How many liters of Na are required to
react completely with 75.0 liters of chlorine?
75. liters Cl2 x 1 mole Cl2
= 3.35moles Cl2
22.4 liters Cl2
3.35 mole Cl2 x 2 mole Na
1 mole Cl2
6.7 mole Na x 22.4 liters Na
1 mole Na
= 6.7 mole Na
= 150. liters Na
PERCENT COMPOSITION FORMULAS
Mass of element in sample of compound X 100= % Element in Compound
Mass of sample of compound
OR
Mass of element in 1 mol of compound X 100= % Element in Compound
Molar mass of compound
PERCENT COMPOSITION CALCULATION
Given: Formula Cu2S
Problem #1: Percent Composition of Sulfur
2 Mol Cu X 63.55 Cu = 127.1 g Cu
mol Cu
1 mol S X 32.07g S = 32.07 g S
mol S
Molar mass of Cu2S = 159.2 g
32.07 g S
159.2 g Cu2S
x 100 = 20.15% S
Problem #2: Calculate the percent composition of
carbon in the following: C6H12O6
6 Mol C x 12.01 g/mole = 72.06 g C
12 Mol H x 1.008 g/mole=12.096 g H
6 Mol O x 16.00 g/mole =96.00 g O
180.2 g C6H12O6
72.06 g C
x 100 = 39.99% C
180.2 g C6H12O6
Problem #3: Calculate the percent composition of
carbon in the following: CO2
1 Mol C x 12.01 g/mole = 12.01 g C
2 Mol O x 16.00 g/mole =32.00 g O
44.01 g CO2
12.01 g C
x 100 = 27.29% C
44.01 g CO26