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Chapter 3
Stoichiometry of Formulas and Equations
ั ันธ์ของสูตรและสมการ
ปริมาณสมพ
3-1
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Mole - Mass Relationships in Chemical Systems
3.1 The Mole
3.2 Determining the Formula of an Unknown Compound
3.3 Writing and Balancing Chemical Equations
3.4 Calculating the Amounts of Reactant and Product
3.5 Fundamentals of Solution Stoichiometry
3-2
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mole(mol) - the amount of a substance that contains
the same number of entities as there are atoms in
exactly 12 g of carbon-12.
This amount is 6.022x1023. The number is called
Avogadro’s number and is abbreviated as N.
One mole (1 mol) contains 6.022x1023 entities
(to four significant figures)
3-3
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Figure 3.1
Counting objects of fixed relative mass.
12 red marbles @ 7g each = 84g
12 yellow marbles @4g each=48g
3-4
55.85g Fe = 6.022 x 1023 atoms Fe
32.07g S = 6.022 x 1023 atoms S
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Figure 3.2
Oxygen
32.00 g
One mole of
common
substances.
CaCO3
100.09 g
Water
18.02 g
Copper
63.55 g
3-5
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Table 3.1
Term
Summary of Mass Terminology
Definition
Isotopic mass
Mass of an isotope of an
element
Atomic mass
Average of the masses of the naturally
occurring isotopes of an element
weighted according to their
abundance
(also called
atomic weight)
Molecular
(or formula) mass
(also called
molecular weight)
Sum of the atomic masses of the atoms
(or ions) in a molecule (or formula unit)
Molar mass (M)
Mass of 1 mole of chemical entities
(atoms, ions, molecules, formula units)
(also called
gram-molecular weight)
3-6
Unit
amu
amu
amu
g/mol
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Information Contained in the Chemical Formula of
Glucose C6H12O6 ( M = 180.16 g/mol)
Table 3.2
Carbon (C)
Hydrogen (H)
Oxygen (O)
Atoms/molecule
of compound
6 atoms
12 atoms
6 atoms
Moles of atoms/
mole of compound
6 moles of
atoms
12 moles of
atoms
6 moles of
atoms
Atoms/mole of
compound
6(6.022 x 1023)
atoms
12(6.022 x 1023)
atoms
6(6.022 x 1023)
atoms
Mass/molecule
of compound
6(12.01 amu)
=72.06 amu
12(1.008 amu)
=12.10 amu
6(16.00 amu)
=96.00 amu
72.06 g
12.10 g
96.00 g
Mass/mole of
compound
3-7
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Interconverting Moles, Mass, and Number of Chemical Entities
no. of grams
Mass (g) = no. of moles x
g
1 mol
1 mol
No. of moles = mass (g) x
no. of grams
No. of entities = no. of moles x
6.022x1023 entities
1 mol
1 mol
No. of moles = no. of entities x
6.022x1023 entities
3-8
M
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Summary of the mass-mole-number relationships
for elements.
Figure 3.3
MASS(g)
of element
M (g/mol)
AMOUNT(mol)
of element
Avogadro’s
number
(atoms/mol)
3-9
ATOMS
of element
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Sample Problem 3.1
PROBLEM:
Calculating the Mass and the Number of Atoms in a
Given Number of Moles of an Element
(a) Silver (Ag) is used in jewelry and tableware but no longer in
U.S. coins. How many grams of Ag are in 0.0342 mol of Ag?
(b) Iron (Fe), the main component of steel, is the most important
metal in industrial society. How many Fe atoms are in 95.8 g of Fe?
PLAN:
(a) To convert mol of Ag to g we have to use
the #g Ag/mol Ag, the molar mass M.
SOLUTION: 0.0342 mol Ag x
107.9 g Ag = 3.69 g Ag
mol Ag
PLAN: (b) To convert g of Fe to atoms we first have
to find the #mols of Fe and then convert
mols to atoms.
mol Fe
SOLUTION: 95.8g Fe x
= 1.72 mol Fe
55.85g Fe
6.022x1023atoms Fe
1.72mol Fe x
mol Fe
3-10
= 1.04x1024 atoms Fe
amount(mol) of Ag
multiply by M of Ag
(107.9g/mol)
mass(g) of Ag
mass(g) of Fe
divide by M of Fe
(55.85g/mol)
amount(mol) of Fe
multiply by 6.022x1023
atoms/mol
atoms of Fe
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Summary of the mass-mole-number relationships for compounds.
