Stoichiometry
AP Chemistry
Mr. Martin
Topics
Law of Conservation of Matter
 Balancing Chem Eq
 Mass Relationships in rxn’s
 Limiting Reagents
 Theoretical, Act and % yld
 Empirical and Molecular formulas

A. Balancing Equations
Law of Conservation of Mass – matter is
neither created or destroyed – during
chem rxn’s some matter is converted to
energy – why we must bal equations
 Stoichiometry – the relationship among
the substances in chem rxn’s.

Chemical Equations

< TARGET="display">
Balancing Equations

< TARGET="display">
Steps to Balance

< TARGET="display">
Practice Problems
Cu + O2 ----> Cu2O
 CaCl2 + AgNO3 ----> AgCl + Ca(NO3)2
 H2 + O2 ----> H2O
 Mg + P4 ---> Mg3P2
 H2SO4 + NaOH ----> H2O + Na2SO4

Solutions
4Cu + O2 ----> 2Cu2O
 CaCl2 + 2AgNO3 ----> 2AgCl + Ca(NO3)2
 2H2 + O2 ----> 2H2O
 6Mg + P4 ---> 2Mg3P2
 H2SO4 + 2NaOH ----> 2H2O + Na2SO4

Additional Practice





Fe2O3 + CO -> Fe3O4 + CO2
Fe3O4 + CO -> FeO + CO2
C12H22O11(s) + O2(g) -> CO2(g) + H2O
Fe(s) + O2(g) -> Fe2O3
Ca(s) + H2O(l) -> Ca(OH)2(aq) + H2(g)
B. Atomic and Molecular Weights

Atomic Mass Scale uses atomic mass units
(AMU’S) relative mass scale originally based
on hydrogen = 1 amu
 Now 1 amu = 1/12 of the carbon-12 isotope
 Average atomic masses (atomic weight)
elements are a mixture of isotopes and
therefore the mass of an element is a
weighted average of the naturally occurring
isotopes.
Calculation of Atomic Mass
AM = sum (abundance) x (isotope mass)
of all the naturally occurring
isotopes
Ex: Ne-20 19.992 amu 90% abund
Ne-22 21.991 amu 10 % abund
(0.9 x 19.992) + (0.1 x 21.991) =
20.192 amu’s atomic mass

Formula Weight
Sum of the average atomic masses of
each atom in it’s chem ormula
 EX C12H22O11
C (12) 12 = 144 amu’s
H (22) 1 = 22 amu’s
O (11) 16 = 176 amu’s
342 amu’s formula weight

Percent Composition by Weight

What is the % comp of Methane CH4?
The Mole
A number 6.02 x 10^23 –
like a dozen = 12 , a pair =
2, or a gross = 144
 A mol is the amount of
matter that contains as
many objects (atoms,
molecules or formula
units) as the # of atoms in
exactly 12g of the carbon12 isotope.

The Mole and Molar Mass



One mole of Carbon =
6.02x10^23 atoms of
carbon
One mole of H2O is
equal to 6.02 x 10^23
H2O molecules.
One mole of HCl is
equal to 6.02 x 10^23
formula units of HCl.

Molar Mass = 12.01 g of
Carbon

Molecular Weight = H2O
H (2) 1 = 2
0 (1) 16 = 16
18 g/mol
Formula Weight = HCl
36.5g/mol

Conversion Problems


Gram moles  atoms, fu’s,
molecules
How many moles are in 200g of sulfuric
acid. ( 2.04 mole of H2SO4)
Conversion Problems

How many formula units are in 10g of
sulfuric acid? ( 6.14 x 10^22 fu’s)
Empirical and Molecular Formulas
Empirical Formulas – gives the lowest
whole # ratio of atoms (moles) in a
compound.
 EF’s are derived from exp info most
notably % comp.

Example Problems

An analysis of sodium dichromate gives
the following mass percentages, 17.5%
Na, 39.7 % Cr, and 42.8% O. Calc the
emp. formula?
Example Problem Mol. Formula
Molecular formula – is a multiple of the
emp. formula. n = exp mm/ef weight
 The % composition of acetaldehyde is
54% C, 9.2% H, and 36.3% O. Its molar
mass is 88 amu’s. What is its molecular
formula.

Combustion Analysis

A method of analysis
where a sample
cont. C,H, O
undergoes complete
combustion. The gas
stream goes through
two tubes one
absorbs CO2, the
other H2O.
Combustion Analysis cont.
The change of mass in the tubes gives
the mass of CO2, and H2O.
 A 4.24mg Acedic Acid sample is
completely burned yielding 6.24mg of
CO2 and 2.54mg of H2O. What is the
mass % of each element?

Quantitive Information from Bal. Eq
grms   mol   mol   grms
Ex. A (moles to moles) How many moles of
H2O will be produced if 17.5 moles of H2

combine with excess
O2? (17.5)
Example Problems Cont.

How many grams of KClO3 must
decompose to produce 185.0 moles of
O2? (1.51 x 10^4 g)
Example Prob Cont.

How many grams of silver chloride can
be produced from the reaction of 17.5 g
of silver nitrate with excess sodium
chloride solution? (14.4g)
Example Problems

How many mg of aluminum nitrate would
be required to completely react with 1.75
kg of barium hydroxide? (1.45 x 10^6mg)
Limiting Reagent & Percent Yld
The reactant that is entirely consumed
when a reaction goes to completion is
the limiting reagent.
 The reactant that is not completely
consumed is said to be in excess.
 The amount of product that is produced
is determined by the limiting reactant.

Ex. Limiting Reagent Prob.

In a lab test reaction 20.0g of CH3CHO
and 10.0g of O2 were put in a reaction
vessel. (a) How many grams of acedic
acid can be produced (b) How many
grams of excess reactant remain after
the reaction is complete.
Theoretical and Percent Yld.
The max amount of product that can be
obtained is the theoretical yield. ( based
on stoichiometric calc.)
 The actual yield is the mass obtained by
the actual reaction.
 % yield = actual yld/theorical yld x 100

Problem

The amount of acedic acid produced
from the previous reaction was 23.8g.
What is the percent yield of the reaction.
(87.2%)