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Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
TEKS 8E Perform stoichiometric calculations, including determination of mass
relationships between reactants and products, calculation of limiting reagents, and
percent yield.
TEKS 8E: Stoichiometric Calculations
How are mole ratios used in stoichiometric calculations?
A balanced chemical equation provides a great deal of quantitative information. It relates particles (atoms,
molecules, formula units), moles of substances, and masses. A balanced chemical equation also is
essential for all calculations involving amounts of reactants and products. For example, suppose you
know the number of moles of one substance. The balanced chemical equation enables you to determine
the number of moles of all other substances in the reaction.
Look at the balanced equation for the production of ammonia.
N2(g) + 3H2(g)  2NH3(g)
You can interpret this equation in a variety of ways, as shown in the table below.

2NH3(g)

2 atoms H and 6 atoms H
3 molecules H2

2 molecules NH3
+
3 mol H2

2 mol NH3
+
3  2.0 g H2

2  17.0 g NH3
34.0 g reactants

34.0 g products
N2(g)
+
2 atoms N2
+
1 molecule N2
+
1 mol N2
28 g N2
@STP:
3H2(g)
6 atoms H
22.4 L
+
3  22.4 L

2  22.4 L
22.4 L N2
+
67.2 L H2

44.8 L NH3
The most important interpretation of this equation is that 1 mol of nitrogen reacts with 3 mol of hydrogen
to form 2 mol of ammonia. Based on this interpretation, you can write ratios that relate moles of reactants
to moles of product. A mole ratio is a conversion factor derived from the coefficients of a balanced
chemical equation interpreted in terms of moles. In chemical calculations, mole ratios are used to convert
between a given number of moles of a reactant or product to moles of a different reactant or product.
Three mole ratios are derived from the balanced equation above:
1 mol N 2 2 mol NH 3 3 mol H 2
3 mol H 2 1 mol N 2 2 mol NH 3
Mole-Mole Calculations In the mole ratio below, W is the unknown, wanted, quantity and G is the
given quantity. The values of a and b are the coefficients from the balanced equation. Thus, a general

 as the next
 Sample Problem, is given in this way:
solution for a mole-mole problem,
such
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Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
Sample Problem: Calculating Moles of a Product How many moles of NH3
are produced when 0.60 mol of nitrogen reacts with hydrogen?
1. Analyze List the known and the unknown. The conversion is mol N2  mol NH3.
Recall the balanced equation for the production of ammonia:
N2(g) + 3H2(g)  2NH3(g)
According to the balanced equation, 1 mol N2 combines with 3 mol H2 to produce 2 mol NH3. To
determine the number of moles of NH3, the given quantity of N2 is multiplied by the form of the mole
ratio from the balanced equation that allows the given unit to cancel.
Known
• moles of nitrogen = 0.60 mol N2
Unknown
• moles of ammonia = ? mol NH3
2. Calculate Solve for the unknown.
Write the mole ratio that will allow you to convert from moles of N2 to moles of NH3.
2 mol NH 3
1 mol N 2
Multiply the given quantity of N2 by the mole ratio in order to find the moles of NH3.

0.60 mol N 2 
2 mol NH 3
 1.2 mol NH 3
1 mol N 2
3. Evaluate Does the result make sense? The ratio of 1.2 mol NH3 to 0.60 mol N2 is 2:1, as predicted by
the balanced equation.

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Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
Mass-Mass Calculations No laboratory balance can measure substances directly
in moles. Instead, the amount of a substance is usually determined by measuring its
mass in grams. From the mass of a reactant or product, the mass of any other reactant
or product in a given chemical equation can be calculated. The mole interpretation of a
balanced equation is the basis for this conversion. If the given sample is measured in
grams, then the mass can be converted to moles by using the molar mass. Then the
mole ratio from the balanced equation can be used to calculate the number of moles of the unknown. If it
is the mass of the unknown that needs to be determined, the number of moles of the unknown can be
multiplied by the molar mass. As in mole-mole calculations, the unknown can be either a reactant or a
product.
Steps for Solving a Mass-Mass Problem Mass-mass problems are solved in basically the same way as
mole-mole problems. The steps for the mass-mass conversion of any given mass (G) to any wanted mass
(W) are outlined below.
1. Change the mass of G to moles of G (mass G  mol G) by using the molar mass of G.
mass G  1 mol G  mol G
molar mass G
2. Change the moles of G to moles of W (mol G  mol W) by using the mole ratio from the balanced
equation.