Figure 3.3
MASS(g)
of compound
M
(g/mol) chemical
formula
AMOUNT(mol)
of compound
Avogadro’s
number
(molecules/mol)
MOLECULES
(or formula units)
of compound
3-11
AMOUNT(mol)
of elements in
compound
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Sample Problem 3.2
PROBLEM:
Calculating the Moles and Number of Formula Units
in a Given Mass of a Compound
Ammonium carbonate is white solid that decomposes with
warming. Among its many uses, it is a component of
baking powder, first extinguishers, and smelling salts. How
many formula unit are in 41.6 g of ammonium carbonate?
PLAN: After writing the formula for the
compound, we find its M by adding the
masses of the elements. Convert the given
mass, 41.6 g to mols using M and then the
mols to formula units with Avogadro’s
number.
SOLUTION:
The formula is (NH4)2CO3.
mass(g) of (NH4)2CO3
divide by M
amount(mol) of (NH4)2CO3
multiply by 6.022x1023
formula units/mol
number of (NH4)2CO3 formula units
M = (2 x 14.01 g/mol N)+(8 x 1.008 g/mol H)
+(12.01 g/mol C)+(3 x 16.00 g/mol O) = 96.09 g/mol
41.6 g (NH4)2CO3 x
3-12
mol (NH4)2CO3
96.09 g (NH4)2CO3
=
x
6.022x1023 formula units (NH4)2CO3
mol (NH4)2CO3
2.61x1023 formula units (NH4)2CO3
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Mass percent from the chemical formula
Mass % of element X =
atoms of X in formula x atomic mass of X (amu)
x 100
molecular (or formula) mass of compound (amu)
Mass % of element X =
moles of X in formula x molar mass of X (amu)
molecular (or formula) mass of compound (amu)
3-13
x 100
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Sample Problem 3.3
Calculating the Mass Percents and Masses of
Elements in a Sample of Compound
PROBLEM: Glucose (C6H12O6) is the most important nutrient in the
living cell for generating chemical potential energy.
(a) What is the mass percent of each element in glucose?
(b) How many grams of carbon are in 16.55g of glucose?
PLAN:
We have to find the total mass of
glucose and the masses of the
constituent elements in order to
relate them.
SOLUTION:
(a)
Per mole glucose there are
6 moles of C
12 moles H
6 moles O
amount(mol) of element
X in 1mol compound
multiply by M (g/mol) of X
mass(g) of X in 1mol of
compound
divide by mass(g) of
1mol of compound
mass fraction of X
multiply by 100
mass % X in compound
3-14
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Sample Problem 3.3
Calculating the Mass Percents and Masses of
Elements in a Sample of Compound
continued
1.008 g H
12.01 g C
6 mol C x
mol C
12 mol H x
= 72.06 g
C
mol H
16.00 g O
6 mol O x
= 12.096 g H
= 96.00 g O
M = 180.16 g/mol
mol O
(b)
mass percent of C =
72.06 g C
180.16 g glucose
= 0.3999 x 100 = 39.99 mass %C
12.096 g H
mass percent of H =
mass percent of O =
3-15
180.16 g glucose
96.00 g O
180.16 g glucose
= 0.06714 x 100 = 6.714 mass %H
= 0.5329 x 100 = 53.29 mass %O
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Empirical and Molecular Formulas
Empirical Formula The simplest formula for a compound that agrees with
the elemental analysis and gives rise to the smallest set
of whole numbers of atoms.
Molecular Formula The formula of the compound as it exists, it may be a
multiple of the empirical formula.
3-16
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Sample Problem 3.4
Determining the Empirical Formula from Masses
of Elements
PROBLEM: Elemental analysis of a sample of an ionic compound
showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are
the empirical formula and name of the compound?
