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3-1 *See PowerPoint Image Slides for all figures and tables pre-inserted into PowerPoint without notes. CHEMISTRY The Molecular Nature of Matter and Change.

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Presentation on theme: "3-1 *See PowerPoint Image Slides for all figures and tables pre-inserted into PowerPoint without notes. CHEMISTRY The Molecular Nature of Matter and Change."— Presentation transcript:

1 3-1 *See PowerPoint Image Slides for all figures and tables pre-inserted into PowerPoint without notes. CHEMISTRY The Molecular Nature of Matter and Change Third Edition Chapter 3 Lecture Outlines* Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 3-2 Table 3.4 Two Compounds with Molecular Formula C 2 H 6 O PropertyEthanolDimethyl Ether M (g/mol) Color Melting Point Boiling Point Density at 20 0 C Use Structural formulas and space-filling model Colorless C C 0.789g/mL(liquid) intoxicant in alcoholic beverages Colorless C C g/mL(gas) in refrigeration

3 3-3 The formation of HF gas on the macroscopic and molecular levels Figure 3.6

4 3-4 A three-level view of the chemical reaction in a flashbulb Figure 3.7

5 3-5 translate the statement Sample Problem 3.7 Balancing Chemical Equations PROBLEM: PLAN:SOLUTION: balance the atomsspecify states of matter Within the cylinders of a car’s engine, the hydrocarbon octane (C 8 H 18 ), 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 C 8 H 18 + O 2 CO 2 + H 2 O 825/29 2C 8 H O 2 16CO H 2 O 2C 8 H 18 ( l ) + 25O 2 ( g ) 16CO 2 ( g ) + 18H 2 O ( g )

6 3-6 Sample Problem 3.8Calculating 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 O 2 find mols SO 2 find g SO 2 find mols Cu 2 O find mols O 2 find kg O 2

7 3-7 SOLUTION: Sample Problem 3.8Calculating Amounts of Reactants and Products continued 2Cu 2 S( s ) + 3O 2 ( g ) 2Cu 2 O( s ) + 2SO 2 ( g ) 3mol O 2 2mol Cu 2 S 10.0mol Cu 2 S= 15.0mol O mol Cu 2 S 2mol SO 2 2mol Cu 2 S 64.07g SO 2 mol SO 2 = 641g SO kg Cu 2 O 10 3 g Cu 2 O kg Cu 2 O = 0.960kg O 2 kg O g O 2 mol Cu 2 O g Cu 2 O = 20.0mol Cu 2 O 20.0mol Cu 2 O 3mol O 2 2mol Cu 2 O 32.00g O 2 mol O 2 (a) (b) (c)

8 3-8 Sample Problem 3.9Calculating 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: 2Cu 2 S( s ) + 3O 2 ( g ) 2Cu 2 O( s ) + 2SO 2 ( g ) Cu 2 O( s ) + C( s ) 2Cu( s ) + CO( g ) 2Cu 2 O( s ) + 2C( s ) 4Cu( s ) + 2CO( g ) 2Cu 2 S( s )+3O 2 ( g )+2C(s) 4Cu( s )+2SO 2 ( g )+2CO(g)

9 3-9 Figure 3.8 Summary of the Mass-Mole-Number Relationships in a Chemical Reaction

10 3-10 Sample Problem 3.10Calculating 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(N 2 H 4 ) and dinitrogen tetraoxide(N 2 O 4 ), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x10 2 g of N 2 H 4 and 2.00x10 2 g of N 2 O 4 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 N 2 divide by M molar ratio mass of N 2 H 4 mol of N 2 H 4 mass of N 2 O 4 mol of N 2 O 4 limiting mol N 2 g N 2 multiply by M

11 3-11 An Ice Cream Sundae Analogy for Limiting Reactions Figure 3.9

12 3-12 Sample Problem 3.10Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant continued SOLUTION: N 2 H 4 ( l ) + N 2 O 4 ( l ) N 2 ( g ) + H 2 O( l ) 1.00x10 2 g N 2 H 4 = 3.12mol N 2 H 4 mol N 2 H g N 2 H mol N 2 H 4 = 4.68mol N 2 3 mol N 2 2mol N 2 H x10 2 g N 2 O 4 = 2.17mol N 2 O 4 mol N 2 O g N 2 O mol N 2 O 4 = 6.51mol N 2 3 mol N 2 mol N 2 O 4 N 2 H 4 is the limiting reactant because it produces less product, N 2, than does N 2 O mol N 2 mol N g N 2 = 131g N 2 243

13 3-13 Sample Problem 3.11Calculating 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: SiO 2 ( s ) + 3C( s ) SiC( s ) + 2CO( g ) 100.0kg SiO 2 mol SiO g SiO g SiO 2 kg SiO 2 = 1664 mol SiO 2 mol SiO 2 = mol SiC = mol SiC 40.10g SiC mol SiC kg 10 3 g = 66.73kg x100=77.0% 51.4kg 66.73kg

14 3-14 Sample Problem 3.12Calculating 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 10 3 mL = 1L = 3.96M

15 3-15 Sample Problem 3.13Calculating 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 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 Na 2 HPO L0.460moles 1 L 0.805mol Na 2 HPO g Na 2 HPO 4 mol Na 2 HPO 4 = 0.805mol Na 2 HPO 4 = 114g Na 2 HPO 4

16 3-16 Sample Problem 3.14Preparing 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 M dil xV dil = #mol solute = M conc xV conc SOLUTION: 0.80L soln = 0.020L soln 0.15mol NaCl L soln 0.12mol NaCl L soln conc 6mol = 0.12mol NaCl

17 3-17 Converting a Concentrated Solution to a Dilute Solution Figure 3.13

18 3-18 Fig. 3.11

19 3-19 Sample Problem 3.15Calculating 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

20 3-20 Sample Problem 3.15Calculating Amounts of Reactants and Products for a Reaction in Solution SOLUTION: continued Mg(OH) 2 ( s ) + 2HCl( aq ) MgCl 2 ( aq ) + 2H 2 O( l ) 0.10g Mg(OH) 2 mol Mg(OH) g 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 HCl 1L 0.10mol HCl = 3.4x10 -2 L HCl

21 3-21 Sample Problem 3.16Solving 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.

22 3-22 SOLUTION: Sample Problem 3.16Solving Limiting-Reactant Problems for Reactions in Solution continued L of Na 2 S mol Na 2 S mol HgS multiply by M mol ratio L of Hg(NO 3 ) 2 mol Hg(NO 3 ) 2 mol HgS multiply by M mol ratio Hg(NO 3 ) 2 ( aq ) + Na 2 S( aq ) HgS( s ) + 2NaNO 3 ( aq ) 0.050L Hg(NO 3 ) 2 x mol/L x 1mol HgS 1mol Hg(NO 3 ) L Hg(NO 3 ) 2 x mol/L x 1mol HgS 1mol Na 2 S = 5.0x10 -4 mol HgS= 2.0x10 -3 mol HgS Hg(NO 3 ) 2 is the limiting reagent. 5.0x10 -4 mol HgS 232.7g HgS 232.7mol HgS = 0.12g HgS

23 3-23 Laboratory Preparation of Molar Solutions Figure 3.12

24 3-24 Figure 3.14


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