Download Stoichiometry: Calculations with Chemical Formulas and Equations

Chapter 3
Stoichiometry:
Calculations with Chemical
Formulas and Equations
Stoichiometry
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Law of Conservation of Mass
“We may lay it down as an
incontestable axiom that, in all
the operations of art and nature,
nothing is created; an equal
amount of matter exists both
before and after the experiment.
Upon this principle, the whole art
of performing chemical
experiments depends.”
--Antoine Lavoisier, 1789
Stoichiometry
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3.1 Chemical Equations
Chemical equations are concise
representations of chemical reactions.
Stoichiometry
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Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Stoichiometry
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Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
Reactants appear on the left
side of the equation.
CO2 (g) + 2 H2O (g)
Stoichiometry
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Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
Products appear on the
right side of the equation.
CO2 (g) + 2 H2O (g)
Stoichiometry
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Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
The states of the reactants and products
are written in parentheses to the right of
each compound.
Stoichiometry
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Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the
equation.
Stoichiometry
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Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of
each element in a molecule.
Stoichiometry
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Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of
each element in a molecule
• Coefficients tell the number of
molecules.
Stoichiometry
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3.1 Chemical Equations
1. Indicating the States of Reactants and
Products
2. • The physical state of each reactant and
product may be added to the equation:
1. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
3. • Reaction conditions occasionally appear
above or below the reaction arrow (e.g.,
"" is often used to indicate the addition of
heat).
Stoichiometry
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3.1 Practice Problems
• Balance the following equations by providing
the missing coefficients:
1. _Fe(s) + _O2(g) → _Fe2O3(s)
• 4, 3, 2
2. _C2H4(g) + _O2(g) → _CO2(g) + _H2O(g)
• 1, 3, 2, 2
3. _Al(s) + _HCl(aq) → _AlCl3(aq) + _H2(g)
• 2, 6, 2, 3
Stoichiometry
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3.2 Reaction Types – Simple
Patterns of Chemical
Reactivity
Stoichiometry
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3.1 Combination and Decomposition
Reactions
1. In combination reactions two or more
substances react to form one product.
2. Combination reactions have more
reactants than products.
Stoichiometry
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Consider the reaction:
• 2Mg(s) + O2(g)  2MgO(s)
Since there are fewer products than reactants,
the Mg has combined with O2 to form MgO.
2. Note that the structure of the reactants has
changed.
3. Mg consists of closely packed atoms and O2
consists of dispersed molecules.
4. MgO consists of a lattice of Mg2+ and O2– ions.
1.
Stoichiometry
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Combination Reactions
• In this type of
reaction two
or more
substances
react to form
one product.
• Examples:
– 2 Mg (s) + O2 (g)  2 MgO (s)
– N2 (g) + 3 H2 (g)  2 NH3 (g)
– C3H6 (g) + Br2 (l)  C3H6Br2 (l)
Stoichiometry
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Decomposition Reactions
• In decomposition reactions one substance
undergoes a reaction to produce two or more other
substances.
• Decomposition reactions have more products than
reactants Examples:
– CaCO3 (s)  CaO (s) + CO2 (g)
– 2 KClO3 (s)  2 KCl (s) + 3 O2 (g)
– 2 NaN3 (s)  2 Na (s) + 3 N2 (g)
Stoichiometry
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Another Decomposition Reaction
• Consider the reaction that occurs in an
automobile air bag:
• 2 NaN3(s)  2Na(s) + 3N2(g)
• Since there are more products than
reactants, the sodium azide has
decomposed into sodium metal and
nitrogen gas.
Stoichiometry
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Combustion Reactions
• Combustion
reactions are rapid
reactions that
produce a flame .
• Most often involve
hydrocarbons
reacting with oxygen
in the air.
• Most combustion reactions involve the reaction of
O2(g) from air.