PLAN:
Once we find the relative number of moles of each element,
we can divide by the lowest mol amount to find the relative
mol ratios (empirical formula).
mol Na
SOLUTION: 2.82 g Na x
= 0.123 mol Na
22.99 g Na
mass(g) of each element
divide by M(g/mol)
amount(mol) of each element
use # of moles as subscripts
mol Cl
4.35 g Cl x 35.45 g Cl = 0.123 mol Cl
mol O
7.83 g O x
16.00 g O
= 0.489 mol O
preliminary formula
change to integer subscripts
empirical formula
3-17
Na1 Cl1 O3.98
NaClO4
NaClO4 is sodium
perchlorate.
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Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
PROBLEM: During physical activity. lactic acid (M = 90.08 g/mol) forms
in muscle tissue and is responsible for muscle soreness.
Elemental analysis shows that this compound contains 40.0
mass% C, 6.71 mass% H, and 53.3 mass% O.
(a) Determine the empirical formula of lactic acid.
(b) Determine the molecular formula.
PLAN:
assume 100g lactic acid and find the
mass of each element
divide each mass by mol mass(M)
amount(mol) of each element
molecular formula
use # mols as subscripts
preliminary formula
convert to integer subscripts
empirical formula
3-18
divide mol mass by
mass of empirical
formula to get a
multiplier
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Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
continued
SOLUTION:
Assuming there are 100 g of lactic acid, the constituents
are
40.0 g C x mol C
12.01g C
= 3.33 mol C
C3.33
3.33
6.71 g H x
mol H
53.3 g O x
1.008 g H
16.00 g O
= 6.66 mol H
= 3.33 mol O
H6.66 O3.33
3.33 3.33
molar mass of lactate
CH2O
empirical
formula
90.08 g
3
mass of CH2O
3-19
mol O
30.03 g
C3H6O3 is the
molecular formula
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Figure 3.4
Combustion train for the determination of the
chemical composition of organic compounds.
m
m
CnHm + (n+ ) O2 = n CO(g) +
H O(g)
2
2 2
3-20
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Sample Problem 3.6
PROBLEM:
Determining a Molecular Formula
from Combustion Analysis
Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in
many natural sources especially citrus fruits. When a 1.000-g sample
of vitamin C is placed in a combustion chamber and burned, the
following data are obtained:
mass of CO2 absorber after combustion
= 85.35g
mass of CO2 absorber before combustion
= 83.85g
mass of H2O absorber after combustion
= 37.96g
mass of H2O absorber before combustion
= 37.55g
What is the molecular formula of vitamin C?
PLAN:
difference (after-before) = mass of oxidized element
find the mass of each element in its combustion product
3-21
find the mols
preliminary
formula
empirical
formula
molecular
formula
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Sample Problem 3.6
Determining a Molecular Formula from
Combustion Analysis
continued
SOLUTION:
CO2
85.35 g - 83.85 g = 1.50 g
H2O
37.96 g - 37.55 g = 0.41 g
12.01 g CO2
1.50 g CO 2 x
There are 12.01 g C per mol CO2
= 0.409 g C
44.01 g CO2
There are 2.016 g H per mol H2O
0.41 g H2O x
2.016 g H2O
= 0.046 g H
18.02 g H2O
O must be the difference:
0.409 g C
= 0.0341 mol C
12.01 g C
C1H1.3O1
3-22
1.000 g - (0.409 + 0.046) = 0.545
0.046 g H
= 0.0456 mol H
16.00 g O
1.008 g H
C3H4O3
0.545 g O
176.12 g/mol
= 2.000
88.06 g
Molecular formular = C6H8O6
= 0.0341 mol O
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Table 3.3 Some Compounds with Empirical Formula CH2O
(Composition by Mass: 40.0% C, 6.71% H, 53.3% O)
M
(g/mol)
Molecular
Formula
Whole-Number
Multiple
formaldehyde
CH2O
1
30.03
disinfectant; biological preservative
acetic acid
C2H4O2
2
60.05
acetate polymers; vinegar (5%soln)
lactic acid
C3H6O3
3
90.09
sour milk; forms in exercising muscle
erythrose
C4H8O4
4
120.10
part of sugar metabolism
ribose
C5H10O5
5
150.13
component of nucleic acids and B2
glucose
C6H12O6
6
180.16
major energy source of the cell
Name
CH2O
3-23
C2H4O2
C3H6O3
C4H8O4
Use or Function
C5H10O5
C6H12O6
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Table 3.4 Two Pairs of Constitutional Isomers
C4H10
Property
Butane
C2H6O
2-Methylpropane
Ethanol
Dimethyl Ether
M(g/mol)
58.12
58.12
46.07
46.07
Boiling Point
-0.50C
-11.060C
78.50C
-250C
Density at 200C 0.579 g/mL 0.549 g/mL
(gas)
(gas)
Structural
formulas
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
H
Space-filling
models
3-24
0.789 g/mL 0.00195 g/mL
(liquid)
(gas)
H
H
C
C
H
H
H
H
H
H
OH H
C
H
O
C
H
H
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The formation of HF gas on the macroscopic and molecular levels.
