Theoretical Yields Power Points

Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 8.3
and 8.4
Book Homework: 25, 27, 29, 31,
33, 37, 43, 47, 49, 51, 53, 55 and 57
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
2 grams Mg + 1 gram O2 makes 2 g MgO
A process in which one or more substances is changed into one
or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what
happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O
8.1 Chemical Equations
A chemical equation uses chemical symbols to denote what occurs
in a chemical reaction.
NH3 + HCl → NH4Cl
Ammonia and hydrogen chloride react to produce ammonium
Each chemical species that appears to the left of the arrow is called a
NH3 + HCl → NH4Cl
Each species that appears to the right of the arrow is called a product.
NH3 + HCl → NH4Cl
Interpreting and Writing Chemical Equations
Labels are used to indicate the physical state:
(g) gas
(l) liquid
(s) solid
(aq) aqueous [dissolved in water]
NH3(g) + HCl(g) → NH4Cl(s)
SO3(g) + H2O(l) → H2SO4(aq)
Balancing Chemical Equations
Chemical equations must be balanced so that the law of conservation
of mass is obeyed.
Balancing is achieved by writing stoichiometric coefficients to the
left of the chemical formulas.
Balancing Chemical Equations
Generally, it will facilitate the balancing process if you do the
1) Change the coefficients of compounds before changing the
coefficients of elements.
2) Treat polyatomic ions that appear on both sides of the equation as
3) Count atoms and/or polyatomic ions carefully, and track their
numbers each time you change a coefficient.
8.3 Calculations with Balanced Chemical Equations
Balanced chemical equations are used to predict how much product
will form from a given amount of reactant.
2 moles of CO combine with 1 mole of O2 to produce 2 moles of CO2.
2 moles of CO is stoichiometrically equivalent to 2 moles of CO2.
Calculations with Balanced Chemical Equations
Consider the complete reaction of 3.82 moles of CO to form CO2.
Calculate the number of moles of CO2 produced.
moles CO2 produced = 3.82 mol CO 
2 mol CO2
= 3.82 mol CO2
2 mol CO
Calculations with Balanced Chemical Equations
Consider the complete reaction of 3.82 moles of CO to form CO2.
Calculate the number of moles of O2 needed.
moles O2 needed = 3.82 mol CO 
1 mol O2
= 1.91 mol O2
2 mol CO
Mass Changes in Chemical Reactions
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the
number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
Stoichiometric Ratio
Moles B
Coefficient B
Coefficient A
Moles A
The stoichiometric ratio is used to determine amounts of compounds
consumed or produced in a balanced chemical reaction
Worked Example 8.5
Urea [(NH2)2CO] is a by-product of protein metabolism. This waste product is
formed in the liver and then filtered from the blood and excreted in the urine by
the kidneys. Urea can be synthesized in the laboratory by the combination of
ammonia and carbon dioxide according to the equation
(a) Calculate the amount of urea that will be produced by the complete reaction of
5.25 moles of ammonia. (b) Determine the stoichiometric amount of carbon
dioxide required to react with 5.25 moles of ammonia.
Strategy Use the balanced chemical equation to determine the correct
stoichiometric conversion factors, and then multiply by the number of moles of
ammonia given.
Worked Example 8.5 (cont.)
(a) moles (NH2)2CO produced = 5.25 mol NH3 ×
(b) moles CO2 produced = 5.25 mol NH3 ×
1 mol (NH2)2CO
= 2.63 mol
2 mol NH3
1 mol CO2
= 2.63 mol CO2
2 mol NH3
Think About It As always, check to be sure that units cancel properly in the
calculation. Also, the balanced equation indicates that there will be fewer moles of
urea produced than ammonia consumed. Therefore, your calculated number of
moles of urea (2.63) should be smaller than the number of moles given in the
problem (5.25). Similarly, the stoichiometric coefficients in the balanced equation
are the same for carbon dioxide and urea, so your answers to this problem should
also be the same for both species.
