CHEMISTRY
The Molecular Nature of Matter and Change
Third Edition
Chapter 3
Lecture Outlines*
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3-1
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Table 3.4 Two Compounds with Molecular Formula C2H6O
Property
3-2
Ethanol
Dimethyl Ether
M(g/mol)
46.07
46.07
Color
Colorless
Colorless
Melting Point
-1170C
-138.50C
Boiling Point
78.50C
Density at 200C
0.789g/mL(liquid)
0.00195g/mL(gas)
Use
intoxicant in alcoholic
beverages
in refrigeration
Structural formulas and
space-filling model
H
H
H
C
C
H
H
-250C
H
O H
H
C
H
H
O
C H
H
The formation of HF gas on the macroscopic and molecular levels
Figure 3.6
3-3
A three-level view of the chemical reaction in a flashbulb
3-4
Figure 3.7
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
3-5
SOLUTION:
2C8H18(l) + 25O2 (g)
16CO2 (g) + 18H2O (g)
Sample Problem 3.8
PROBLEM:
Calculating Amounts of Reactants and Products
In a lifetime, the average American uses 1750lb(794g) of copper
in coins, plumbing, and wiring. Copper is obtained from sulfide
ores, such as chalcocite, or copper(I) sulfide, by a multistage
process. After an initial grinding step, the first stage 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 10.0mol of
copper(I) sulfide?
(b) How many grams of sulfur dioxide are formed when 10.0mol
of copper(I) sulfide is roasted?
(c) How many kilograms of oxygen are required to form 2.86Kg
of copper(I) oxide?
PLAN:
write and balance equation
find mols O2
find mols SO2
find g SO2
3-6
find mols Cu2O
find mols O2
find kg O2
Sample Problem 3.8
Calculating Amounts of Reactants and Products
continued
SOLUTION:
2Cu2S(s) + 3O2(g)
(a) 10.0mol Cu2S
3mol O2
2Cu2O(s) + 2SO2(g)
= 15.0mol O2
2mol Cu2S
(b) 10.0mol Cu2S
(c) 2.86kg Cu2O
2mol SO2
64.07g SO2
2mol Cu2S
mol SO2
103g Cu2O
= 641g SO2
mol Cu2O
kg Cu2O 143.10g Cu2O
20.0mol Cu2O
3mol O2
2mol Cu2O
3-7
32.00g O2
mol O2
= 20.0mol Cu2O
kg O2
103g
O2
= 0.960kg O2
Sample Problem 3.9
PROBLEM:
Calculating Amounts of Reactants and Products in
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)
Cu2O(s) + C(s)
cancel reactants and products common
to both sides of the equations
sum the equations
3-8
2Cu(s) + CO(g)
2Cu2O(s) + 2C(s)
2Cu2S(s)+3O2(g)+2C(s)
2Cu2O(s) + 2SO2(g)
4Cu(s) + 2CO(g)
4Cu(s)+2SO2(g)+2CO(g)
Summary of the Mass-Mole-Number
Relationships in a Chemical Reaction
Figure 3.8
3-9
Sample Problem 3.10 Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
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 1.00x102g of N2H4
and 2.00x102g 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
3-10
limiting mol N2
mol of N2
g N2
An Ice Cream Sundae Analogy for Limiting Reactions
Figure 3.9
3-11
Sample Problem 3.10 Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
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
3-12
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
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 100.0kg of sand are processed, 51.4kg of SiC
are recovered. What is the percent yield of SiC in this process?
PLAN:
write balanced equation
find mol reactant & product
SOLUTION:
SiO2(s) + 3C(s)
100.0kg SiO2
SiC(s) + 2CO(g)
103g SiO2
mol SiO2
kg SiO2
60.09g SiO2
= 1664 mol SiO2
mol SiO2 = mol SiC = 1664
find g product predicted
1664mol SiC
actual yield/theoretical yield x 100
percent yield
mol SiC
51.4kg
66.73kg
3-13
40.10g SiC kg
103g
x100 =77.0%
= 66.73kg
Sample Problem 3.12 Calculating the Molarity of a Solution
PROBLEM:
PLAN:
Hydrobromic acid(HBr) is a solution of hydrogen bromide gas in
water. Calculate the molarity of hydrobromic acid solution if
455mL contains 1.80mol of hydrogen bromide.
Molarity is the number of moles of solute per liter of solution.
mol of HBr
divide by volume
concentration(mol/mL) HBr
103mL = 1L
molarity(mol/L) HBr
3-14
SOLUTION: 1.80mol HBr 1000mL
= 3.96M
455 mL soln
1L
Sample Problem 3.13 Calculating Mass of Solute in a Given Volume of
Solution
PROBLEM:
PLAN:
How many grams of solute are in 1.75L of 0.460M 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:
1.75L 0.460moles
1L
= 0.805mol Na2HPO4
0.805mol Na2HPO4 141.96g Na2HPO4
mol Na2HPO4
= 114g Na2HPO4
3-15
Sample Problem 3.14 Preparing a Dilute Solution from a Concentrated
Solution
PROBLEM:
“Isotonic saline” is a 0.15M 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
0.80L of isotomic saline from a 6.0M 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
SOLUTION:
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
3-16
0.80L soln 0.15mol NaCl = 0.12mol NaCl
L soln
L solnconc
0.12mol NaCl
= 0.020L soln
6mol
Converting a Concentrated Solution to a Dilute Solution
Figure 3.13
3-17
Fig. 3.11
3-18
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.
L HCl
mass Mg(OH)2
divide by M
mol HCl
mol Mg(OH)2
mol ratio
3-19
divide by M
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
mol Mg(OH)2
= 1.7x10-3 mol Mg(OH)2
58.33g Mg(OH)2
2 mol HCl
1.7x10-3
mol Mg(OH)2
3.4x10-3
= 3.4x10-3 mol HCl
1 mol Mg(OH)2
1L
mol HCl
0.10mol HCl
3-20
MgCl2(aq) + 2H2O(l)
= 3.4x10-2 L HCl
Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions
in Solution
PROBLEM:
PLAN:
3-21
Mercury and its compounds have many uses, from filling 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 in Sample Problem
3.10 and find the amount of product which would be made from each
reactant. We then chose the reactant which gives the lesser amount
of product.
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 0. 10 mol/L
x 1mol HgS
x 1mol HgS
1mol Hg(NO3)2
1mol Na2S
= 5.0x10-4 mol HgS
mol HgS
HgS(s) + 2NaNO3(aq)
5.0x10-4
3-22
mol HgS
232.7g HgS
232.7mol HgS
multiply by M
= 2.0x10-3 mol HgS
Hg(NO3)2 is the limiting reagent.
= 0.12g HgS
L of Na2S
mol Na2S
mol ratio
mol HgS
Laboratory Preparation of Molar Solutions
Figure 3.12
3-23
Figure 3.14
3-24