1 Gilbert: Chapter 3Slide 1of 44 Chapter 3: Chemical Reactions and Earth?s Composition Dr. Charlene Hayden Oakland University CHM 157 / CHM 143 Gilbert: Chapter 3Slide 2of 44 The Mole Concept ? Since we can?t count atoms in the lab, we use a weight scale to measure macroscopic amounts. ? 1 mole = the # of atoms in exactly 12 g of pure carbon-12 ? Avogadro experimentally determined the # to be 6.022 x 1023 (known as ?Avogadro?s number?, NA) ? Think of this as a ?chemist?s dozen?: square4 1 dozen of anythingcontains 12 items square4 1 mole of anythingcontains 6.022 x 1023 items square4 Abbreviation for mole is mol Gilbert: Chapter 3Slide 3of 44 Conversions ? Avogadro?s number can be used as a conversion factor to relate number of particles (e.g., atoms, molecules, formula units, ions, etc.) and number of moles. Eggs Dozens of Eggs 2 Gilbert: Chapter 3Slide 4of 44 Terminology ? Molecular mass = the mass of one molecule of a molecular compound (in amu) ? Formula mass = the mass of one formula unit of a compound (in amu) ? Molar mass (M) = the mass of one mole of the particles that comprise a substance (in grams) ?Particle?scale masses are measured in amu, and macroscopic quantities are in grams. Gilbert: Chapter 3Slide 5of 44 Molar Mass ? Molar mass (M) = the mass of one mole of the particles that comprise a substance ? To calculate the molar mass (in grams / mole), add up the atomic mass values and change the units from amu to grams. square4 Example: 1 atom of He = 4.003 amu 1 mole of He = 4.003 g Themolar mass (M) of helium is 4.003 g/mol. Avogardro?s number is the # of amu in 1 gram. Gilbert: Chapter 3Slide 6of 44 Calculation of Molar Mass ? Count the # of each type of atom in a formula, and multiply byits atomic mass. ? Example: CO2 1C + 2O 12.01 + 2(16.00) = 44.01g/mol ? Molar mass is a conversion factor relating mass of a substance to moles of a substance. 3 Gilbert: Chapter 3Slide 7of 44 Summary of Conversions Gilbert: Chapter 3Slide 8of 44 Examples 1. What is the mass of 1.06 x 1024 molecules of carbon tetrachloride? 2. How many atoms of carbon are present in a teaspoon of table sugar (C12H22O11) with a mass of 4.16 g? Gilbert: Chapter 3Slide 9of 44 Practice 1. How many moles of Au atoms are present in 10.0 g of gold? 2. How many molecules are present in 50.0 g of methane, CH4 ? 3. How many oxygen atoms are present in 5.32 moles of chalk (CaCO3)? 4 Gilbert: Chapter 3Slide 10of 44 Chemical Equations ? Chemical equations describe the proportions of substances participating in a chemical reaction. square4 Reactants= initial substances (consumed) square4 Products= final substances (formed) Gilbert: Chapter 3Slide 11of 44 Other Symbols in Chemical Equations ? States of matter square4 Gas (g) square4 Liquid (l) square4 Solid (s) square4 Aqueous (aq) = dissolved in water ? Reaction conditions square4 High temperature (?) square4 Pressure, catalysts, etc. (written with arrow) Gilbert: Chapter 3Slide 12of 44 ? Relationships in a balanced chemical reaction: 2 H2 + O2 barb2right 2 H2O 2 molecules H2 + 1 molecule O2 barb2rightbarb2rightbarb2rightbarb2right 2 molecules H2O 2 mol H2 + 1 mol O2 barb2rightbarb2rightbarb2rightbarb2right 2 mol H2O (2 X 2.02) g H2 + 32.0 g O2 barb2rightbarb2rightbarb2rightbarb2right (2 X 18.0) g H2O square4 molecular interpretation square4 molar interpretation square4 mass interpretation Interpretation of Chemical Equations 5 Gilbert: Chapter 3Slide 13of 44 Law of Conservation of Mass ? The sum of the masses of the reactants of a chemical equation is equal to the sum of the masses of the products. ? Mass is neither created nor destroyed in