Chemical Reactions, Chemical
Equations, and Stoichiometry
Brown, LeMay Ch 3
AP Chemistry
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3.2: Types of reactions
http://web.fuhsd.org/kavita_gupta/July.html
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3.3: Atomic, Molecular & Formula Weights
Atomic mass units:
 1 amu = 1.66054 x 10-24 g
or 1 g = 6.02214 x 1023 amu
 1 atom of 12C isotope defined as weighing
exactly 12 amu
Average atomic mass (or atomic weight):
11.867
 12C: 98.892% x 12 amu =
13C: 1.108% x 13.00335 amu = + 0.1441
AW = 12.011 amu…....
Therefore 12.011 g C = 1 mol C
3

Molecular mass or weight (MM or MW):
sum of atomic masses of each atom in the
chemical formula. Ex: H2SO4
MW = (2)(1.0079) + 32.06 + (4)(16.00)
= 98.08 amu

Formula weight (FW): same as MW, except
for ionic substances in which no “molecule”
exists. Then, FW is the simplest integer ratio
of moles of each element (ions) present. Ex:
NaCl is a 3D array of ions
FW = 22.99 + 35.453 = 58.44 amu
l
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3.4 – 3.7: Mass Relationships

Percent Composition:
Ex: What is the percentage of oxygen in
phosphoric acid?
4( AW O )
%O 
 100 
MW H 3 PO4
4(16.00)
100  65.31%
3(1.0079)  30.974  4(16.00)
Ex: What is the percent of O in CuSO4.5H2O?
57.7%
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
The Mole and Molar Mass (MM)
◦ From Latin moles: “a mass”
◦ Avogadro’s Number = 6.022 x 1023 atoms in
exactly 12.000 g of 12C = 1 mol
◦ Moles important because they can be used
for ratios. Mass can not be used for ratios.
Ex. CO2 etc.
◦ Converting: grams → moles → molecules
Ex: How many molecules of H2O in 100.0 g
H2O?
23
 100.0 g H 2O   1 mol H 2O  6.022 10 molecules H 2O


 
1
1 mol H 2O

  18.0 g H 2O 
molar mass
Avogadro' s number
= 3.343 x 1024 molecules H2O

 

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Ex: What is the empirical formula of a compound that
is composed of 80.% Carbon and 20.% Hydrogen? If
the molar mass is found to be 30 g/mol, what is the
molecular formula?
Strategy: Assume 100. g of unknown
 80. g C  1 mol C 
  6.7 mol C


 1  12.011 g C 
 20. g H  1 mol H 
  20. mol H


 1  1.0079 g H 
C:H mole ratio is 6.7:20. ≈ 1:3. So, empirical
formula is CH3,
and the molecular formula is C2H6.
Percent Composition Tutorial
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Practice Problem 1:
One of the most commonly used white
pigments in the paint is a compound of
Titanium and Oxygen that contains 59.9% Ti
by mass. Determine the empirical formula of
this compound.
Ans; TiO2
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Practice problem2:
Benzene contains only C and H and is
7.74% H by mass; the molar mass of benzene is 184
g/mol. What are the
empirical and molecular formula of this compound?
Ans: CH, C14H14
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EF from Combustion Data: To find out the
EF of a compound, the compound is burned in the air. The
equation for combustion of these compounds is all
follows.
For Hydrocarbons: CxHy + O2 CO2 + H2O
For Compounds Containing CHN:
CxHyNz+ O2 CO2 + H2O+ NO2
For Compounds containing C, H and O
CxHyOz + O2 CO2 + H2O
In such cases to figure out mass of O,
Mass of O = total mass of compound – (mass of C + mass
of H)
Animation: Must See!!
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Important Fact to remember for
Combustion:

Complete combustion produces CO2 as
opposed to incomplete combustion which
produces CO.
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Practice Problem on EF from
Combustion Data:
Many homes in rural America are heated by
propane gas, a compound that contains only
carbon and hydrogen. Complete combustion
of a sample of propane produced 2.641 g of
carbon dioxide and 1.442 g of water as the
only products. Find the empirical formula of
propane.
Answer: C3H8
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Practice Problem Level 2: EF from
Combustion Data
Cumene is a compound containing only
carbon and hydrogen that is used in the
production of acetone and phenol in the
chemical industry. Combustion of 47.6 mg
cumene produces some CO2 and 42.8 mg
water. The molar mass of cumene
produces between 115 and 125 g/mol.
Determine the empirical and molecular
formulas.
Answer: C3H4, C9H12
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Practice Problem 3:
A 0.2500 g sample of a compound
known to contain carbon, hydrogen
and oxygen undergoes complete
combustion to produce 0.3664 g of
CO2 and 0.1500 g of H2O. What is
the empirical formula of this
compound?
Ans: CH2O
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
Limiting Reagent (or reactant):
Ex: If 6.0 g hydrogen gas reacts with 40.0 g oxygen gas, what
mass of water will be produced?
2 H2 (g) + O2 (g) → 2 H2O (g)
 6.0 g H 2  1 mol H 2 
  2.976  3.0 mol H 2


1

 2.0158 g H 2 
 40.0 g O 2  1 mol O 2 
  1.25 mol O 2


1

 32.00 g O 2 
O2 is limiting reagent, H2 is excess.
 1.25 mol O 2  2 mol H 2O  18.0158 g H 2O 

  45.0 g H 2O


1

 1 mol O 2  1 mol H 2O 
Limiting Reactant Movie
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
Percent yield & Percent error:
Ex: If in the previous example, only 40.0 g
water were formed, what is the percent yield
and percent error?
Actual yield
% Yield 
100
Theoretica l yield
40.0 g H 2 O
100  88.9 %
45.0 g H 2 O
% Error 
Accepted value - Measured result
Accepted value
45.0 g H 2 O - 40.0 g H 2 O
45.0 g H 2O
100 
5.0 g H 2O
45.0 g H 2O
100
100  11.%
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Determining Formula of a Hydrate: To determine
the formula of a hydrate, a certain mass of hydrate is heated
to drive off the water. Then the mass of water driven-off is
calculated. Now the mole ratio of water to the compound is
calculated. For more information visit the following website:
Movie on Determining formula of a Hydrate
Questions to ponder over:
1. What might be a good procedure to calculate
percent of water in a hydrate?
2. How do you identify hydrates? Do you remember
how to name hydrates?
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Practice Problem:
A student is given three unknowns labeled A, B,
and C. The student is told they are Na2CO3.H2O;
Na2CO3.7H2O; and Na2CO3.10H2O, not
necessarily in that order.
a. When 2.000 gram of A was heated, 0.914 gram
of anhydrous residue remained.
Formula for sample A.
Ans: Na2CO3.7H2O Sodium carbonate
heptahydrate
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Practice Problem 2
b. When 4.000 grams of B was heated, 1.48 gram of
anhydrous residue remained.
Formula for sample B.
Chemical name for sample B.
Ans: Na2CO3.10H2O Sodium carbonate
decahydrate
c. When 1.000 gram of C was heated, 0.855 gram of
anhydrous residue remained.
Formula for sample C.
Chemical name for sample C.
Ans: Na2CO3.H2O Sodium carbonate monohydrate
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