Worked Example 8.6
Dinitrogen monoxide (N2O), also known as nitrous oxide or “laughing gas,” is
used as an anesthetic in dentistry. It is manufactured by heating ammonium
nitrate. The balanced equation is
NH4NO3(s) →
N2O(g) + 2H2O(l)
(a) Calculate the mass of ammonium nitrate that must be heated in order to
produce 10.0 g of nitrous oxide. (b) Determine the corresponding mass of water
produced in the reaction.
Strategy For part (a), use the molar mass of nitrous oxide to convert the given
mass of nitrous oxide to moles, use the appropriate stoichiometric conversion
factor to convert to moles of ammonium nitrate, and then use the molar mass of
ammonium nitrate to convert to grams of ammonium nitrate. For part (b), use the
molar mass of nitrous oxide to convert the given mass of nitrous oxide to moles,
use the stoichiometric conversion factor to convert from moles of nitrous oxide to
moles of water, and then use the molar mass of water to convert to grams of
Worked Example 8.6 (cont.)
(a) 10.0 g N2O ×
1 mol N2O
= 0.227 mol N2O
44.02 g N2O
0.227 mol N2O ×
1 mol NH4NO3
1 mol N2O
= 0.227 mol NH4NO3
Think About It Use80.05
the law
of 4conservation
of mass to check your
g NH
= 18.2 g NH4NO3 is
mol NH
the combined
3 × that
1 mol
NH4NO3 mass of both products
equal to the mass of reactant you determined in part (a). In this case
Thus, 18.2
g of ammonium
nitrate number
must beof
in order
to produce
to the appropriate
10.0 10.0
g + g of
g = 18.2 g. Remember that small differences may arise as the
result of rounding.
(b) Starting with the number of moles of nitrous oxide determined in the first step
of (a),
2 mol H2O
0.227 mol N2O ×
= 0.454 mol H2O
1 mol N2O
0.454 mol H2O ×
18.02 g H2O
= 8.18 g H2O
1 mol H2O
8.4 Limiting Reactants
The reactant used up first in a reaction is called the limiting reactant.
Excess reactants are those present in quantities greater than
necessary to react with the quantity of the limiting reactant.
CO(g) + 2H2(g) → CH3OH(l)
Limiting Reactants
Consider the reaction between 5 moles of CO and 8 moles of H2 to
produce methanol.
CO(g) + 2H2(g) → CH3OH(l)
How many moles of H2 are necessary in order for all the CO to react?
moles of H2 = 5 mol CO 
2 mol H2
= 10 mol H2
1 mol CO
How many moles of CO are necessary in order for all of the H2 to
moles of CO = 8 mol H2 
1 mol CO
= 4 mol CO
2 mol H2
10 moles of H2 required; 8 moles of H2 available; limiting reactant.
4 moles of CO required; 5 moles of CO available; excess reactant.
Worked Example 8.7
Alka-Seltzer tablets contain aspirin, sodium bicarbonate, and citric acid. When
they come into contact with water, the sodium bicarbonate (NaHCO3) and citric
acid (H3C6H5O7) react to form carbon dioxide gas, among other products.
3NaHCO3(aq) + H3C6H5O7(aq) → 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
The formation of CO2 causes the trademark fizzing when the tablets are dropped
into a glass of water. An Alka-Seltzer tablet contains 1.700 g of sodium
bicarbonate and 1.000 g citric acid. Determine, for a single tablet dissolved in
water, (a) which ingredient is the limiting reactant, (b) what mass of the excess
reactant is left over when the reaction is complete, and (c) what mass of CO2
Strategy Convert each of the reactant masses to moles. Use the balanced
equation to write the stoichiometric conversion factor and determine which
reactant is limiting. Next, determine the number of moles of excess reactant
remaining and the number of moles of CO2 produced. Finally, use the appropriate
molar masses to convert moles of excess reactant and moles of CO2 to grams.
Worked Example 8.7 (cont.)
1 mol NaHCO3
1.700 g NaHCO3 ×
= 0.02024 mol NaHCO3
84.01 g NaHCO3
1.000 g H3C6H5O7 ×
1 mol H3C6H5O7
192.12 g
= 0.005205 mol H3C6H5O7
(a) To determine which reactant is limiting, calculate the amount of citric acid
necessary to react completely with 0.02024 mol sodium bicarbonate.
