Chapter 4: Chemical Equations and Stoichiometry
Chapter 4 Problem Set
Pages 172-178
7b, 9c, 11b, 17, 19, 23, 29, 33, 35, 39, 59(opt)
4.1
CHEMICAL EQUATIONS
The basis for a balanced chemical equation is the law of conservation of matter.
The relationship between the quantities of chemical reactants and products is called
stoichiometry.
EXERCISE 4.1 Page 149 Chemical Reactions
The reaction of iron with oxygen is shown in Figure 4.2. The equation for the reaction is
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
1. What are the stoichiometric coefficients?
2. If you were to use 8000 atoms of Fe, how many molecules of O2 are required to consume the
iron completely?
4.2
BALANCING CHEMICAL EQUATIONS
In a balanced equation there are the same number of atoms of each element of both sides of
the equation
Hints for balancing equations
1. Equations are balanced by adjusting coefficients in front of formulas.
2. It is best to start with an element that appears in only one species on each side of the
equation.
3. Usually the simplest whole number coefficients are used in balancing an equation but
decimal numbers may also be used. Make sure all coefficients are in lowest ratio possible
4. For Combustion reactions, balance carbon first, then hydrogen, then oxygen
5. The 11 types of reactions on the help sheet are just a few of the types that you will need to
memorize this year. Chemical Reactions and the equations that represent them are the
“guts” of chemistry
EXERCISE 4.2 Page 153 Balancing the Equation for a Combustion Reaction
1. Pentane, C5H12, can burn completely in air to give carbon dioxide and water. Write the
balanced equation for this combustion reaction.
2. Write a balanced chemical equation for the complete combustion of tetraethyllead,
Pb(C2H5)4 (which was once used as a gasoline additive). The products are PbO(s), H2O(  ), and
CO2(g)
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4.3
MASS RELATIONSHIPS IN CHEMICAL REACTIONS: STOICHIOMETRY
A balanced chemical equation shows the quantitative relationship between reactants and
products in a chemical reaction. The coefficients of a balanced equation represent the
number of moles or molecules of reactants and products.
CS2(  l) + 3 O2(g)  CO2(g) + 2 SO2(g)
We can also use the coefficients as conversion factors or as stoichiometric factors.
EXERCISE 4.3 Page 157 Mass Relations in Chemical Reactions
1. What mass of carbon, in grams, can be consumed by 454 g O2 in a combustion to give
carbon monoxide? (incomplete combustion)
2. What mass of CO is produced
EXAMPLE 4L.3 Mass Relations in Reactions
The ammonium sulfate which is used as a fertilizer on lawns and gardens is made by reacting
ammonia with sulfuric acid.
2 NH3(g) + H2SO4(  )  (NH4)2SO4(s)
Determine
a. the mass in grams of (NH4)2SO4 when 1.43 mols NH3 is reacted
b. the mass in grams of H2SO4 required to form 1.00 kg (NH4)2SO4
c. determine mass in grams of NH3 required to react with 35.0 g of H2SO4
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4.4
REACTANTS IN WHICH ONE REACTANT IS PRESENT IN LIMITED SUPPLY
To determine which reactant is the limiting reagent
1. Calculate the number of moles of each reactant. Then divide this number by the coefficient
that appears before the reactant in the balanced equation. The smallest number is the
limiting reagent.
2. Or use method given in text
3. Or find moles of product that each reactant can produce. Limiting reactant is the one that
produces the least amount of product
EXERCISE 4.5 Page 162 A Reaction Involving a Limiting Reactant
Titanium tetrachloride, TiCl4, is an important industrial chemical. For example, TiO2, the
material used a s a white pigment in paper and paints, can be make from it. Titanium
tetrachloride can be made to combining titanium-containing ore (which is often impure TiO2)
with carbon and chlorine.
TiO2(s) + 2 Cl2(g) + C(s)  TiCl4(  ) + CO2(g)
1. If one begins with 125 grams each of Cl2 and C, but plenty of TiO2 containing ore, which is
the limiting reactant in this reaction?
2. What quantity of TiCl4, in grams, can be produced?
4.5
PERCENT YIELD
The maximum quantity of product that can be obtained from a chemical reaction is the
theoretical yield, usually found by doing a stoichiometric calculation.
actual yield (the amount of product produced in the laboratory or chemical plant, usually less
than the theoretical yield)
The percent yield is a comparison of the actual to the theoretical yield. In equation form we
have
exp erimental yield
percent yield 
x 100
theoretica l yield
Sometimes a % yield question will first involve finding the limiting reactant.
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EXERCISE 4.6 Page 164 Percent Yield
Methanol, CH3OH, can be burned in oxygen to provide energy, or it can be decomposed to
form hydrogen gas, which can then be used as a fuel (see Figure 4.8)
CH3OH(  )  2 H2(g) + CO(g)
1. If you decompose 125 g of methanol, what is the theoretical yield of hydrogen?
2. If you obtained only 13.6 g of hydrogen, what is the percent yield?
4.6
CHEMICAL EQUATIONS AND CHEMICAL ANALYSIS
Analysis of a Mixture – can be very challenging
EXERCISE 4.7 Page 167 Analysis of a Mixture
You have 2.357 g of a mixture of BaCl2 and BaCl2  2 H2O. If experiment shows that the
mixture has a mass of only 2.108 g after heating to drive off all the water of hydration in
BaCl2  2 H2O, what is the weight percent of BaCl2  2 H2O in the original mixture?
Determining the Formula of a Compound
The empirical formula of a compound can be determined if the percent composition of the
compound is known.
All formulas for ionic compounds are empirical (lowest whole number ratio)
Formulas for molecular compounds may be empirical or they may be a whole number multiple
greater than the empirical formula, which is called the molecular formula
Ex: The molecular formula for octane is C8H18. Its empirical formula is C4H9 (which may or
may not exist). The molecular formula C12H27 also has the empirical formula C4H9.
The molecular formula for propane, C3H8, has the same empirical formula, C3H8.
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EXERCISE 4.8 Page 169 Determining the Empirical and Molecular Fomulas for a
Hydrocarbon
A 0.523-g sample of the unknown compound CxHy was burned in air to give 1.612 g of CO2
and 1.7425 g of H2O. A separate experiment gave a molar mass for CxHy of 114 g/mol.
Determine the empirical and molecular formulas for the hydrocarbon.
EXERCISE 4.9 Page 172 Formula Determination from Combustion Analysis
Vitamin C is composed of C, H, and O. Determine the empirical formula of vitamin C from the
following data: burning 0.400 g of solid vitamin C in pure oxygen gives 0.600 g of CO2 and
0.163 g of H2O.
EXERCISE 4.10 Page 172 Formula Determination from Combustion Analysis
A compound is composed of only C, H, and Cr. When 0.178 g of the compound is burned in
air, the products are CO2 (0.452 g), H2O (0.0924 g), and Cr2O3.
CxHyCrz (s) + some O2  x CO2 + y/2 H2O + z/2 Cr2O3
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