Figure 3.6
3-25
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Figure 3.7
3-26
A three-level view of the chemical reaction in a flashbulb.
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translate the statement
balance the atoms
adjust the coefficients
check the atom balance
specify states of matter
3-27
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Sample Problem 3.7
PROBLEM:
PLAN:
Balancing Chemical Equations
Within the cylinders of a car’s engine, the hydrocarbon
octane (C8H18), one of many components of gasoline, mixes
with oxygen from the air and burns to form carbon dioxide
and water vapor. Write a balanced equation for this reaction.
SOLUTION:
translate the statement
C8H18 +
balance the atoms
adjust the coefficients
check the atom balance
O2
3-28
H2O
C8H18 + 25/2 O2
8 CO2 + 9 H2O
2C8H18 + 25O2
16CO2 + 18H2O
2C8H18 + 25O2
16CO2 + 18H2O
2C8H18(l) + 25O2 (g)
specify states of matter
CO2 +
16CO2 (g) + 18H2O (g)
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Figure 3.8
Summary of the mass-mole-number relationships
in a chemical reaction.
MASS(g)
of compound A
MASS(g)
of compound B
M (g/mol) of
compound A
AMOUNT(mol)
of compound A
molar ratio from
balanced equation
Avogadro’s
number
(molecules/mol)
MOLECULES
(or formula units)
of compound A
3-29
M (g/mol) of
compound B
AMOUNT(mol)
of compound B
Avogadro’s
number
(molecules/mol)
MOLECULES
(or formula units)
of compound B
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Sample Problem 3.8
Calculating Amounts of Reactants and Products
PROBLEM: In a lifetime, the average American uses 1750 lb(794 g) of copper in
coins, plumbing, and wiring. Copper is obtained from sulfide ores,
such as chalcocite, or copper(I) sulfide, by a multistep process.
After an initial grinding, the first step is to “roast” the ore (heat it
strongly with oxygen gas) to form powdered copper(I) oxide and
gaseous sulfur dioxide.
(a) How many moles of oxygen are required to roast 10.0 mol of
copper(I) sulfide?
(b) How many grams of sulfur dioxide are formed when 10.0 mol of
copper(I) sulfide is roasted?
(c) How many kilograms of oxygen are required to form 2.86 kg of
copper(I) oxide?
PLAN:
write and balance equation
find mols O2
3-30
find mols SO2
find mols Cu2O
find g SO2
find mols O2
find kg O2
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Sample Problem 3.8
Calculating Amounts of Reactants and Products
continued
SOLUTION:
2Cu2S(s) + 3O2(g)
2Cu2O(s) + 2SO2(g)
(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?
3mol O2
(a) 10.0mol Cu2S x
= 15.0mol O2
2mol Cu2S
(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I)
sulfide is roasted?
(b) 10.0mol Cu2S x 2mol SO2 x 64.07g SO2
2mol Cu2S
mol SO2
= 641g SO2
(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?
