Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 3
Stoichiometry of Formulas and Equations
3-1
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 3.1
Counting objects of fixed relative mass.
12 red marbles @ 7g each = 84g
12 yellow marbles @4g each = 48g 55.85g Fe = 6.022 x 1023 atoms Fe
32.07g S = 6.022 x 1023 atoms S
3-4
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Information Contained in the Chemical Formula of Glucose
C6H12O6 ( M = 180.16 g/mol)
Table 3.1
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
12.10 g
96.00 g
Mass/mole of
compound
3-6
72.06 g
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Interconverting Moles, Mass, and Number of Chemical Entities
no. of grams
Mass (g) = no. of moles x
g
1 mol
No. of moles = mass (g) x
1 mol
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
3-7
M
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 3.3
MASS(g)
of element
Summary of the mass-molenumber relationships for
elements.
M (g/mol)
AMOUNT(mol)
of element
Avogadro’s
number
(atoms/mol)
ATOMS
of element
3-8
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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 0.0342mol of Ag?
(b) Iron (Fe), the main component of steel, is the most important
metal in industrial society. How many Fe atoms are in 95.8g 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: 0.0342mol Ag x 107.9 g Ag = 3.69g 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
= 1.72mol Fe
55.85g Fe
6.022x1023atoms Fe = 1.04x1024 atoms
1.72mol Fe x
Fe
mol Fe
3-9
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
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 3.4
MASS(g)
of compound
Summary of the mass-molenumber relationships for
compounds.
M (g/mol)
AMOUNT(mol)
of compound
chemical
formula
Avogadro’s
number
(molecules/mol)
MOLECULES
(or formula units)
of compound
3-10
AMOUNT(mol)
of elements in
compound
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.2
Calculating the Moles and Number of Formula Units
in a Given Mass of a Compound
PROBLEM:
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
mass(g) of (NH4)2CO3
compound, we find its M by adding the
masses of the elements. Convert the given
divide by M
mass, 41.6 g to mols using M and then the
amount(mol) of (NH4)2CO3
mols to formula units with Avogadro’s
multiply by 6.022x1023
formula units/mol
number of (NH4)2CO3 formula units
number.
SOLUTION: The formula is (NH4)2CO3.
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
mol (NH4)2CO3
41.6 g (NH4)2CO3 x
x
96.09 g (NH4)2CO3
6.022x1023 formula units (NH4)2CO3
mol (NH4)2CO3
2.61x1023 formula units (NH4)2CO3
3-11
=
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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)
3-12
x 100
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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:
Per mole glucose there are
(a)
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
3-13
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.3
Calculating the Mass Percents and Masses of
Elements in a Sample of Compound
continued
6 mol C x
12.01 g C
= 72.06 g C
12 mol H x
16.00 g O
= 96.00 g O
= 12.096 g H
mol H
mol C
6 mol O x
1.008 g H
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 =
3-14
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
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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.
3-15
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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).
SOLUTION: 2.82 g Na
mass(g) of each element
divide by M(g/mol)
4.35 g Cl
amount(mol) of each element
use # of moles as subscripts
preliminary formula
change to integer subscripts
empirical formula
3-16
7.83 g O
mol Na
22.99 g Na
mol Cl
35.45 g Cl
mol O
16.00 g O
Na1 Cl1 O3.98
= 0.123 mol Na
= 0.123 mol Cl
= 0.489 mol O
NaClO4
NaClO4 is sodium perchlorate.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
3-17
divide mol mass by
mass of empirical
formula to get a
multiplier
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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 mol C
12.01g C
3.33 mol C
C3.33
3.33
6.71 g H mol H
53.3 g O mol O
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
3-18
30.03 g
C3H6O3 is the
molecular formula
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 3.5
Combustion apparatus for determining formulas
of organic compounds.