an ?ordinary? chemical reaction. Before Reaction After Reaction Gilbert: Chapter 3Slide 14of 44 Balanced Chemical Equations ? Chemical equations should be ?balanced? to follow the law of conservation of mass. square4 Total mass of each element on reactant side must equal the total mass of each element on the product side. square4 Same total number of atoms of each element on each side. square4 Total charge of reactant side must equal the total charge of product side. Gilbert: Chapter 3Slide 15of 44 Rules for Balancing Equations ? Never introduce extraneous chemical speciesto balance. NO + O 2? NO2 + O ? Never change a formulafor the purpose of balancing an equation (i.e., you can?t change the subscripts!). NO + O2? NO3 ? An equation can be balanced only by adjusting the coefficientsof formulas (#?s in front of species). Given: NO + O2? NO2 6 Gilbert: Chapter 3Slide 16of 44 Balancing Equation Strategy ? Balance elements that occur in only one compoundon each side first. ? Balance free elements last (e.g. O2) ? Balance unchanged polyatomic ions (or other groups of atoms) as groups. ? Fractional coefficientsare acceptable and can be clearedat the end by multiplication. Gilbert: Chapter 3Slide 17of 44 Combustion Reactions ? A combustion reaction is one that occurs between oxygen (O2)and another substance. ? Hydrocarbons = molecular compounds composed of only hydrogen & carbon. square4 Generic chemical formula is: CXHY ? If the compound is a hydrocarbon, then theproducts of completecombustion are carbon dioxide and water. Gilbert: Chapter 3Slide 18of 44 Examples Balance the following chemical equations: CH3OH + O2 barb2right CO2 + H2O K3PO4 + Pb(C2H3O2)2 barb2right KC2H3O2 + Pb3(PO4)2 7 Gilbert: Chapter 3Slide 19of 44 Practice Balance the following chemical equations: C4H10 + O2 barb2right CO2 + H2O (NH4)2CO3 + AgNO3 barb2right NH4NO3 + Ag2CO3 Gilbert: Chapter 3Slide 20of 44 Stoichiometry ? Stoichiometry includes all the quantitative relationships involving: square4 atomic and formula masses square4 chemical formulas square4 chemical equations. ? Stoichiometryasks questions of ?how much?or ?how many?. ? Mole-to-mole ratio is a keyconversion factor(called ?stoichiometric factor?). Gilbert: Chapter 3Slide 21of 44 Stoichiometric Calculations ? Step 1: Convert massof chemical species to molesusing its molar mass. ? Step 2: Convert molesof one chemical species to molesof another using a molar ratio. Use the balanced equation to derive this! ? Step 3: Convert molesof chemical species back to massusing its molar mass. 8 Gilbert: Chapter 3Slide 22of 44 Relating the Masses of Two Reactants to Each Other. What mass of O2 is consumed in the complete combustion of 6.86 g of triethylene glycol (TEG), C6H14O4? Calculate the molar mass of triethyleneglycol (TEG): 2C6H14O4 + 15O2?12 CO2 + 14H2O Use dimensional analysisto solve the problem: (6 x 12.01) + (14 x 1.008) + (4 x 16.00) = 150.2 g/mole 6.86 g TEG × × × =11.0 g O215mol O22 mol TEG150.2 g TEG1 mol TEG . 32.00 g O2 .1 mol O 2 Example Stoichiometric Calculation Gilbert: Chapter 3Slide 23of 44 Practice Sodium carbonate reacts with hydrochloric acid to produce sodium chloride, water, and carbon dioxide. How much hydrochloric acid is required to produce 10.0 g of carbon dioxide? Gilbert: Chapter 3Slide 24of 44 Why does anyone care?? ? Example: The Chemistry of a Car?s Air Bag 6NaN3 + Fe2O3 barb2right 3 Na2O + 2 Fe + 9N2 square4 The air bag contains solid sodium azide (NaN3) and solid iron oxide (i.e., rust!). square4 A sensor detects impact & initiates reaction by a spark. square4 The nitrogen gas (N2) produced inflates the