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Chapter 3 Stoichiometry 3-1 Dr. Wolf’s CHM 101 Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating the Amounts of Reactant and Product 3.5 Fundamentals of Solution Stoichiometry 3-2 Dr. Wolf’s CHM 101 The Mole mole - the amount of a substance that contains the same number of entities as there are atoms in exactly 12g of carbon-12, i.e. the numerical value of the atom’s mass in grams This amount is 6.022x1023. The number is called Avogadro’s number and is abbrieviated as N. One mole (1 mol) contains 6.022x1023 entities (to four significant figures) 3-3 Dr. Wolf’s CHM 101 Counting Objects of Fixed Relative Mass 12 red marbles @ 7g each = 84g 12 yellow marbles @4g each=48g 55.85g Fe = 6.022 x 1023 atoms Fe 32.07g S = 6.022 x 1023 atoms S So equal numbers will always have the same 7:4 ratio = 84:48 These values come from the atomic mass values for Fe and S in the Periodic Table 3-4 Dr. Wolf’s CHM 101 One Mole of Common Substances CaCO3 100.09 g Oxygen, O2 32.00 g Water, H2O 18.02 g Copper 63.55 g 3-5 Dr. Wolf’s CHM 101 Table 3.1 Summary of Mass Terminology Term Definition Unit Isotopic mass Mass of an isotope of an element amu Atomic mass Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance amu Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) amu (also called atomic weight) Molecular (or formula) mass (also called molecular weight) Molar mass (M) Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units) (also called gram-molecular weight) 3-6 Dr. Wolf’s CHM 101 g/mol Calculating the Molar Mass of a Substance For monatomic elements, the molar mass is the numerical value on the periodic table expressed in g/mol For molecules, the molar mass is the sum of the molar masses of each of the atoms in the molecular formula. 3-7 Dr. Wolf’s CHM 101 Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/ mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms 6(6.022 x 1023) atoms 12(6.022 x 1023) atoms 6(6.022 x 1023) atoms 6(16.00 amu) =96.00 amu Atoms/mole of compound Mass/molecule of compound 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu Mass/mole of compound 72.06 g 12.10 g 96.00 g So for glucose with 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atom, the molar mass is 72.06 + 12.10 + 96.00 = 180.16 g/mol 3-8 Dr. Wolf’s CHM 101 Interconverting Moles, Mass, and Number of Chemical Entities no. of grams Mass (g) = no. of moles x 1 mol No. of moles = mass (g) x 1 mol no. of grams No. of entities = no. of moles x 6.022x1023 entities 1 mol 1 mol No. of moles = no. of entities x 6.022x1023 entities 3-9 Dr. Wolf’s CHM 101 3-10 Dr. Wolf’s CHM 101 Sample Problem 3.1 PROBLEM: Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element (a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe? PLAN: (a) To convert mol of Ag to g we have to use the #g Ag/mol Ag, the molar mass M. SOLUTION: 0.0342mol Ag x 107.9g Ag = 3.69g Ag mol Ag PLAN: (b) To convert g of Fe to atoms we first have to find the #mols of Fe and then convert mols to atoms. SOLUTION: 95.8g Fe x mol Fe = 1.72mol Fe 55.85g Fe 6.022x1023atoms Fe = 1.04x1024 atoms 1.72mol Fe x Fe mol Fe 3-11 Dr. Wolf’s CHM 101 amount(mol) of Ag multiply by M of Ag (107.9g/mol) mass(g) of Ag mass(g) of Fe divide by M of Fe (55.85g/mol) amount(mol) of Fe multiply by 6.022x1023 atoms/mol atoms of Fe Sample Problem 3.2 PROBLEM: Calculating the Moles and Number of Formula Units in a Given Mass of a Compound Ammonium carbonate is white solid that decomposes with warming. Among its many uses, it is a component of baking powder, first extinguishers, and smelling salts. How many formula units are in 41.6g of ammonium carbonate? PLAN: After writing the formula for the compound, we find its M by adding the masses of the elements. Convert the given mass, 41.6g to mols using M and then the mols to formula units with Avogadro’s number. SOLUTION: The formula is (NH4)2CO3. mass(g) of (NH4)2CO3 divide by M amount(mol) of (NH4)2CO3 multiply by 6.022x1023 formula units/mol number of (NH4)2CO3 formula units M = (2 x 14.01g/mol N)+(8 x 1.008g/mol H) +(12.01g/mol C)+(3 x 16.00g/mol O)= 96.09g/mol 41.6g (NH4)2CO3 x mol (NH4)2CO3 96.09g (NH4)2CO3 x 6.022x1023 formula units (NH4)2CO3 mol (NH4)2CO3 2.61x1023 formula units (NH4)2CO3 3-12 Dr. Wolf’s CHM 101 = Mass % of element X = atoms of X in formula x atomic mass of X (amu) x 100 molecular (or formula) mass of compound(amu) Mass % of element X = moles of X in formula x molar mass of X (g) molecular (or formula) mass of compound (g) 3-13 Dr. Wolf’s CHM 101 x 100 Flow Chart of Mass Percentage Calculation moles of X in one mol of compound M = (g/mol) of X mass (g) of X in one mol of compound Divide by mass(g) of one mol of compound Mass fraction of X Multiply by 100 Mass % of X 3-14 Dr. Wolf’s CHM 101 So for glucose with 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atom, the molar mass is 72.06 + 12.10 + 96.00 = 180.16 g/mol So to calculate the mass percent of carbon in glucose, we just use the mass of carbon in one mole of glucose over the mass of one mole of glucose to get the mass fraction and then multiply by 100. Mass % C = (76.06g/mol / 180.16 g/mol) x 100 = 40.00 % 3-15 Dr. Wolf’s CHM 101 Empirical and Molecular Formulas Empirical Formula The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. Molecular Formula The formula of the compound as it exists, it may be a multiple of the empirical formula. 