topic 9: stoichiometry · topic 9: stoichiometry (chapter 9 in modern chemistry p. 298)...
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HN Chemistry Page 1
Topic 9: Stoichiometry (Chapter 9 in Modern Chemistry p. 298)
Introduction to Stoichiometry
In this topic we will focus on the quantitative aspects of chemical reactions. That’s right; you’ll
need your calculators. Composition Stoichiometry (Topic 7) deals with the mass relationships
of elements in compounds. Reaction Stoichiometry involves the mass relationships between
reactants and products in a chemical reaction. Reaction stoichiometry is the subject of this topic
and it is based on chemical equations and the law of conservation of mass. All reaction
stoichiometry calculations must start with a balanced chemical equation.
Reaction Stoichiometry Problems
Problems in reaction stoichiometry will be classified by the given and the unknown.
Problem Type 1: Given & unknown quantities are amounts in moles.
amount of given
substance (mol)
amount of unknown
substance (mol)
Problem Type 2: Given amount is in moles and unknown amount is mass in grams.
amount of given
substance (mol)
amount of unknown
substance (mol)
mass of unknown
substance (g)
Problem Type 3: Given amount is mass in grams and unknown amount is in moles.
mass of given
substance (g)
amount of given
substance (mol)
amount of unknown
substance (mol)
Problem Type 4: Given & unknown amounts are mass in grams.
mass of given
substance (g)
amount of given
substance (mol)
amount of unknown
substance (mol)
mass of unknown
substance (g)
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Mole Ratio
Solving any reaction stoichiometry problem requires the use of a
mole ratio to convert from mores or grams of one substance in a reaction to moles or grams of
another substance. A mole ratio is a conversion factor that relates the amounts in moles of any
two substances involved in a chemical reaction. This information is obtained directly from the
balanced chemical equation.
For example, here is the equation for the electrolysis of melted aluminum oxide to produce
aluminum and oxygen.
2 Al2O3(l) 4 Al(s) + 3 O2(g)
This means: 2 moles of Al2O3 can produce 4 moles of Al and 3 moles of O2
This can be represented as a mole ratio like:
2 mol Al2O3 or
4 mol Al
4 mol Al 2 mol Al2O3
2 mol Al2O3 or
3 mol O2
3 mol O2 2 mol Al2O3
4 mol Al or
3 mol O2
3 mol O2 4 mol Al
Can you see that this would also mean:
4 moles of Al2O3 can produce 8 moles of Al and 6 moles of O2
Example Problem 1:
Determine the amount in moles of aluminum that can be produced from 13.0 mol of
aluminum oxide.
What is the mole ratio that you will need? Al to Al2O3
The Al produce is 2x as many moles as aluminum oxide. So the answer is simply 26.0 moles of
Al.
It’s not always so easy to do the math in your head. If you need to you can use dimensional
analysis.
13.0 mol Al2O3 x 4 mol Al
= 26.0 mol Al 2 mol Al2O3
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Note: You will always use a mole ratio when you need to change from one substance in an
equation to another substance in that equation.
Molar Mass
Remember in Topic 7 that the molar mass is the mass, in grams, of one mole of a substance.
Recall that the molar mass can also be used as a conversion factor that relates the mass of a
substance to the amount in moles of that same substance. You can use molar mass to go from
grams to moles or to go from moles to grams.
Again using the equation on p. 2:
1 mol Al2O3 = 101.96 g 1 mol Al = 26.98 g 1 mol O2 = 32.00 g
These molar masses can be expressed by the following conversion factors.
101.96 g Al2O3 or 1 mol Al2O3
1 mol Al2O3 101.96 g Al2O3
26.98 g Al or
1 mol Al
1 mol Al 26.98 g Al
32.00 g O2 or 1 mol O2
1 mol O2 32.00 g O2
Example Problem 2:
Determine the grams of aluminum equivalent to 26.0 moles of aluminum.
26.0 mol Al x 26.98 g Al
= 701 g Al 1 mol Al
Ideal Stoichiometric Calculations Chemical equations help us make predictions about chemical reactions without having to run the
reactions in the laboratory. These reactions are theoretical. They tell us the amounts of reactants
and products for a given chemical reaction under ideal conditions, in which all reactants are
completely converted into products. Theoretical stoichiometric calculations allow us to
determine the maximum amount of product that could be obtained in a reaction.