1 mol H3C6H5O7
= 0.006745 mol H3C6H5O7
0.02024 mol NaHCO3× 3 mol NaHCO
The amount of H3C6H5O7 required to react with 0.02024 mol of NaHCO3 is more
than a tablet contains. Therefore, citric acid is the limiting reactant and sodium
bicarbonate is the excess reactant.
Worked Example 8.7 (cont.)
(b) To determine the mass of excess reactant (NaHCO3) left over, first calculate
the amount of NaHCO3 that will react:
0.005205 mol H3C6H5O7 ×
3 mol NaHCO3
1 mol H3C6H5O7
= 0.01562 mol NaHCO3
Thus, 0.01562 mole of NaHCO3 will be consumed, leaving 0.00462 mole
unreacted. Convert the unreacted amount to grams as follows:
84.01 g NaHCO
0.00462 mol NaHCO3 × 1 mol NaHCO 3 = 0.388 g NaHCO3
Worked Example 8.7 (cont.)
(c) To determine the mass of CO2 produced, first calculate the moles of CO2
produced from the number of moles of limiting reactant (H3C6H5O7) consumed.
3 mol CO2
CO2 your
Think About It 3 In6 a 5problem
mol as
work by calculating the amounts of the other products in the reaction.
this amount
as follows: of mass, the combined starting mass
to the to
of conservation
of the two reactants (1.700
1.000 g = 2.700 g) should equal the sum of
44.01g g+CO
= 0.6874 g CO2 this case, the
the masses of products
CO2 excess reactant. In
masses of H2O and Na3C6H5O7 produced are 0.2815 g and 1.343 g,
To summarize
acid is theislimiting
(b) (c)]
of CO
0.6874 greactant,
[from part
and (c) 0.6874 g carbon dioxide is
the amount
of excess
3 is 0.388 g [from part (b)]. The total,
0.2815 g + 1.343 g + 0.6874 g + 0.388 g, is 2.700 g, identical to the total
mass of reactants.
Limiting Reagent
The reactant that is completely consumed by the reaction
The number of bicycles that can be assembled is limited by whichever part
runs out first. In the inventory shown in this figure, wheels are that part.
Limiting Reactants
The theoretical yield is the amount of product that forms when all
the limiting reactant reacts to form the desired product.
The actual yield is the amount of product actually obtained from a
The percent yield tells what percentage the actual yield is of the
theoretical yield.
% yield =
actual yield
 100%
theoretical yield
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
x 100
Theoretical Yield
Worked Example 8.8
Aspirin, acetylsalicylic acid (C9H8O4), is the most commonly used pain reliever in
the world. It is produced by the reaction of salicylic acid (C7H6O3) and acetic
anhydride (C4H6O3) according to the following equation:
salicylic acid
acetic anhydride
+ HC2H3O2
acetylsalicylic acid
acetic acid
In a certain aspirin synthesis, 104.8 g of salicylic acid and 110.9 g of acetic
anhydride are combined. Calculate the percent yield if 105.6 g of aspirin are
Strategy Convert reactant grams to moles, and determine which is the limiting
reactant. Use the balanced equation to determine the moles of aspirin that can be
produced and convert to grams for the theoretical yield. Use this and the actual
yield given to calculate the percent yield.
Worked Example 8.8 (cont.)
104.8 g C7H6O3×
1 mol C7H6O3
138.12 g C7H6O3 = 0.7588 mol C7H6O3
110.9 g C4H6O3×
1 mol C4H6O3
102.09 g C4H6O3 = 1.086 mol C4H6O3
Because the two reactants combine in a 1:1 mole ratio, the reactant present in the
smallest number of moles (in this case, salicylic acid) is the limiting reactant.
According to the balanced equation, one mole of aspirin is produced for every
mole of salicylic acid consumed.
Therefore, the theoretical yield of aspirin is 0.7588 mol. We convert this to grams
using the molar mass of aspirin:
180.15 g C9H8O4
0.7588 mol C9H8O4×
1 mol C9H8O4 = 136.7 g C9H8O4
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
209 g CH3OH x
moles H2O
grams H2O
molar mass
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
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