bag. ? Stoichiometryis needed to calculate the proper amount of NaN3 to fill the bag with N2. square4 too much barb2right bursts; too little barb2rightunderinflated 9 Gilbert: Chapter 3Slide 25of 44 Percent Composition ? (Mass) Percent Composition = mass percent of an element in a compound Mass % = mass of element in compound x 100 molar mass of compound ? Example: Calculate the mass % O in CO2. Mass % O = (2 x 16.00)g x 100 = 72.71% O 12.01 + (2 x 16.00) Gilbert: Chapter 3Slide 26of 44 % Composition and Empirical Formulas 1. Assumethere is 100 gof the sample, so that the% values will equal the #of grams of each element. 2. Convert gramsintomolesof each element using its atomic mass. 3. Write a ?tentative formula? using these mole values. 4. Divideeach mole value by the smallest onefound. 5. Convertany fractions to wholenumbersby multiplying all the mole values bythe same number. 6. When you have all whole numbers with no common factor, you have the ?empirical formula?. NOTE: Sig figs don?t matter here, but round off errors will lead to wrong answer. KEEP extra digits! Gilbert: Chapter 3Slide 27of 44 Determining the Empirical Formulaof a Compound from Its Mass Percent Composition Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C,9.63% Hand27.79% O. Whatisthe empirical formulaof dibutyl succinate? Step 1: Determine the mass of each element in a 100g sample. C 62.58 g H 9.63 g O 27.79 g EmpiricalFormula from Mass % 10 Gilbert: Chapter 3Slide 28of 44 Step 2: Convert masses to amounts in moles. OmolOgOmolOgn HmolHgHmolHgn CmolCgCmolCgn O H C 737.1999.15179.27 55.9008.1163.9 210.5011.12158.62 =×= =×= =×= Step 3: Write a ?tentative?formula. Step 4: Divide by the smallest mole value. C5.21H9.55O1.74 C2.99H5.49O C5.21H9.55O1.74 1.74 1.74 1.74 EmpiricalFormula from Mass % Gilbert: Chapter 3Slide 29of 44 Step 5: Convert to a small whole number ratio. Multiply ×2to get C5.98H10.98O2 The empirical formula is C6H11O2 [ C2.99H5.49O ] x 2 Since 5.49 is ~5.5, that suggests a fraction of 1/2. Step 6: Check for whole numbers and no common factors. EmpiricalFormula from Mass % Gilbert: Chapter 3Slide 30of 44 Practice Asbestos is a mineral containing magnesium, silicon, oxygen, and hydrogen. One form of asbestos, chrysotile (520.27 g/mol), has the composition 28.03% magnesium, 21.60% silicon, 1.16% hydrogen. Determine the empirical formula of chrysotile. 11 Gilbert: Chapter 3Slide 31of 44 Mass Spectrometry &Molecular Mass ? Mass spectrometers are instruments used to determine the mass of substances. ? Mass spectrometers convert atoms and molecules into ions & then separate those ions based on their mass-to-charge ratio. ? For organic compounds, can also be used as a ?fingerprint?type identification. Gilbert: Chapter 3Slide 32of 44 Mass Spectrometer Gilbert: Chapter 3Slide 33of 44 Mass Spectra 12 Gilbert: Chapter 3Slide 34of 44 Determining the Molecular Formula ? Given the molecular mass (e.g. from a mass spectrometer), then the molecular formula can be derived from an empirical formula. ? The molecular formula will always be a whole number multiple of the empirical formula. ? Compare the ?molar mass? of the empirical formula to that experimental molecular mass. Example: CH2O & molecular mass = 180.2 12.01 + (2x1.008) + 16.00 = 30.03 (180.2 / 30.03) = 6.00 barb2right C6H12O6 Gilbert: Chapter 3Slide 35of 44 Combustion Analysis CaHb + excess O2 ---> aCO2(g) + b/2H2O(l) The masses ofcarbon and hydrogen in CaHb can be determined from the mass of H2O and CO2 produced. Gilbert: Chapter 3Slide 36of 44 Empirical Formula from Combustion Data 1. Use mass of CO2 to calculate mass and moles of C. 2. Use mass of H2Oto calculate mass and moles of H. 3. Subtract masses of C and H from mass of sample to find mass of ?other? element (often O). 