These problems are extensions of the composition stoichiometry problems that were solved in
Topic 7 (mole conversions). It is important to use a logical, systematic approach to successfully
solve these problems. We will use dimensional analysis.
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Conversion of Quantities in Moles (Mole-Mole stoichiometry)
The Plan:
amount of amount of
given substance (mol) unknown substance (mol)
To solve this plan requires one conversion factor – the mole ratio of the unknown to the given
from the balanced equation.
Mole ratio
(Balanced equation)
Amount of given
substance (mol)
Amount of unknown substance
(mol)
x mol unknown
= mol given
GIVEN IN THE PROBLEM
CONVERSION FACTOR
CALCULATED
Example Problem 3 (p. 305):
In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction
with lithium hydroxide, LiOH, according to the following chemical equation.
CO2(g) + 2 LiOH(s) Li2CO3(s) + H2O(l)
How many moles of lithium hydroxide are required to react with 20 moles of CO2, the
average amount exhaled by a person each day?
20 mol CO2 x 2 mol LiOH
= 40 mol LiOH 1 mol CO2
Task 9a
1. Ammonia, NH3, is widely used as a fertilizer and in many household cleaners. How many
moles of ammonia are produced when 6 mol of hydrogen gas react with an excess of nitrogen
gas?
2. The decomposition of potassium chlorate is used as a source of oxygen in the laboratory.
How many moles of potassium chlorate are needed to produce 15 mole of oxygen gas?
Given Conversion factor from
balanced equation Calculated
answer
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Conversions of Amounts in Moles to Mass (Mole-Mass Stoichiometry)
The Plan:
amount of amount of mass of
given substance unknown substance unknown substance
(mol) (mol) (g)
To solve this plan two conversion factors are required – the mole ratio of the unknown to the
given from the balanced equation and the molar mass of the unknown.
Mole ratio
Molar mass factor
(Balanced equation)
(Periodic table)
Amount of given
substance (mol)
Mass of
unknown substance (g)
x mol unknown
x Molar mass of unknown (in g)
= mol given 1 mol of unknown
GIVEN IN THE
PROBLEM
CONVERSION FACTORS
CALCULATED
Example Problem 4 (p. 306):
In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6, and oxygen
from the reaction of carbon dioxide and water. What mass, in grams, of glucose is
produced when 3.00 mol of water react with carbon dioxide?
First, write a balanced equation
6 CO2(g) + 6 H2O(l) C6H12O6(s) + 6 O2(g)
3.00 mol H2O
x 1 mol C6H12O6
x 180.18 g C6H12O6
= 90.1 g
C6H12O6 6 mol H2O 1 mol C6H12O6
Task 9b
1. When magnesium burns in air, it combines with oxygen to form magnesium oxide. What
mass in grams of magnesium oxide is produced from 2.00 mol of magnesium?
2. What mass of glucose can be produced from a photosynthesis reaction that occurs using 10
mol CO2?
Given Molar mass from the
periodic table
Conversion factor from
balanced equation Calculated
answer
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Conversion of Mass to Amounts in Moles (Mass-Mole Stoichiometry)
The Plan:
mass of amount of amount of
given substance given substance unknown substance
(g) (mol) (mol)
To solve this plan two conversion factors are required – the molar mass of the unknown and the
mole ratio.
Molar mass factor
Mole ratio
(Periodic table)
(Balanced equation)
Mass of given
substance (g)
Amount of unknown substance
(mol)
x 1 mol given
x mol unknown
= Molar mass of given (in g) mol given
GIVEN IN THE
PROBLEM
CONVERSION FACTORS
CALCULATED
Example Problem 5 (p. 309):
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of
ammonia.
NH3(g) + O2(g) NO(g) + H2O(g)
The reaction is run using 824 g ammonia and excess oxygen. How many moles of H2O are
formed?
Even though we were given the equation, it is not balanced. Balance the equation.
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
824 g NH3 x 1 mol NH3 x
6 mol H2O = 72.5 mol H2O
17.04 g NH3 4 mol NH3
Task 9c
1. How many moles or mercury(II) oxide, HgO, is required to decompose to produce 124 g of
oxygen?
Given Conversion factor from
balanced equation Molar mass from the
periodic table
Calculated
answer
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Mass-Mass Calculations
Mass-mass calculations are more practical than other mole calculations. You can never measure
moles directly but you can measure mass directly. Mass-mass calculations can be viewed as the
combination of the other types of problems.