m
m
CnHm + (n+ ) O2 = n CO(g) +
H O(g)
2
2 2
3-19
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.6
Determining a Molecular Formula from Combustion
Analysis
PROBLEM:
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
find the mols
3-20
preliminary
formula
empirical
formula
molecular
formula
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.6
Determining a Molecular Formula from Combustion
Analysis
continued
SOLUTION:
CO2
H2O
85.35 g-83.85 g = 1.50 g
There are 12.01 g C per mol CO2
1.50 g CO2
37.96 g-37.55 g = 0.41 g
12.01 g CO2
= 0.409 g C
44.01 g CO2
There are 2.016 g H per mol H2O. 0.41 g H2O 2.016 g H2O
18.02 g H2O
O must be the difference:
0.409 g C
= 0.0341 mol C
12.01 g C
C1H1.3O1
1.000 g - (0.409 + 0.046) = 0.545
0.046 g H
= 0.0456 mol H
1.008 g H
C3H4O3
176.12 g/mol
88.06 g
3-21
= 0.046 g H
0.545 g O
= 0.0341 mol O
16.00 g O
= 2.000
C6H8O6
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Table 3.2 Constitutional Isomers of C2H6O
Ethanol
Property
Dimethyl Ether
M(g/mol)
46.07
46.07
Boiling Point
78.50C
-250C
Density at 200C
0.789 g/mL
(liquid)
Structural
formulas
Space-filling
models
3-22
H
H
H
C
C
H
H
0.00195 g/mL
(gas)
H
H
OH
H
C
H
O
C
H
H
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 3.6
The formation of HF gas on the macroscopic and molecular levels.
3-23
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 3.7 A three-level view of the chemical reaction in a flashbulb.
3-24
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
translate the statement
balance the atoms
adjust the coefficients
check the atom balance
specify states of matter
3-25
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.7
PROBLEM:
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.
PLAN:
translate the statement
balance the atoms
C8H18 +
O2
CO2 +
H2O
C8H18 + 25/2 O2
8 CO2 + 9 H2O
adjust the coefficients
2C8H18 + 25O2
16CO2 + 18H2O
check the atom balance
2C8H18 + 25O2
16CO2 + 18H2O
specify states of matter
3-26
SOLUTION:
2C8H18(l) + 25O2 (g)
16CO2 (g) + 18H2O (g)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
3-27
M (g/mol) of
compound B
AMOUNT(mol)
of compound B
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound B
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.8
PROBLEM:
Calculating Amounts of Reactants and Products
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?
write and balance equation
PLAN:
find mols O2
find mols SO2
find g SO2
3-28
find mols Cu2O
find mols O2
find kg O2
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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?
(a)
10.0mol Cu2S
3mol O2
= 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
2mol SO2
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?
(c) 2.86kg Cu2O
103g Cu2O
mol Cu2O
kg Cu2O 143.10g Cu2O
20.0mol Cu2O
3mol O2
2mol Cu2O
3-29
32.00g O2
mol O2
= 20.0mol Cu2O
kg O2
103g
O2
= 0.959kg O2
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Table 3.3 Information Contained in a Balanced Equation
Viewed in
Terms of
molecules
amount (mol)
mass (amu)
mass (g)
total mass (g)
3-30
Reactants
C3H8(g) + 5O2(g)
Products
3CO2(g) + 4H2O(g)
1 molecule C3H8 + 5 molecules O2
3 molecules CO2 + 4 molecules H2O
1 mol C3H8 + 5 mol O2
3 mol CO2 + 4 mol H2O
44.09 amu C3H8 + 160.00 amu O2
132.03 amu CO2 + 72.06 amu H2O
44.09 g C3H8 + 160.00 g O2
132.03 g CO2 + 72.06 g H2O
204.09 g
204.09 g
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.9
Using Molecular Depictions to Solve a LimitingReactant Problem
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.
3-31
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.9
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
0.750 mol Cl2
3-32
4
=
1
Since we find that the ratio is 4/1,
that means F2 is in excess and
Cl2 is the limiting reactant.
