Chapter 9 - Stoichiometry
9-1 Introduction to
Stoichiometry
 Carrying out chemical reactions is like
baking – you need to combine the
ingredients in the right amounts.
 Reaction stoichiometry – study of mass
relationships between reactants and
products in a chemical reaction
9-1 ReactionStoichiometry Problems
 What is given?
 What is unknown?
 All can be solved using mole ratios
and/or molar masses.
9-1 Mole Ratio
 A conversion factor
that relates the
amounts in moles of
any two substances
involved in a
chemical reaction.
 Mole ratios are
obtained from a
balanced chemical
equation.
9-1 Mole Ratios
 What mole ratios can be written for the
balanced equation 2H2 + O2  2H2O ?
9-1 Mole Ratios
 What mole ratios can be written for the
balanced equation
H2SO4 + 2KOH  2H2O + K2SO4
9-1 Mole Ratio
 If the number of
moles of a reactant
or product is known,
the moles of any
other reactant or
product can be found
using the mole ratio.
9-1 Mole Ratio
 Consider the synthesis of water from
hydrogen and oxygen. If 3.2 moles of
oxygen are available, how many moles of
hydrogen are required to react with it?
9-1 Mole Ratio
 Consider the formation of ammonia from
hydrogen and nitrogen. If 1.5 moles of
hydrogen react with excess nitrogen, how
many moles of ammonia will form?
9-1 Molar Mass
 Molar mass is the mass, in grams, of one
mole of a substance.
 Molar mass is a conversion factor that
relates mass to moles.
 Molar mass is expressed in grams per
mole (g/mol).
9-1 Molar Mass
 To calculate the molar mass of a
substance, add up the atomic masses of
the atoms in the formula.
 What is the molar mass of ammonium
chloride?
9-1 Molar Mass
 How many moles are there in 242.5 g of
sodium chloride?
9-1 Molar Mass
 If you have 1.72 moles of barium sulfate,
what is the mass of this sample?
9-2 Ideal Stoichiometric
Calculations
 Remember, if you know
the amount of a reactant
or product in moles, you
can find the amount of
any other reactant or
product.
 The only conversion
factor required is the
mole ratio.
 The mole ratio must be
obtained from the
balanced chemical
equation.
 How many moles of
ammonia are produced
when six moles of
hydrogen gas react with
an excess of nitrogen
gas?
9-2 Amount in Moles to
Mass
 In photosynthesis, plants
use energy from the sun
to produce glucose,
C6H12O6, and oxygen
from carbon dioxide and
water. What mass, in
grams, of glucose is
produced when 3 mol of
water react with excess
carbon dioxide?
9-2 Amount in Mass to
Moles
 When magnesium
burns in air, it
combines with
oxygen to form
magnesium oxide. If
1.52 g of magnesium
are burned in excess
oxygen, how many
moles of magnesium
oxide will form?
9-2 Mass-Mass
Calculations
 Given the mass of any reactant or
product, determine the mass of another
reactant or product.
 mass given  moles given  moles unknown  mass unknown
9-2 Mass-Mass
Calculations
 Tin (II) fluoride is used in
some toothpastes. It is
made by the reaction of
tin with hydrogen
fluoride.
 Sn(s) + 2HF(g) 
SnF2(s) + H2(g)
 How many grams of tin
(II) fluoride are produced
from the reaction of 30.0
g of hydrogen fluoride
with excess tin?
Molar Volume
 Sometimes amounts of reactants or
products are given as volumes, usually in
liters.
 The volume of one mole of ANY gas at
STP is 22.4L.
 STP stands for standard temperature and
pressure (0°C, 1 atm).
Mass-Volume
 What volume of carbon dioxide gas will
form from the combustion of 12.4g of
butane at STP?
Volume-Mass
 If 3 L of nitrogen gas combine with
excess hydrogen gas at STP, how many
grams of ammonia will form?
Volume-Volume
 What volume of hydrogen gas is required
to completely react with 3.5 L of oxygen
gas at STP?
9-3 Limiting Reactants
 limiting reactant – the reactant that limits the amount
of product that forms – will be completely consumed
 excess reactant – the substance that is not completely
consumed – more that enough is available
9-3 Limiting Reactant
 Silicon dioxide (quartz) is usually quite unreactive but
reacts readily with hydrogen fluoride accoring to the
following equation.
SiO2 + 4HF  SiF4 + 2H2O
If 2.0 mol of HF are exposed to 4.5 mol of SiO2, which
is the limiting reactant?
9-3 Limiting Reactant
 If 20.5 g of chlorine is reacted with 20.5 g
of sodium, which reactant is limiting?
9-3 Limiting Reactant
 A black oxide of iron, Fe3O4, can be formed in the lab by the
reaction between red hot iron and steam according to the following
equation.
3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)
When 36.0 g of water is mixed with 167 g of Fe, which is the
limiting reactant? How many grams of black iron oxide will be
produced? What mass of excess reactant will remain when the
reaction is complete?
9-3 Percent Yield
 The amounts of products calculated in
stoichiometric problems represent theoretical
yields.
 theoretical yield – the maximum amount of
product that can be produced from a given
amount of reactant
 Usually, the amount of product obtained in a
chemical reaction is less than the theoretical
yield.
 Reasons include competing side reactions and
loss of product in purification process.
 actual yield – the measured amount of
product obtained from a reaction in the lab –
less than theoretical yield
9-3 Percent Yield
 Chemists are interested in the efficiency
of their reactions.
 Express as percentage by comparing
actual and theoretical yield.
percent yield = actual yield/theoretical yield x 100
9-3 Percent Yield
theoretical yield
actual yield
20.0 g
15.0 g
1.0 g
5.00 g
3.45 g
percent yield
90.0%
4.75 g
48.0%
9-3 Percent Yield
 Chlorobenzene, C6H5Cl, is used in the production of
many important chemicals, such as aspirin, dyes, and
disinfectants. It can be prepared by the following
reaction.
C6H6(l) + Cl2(g)  C6H5Cl(s) + HCl(g)
When 36.8g of C6H6 react with excess chlorine, the
actual yield of C6H5Cl is 38.8 g. What is the percent
yield?
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