mol G  b mol W  mol W
a mol G
3. Change the moles of W to grams of W (mol W  mass W) by using the molar mass of W.

mol W  molar mass W  mass W
1 mol W
The diagram below shows another way to represent the steps for doing mole-mass and mass-mole
stoichiometric calculations. For a mole-mass problem, the first conversion (from mass to moles) is
skipped. For a mass-moleproblem, the last conversion (from moles to mass) is skipped. You can use
parts of the three-step process shown below as they are appropriate to the problem you are solving.
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Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
Sample Problem: Calculating the Mass of a Product Calculate the number of
grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of
nitrogen. The balanced equation is as follows:
N2(g) + 3H2(g)  2NH3(g)
1. Analyze List the knowns and the unknown. The mass of hydrogen will be used to find the mass of
ammonia: g H2  g NH3. The coefficients of the balanced equation show that 3 mol H2 reacts with
1 mol N2 to produce 2 mol NH3. The following steps are necessary to determine the mass of ammonia:
g H2  mol H2  mol NH3  g NH3
Knowns
• mass of hydrogen = 5.40 g H2
• 2 mol NH3/3 mol H2 (from balanced equation)
• 1 mol H2 = 2.0 g H2 (molar mass)
• 1 mol NH3 = 17.0 g NH3 (molar mass)
Unknown
• mass of ammonia = ? g NH3
2. Calculate Solve for the unknown.
Start with the given quantity, and convert from mass to moles.
5.40 g H2 
1 mol H 2
2.0 g H2
Then convert from moles of reactant to moles of product by using the correct mole ratio.

5.40 g H2 
1 mol H 2 2 mol NH 3

2.0 g H2
3 mol H 2
Finish by converting from moles of product to mass of product, using the molar mass of NH3.

5.40 g H2 
1 mol H 2 2 mol NH 3 17.0 g NH3


 31 g NH3
2.0 g H2
3 mol H 2
1 mol NH 3
g H2  mol H2  mol NH3 
g NH3
3. Evaluate Does the result make sense? Because three conversion factors are involved in this solution, it
is more difficult to estimate an answer. However, because the molar mass of NH3 is substantially greater
than the molar mass of H2, the answer should have a larger mass than the given mass. The answer should
have two significant figures.
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Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
What is the general procedure for solving a
stoichiometric problem?
As you already know, you can obtain mole ratios from a balanced chemical equation.
From the mole ratios, you can calculate any measurement unit that is related to the
mole. The given quantity can be expressed in numbers of representative particles, units
of mass, or volumes of gases at STP. The problems can include mass-volume, particle-mass, and volumevolume calculations.
In a typical stoichiometric problem, the given quantity is first converted to moles. Then, the mole ratio
from the balanced equation is used to calculate the number of moles of the wanted substance. Finally, the
moles are converted to any other unit of measurement related to the unit mole, as the problem requires.
Thus far, you have learned how to use the relationship between moles and mass (1 mol = molar mass) in
solving mass-mass, mass-mole, and mole-mass stoichiometric problems. The mole-mass relationship
gives you two conversion factors.
1 mol
and molar mass
molar mass
1 mol
Recall that the mole can be related to other quantities as well. For example, 1 mol = 6.02 × 1023
representative particles, and 
1 mol of a gas = 22.4
 L at STP. These two relationships provide four more
conversion factors that you can use in stoichiometric calculations.
6.02  1023 particles
1 mol
and
1 mol
6.02  1023 particles
1 mol and 22.4 L
22.4 L
1 mol


The figure below summarizes the steps for a typical stoichiometric problem. Notice that the units of the
given quantity will not necessarily be the same as the units of the wanted quantity. For example, given the
mass of G, you might be asked to
calculate thevolume of W at STP.
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Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
Sample Problem: Calculating Molecules of a Product How many molecules
of oxygen are produced when 29.2 g of water is decomposed by electrolysis according
to this balanced equation?
electricity
2H2O(l) 

