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CHEMISTRYThe Molecular Nature of Matter and Change

Third Edition

Chapter 3

Lecture Outlines*

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Table 3.4 Two Compounds with Molecular Formula C2H6O

Property Ethanol Dimethyl Ether

M(g/mol)

Color

Melting Point

Boiling Point

Density at 200C

Use

Structural formulas and space-filling model

46.07

Colorless

-1170C

78.50C

0.789g/mL(liquid)

intoxicant in alcoholic beverages

46.07

Colorless

-138.50C

-250C

0.00195g/mL(gas)

in refrigeration

C CH

H

H

H

H

O H C OH

H

H

C H

H

H

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The formation of HF gas on the macroscopic and molecular levels

Figure 3.6

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A three-level view of the chemical reaction in a flashbulb

Figure 3.7

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translate the statement

Sample Problem 3.7 Balancing Chemical Equations

PROBLEM:

PLAN: SOLUTION:

balance the atoms

specify states of matter

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.

adjust the coefficients

check the atom balance

C8H18 + O2 CO2 + H2O

C8H18 + O2 CO2 + H2O825/2 9

2C8H18 + 25O2 16CO2 + 18H2O

2C8H18 + 25O2 16CO2 + 18H2O

2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g)

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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 g SO2

find mols Cu2O

find mols O2 find kg O2

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SOLUTION:

Sample Problem 3.8 Calculating Amounts of Reactants and Products

continued

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

3mol O2

2mol Cu2S10.0mol Cu2S = 15.0mol O2

10.0mol Cu2S2mol SO2

2mol Cu2S

64.07g SO2

mol SO2

= 641g SO2

2.86kg Cu2O103g Cu2O

kg Cu2O

= 0.960kg O2

kg O2

103g O2

mol Cu2O

143.10g Cu2O= 20.0mol Cu2O

20.0mol Cu2O3mol O2

2mol Cu2O

32.00g O2

mol O2

(a)

(b)

(c)

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Sample Problem 3.9 Calculating Amounts of Reactants and Products in a Reaction Sequence

PROBLEM: 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:

write balanced equations for each step

cancel reactants and products common to both sides of the equations

sum the equations

SOLUTION:

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

Cu2O(s) + C(s) 2Cu(s) + CO(g)

2Cu2O(s) + 2C(s) 4Cu(s) + 2CO(g)

2Cu2S(s)+3O2(g)+2C(s) 4Cu(s)+2SO2(g)+2CO(g)

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Figure 3.8

Summary of the Mass-Mole-NumberRelationships in a Chemical Reaction

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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.

mol of N2 mol of N2

divide by M

molar ratio

mass of N2H4

mol of N2H4

mass of N2O4

mol of N2O4

limiting mol N2

g N2

multiply by M

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An Ice Cream Sundae Analogy for Limiting Reactions

Figure 3.9

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Sample Problem 3.10 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant

continued

SOLUTION: N2H4(l) + N2O4(l) N2(g) + H2O(l)

1.00x102g N2H4 = 3.12mol N2H4

mol N2H4

32.05g N2H4

3.12mol N2H4 = 4.68mol N2

3 mol N2

2mol N2H4

2.00x102g N2O4 = 2.17mol N2O4

mol N2O4

92.02g N2O4

2.17mol N2O4 = 6.51mol N2

3 mol N2

mol N2O4

N2H4 is the limiting reactant

because it produces less product, N2, than does N2O4.

4.68mol N2mol N2

28.02g N2 = 131g N2

2 43

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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:

write balanced equation

find mol reactant & product

find g product predicted

percent yield

actual yield/theoretical yield x 100

SOLUTION:

SiO2(s) + 3C(s) SiC(s) + 2CO(g)

100.0kg SiO2 mol SiO2

60.09g SiO2

103g SiO2

kg SiO2 = 1664 mol SiO2

mol SiO2 = mol SiC = 1664

1664mol SiC40.10g SiC

mol SiC

kg

103g= 66.73kg

x100 =77.0%51.4kg

66.73kg

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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

concentration(mol/mL) HBr

molarity(mol/L) HBr

SOLUTION:

PLAN: Molarity is the number of moles of solute per liter of solution.

1.80mol HBr

455 mL soln

1000mL

1 L

divide by volume

103mL = 1L

= 3.96M

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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?

volume of soln

moles of solute

grams of solute

multiply by M

multiply by M

PLAN:

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.

1.75L 0.460moles

1 L

0.805mol Na2HPO4 141.96g Na2HPO4

mol Na2HPO4

= 0.805mol Na2HPO4

= 114g Na2HPO4

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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.

volume of dilute soln

moles of NaCl in dilute soln = mol NaCl in concentrated soln

L of concentrated soln

multiply by M of dilute solution

divide by M of concentrated soln

MdilxVdil = #mol solute = MconcxVconc

SOLUTION:

0.80L soln

= 0.020L soln

0.15mol NaCl

L soln

0.12mol NaCl L solnconc

6mol

= 0.12mol NaCl

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Converting a Concentrated Solution to a Dilute SolutionFigure 3.13

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Fig. 3.11

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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.

mass Mg(OH)2

divide by M

mol Mg(OH)2

mol ratio

mol HCl

L HCl

divide by M

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Sample Problem 3.15 Calculating Amounts of Reactants and Products for a Reaction in Solution

SOLUTION:

continued

Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)

0.10g Mg(OH)2

mol Mg(OH)2

58.33g Mg(OH)2

= 1.7x10-3 mol Mg(OH)2

1.7x10-3 mol Mg(OH)2

2 mol HCl

1 mol Mg(OH)2

= 3.4x10-3 mol HCl

3.4x10-3 mol HCl1L

0.10mol HCl= 3.4x10-2 L HCl

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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.

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SOLUTION:

Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions in Solution

continued

L of Na2S

mol Na2S

mol HgS

multiply by M

mol ratio

L of Hg(NO3)2

mol Hg(NO3)2

mol HgS

multiply by M

mol ratio

Hg(NO3)2(aq) + Na2S(aq) HgS(s) + 2NaNO3(aq)

0.050L Hg(NO3)2

x 0.010 mol/L

x 1mol HgS

1mol Hg(NO3)2

0.020L Hg(NO3)2

x 0. 10 mol/L

x 1mol HgS

1mol Na2S

= 5.0x10-4 mol HgS = 2.0x10-3 mol HgS

Hg(NO3)2 is the limiting reagent.

5.0x10-4 mol HgS232.7g HgS

232.7mol HgS= 0.12g HgS

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Laboratory Preparation of Molar Solutions

Figure 3.12

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Figure 3.14