Chapter

1
Table 3.4 Two Compounds with Molecular Formula
C2H6O
Property
Ethanol
Dimethyl Ether
46.07
M(g/mol)
46.07
Colorless
Color
Colorless
-138.50C
Melting Point
-1170C
Boiling Point
78.50C
-250C
Density at 200C
0.789g/mL(liquid)
0.00195g/mL(gas)
Use
intoxicant in alcoholic beverages
in refrigeration
Structural formulas and space-filling model
2
The formation of HF gas on the macroscopic and
molecular levels
Figure 3.6
3
A three-level view of the chemical reaction in a
flashbulb
Figure 3.7
4
Sample Problem 3.7
Balancing Chemical Equations
PROBLEM
Within the cylinders of a cars 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.
translate the statement
8
25/2
9
5
Sample Problem 3.8
Calculating Amounts of Reactants and Products
PROBLEM
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 mols Cu2O
find g SO2
find mols O2
find kg O2
6
Sample Problem 3.8
Calculating Amounts of Reactants and Products
continued
SOLUTION
10.0mol Cu2S
15.0mol O2
(a)
10.0mol Cu2S
(b)
641g SO2
2.86kg Cu2O
(c)
20.0mol Cu2O
20.0mol Cu2O
0.960kg O2
7
Sample Problem 3.9
Calculating Amounts of Reactants and Products in
a Reaction Sequence
PLAN
SOLUTION
write balanced equations for each step
cancel reactants and products common to both
sides of the equations
sum the equations
8
Summary of the Mass-Mole-Number Relationships in
a Chemical Reaction
Figure 3.8
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.
divide by M
molar ratio
mol of N2
mol of N2
10
An Ice Cream Sundae Analogy for Limiting Reactions
Figure 3.9
11
Sample Problem 3.10
Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
continued
SOLUTION
2
4
3
1.00x102g N2H4
3.12mol N2H4
N2H4 is the limiting reactant because it produces
less product, N2, than does N2O4.
3.12mol N2H4
4.68mol N2
4.68mol N2
131g N2
2.00x102g N2O4
2.17mol N2O4
2.17mol N2O4
6.51mol N2
12
Sample Problem 3.11
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 100.0kg of sand are processed,
51.4kg of SiC are recovered. What is the percent
yield of SiC in this process?
PLAN
SOLUTION
write balanced equation
100.0kg SiO2
1664 mol SiO2
find mol reactant product
mol SiO2 mol SiC 1664
find g product predicted
1664mol SiC
66.73kg
actual yield/theoretical yield x 100
percent yield
x100
77.0
13
Sample Problem 3.12
Calculating the Molarity of a Solution
PROBLEM
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.
mol of HBr
SOLUTION
divide by volume
3.96M
concentration(mol/mL) HBr
103mL 1L
molarity(mol/L) HBr
14
Sample Problem 3.13
Calculating Mass of Solute in a Given Volume of
Solution
PROBLEM
How many grams of solute are in 1.75L of 0.460M
sodium monohydrogen phosphate?
PLAN
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
SOLUTION
multiply by M
1.75L
moles of solute
0.805mol Na2HPO4
multiply by M
0.805mol Na2HPO4
grams of solute
114g Na2HPO4
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
0.80L soln
0.12mol NaCl
divide by M of concentrated soln
0.12mol NaCl
0.020L soln
L of concentrated soln
16
Converting a Concentrated Solution to a Dilute
Solution
Figure 3.13
17
Fig. 3.11
18
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.
19
Sample Problem 3.15
Calculating Amounts of Reactants and Products for
a Reaction in Solution
continued
SOLUTION
0.10g Mg(OH)2
1.7x10-3 mol Mg(OH)2
1.7x10-3 mol Mg(OH)2
3.4x10-3 mol HCl
3.4x10-3 mol HCl
3.4x10-2 L HCl
20
Sample Problem 3.16
Solving Limiting-Reactant Problems for Reactions
in Solution
PROBLEM
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?
PLAN
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.
21
Sample Problem 3.16
Solving Limiting-Reactant Problems for Reactions
in Solution
continued
SOLUTION
0.050L Hg(NO3)2
0.020L Hg(NO3)2
x 0.010 mol/L
x 0. 10 mol/L
5.0x10-4 mol HgS
2.0x10-3 mol HgS
Hg(NO3)2 is the limiting reagent.
5.0x10-4 mol HgS
0.12g HgS
22
Laboratory Preparation of Molar Solutions
Figure 3.12
23
Figure 3.14