• Examples:
– CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
– C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Stoichiometry
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Practice exercise for 3.2
• Write balanced chemical equations for the
following reactions:
• (a) Solid mercury(II) sulfide decomposes
into its component elements when heated.
• HgS(s) → Hg(l) + S(s)
• (b) The surface of aluminum metal
undergoes a combination reaction with
oxygen in the air.
• 4 Al(s) + 3 O2(g) → 2 Al2O3(s) Stoichiometry
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Practice exercise for 3.2
•
Write the balanced equation for the reaction
that occurs when ethanol, C2H5OH(l), is burned
in air.
•
Answer: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)
Stoichiometry
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3.3 Formula
Weights
Stoichiometry
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Formula Weight (FW)
• A formula weight is the sum of the
atomic weights for the atoms in a
chemical formula.
• So, the formula weight of calcium
chloride, CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
• Formula weights are generally reported
for ionic compounds.
Stoichiometry
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3.3 Examples: Formula Weight
•
•
•
•
Example: FW (H2SO4)
= 2 (A.W.of H) + 1(A.W. of S) + 4(A.W. of O)
= 2(1.0 amu) + 32.1 amu + 4(16.0 amu)
= 98.1 amu.
• Example: MW (C6H12O6)
• = 6(12.0 amu) + 12 (1.0 amu) + 6 (16.0 amu)
•
= 180.0 amu.
Stoichiometry
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3.3 Practice Formula Weight
• Calculate the formula weight of
• (a) Al(OH)3
• Answer: (a) 78.0 amu
• (b) CH3OH.
• Answer:(b) 32.0 amu
Stoichiometry
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Molecular Weight (MW)
• A molecular weight is the sum of the
atomic weights of the atoms in a
molecule.
• For the molecule ethane, C2H6, the
molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu
Stoichiometry
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3.3 Examples: Molecular Mass
• Example: MW (C6H12O6)
• = 6(12.0 amu) + 12 (1.0 amu) + 6 (16.0 amu)
•
= 180.0 amu.
• Note: Formula weight of the repeating unit
(formula unit) is used for ionic substances.
• Example: FW (NaCl)
•
= 23.0 amu + 35.5 amu
•
= 58.5 amu.
Stoichiometry
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3.3 Percent Composition
Percentage composition is obtained by
dividing the mass contributed by each
element (number of atoms times AW) by the
formula weight of the compound and
multiplying by 100 :
(number of atoms)(atomic weight)
% element =
(FW of the compound)
x 100
Stoichiometry
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Percent Composition
So the percentage of carbon in ethane
is…C2H6
(2)(12.0 amu)
%C =
(30.0 amu)
24.0 amu
x 100
=
30.0 amu
= 80.0%
Stoichiometry
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% Mass Problem
• Calculate the percentage of nitrogen, by
mass, in Ca(NO3)2.
• %N = 28/164.1 x 100 %
• Answer: 17.1%
Stoichiometry
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3.4 Avogadro’s
Number Moles
Stoichiometry
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Avogadro’s Number
• 6.02 x 1023
• 1 mole of 12C has a
mass of 12 g.
Stoichiometry
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The Mole
• The mole (abbreviated "mol") is a
convenient measure of chemical
quantities.
• 1 mole of something = 6.0221421 x 1023
of that thing.
• This number is called Avogadro’s
number.
Stoichiometry
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Molar Mass
• By definition, the molar mass is the mass of 1
mol of a substance (i.e., g/mol).
– The molar mass of an element is the mass
number for the element that we find on the
periodic table.
– The mass of 1 mole of 12C = 12 g. Exactly.
– The molar mass of a molecule is the sum of
the molar masses of the atoms:
– Example: The molar mass of N2 = 2 x 14
(molar mass of N). = 28 g
– The formula weight (in amu’s) will be the Stoichiometry
same number as the molar mass (in g/mol).
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Using Moles
Moles provide a bridge from the molecular
scale to the real-world scale.
Stoichiometry
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Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles.