2H2 (g) + O2 (g)
1. Analyze List the knowns and the unknown. The following calculations need to be performed:
g H2O  mol H2O  mol O2  molecules O2

The appropriate mole ratio relating moles of O2 to moles of H2O from the balanced equation is
1 mol O2/2 mol H2O.
Knowns
• mass of water = 29.2 g H2O
• 1 mol O2/2 mol H2O (from balanced equation)
• 1 mol H2O = 18.0 g H2O (molar mass)
• 1 mol O2 = 6.02 × 1023 molecules O2
Unknown
• molecules of oxygen = ? molecules O2
2. Calculate Solve for the unknown.
Start with the given quantity, and convert from mass to moles.
29.2 g H2O 
1 mol H 2O
18.0 g H2O
Then, convert from moles of reactant to moles of product.

29.2 g H2O 
1 mol H 2O 1 mol O 2

18.0 g H2O 2 mol H 2O
Finish by converting from moles to molecules.
29.2 g H O  1 mol H 2O  1 mol O 2  6.02  1023 molecules O2  4.88  1023 molecules O
2
2
18.0 g H2O 2 mol H 2O
1 mol O 2
g H2O  mol H2O 

mol O2 
molecules O2
3. Evaluate Does the result make sense? The given mass of water should produce a little less than
1 mole of oxygen, or a little less than Avogadro’s number of molecules. The answer should have three
significant figures.
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Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
How do you determine limiting reagents?
Many cooks follow a recipe when making a new dish. They know that sufficient
quantities of all the ingredients must be available in order to follow the recipe.
Suppose, for example, that you are preparing to make tacos. You happen to have more
than enough meat, cheese, lettuce, tomatoes, and salsa on hand. However, you have
only two taco shells. In this case, the quantity of taco shells you have will limit the number of tacos you
can make. Thus, the taco shells are the limiting ingredient in this cooking venture. A chemist often faces a
similar situation. In a chemical reaction, an insufficient quantity of any of the reactants will limit the
amount of product that forms.
As you know, a balanced chemical equation is a chemist’s recipe. You can interpret the recipe on a
microscopic scale (interacting particles) or on a macroscopic scale (interacting moles). The coefficients
used to write the balanced equation give both the ratio of representative particles and the mole ratio.
Recall the equation for the preparation of ammonia:
N2(g) + 3H2(g)  2NH3(g)
When one molecule (mole) of N2 reacts with three molecules (moles) of H2, two molecules (moles) of
NH3 are produced. What would happen if two molecules (moles) of N2 reacted with three molecules
(moles) of H2? Would more than two molecules (moles) of NH3 be formed? The table below shows both
the particle and the mole interpretations of this problem.
Chemical Equations
N2(g)
+
3H2(g)

2NH3(g)
Microscopic recipe: 1 molecule N2 + 3 molecules H2  2 molecules NH3
Macroscopic recipe: 1 mol N2
+
3 mol H2
7

2 mol NH3
Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
Before the reaction takes place, nitrogen and hydrogen are present in a 2:3 molecule
(mole) ratio. The reaction takes place according to the balanced equation. One
molecule (mole) of N2 reacts with three molecules (moles) of H2 to produce two
molecules (moles) of NH3. At this point, all the hydrogen has been used up, and the
reaction stops. One molecule (mole) of unreacted nitrogen is left in addition to the two
molecules (moles) of NH3 that have been produced by the reaction.
In this reaction, only the hydrogen is completely used up. This reactant is the limiting reagent, or the
reactant that determines the amount of product that can be formed by a reaction. The reaction occurs only
until the limiting reagent is used up. By contrast, the reactant that is not completely used up in a reaction
is called the excess reagent. In this example, nitrogen is the excess reagent because some nitrogen
remains unreacted.
Sometimes in stoichiometric problems, the given quantities of reactants are expressed in units other than
moles. In such cases, the first step in the solution is to convert the quantity of each reactant to moles.
Then the limiting reagent can be identified. The amount of product formed in a reaction can be
determined from the given amount of limiting reagent.
Sample Problem: Determining the Limiting Reagent Copper reacts with sulfur to form copper(I)
sulfide according to the following balanced equation:
2Cu(s) + S(s)  Cu2S(s)
What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?
1. Analyze List the knowns and the unknown. The number of moles of each reactant must be found first.
The balanced equation is used to calculate the number of moles of one reactant needed to react with the
given amount of the other reactant.
Knowns
• mass of copper = 80.0 g Cu
• mass of sulfur = 25.0 g S
• 1 mol S/2 mol Cu
Unknown
• limiting reagent = ?
2. Calculate Solve for the unknown.
Start with one of the reactants and convert from mass to moles.
80.0 g Cu  1 mol Cu  1.26 mol Cu
63.5 g Cu
Then convert the mass of the other reactant to moles.