3
(c) 2.86kg Cu2O x 10 g Cu2O x mol Cu2O
= 20.0mol Cu2O
kg Cu2O 143.10g Cu2O
20.0mol Cu2O x
3-31
3mol O2
x
2mol Cu2O
32.00g O2
mol O2
x
kg O2
103g
O2
= 0.959kg O2
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Sample Problem 3.9
PROBLEM:
Writing an Overall Equation for a Reaction Sequence
Roasting is the first step in extracting copper from chalcocite,
the ore used in the previous problem. In the next step,
copper(I) oxide reacts with powdered carbon to yield copper
metal and carbon monoxide gas. Write a balanced overall
equation for the two-step process.
PLAN:
SOLUTION:
write balanced equations for each step
2Cu2S(s) + 3O2(g)
cancel reactants and products common
to both sides of the equations
2Cu2O(s) + 2C(s)
sum the equations
3-32
2Cu2S(s)+3O2(g)+2C(s)
2Cu2O(s) + 2SO2(g)
4Cu(s) + 2CO(g)
4Cu(s)+2SO2(g)+2CO(g)
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Figure 3.10
3-33
An ice cream sundae analogy for limiting reactions.
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Table 3.5 Information Contained in a Balanced Equation
Viewed in
Terms of
Reactants
C3H8(g) + 5O2(g)
molecules 1 molecule C3H8 + 5 molecules O2
amount (mol)
1 mol C3H8 + 5 mol O2
mass (amu) 44.09 amu C3H8 + 160.00 amu O2
mass (g)
total mass (g)
3-34
44.09 g C3H8 + 160.00 g O2
204.09 g
Products
3CO2(g) + 4H2O(g)
3 molecules CO2 + 4 molecules H2O
3 mol CO2 + 4 mol H2O
132.03 amu CO2 + 72.06 amu H2O
132.03 g CO2 + 72.06 g H2O
204.09 g
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Sample Problem 3.10
PROBLEM:
Using Molecular Depictions to Solve a LimitingReactant Problem
Nuclear engineers use chlorine trifluoride in the processing
of uranium fuel for power plants. This extremely reactive
substance is formed as a gas in special metal containers by
the reaction of elemental chlorine and fluorine.
(a) Suppose the box shown at left represents a container of
the reactant mixture before the reaction occurs (with
chlorine colored green). Name the limiting reactant, and
draw the container contents after the reaction is complete.
(b) When the reaction is run again with 0.750 mol of Cl2 and 3.00
mol of F2, what mass of chlorine trifluoride will be prepared?
PLAN: Write a balanced chemical equation. Compare the number of
molecules you have to the number needed for the products.
Determine the reactant that is in excess. The other reactant is the
limiting reactant.
3-35
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Sample Problem 3.10 Using Molecular Depictions to Solve a LimitingReactant Problem
continued
SOLUTION: Cl2(g) + 3F2(g)
2ClF3(g)
(a) You need a ratio of 2 Cl and 6 F for the reaction.
You have 6 Cl and 12 F.
6 Cl would require 18 F.
12 F need only 4 Cl (2 Cl2 molecules).
There isn’t enough F, therefore it must be the limiting reactant.
F2
Cl2
You will make 4 ClF2 molecules (4 Cl, 12 F) and have
2 Cl2 molecules left over.
(b) We know the molar ratio of F2/Cl2 should be 3/1.
3.00 mol F2
0.750 mol Cl2
4
=
1
Since we find that the ratio is 4/1,
that means F2 is in excess and Cl2
is the limiting reactant.
0.750 mol Cl2 x 2 mol ClF3 x 92.5 g ClF3 =
1 mol Cl
1 mol ClF3
3-36
139 g ClF3
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Sample Problem 3.11
PROBLEM:
Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
A fuel mixture used in the early days of rocketry is composed
of two liquids, hydrazine(N2H4) and dinitrogen
tetraoxide(N2O4), which ignite on contact to form nitrogen gas
and water vapor. How many grams of nitrogen gas form when
1.00x102 g of N2H4 and 2.00x102 g of N2O4 are mixed?
PLAN: We always start with a balanced chemical equation and find the number
of mols of reactants and products which have been given.