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms
or ions of each element in the compound.
Stoichiometry
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3.4 Practice Problem
• Without using a calculator, arrange the
following samples in order of increasing
number of O atoms: 1 mol H2O, 1 mol CO2, 3
 1023 molecules O3.
• Answer:
• 1 mol H2O (6  1023 O atoms)
• 3  1023 molecules O3 (9  1023 O atoms)
• 1 mol CO2 (12  1023 O atoms)
Stoichiometry
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3.4 Practice Problem
Converting Moles to Atoms
• Calculate the number of H atoms in 0.350 mol of
C6H12O6.
• 0.350 mol of C6H12O6 X 6.2 x 1023 molecule C6H12O6 X 12 H atoms
•
1 mol of C6H12O6.
1 molecule C6H12O6.
• = 2.53 x 1024 H atoms
•
•
•
•
•
How many oxygen atoms are in
(a) 0.25 mol Ca(NO3)2
Answer: (a) 9.0  1023
(b) 1.50 mol of sodium carbonate?
Answer (b) 2.71  1024
Stoichiometry
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Calculating Molar Mass
What is the mass in grams of 1.000 mol of glucose,
C6H12O6?
Solution
Analyze We are given a chemical formula and asked to
determine its molar mass.
Plan The molar mass of a substance is found by adding the
atomic weights of its component atoms.
Because glucose has a formula weight of 180.0 amu, one
mole of this substance has a mass of 180.0 g. In other words,
C6H12O6 has a molar mass of 180.0 g/mol.
Practice Exercise
Calculate the molar mass of Ca(NO3)2.
Answer: 164.1 g/mol
Stoichiometry
3.4 Practice Problem
Sample Exercise 3.10 Converting Grams to Moles
• Calculate the number of moles of glucose (C6H12O6) in
5.380 g of C6H12O6.
• The molar mass of a substance provides the factor for
converting grams to moles. The molar mass of C6H12O6
is 180.0 g/mol
• 5.380 x 1mol/180.0 g =
• Answer: .02989 mol of C6H12O6
Practice Exercise
• How many moles of sodium bicarbonate (NaHCO3) are in
508 g of NaHCO3?
Stoichiometry
• Answer: 6.05 mol NaHCO3
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3.4 Practice Problem
Converting Moles to Grams
• Calculate the mass, in grams, of 0.433 mol of calcium
nitrate
• .433 mol Ca(NO3)2 x 164.1 g Ca(NO3)2
» 1mol Ca(NO3)2
= 71.1 grams Ca(NO3)2
Practice Exercise
• What is the mass, in grams, of
• (a) 6.33 mol of NaHCO3
• (b) 3.0  10–5 mol of sulfuric acid?
• Answer: (a) 532 g, (b) 2.9  10–3 g
Stoichiometry
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3.4 Calculating the Number of Molecules and
Number of Atoms from Mass
• (a) How many glucose molecules are in 5.23 g of C6H12O6?
• 5.23 g C6H12O6 X 1 mol C6H12O6 X 6.02 x 1023 molecules
180.0 g C6H12O6 1 mol C6H12O6
• =
1.75 x 1022 molecules of C6H12O6
• (b) How many oxygen atoms are in this sample?
• 1.75 x 1022 molecules of C6H12O6 x 6 Oxygen atom
•
1molecule of C6H12O6
• = 1.05 x 1023 Oxygen atoms
Stoichiometry
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3.4 Practice Probem
Stoichiometry
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3.4 Practice Probem
Stoichiometry
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3.4 Practice Problems
Stoichiometry
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3.5 Empirical
Formulas from
Analyses
Stoichiometry
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Calculating Empirical Formulas
• Recall that the empirical formula gives the relative
number of atoms of each element in the molecule.
Finding empirical formula from mass percent data:
• We start with the mass percent of elements (i.e., empirical data)
and calculate a formula.
• Assume we start with 100 g of sample.