The Plan: mass of amount of amount of amount of
given substance given substance unknown substance unknown substance
(g) (mol) (mol) (g)
To solve this plan two conversion factors are required – the molar mass of the unknown and the
mole ratio.
Molar mass factor
Mole ratio
Molar mass factor
(Periodic table)
(Balanced equation)
(Periodic table)
Mass of given
substance (g)
Amount of unknown substance
(mol)
x 1 mol given
x mol unknown
x
Molar mass of unknown (in g)
= Molar mass of given
(in g) mol given 1 mol unknown
GIVEN IN THE
PROBLEM CONVERSION FACTORS
CALCULATED
Example Problem 6 (p. 310):
Tin(II) fluoride, SnF2, is used in some toothpastes. It is made by the reaction of tin with
hydrogen fluoride according to the following equation.
Sn(s) + 2 HF(g) SnF2(s) + H2(g)
How many grams of SnF2 are produced from the reaction of 30.00 g HF with Sn?
30.00 g HF x 1 mol HF
x 1 mol SnF2 x
156.71 g SnF2 =
117.5 g SnF2 20.00 g HF 2 mol HF 1 mol SnF2
Given
Molar mass from the
periodic table Molar mass from the
periodic table
Conversion factor from
balanced equation Calculated
answer
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Task 9d
1. When copper metal is added to silver nitrate in solution, silver metal and copper(II) nitrate are
produced. What mass of silver is produced from 100.0 g Cu?
2. What mass of aluminum is produced by the decomposition of 5.0 kg Al2O3?
Other conversion factors
Since all of these stoichiometric calculations must go through a mole ratio step, any conversion
factor that can be used to change one to another can be use in a problem.
For example: molar volume of a gas 22.4 L of gas at STP
mole of gas
Example Problem 7:
32. 7 g of potassium chlorate decomposes to produce potassium chloride and oxygen. What
volume of oxygen gas will be produced?
2 KClO3(s) 2 KCl(s) + 3 O2(g)
32.7 g KClO3 x 1 mol KClO3 x
1 mol O2 x 22.4 L O2
= 5.97 L O2 122.6 g KClO3 1 mol KClO3 1 mol O2
Limiting Reactants and Percentage Yield
Limiting Reactants
In the laboratory, a reaction is rarely carried out with exactly the required amount of each of the reactants. In many cases, one or more reactants is present in excess; that is, there is more than
the exact amount required to react.
Once one of the reactants is used up, no more products can be formed. The substance that is
completely used up first in a reaction is called the limiting reactant. The limiting reactant is the
reactant that limits the amount of the other reactant that can combine and the amount of product
that can form in a chemical reaction. The substance that is not used up completely in a reaction
is called the excess reactant. A limiting reactant may also be referred to as a limiting reagent.
Molar volume of a
gas at STP
Molar mass from the
periodic table Conversion factor
from balanced
equation
Given Calculated
answer
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If I want to make 3 batches of chocolate chip cookies, but only have enough sugar to make 1
batch, then I can only make 1 batch. The sugar is the limiting reactant. The other ingredients are
excess reactants.
Example Problem 8 (p. 313):
Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen
fluoride according to the following equation.
SiO2(s) + 4 HF(g) SiF4(g) + 2 H2O(l)
If 6.0 mol HF is added to 4.5 mol SiO2, which one is the limiting reactant?
There are 2 ways you can approach this problem.
Method 1: Determine the amount of product that each of these reactants will produce. The
reactant that produces the least amount of product is the limiting reactant.
6.0 mol HF 1 mol SiF4 = 1.5 mol SiF4 produced
4 mol HF
4.5 mol SiO2 1 mol SiF4 = 4.5 mol SiF4 produced
1 mol SiO2
Since HF produced the least amount of product, it is the limiting reactant.
Method 2: Compare the two reactants. Determine from that information which is limiting.
6.0 mol HF 1 mol SiF2 = 1.5 mol SiF2 produced
4 mol HF
Since SiF2 is in excess, HF is limiting.
From here you could find out the mass of SiF4 or H2O produced.
You could also find out how much SiF2 is leftover after the limiting reactant is used up.
This is the lowest #
of moles so this
reaction stopped first.
6.0 mol of HF needs 1.5 mol of SiF2. The
problem gives 4.5 mol of SiF2 so there is
plenty of SiF2.