2 mol ClF3 92.5 g ClF3
1 mol Cl
1 mol ClF3
=
139 g ClF3
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.10 Calculating Amounts of Reactant and Product in a
Limiting-Reactant Problem
PROBLEM:
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
3-33
limiting mol N2
mol of N2
g N2
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.10 Calculating Amounts of Reactant and Product in a
Limiting-Reactant Problem
continued
SOLUTION:
1.00x102g N2H4
2 N2H4(l) + N2O4(l)
mol N2H4
32.05g N2H4
3.12mol N2H4
3 mol N2
= 3.12mol N2H4
2.17mol N2O4
mol N2O4
3-34
4.68mol N2
mol
N2O4 N O
2 4
92.02g N2O4
3 mol N2
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
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 3.9
The effect of side reactions on yield.
A +B
C
(reactants)
(main product)
D
(side products)
3-35
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.11
PROBLEM:
Calculating Percent Yield
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:
write balanced equation
find mol reactant & product
SOLUTION:
SiO2(s) + 3C(s)
100.0 kg SiO2
SiC(s) + 2CO(g)
103 g SiO2
mol SiO2
kg SiO2
60.09 g SiO2
= 1664 mol SiO2
mol SiO2 = mol SiC = 1664
find g product predicted
1664 mol SiC
mol SiC
actual yield/theoretical yield x 100
51.4 kg
percent yield
3-36
40.10 g SiC kg
66.73 kg
103g
x 100 =77.0%
= 66.73 kg
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.12 Calculating the Molarity of a Solution
PROBLEM:
PLAN:
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
3-37
SOLUTION:
0.715 mol glycine
1000mL
495 mL soln
1L
= 1.44 M glycine
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 3.10
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
3-38
M (g/mol)
VOLUME (L)
of solution
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.13 Calculating Mass of Solute in a Given Volume
PROBLEM:
PLAN:
of Solution
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?
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.
volume of soln
multiply by M
moles of solute
multiply by M
grams of solute
SOLUTION:
1.75 L 0.460 moles
1L
= 0.805 mol Na2HPO4
0.805 mol Na2HPO4 141.96 g Na2HPO4
mol Na2HPO4
3-39
= 114 g Na2HPO4
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 3.11 Converting a concentrated solution to a dilute solution.
3-40
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.14 Preparing a Dilute Solution from a Concentrated
Solution
PROBLEM:
“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?
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
SOLUTION:
multiply by M of dilute solution
PLAN:
moles of NaCl in dilute soln = mol NaCl
in concentrated soln
divide by M of concentrated soln
L of concentrated soln
3-41
0.80 L soln 0.15 mol NaCl = 0.12 mol NaCl
L soln
L solnconc
0.12 mol NaCl
= 0.020 L soln
6 mol
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.15 Calculating Amounts of Reactants and Products for
a Reaction in Solution
PROBLEM:
PLAN:
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?
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.
mass Mg(OH)2
L HCl
divide by M
mol Mg(OH)2
mol HCl
mol ratio
3-42
divide by M
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
1.7x10-3
mol Mg(OH)2
3.4x10-3
2 mol HCl
1 mol Mg(OH)2
1L
mol HCl
0.10mol HCl
3-43
= 1.7x10-3 mol Mg(OH)2
58.33g Mg(OH)2
mol Mg(OH)2
MgCl2(aq) + 2H2O(l)
= 3.4x10-3 mol HCl
= 3.4x10-2 L HCl
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions
in Solution
PROBLEM:
PLAN:
3-44
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.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
0.050L Hg(NO3)2
multiply by M
mol Hg(NO3)2
mol ratio
x 0.010 mol/L
0.020L Hg(NO3)2
x 1mol HgS
1mol Hg(NO3)2
1mol Na2S
= 2.0x10-3 mol HgS
Hg(NO3)2 is the limiting reagent.
5.0x10-4
3-45
mol HgS
232.7g HgS
1 mol HgS
L of Na2S
x 0. 10 mol/L multiply by M
x 1mol HgS
= 5.0x10-4 mol HgS
mol HgS
HgS(s) + 2NaNO3(aq)
= 0.12g HgS
mol Na2S
mol ratio
mol HgS