• The mass percent then translates as the number of grams of
each element in 100 g of sample.
• From these masses, the number of moles can be calculated
(using the atomic weights from the periodic table).
• The lowest whole-number ratio of moles is the empirical formula
Stoichiometry
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Calculating Empirical Formulas
One can calculate the empirical formula from
the percent composition.
Stoichiometry
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Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the
empirical formula of PABA.
Stoichiometry
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Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C:
H:
N:
O:
1 mol
12.01 g
1 mol
5.14 g x
1.01 g
1 mol
10.21 g x
14.01 g
1 mol
23.33 g x
16.00 g
61.31 g x
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
Stoichiometry
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Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
C:
5.105 mol
0.7288 mol
= 7.005  7
H:
5.09 mol
0.7288 mol
= 6.984  7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001  2
Stoichiometry
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Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
Stoichiometry
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Combustion Analysis
• Compounds containing C, H and O are routinely
analyzed through combustion in a chamber like this.
– C is determined from the mass of CO2 produced.
– H is determined from the mass of H2O produced.
– O is determined by difference after the C and H have been
determined.
Stoichiometry
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Elemental Analyses
Compounds
containing other
elements are
analyzed using
methods analogous
to those used for C,
H and O.
Stoichiometry
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3.5 Calculating Empirical Formula
• Ascorbic acid (vitamin C) contains
40.92% C, 4.58% H, and 54.50% O by
mass. What is the empirical formula of
ascorbic acid?
Stoichiometry
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3.5 Calculating Empirical Formula
Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H,
and 54.50% O by mass. What is the empirical formula of
ascorbic acid?
Solve We first assume, for simplicity, that we have exactly 100 g of material (although
any mass can be used). In 100 g of ascorbic acid, therefore, we have:
40. 92 g C, 4.58 g H, 54,50 g O2 .
Second, we calculate the number of moles of each element:
Third, we determine the simplest whole-number ratio of moles by dividing each
number of moles by the smallest number of moles, 3.406:
Stoichiometry
3.5 Calculating Empirical Formula
Solution (continued)
The ratio for H is too far from 1 to
attribute the difference to experimental
error; in fact, it is quite close to 1 1/3. This
suggests that if we multiply the ratio by 3,
we will obtain whole numbers:
The whole-number mole ratio gives us
the subscripts for the empirical formula
PA 5.325-g sample of methyl benzoate, a compound used in the
manufacture of perfumes, contains 3.758 g of carbon, 0.316 g of
hydrogen, and 1.251 g of oxygen. What is the empirical formula of this
substance?
Answer: C4H4O
Stoichiometry
3.5.1 Sample Problem
• An organic compound was found to contain only C, H, & Cl. When
a 1.50 sample of the compound was completely combusted in air,
3.52 g of CO2 was formed
• In a separate experiment the chlorine in a 1.00 gram sample was
converted to 1.27 g of AgCl.
• Determine the empirical formula for this compound.
• Plan: Because different sample sizes were used to analyze the
different elements, calculate mass % of each element in the sample.
– Calculate mass % C from CO2.
– Calculate Mass % Cl from AgCl.
– Get Mass % H by subtraction.
– Calculate mole ratios & the empirical formula
Stoichiometry
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3.5.1 Sample Problem(cont)
• Calculate mass % C from CO2.
• 3.52 g CO2 x 1 mol CO2 x 1 mol C x 12.01g C
•
44.01 g CO2 1 mol CO2 1 mol C =.961 g C
•
•
.961 g C
1.50 g sample
x 100 %
=
• Calculate Mass % Cl from AgCl.
• 1.27 g AgCl x 1 mol AgCl x 1 mol Cl x 35.45 g Cl
•
143.3 g AgCl 1 mol AgCl 1 mol Cl = .314 g Cl
•
•
.314 g Cl
x 100 %
= 31.4% C
Stoichiometry
•
1.00 g sample
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3.5.1 Sample Problem(cont)
•
•
•
•
Get Mass % H by subtraction.