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Example Problem 9 (p. 314):
The black oxide of iron, Fe3O4, occurs in nature as the mineral magnetite. This substance
can also be made in the laboratory by the reaction between red-hot iron and steam
according to the following equation.
3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g)
When 36.0 g H2O is mixed with 67.0 g Fe, what mass in grams of black iron oxide is
produced? What reactant is limiting? How much of the excess reactant is left over?
Since we do not know what the limiting reactant is, let’s find that first.
36.0 g H2O 1 mol H2O 3 mol Fe 55.8 g Fe = 83.7 g Fe
18.0 g H2O 4 mol H2O 1 mol Fe
This means that 36.0 g of H2O need 83.7 g of Fe. The problem only provided 67.0 g of Fe
therefore Fe is the limiting reactant.
Now that we know that Fe is the limiting reactant, we will start the rest of the problems with the
67.0 g of Fe.
Calculation of the mass of Fe3O4
67.0 g Fe 1 mol Fe 1 mol Fe3O4 231 g Fe3O4 = 92.5 g Fe3O4
55.8 g Fe 3 mol Fe 1 mol Fe3O4
Now we need to find how much H2O is used up by the 67.0 g of Fe. Then we can subtract from
36.0 g that was given in the problem.
67.0 g Fe 1 mol Fe 4 mol H2O 18.0 g H2O = 28.8 g H2O
55.8 g Fe 3 mol Fe 1 mol H2O
This means that 67.0 g of Fe reacts with 28.8 g of H2O. The problem gave us 36.0 g of H2O. We
need to subtract.
36.0 g H2O given – 28.8 g needed = 7.2 g H2O left over.
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Task 9e
1. Zinc & sulfur react to form zinc sulfide according to the following equation.
8 Zn(s) + S8(s) 8 ZnS(s)
a. If 2.00 mol of Zn are heated with 1.00 mol of S8, identify the limiting reactant.
b. How many moles of excess reactant remain?
c. How many moles of the product are formed?
2. 90.0 g of FeCl3 reacts with 52.0 g of H2S. What is the limiting reactant? What is the mass of
HCl produced? What mass of excess reactant remains after the reaction?
Percentage Yield
So far the amounts of products calculated represent theoretical yields. The theoretical yield is
the maximum amount of product that can be produced from a given amount of reactant. This
would be the amount produced if everything went exactly as the balanced chemical equation is
written, with absolutely no waste. Of course, in most chemical reactions, the amount of product
obtained is less than the theoretical yield. There may be impurities in the reactants, or
byproducts could be produced in competing side reactions. Also many reactions do not go to
completion. This measured amount of a product obtained from a reaction is called the actual
yield of that product.
Chemists are interested in the efficiency of a reaction. The efficiency is expressed by comparing
the actual and theoretical yields. The percentage yield is the ration of the actual yield to the
theoretical yield, multiplied by 100.
percentage yield = actual yield
x 100 theoretical yield
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Example Problem 10 (p. 317):
Chlorobenzene, C6H5Cl, is used in the production of many important chemicals, such as
aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to
react benzene, C6H6, with chlorine, as represented by the following equation.
C6H6(l) + Cl2(g) C6H5Cl(l) + HCl(g)
When 36.8 g C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g. What is
the percentage yield of C6H5Cl?
You have two givens in this problem: 36.8 g of C6H6 (a reactant) and 38.8 g of C6H5Cl (a
product). You can use the 36.8 g of C6H6 to determine the theoretical yield of the problem.
36.8 g C6H6 x 1 mol C6H6 x
1 mol C6H5Cl x
112.56 g C6H5Cl = 53.0 g C6H5Cl
78.12 g C6H6 1 mol C6H6 1 mol C6H5Cl
This is the
theoretical
yield
Now that we have the theoretical yield, we can use the percentage yield formula.
percentage yield = actual yield
x 100 theoretical yield
percentage
yield =
38.8 g x 100 = 73.2%
53.0
Percent Yield
Task 9f
1. Methanol can be produced through the reaction of CO and H2 in the presence of a catalyst.
CO(g) + 2 H2(g) CH3OH(l)
If 75.0 g of CO reacts to produce 68.4 g CH3OH, what is the percentage yield of CH3OH?
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2. Aluminum reacts with excess copper(II) sulfate according to the reaction given below. If
1.85 g of Al react and the percentage yield of Cu is 56.6%, what mass of Cu is produced?
Al(s) + CuSO4(aq) Al2(SO4)3(aq) + Cu(s) (unbalanced)