In this case one of the reactants is in molar excess and the other will
limit the extent of the reaction.
mass of N2H4
mass of N2O4
divide by M
mol of N2H4
multiply by M
mol of N2O4
molar ratio
mol of N2
3-37
limiting mol N2
mol of N2
g N2
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Sample Problem 3.11
Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
continued
SOLUTION:
2 N2H4(l) + N2O4(l)
mol N2H4
1.00x102g N2H4 x
32.05g N2H4
3 mol N2
3.12mol N2H4 x
= 3.12 mol N2H4
N2O4 x
4.68mol N2 x
mol N2O4
92.02g N2O4
2.17mol N2O4 x
3 mol N2
mol N2O4
3-38
N2H4 is the limiting reactant
because it produces less
product, N2, than does N2O4.
= 4.68mol N2
2mol N2H4
2.00x102g
3 N2(g) + 4 H2O(l)
= 2.17mol N2O4
= 6.51mol N2
28.02g N2
mol N2
= 131g N2
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Figure 3.11
The effect of side reactions on yield.
A +B
C
(reactants)
(main product)
D
(side products)
3-39
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Sample Problem 3.12
Calculating Percent Yield
PROBLEM: Silicon carbide (SiC) is an important ceramic material that is made
by allowing sand(silicon dioxide, SiO2) to react with powdered
carbon at high temperature. Carbon monoxide is also formed.
When 100.0 kg of sand is processed, 51.4 kg of SiC is recovered.
What is the percent yield of SiC from this process?
PLAN:
SOLUTION:
write balanced equation
SiO2(s) + 3C(s)
SiC(s) + 2CO(g)
103 g SiO2
mol SiO2
x
= 1664 mol SiO2
100.0
kg
SiO
x
2
find mol reactant & product
kg SiO2 60.09 g SiO2
mol SiO2 = mol SiC = 1664
find g product predicted
1664 mol SiC x
mol SiC
actual yield/theoretical yield x 100
percent yield
51.4 kg
66.73 kg
3-40
40.10 g SiC
x 100
x
=77.0%
kg
103g
= 66.73 kg
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Sample Problem 3.13
PROBLEM:
PLAN:
Calculating the Molarity of a Solution
Glycine (H2NCH2COOH) is the simplest amino acid. What is
the molarity of an aqueous solution that contains 0.715 mol
of glycine in 495 mL?
Molarity is the number of moles of solute per liter of solution.
mol of glycine
divide by volume
concentration(mol/mL) glycine
103mL = 1L
molarity(mol/L) glycine
3-41
SOLUTION:
0.715 mol glycine
495 mL soln
x
1000mL
1L
= 1.44 M glycine
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Figure 3.12
Summary of
mass-mole-number-volume
relationships in solution.
MASS (g)
of compound
in solution
M (g/mol)
AMOUNT (mol)
of compound
in solution
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound
in solution
3-42
M (g/mol)
VOLUME (L)
of solution
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Sample Problem 3.14
A “buffered” solution maintains acidity as a reaction occurs.
In living cells phosphate ions play a key buffering role, so
biochemistry often study reactions in such solutions. How
many grams of solute are in 1.75 L of 0.460 M sodium
monohydrogen phosphate?
PROBLEM:
PLAN:
volume of soln
Calculating Mass of Solute in a Given Volume of
Solution
Molarity is the number of moles of solute per liter of solution.
Knowing the molarity and volume leaves us to find the # moles
and then the # of grams of solute. The formula for the solute is
Na2HPO4.
multiply by M
moles of solute
SOLUTION:
1.75 L x
0.460 moles
1L
multiply by M
grams of solute
3-43
0.805 mol Na2HPO4 x
= 0.805 mol Na2HPO4
141.96 g Na2HPO4
mol Na2HPO4
= 114 g Na2HPO4
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Figure 3.12
Laboratory preparation of molar solutions.
A
•Weigh the solid needed.
•Transfer the solid to a
volumetric flask that
contains about half the
final volume of solvent.
3-44
C Add solvent until the solution
reaches its final volume.
B Dissolve the solid
thoroughly by
swirling.
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Figure 3.13
3-45
Converting a concentrated solution to a dilute solution.