% H = 100 – (64.04 + 31.42) = 4.54% H
Calculate mole ratios & the empirical formula
Assume 100% g Sample
• 64.04 g C x 1 mol C/12.01 g C = 5.33 mol C
• 31.24 g Cl x 1 mol Cl/35.45 g Cl = 0.886 mol Cl
• 4.54 g H x 1 mol H /1.008 g H = 4.50 mol H
5.33/0.886 = 6.02
0.886/0.886 = 1.00
4.50/ 0.886 = 5.08
• The empirical formula is C6H5Cl
Stoichiometry
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3.6 Quantitative Information from
Balanced Equations
1. The coefficients in a balanced chemical equation give
the relative numbers of molecules (or formula units)
involved in the reaction.
2. The stoichiometric coefficients in the balanced
equation may be interpreted as:
1. the relative numbers of molecules or formula units involved in
the reaction or
2. the relative numbers of moles involved in the reaction.
Stoichiometry
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3.6 Stoichiometric Calculations
The coefficients in the balanced equation give
the ratio of moles of reactants and products.
Stoichiometry
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3.6 Quantitative Information from
Balanced Equations
• The molar quantities indicated by the coefficients in
a balanced equation are called stoichiometrically
equivalent quantities.
• Stoichiometric factors (or molar ratios) may be used
to convert between quantities of reactants and
products in a reaction.
• It is important to realize that the stoichiometric
ratios are the ideal proportions in which reactants
are needed to form products.
Stoichiometry
© 2009, Prentice-Hall, Inc.
3.6 Quantitative Information from
Balanced Equations
1. A balanced reaction equation often provides more
stoichiometric factors (or molar ratios) than needed to
solve any particular stoichiometric problem. Often only
one or two of them are relevant in a given problem.
2. The number of grams of reactant cannot be directly
related to the number of grams of product.
3. To get grams of product from grams of reactant:
1. convert grams of reactant to moles of reactant (use molar mass),
2. convert moles of one reactant to moles of other reactants and
products (use the stoichiometric ratio from the balanced chemical
equation),
3. convert moles back into grams for desired product (use molar
Stoichiometry
mass).
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3.6 Stoichiometric Calculations
Starting with the
mass of Substance
A you can use the
ratio of the
coefficients of A and
B to calculate the
mass of Substance
B formed (if it’s a
product) or used (if
it’s a reactant).
Stoichiometry
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Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams.
Stoichiometry
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Sample Exercise 3.16 Calculating Amounts of Reactants and Products
How many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6?
C6H12O6(s) + 6 O2(g)→6 CO2(g) + 6 H2O(l)
Solution
Analyze We are given the mass of a reactant and are asked to determine the mass of a product in
the given equation.
Plan The general strategy, as outlined in Figure 3.13, requires three steps. First, the amount of
C6H12O6 must be converted from grams to moles. Second, we can use the balanced equation,
which relates the moles of C6H12O6 to the moles of H2O: 1 mol C6H12O6
6 mol H2O. Third, we
must convert the moles of H2O to grams.
Solve First, use the molar mass of C6H12O6 to
convert
from grams C6H12O6 to moles C6H12O6:
Second, use the balanced equation to convert moles
of C6H12O6 to moles of H2O:
Third, use the molar mass of H2O to convert from
moles of H2O to grams of H2O:
The steps can be summarized in a diagram like that
in Figure 3.13:
Stoichiometry
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3.7 Limiting
Reactants
Stoichiometry
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How Many Cookies Can I Make?
• You can make cookies
until you run out of one
of the ingredients.
• Once this family runs
out of sugar, they will
stop making cookies
(at least any cookies
you would want to eat).
Stoichiometry
© 2009, Prentice-Hall, Inc.
How Many Cookies Can I Make?