25.0 g S  1 mol S  0.779 mol S
32.1 g S

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Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
Now convert moles of Cu to moles of S needed to react with 1.26 moles of Cu.
1.26 mol Cu  1 mol S  0.630 mol S
2 mol Cu
Compare the amount of sulfur needed with the given amount of sulfur.

0.630 mol S (amount needed to react) < 0.779 mol S (given amount)
Sulfur is in excess, so copper is the limiting reagent.
3. Evaluate Does the result make sense? Since the ratio of the given mol Cu to mol S was less than the
ratio (2:1) from the balanced equation, copper should be the limiting reagent.
How do you calculate percent yield?
When a teacher gives an exam to the class, every student could get a grade of 100 percent. However, this
outcome generally does not occur. Instead, the performance of the class is usually spread over a range of
grades. Your exam grade, expressed as a percentage, is a ratio of two items. The first item is the number
of questions you answered correctly. The second is the total number of questions. The grade compares
how well you performed with how well you could have performed if you had answered all the questions
correctly. Chemists perform similar calculations in the laboratory when the product from a chemical
reaction is less than expected, based on the balanced chemical equation.
When a balanced chemical equation is used to calculate the amount of product that will form during a
reaction, the calculated value represents the theoretical yield. The theoretical yield is the maximum
amount of product that could be formed from given amounts of reactants. In contrast, the amount of
product that actually forms when the reaction is carried out in the laboratory is called the actual yield.
The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percent.
Percent yield =
actual yield
 100%
theoretical yield
Because the actual yield of a chemical reaction is often less than the theoretical yield, the percent yield is
often less than 100 percent. The percent yield is a measure of the efficiency of a reaction carried out
yield is similar to an exam score measuring your efficiency of learning or a
in the laboratory. This
batting average measuring your efficiency of hitting a baseball.
Stoichiometry and conservation of mass dictate that yields of greater than 100 percent are not possible.
However, errors and lack of knowledge in a process can cause a reaction to appear to have a yield that is
more than 100 percent. For example, if air or water leaks into a system, then more product may be formed
than expected.
Many factors can cause percent yields to be less than 100 percent. Reactions do not always go to
completion; when a reaction is incomplete, less than the calculated amount of product is formed. Impure
reactants and competing side reactions may cause unwanted products to form. Actual yield can also be
lower than the theoretical yield due to a loss of product during filtration or in transferring between
containers. In addition, if reactants or products have not been carefully measured, a percent yield of 100
percent is unlikely.
9
Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
Sample Problem: Calculating the Theoretical Yield of a Reaction Calcium
carbonate, which is found in seashells, is decomposed by heating. The balanced
equation for this reaction is

CaCO3 (s) 
CaO(s) + CO2 (g)
What is the theoretical yield of CaO if 24.8 g CaCO3 is heated?
1. Analyze List
the known and the unknown. Calculate the theoretical yield using the mass of the
reactant. The following calculations need to be performed:
g CaCO3  mol CaCO3  mol CaO  g CaO
Knowns
• mass of calcium carbonate = 24.8 g CaCO3
• 1 mol CaO/1 mol CaCO3 (from balanced equation)
Unknown
• theoretical yield = ? g CaO
2. Calculate Solve for the unknown.
Start with the mass of the reactant and convert to moles of the reactant.
24.8 g CaCO3 
1 mol CaCO3
100.1 g CaCO3
Next, convert to moles of the product using the mole ratio.

24.8 g CaCO3 
1 mol CaCO3
 1 mol CaO
100.1 g CaCO3 1 mol CaCO3
Finish by converting from moles to mass of the product.