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Sample Problem 3.14
PROBLEM:
Preparing a Dilute Solution from a Concentrated
Solution
“Isotonic saline” is a 0.15 M aqueous solution of NaCl that
simulates the total concentration of ions found in many cellular
fluids. Its uses range from a cleaning rinse for contact lenses to
a washing medium for red blood cells. How would you prepare
0.80 L of isotomic saline from a 6.0 M stock solution?
PLAN:
It is important to realize the number of moles of solute does not
change during the dilution but the volume does. The new
volume will be the sum of the two volumes, that is, the total final
volume.
MdilxVdil = #mol solute = MconcxVconc
volume of dilute soln
multiply by M of dilute solution
moles of NaCl in dilute soln = mol NaCl
in concentrated soln
divide by M of concentrated soln
L of concentrated soln
3-46
SOLUTION:
0.80 L soln x 0.15 mol NaCl = 0.12 mol NaCl
L soln
0.12 mol NaCl x L solnconc = 0.020 L soln
6 mol
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Sample Problem 3.15
Calculating Amounts of Reactants and Products for
a Reaction in Solution
PROBLEM: Specialized cells in the stomach release HCl to aid digestion. If
they release too much, the excess can be neutralized with
antacids. A common antacid contains magnesium hydroxide,
which reacts with the acid to form water and magnesium
chloride solution. As a government chemist testing commercial
antacids, you use 0.10M HCl to simulate the acid concentration
in the stomach. How many liters of “stomach acid” react with a
tablet containing 0.10g of magnesium hydroxide?
PLAN:
Write a balanced equation for the reaction; find the grams of
Mg(OH)2; determine the mol ratio of reactants and products;
use mols to convert to molarity.
L HCl
mass Mg(OH)2
divide by M
mol Mg(OH)2
mol HCl
mol ratio
3-47
divide by M
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Sample Problem 3.15
Calculating Amounts of Reactants and Products for
a Reaction in Solution
continued
SOLUTION:
Mg(OH)2(s) + 2HCl(aq)
0.10g Mg(OH)2 x
1.7x10-3
mol Mg(OH)2
58.33g Mg(OH)2
mol Mg(OH)2 x
MgCl2(aq) + 2H2O(l)
= 1.7x10-3 mol Mg(OH)2
2 mol HCl
= 3.4x10-3 mol HCl
1 mol Mg(OH)2
3.4x10-3 mol HCl x
1L
0.10mol HCl
3-48
= 3.4x10-2 L HCl
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Sample Problem 3.16
PROBLEM:
PLAN:
3-49
Solving Limiting-Reactant Problems for Reactions
in Solution
Mercury and its compounds have many uses, from fillings for
teeth (as an alloy with silver, copper, and tin) to the industrial
production of chlorine. Because of their toxicity, however,
soluble mercury compounds, such mercury(II) nitrate, must
be removed from industrial wastewater. One removal method
reacts the wastewater with sodium sulfide solution to
produce solid mercury(II) sulfide and sodium nitrate solution.
In a laboratory simulation, 0.050L of 0.010M mercury(II) nitrate
reacts with 0.020L of 0.10M sodium sulfide. How many grams
of mercury(II) sulfide form?
As usual, write a balanced chemical reaction. Since this is a problem
concerning a limiting reactant, we proceed as we would for a limiting
reactant problem. Find the amount of product which would be made
from each reactant. Then choose the reactant that gives the lesser
amount of product.
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Sample Problem 3.16
Solving Limiting-Reactant Problems for Reactions
in Solution
continued
SOLUTION:
Hg(NO3)2(aq) + Na2S(aq)
L of Hg(NO3)2
0.050 L Hg(NO3)2
multiply by M
mol Hg(NO3)2
x
mol ratio
mol HgS
HgS(s) + 2NaNO3(aq)
0.020 L Hg(NO3)2
x 0.010 mol/L
x 0.10 mol/L multiply by M
1 mol HgS
1 mol HgS
1 mol Hg(NO3)2
= 5.0x10-4 mol HgS
x
1 mol Na2S
= 2.0x10-3 mol HgS
Hg(NO3)2 is the limiting reagent.
5.0x10-4
3-50
mol HgS x
L of Na2S
232.7g HgS
1 mol HgS
= 0.12 g HgS
mol Na2S
mol ratio
mol HgS
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Figure 3.15
3-51
An overview of the key mass-mole-number
stoichiometry relationships.