• In this example the
sugar would be the
limiting reactant,
because it will limit the
amount of cookies you
can make.
Stoichiometry
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Limiting Reagent
• The limiting reactant (or limiting
reagent) is the reactant that is entirely
consumed when the reaction goes to
completion.
Stoichiometry
Limiting Reactants
• The limiting reactant is the reactant present in
the smallest stoichiometric amount.
– In other words, it’s the reactant you’ll run out of first (in
this case, the H2).
Stoichiometry
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Limiting Reactants
In the example below, the O2 would be the
excess reagent.
Stoichiometry
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Limiting Reagent
• Zinc metal reacts with hydrochloric acid
by the following reaction.
Zn(s)  2 HCl(aq)  ZnCl 2 (aq)  H 2 ( g )
– If 0.30 mol Zn is added to hydrochloric acid
containing 0.52 mol HCl, how many moles of H2
are produced?
Stoichiometry
Limiting Reagent
• Take each reactant in turn and ask how much
product would be obtained if each were
totally consumed. The reactant that gives the
smaller amount is the limiting reagent.
1 mol H 2
0.30 mol Zn 
 0.30 mol H 2
1 mol Zn
1 mol H 2
0.52 mol HCl 
 0.26 mol H 2
2 mol HCl
• Since HCl is the limiting reagent, the amount of H2
produced must be 0.26 mol.
Stoichiometry
Theoretical and Percent Yield
• The theoretical yield of product is the
maximum amount of product that can
be obtained from given amounts of
reactants.
– The percentage yield is the actual yield
(experimentally determined) expressed as a
percentage of the theoretical yield (calculated).
actual yield
%Yield 
 100%
theoretical yield
Stoichiometry
Theoretical Yield
• The theoretical yield is the maximum
amount of product that can be made.
– In other words it’s the amount of product
possible as calculated through the
stoichiometry problem.
• This is different from the actual yield,
which is the amount one actually
produces and measures.
Stoichiometry
© 2009, Prentice-Hall, Inc.
Sample Exercise 3.18 Calculating the Amount of Product Formed from
a Limiting Reactant
The most important commercial process for converting N2 from the air into nitrogen-containing
compounds is based on the reaction of N2 and H2 to form ammonia (NH3):
N2(g) + 3 H2(g)→2 NH3(g)
How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?
Solution
Analyze We are asked to calculate the number of moles of product, NH3, given the quantities of
each reactant, N2 and H2, available in a reaction. Thus, this is a limiting reactant problem.
Plan If we assume that one reactant is completely consumed, we can calculate how much of the
second reactant is needed in the reaction. By comparing the calculated quantity with the available
amount, we can determine which reactant is limiting. We then proceed with the calculation, using
the quantity of the limiting reactant.
Solve The number of moles of H2 needed for
complete consumption of 3.0 mol of N2 is:
Because only 6.0 mol H2 is available, we will run out
of
H2 before the N2 is gone, and H2 will be the limiting
reactant. We use the quantity of the limiting reactant,
H2, to calculate the quantity of NH3 produced:
Comment The table on the right summarizes
this example:
Stoichiometry
Sample Exercise 3.18 Calculating the Amount of Product Formed from
a Limiting Reactant
Consider the reaction 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s). A mixture
of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react.
(a) Which is the limiting reactant?
(b) How many moles of AlCl3 are formed?
(c) How many moles of the excess reactant remain at the end of
the reaction?
Stoichiometry
Sample Exercise 3.19 Calculating the Amount of Product Formed from
a Limiting Reactant
Consider the following reaction that occurs in a fuel cell:
2 H2(g) + O2 (g) → 2 H2O (g)
This reaction, properly done, produces energy in the form of electricity and water. Suppose a fuel cell is set
up with 150 g of hydrogen gas and 1500 grams of oxygen gas (each measurement is given with two
significant figures). How many grams of water can be formed?