24.8 g CaCO3 
56.1 g CaO
1 mol CaCO 3
 1 mol CaO 
 13.9 g CaO
100.1 g CaCO3 1 mol CaCO3 1 mol CaO
g CaCO3 
mol CaCO3 
mol CaO 
g CaO
3. Evaluate Does the result make sense? The mole ratio of CaO to CaCO3 is 1:1. The ratio of their
masses in the reaction should be the same as the ratio of their molar masses, which is slightly greater
than 1:2. The result of the calculations shows that the mass of CaO is slightly greater than half the mass
of CaCO3.
10
Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
Sample Problem: Calculating the Percent Yield of a Reaction What is the
percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO3 is heated?

CaCO3 (s) 
CaO(s) + CO2 (g)
1. Analyze List the knowns and the unknown. Use the equation for percent yield. The
theoretical yield for this problem was calculated in the previous Sample Problem.

Known
• actual yield = 13.1 g CaO
• theoretical yield = 13.9 g CaO (from previous Sample Problem)
Unknown
• percent yield = ? %
2. Calculate Solve for the unknown.
Substitute the values for actual yield and theoretical yield into the equation for percent yield.
percent yield =
actual yield
 100%
theoretical yield
percent yield =
13.1 g CaO
 100%  94.2%
13.9 g CaO

3. Evaluate Does the result make sense? In this example, the actual yield is slightly less than the
theoretical yield. Therefore, the percent yield should be slightly less than 100 percent.

11
Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
Lesson Check
1. Calculate Mole Relationships Between Reactants and Products The following equation
describes the formation of aluminum oxide, which is found on the surface of aluminum objects
exposed to the air.
4Al(s) + 3O2(g) → 2Al2O3(s)
a. Write the six mole ratios that can be derived from this equation.
b. How many moles of aluminum are needed to form 3.7 mol Al2O3?
2. Calculate Mass Relationships Between Reactants and Products Acetylene gas (C2H2) is
produced by adding water to calcium carbide (CaC2).
CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq)
How many grams of acetylene are produced by adding water to 5.00 g CaC2?
12
Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
3. Calculate Mass Relationships Between Reactants and Products Iron metal
(Fe) can be obtained from iron ore (Fe2O3) by the following reaction:
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
How much iron ore is needed to obtain 92.8 grams of iron metal?
4. Calculate Mass Relationships Between Reactants and Products Methanol (CH3OH) is used in
the production of many chemicals. Methanol is made by reacting carbon monoxide and hydrogen at
high temperature and pressure.
CO(g) + 2H2(g)  CH3OH(g)
Calculate the number of grams of each reactant needed to produce 64.1 g CH3OH.
5. Define What is a limiting reagent? What is an excess reagent?
_______________________________________________________________________________
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13
Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
6. Calculate the Limiting Reagent The following equation describes the complete
combustion of ethene (C2H4):
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(g)
If 2.70 mol C2H4 reacts with 6.30 mol O2, calculate the limiting reagent.
7. Calculate the Limiting Reagent In a reaction chamber, 3.0 mol of aluminum is mixed with
5.3 mol Cl2 and then reacts. The following balanced chemical equation describes the reaction:
2Al(s) + 3Cl2(g)  2AlCl3(s)
Identify the limiting reagent for the reaction. Then calculate the number of moles of excess reagent
remaining after the reaction.
8. Calculate the Limiting Reagent Hydrogen gas can be produced by the reaction of magnesium
metal with hydrochloric acid.
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
Calculate the limiting reagent when 6.00 g HCl reacts with 5.00 g Mg.
14
Name ___________________________ Class ________ Date ___________
TEKS
Chemistry
Lesson 8E
9. Calculate Percent Yield If 50.0 g of silicon dioxide is heated with an excess of
carbon, 27.9 g of silicon carbide is produced.

SiO2(s) + 3C(s) 
 SiC(s) + 2CO(g)
What is the percent yield of this reaction?

10. Calculate Percent Yield If 15.0 g of nitrogen reacts with 15.0 g of hydrogen, 10.5 g of ammonia is
produced. What is the percent yield of this reaction?
11. Calculate Percent Yield What is the percent yield if 4.65 g of copper is produced when 1.87 g of
aluminum reacts with an excess of copper(II) sulfate?
2Al(s) + 3CuSO4(aq)  Al2(SO4)3(aq) + 3Cu(s)
15