4. Convertmass of 3rd element intomoles. 5. Write a ?tentative formula? using the molevalues. 6. Divideeach mole value by the smallest onefound. 7. Convertany fractions to wholenumbersby multiplying all the mole values bythe same number. 8. When you have all whole numbers with no common factor, you have the ?empirical formula?. NOTE: Sig figs don?t matter here, but round off errors will lead to wrong answer. KEEP extra digits! 13 Gilbert: Chapter 3Slide 37of 44 Dimethylhydrazine is a carbon-hydrogen-nitrogen compound used in rocket fuels. When burned completely, a 0.312 g sample yields 0.458 g CO2 and 0.374 g H2O. What is the empirical formula of diemthylhydrazine? Example Gilbert: Chapter 3Slide 38of 44 The Limiting Reactant Concept ? Simple Food Analogy: Salami Sandwiches 2 Bread + 1 Salami barb2right 1 Sandwich (SW) 30 slices 20 slices ??? ? Solution: square4 Step 1:Calculate the maximum number of sandwiches each ingredient could make. square4 Step 2:Compare the two numbers. The ingredient which produces fewer sandwiches is the ?limiting reactant?. 15 SW < 20 SW . .Breadis the ?limiting reactant?! 30 bread 1 SW 2 bread= 15 SW 20 salami 1 SW 1 salamiX = 20 SW . X Gilbert: Chapter 3Slide 39of 44 Some Formal Definitions ? The ?limiting reactant? is the reactant (reagent) that is entirely consumed when a reaction goes to completion. square4 Any other reactant that is not completely consumed is called an ?excess reactant?. square4 The maximum amount of product(s) which can be formed is determined by the amount of limiting reactant. Example: The maximum number of sandwiches made was determined by the slices of bread. 14 Gilbert: Chapter 3Slide 40of 44 Determining the Limiting Reactant in a Reaction. Phosphorus trichloride, PCl3, is a commercially important compound used in the manufacture of pesticides, gasoline additives, and a number of other products. It is made by the direct combination of phosphorus and chlorine P4 (s) + 6 Cl2 (g) ?4 PCl3(l) What mass of PCl3forms in the reaction of 125 g P4with 323 g Cl2? Strategy: Calculate the amount of PCl3product each reactant couldproduce (i.e. if not limited). Limiting Reactant Example Gilbert: Chapter 3Slide 41of 44 Thus, chlorine gas is the limiting reagent, and the amount of product would be 417 g PCl3. 323 g Cl2× × × = 417 g PCl31 mol Cl270.91 g Cl 2 4 mol PCl3 6 mol Cl2 1 mol PCl3 137.3 g PCl3 125 g P4× × × = 554 g PCl31 mol P4123.9 g P 4 4 mol PCl3 1 mol P4 1 mol PCl3 137.3 g PCl3 Compare the two values: 417 g PCl3 < 554 g PCl3 Limiting Reactant Example Gilbert: Chapter 3Slide 42of 44 Practice 10.0 g of methane (CH4) is burned in 20.0 g of oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). a. What is the limiting reactant? b. How many grams of water will be produced? c. How many grams of which reactant are in excess? 15 Gilbert: Chapter 3Slide 43of 44 Percent Yield ? Theoretical Yield: the stoichiometric amount of product expected. ? Actual Yield: the experimentally determined amountof product formed Percent Yield = Actual Yield x 100% Theoretical Yield Gilbert: Chapter 3Slide 44of 44 Percent Yield Example Aluminum burns in bromine liquid, producing aluminum bromide. In a certain experiment, 6.0 g of aluminum reacted with an excess of bromine to yield 50.3 g aluminum bromide. Calculate the theoretical and percent yields for this experiment. hayden2 Microsoft PowerPoint - Gilbert_Chapter3_CH_W11.ppt
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