Solution
Analyze We are asked to calculate the amount of a product, given the amounts of two reactants, so this is a
limiting reactant problem.
Plan We must first identify the limiting reagent. To do so, we can calculate the number of moles of each
reactant and compare their ratio with that required by the balanced equation. We then use the quantity of the
limiting reagent to calculate the mass of water that forms.
Solve From the balanced equation, we have the following stoichiometric relations:
Using the molar mass of each substance, we can calculate the number of moles of each reactant:
Stoichiometry
Sample Exercise 3.19 Calculating the Amount of Product Formed from
a Limiting Reactant
Solution (continued)
Thus, there are more moles of H2 than O2. The coefficients in the balanced equation indicate,
however, that the reaction requires 2 moles of H2 for every 1 mole of O2. Therefore, to completely
react all the O2, we would need 2  47 = 94 moles of H2. Since there are only 75 moles of H2, H2 is
the limiting reagent. We therefore use the quantity of H2 to calculate the quantity of product
formed. We can begin this calculation
with the grams of H2, but we can save a step by starting with the moles of H2 that were calculated
previously in the exercise:
Check The magnitude of the answer seems reasonable. The units are correct, and the number of
significant figures (two) corresponds to those in the numbers of grams of the starting materials.
Comment The quantity of the limiting reagent, H2, can also be used to determine the quantity of O2
used (37.5 mol = 1200 g). The number of grams of the excess oxygen remaining at the end of the
reaction equals the starting amount minus the amount consumed in the reaction, 1500 g – 1200 g
= 300 g.
Stoichiometry
Sample Exercise 3.19 Practice Problem
Practice Exercise
A strip of zinc metal with a mass of 2.00 g is placed in an
aqueous solution containing 2.50 g of silver nitrate, causing
the following reaction to occur:
Zn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq)
(a) Which reactant is limiting?
(b) How many grams of Ag will form?
(c) How many grams of Zn(NO3)2 will form?
(d) How many grams of the excess reactant will be left at
Stoichiometry
the end of the reaction?
Sample Exercise 3.20 Calculating the Theoretical Yield and the Percent
Yield for a Reaction
Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made
commercially by a controlled reaction between cyclohexane (C6H12) and O2:
2 C6H12(l) + 5 O2(g) → 2 H2C6H8O4(l) + 2 H2O(g)
(a) Assume that you carry out this reaction starting with 25.0 g of
cyclohexane and that cyclohexane is the limiting reactant. What is the
theoretical yield of adipic acid?
(b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent
yield of adipic acid?
Solution
Analyze We are given a chemical equation and the quantity of the limiting
reactant (25.0 g of C6H12). We are asked first to calculate the theoretical
yield of a product (H2C6H8O4) and then to calculate its percent yield if only
33.5 g of the substance is actually obtained.
Plan (a) The theoretical yield, which is the calculated quantity of adipic acid
formed in the reaction, can be calculated using the following sequence of
conversions:
g C6H12 → mol C6H12 → mol H2C6H8O4 → g H2C6H8O4
Stoichiometry
Sample Exercise 3.20 Calculating the Theoretical Yield and the Percent
Yield for a Reaction
Solution (continued)
(b) The percent yield is calculated by comparing the actual yield (33.5 g) to the theoretical yield
using Equation 3.14.
Solve
Check Our answer in (a) has the appropriate magnitude, units, and significant figures. In (b) the
answer is less than 100% as necessary.
Stoichiometry
Exercise 3.20 Practice Problem
Imagine that you are working on ways to improve the
process by which iron ore containing Fe2O3 is converted
into iron. In your tests you carry out the following reaction
on a small scale:
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
(a)If you start with 150 g of Fe2O3 as the limiting reagent,
what is the theoretical yield of Fe?
(b)(b) If the actual yield of Fe in your test was 87.9 g, what
was the percent yield?
Stoichiometry
© 2009, Prentice-Hall, Inc.