3-16 Dr. Wolf’s CHM 101 Sample Problem 3.4 Determining the Empirical Formula from Masses of Elements PROBLEM: Elemental analysis of a sample of an ionic compound gave the following results: 2.82g of Na, 4.35g of Cl, and 7.83g of O. What are the empirical formula and name of the compound? PLAN: Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula). SOLUTION: 2.82g Na mol Na = 0.123 mol Na 22.99g Na mass(g) of each element mol Cl divide by M(g/mol) 4.35g Cl = 0.123 mol Cl 35.45g Cl amount(mol) of each element mol O 7.83g O = 0.489 mol O use # of moles as subscripts 16.00 O preliminary formula Na1 Cl1 O3.98 NaClO4 change to integer subscripts empirical formula 3-17 Dr. Wolf’s CHM 101 NaClO4 is sodium perchlorate. Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: During physical activity. lactic acid (M=90.08g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental anaylsis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. PLAN: assume 100g lactic acid and find the mass of each element divide each mass by mol mass(M) amount(mol) of each element molecular formula use #mols as subscripts preliminary formula convert to integer subscripts empirical formula 3-18 Dr. Wolf’s CHM 101 divide mol mass by mass of empirical formula to get a multiplier Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass continued SOLUTION: Assuming there are 100.g of lactic acid, the constituents are 40.0g C mol C 12.01g C 3.33mol C C3.33 3.33 6.71g H mol H 53.3g O 1.008g H 16.00g O 6.66mol H 3.33mol O H6.66 O3.33 3.33 3.33 molar mass of lactate CH2O empirical formula 90.08g 3 mass of CH2O 3-19 mol O Dr. Wolf’s CHM 101 30.03g C3H6O3 is the molecular formula Combustion Analysis Determining Composition by Looking at amounts of Products of Combustion Combustion Train for the Determination of the Chemical Composition of Organic Compounds. m m CnHm + (n+ ) O2 = n CO2(g) + H O(g) 4 2 2 3-20 Dr. Wolf’s CHM 101 Sample Problem 3.6 PROBLEM: Determining a Molecular Formula from Combustion Analysis Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in many natural sources especially citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained: mass of CO2 absorber after combustion =85.35g mass of CO2 absorber before combustion =83.85g mass of H2O absorber after combustion =37.96g mass of H2O absorber before combustion =37.55g What is the molecular formula of vitamin C? PLAN: difference (after-before) = mass of oxidized element find the mass of each element in its combustion product find the mols 3-21 Dr. Wolf’s CHM 101 preliminary formula empirical formula molecular formula Sample Problem 3.6 continued Determining a Molecular Formula from Combustion Analysis SOLUTION: CO2 85.35g-83.85g = 1.50g There are 12.01g C per mol CO2 . 37.96g-37.55g = 0.41g H2O 12.01g C 1.50g CO2 = 0.409g C 44.01g CO2 There are 2.016g H per mol H2O . 0.41g H2O 2.016g H = 0.046g H 18.02g H2O O must be the difference: 0.409g C = 0.0341mol C 12.01g C C1H1.3O1 1.000g - (0.409 + 0.049) = 0.545 0.046g H = 0.0461mol H 1.008g H C3H4O3 Dr. Wolf’s CHM 101 = 0.0341mol O 16.00g O 176.12g/mol 88.06g 3-22 0.545g O = 2.000 C6H8O6 Writing and Balancing Chemical Equations A chemical equation shows reactants going to products. In addition, it must be balanced....meaning the same number of each kind of atom must appear on both sides of the equation. Example: hydrogen and fluorine to give hydrogen fluoride 3-23 Dr. Wolf’s CHM 101 H2 3-24 + Dr. Wolf’s CHM 101 F2 2 HF A three-level view of the chemical reaction in a flashbulb 3-25 Dr. Wolf’s CHM 101 Sample Problem 3.7 PROBLEM: PLAN: 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. SOLUTION: 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 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g) The mole ratios are the same as the molecular coefficients in the balanced reaction. 3-26 Dr. Wolf’s CHM 101 Calculation Amounts of Reactants and Products in a Chemical Reaction 3-27 Dr. Wolf’s CHM 101 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 3-28 Dr. Wolf’s CHM 101 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) 3mol O2 (a) 10.0mol Cu2S 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 3-29 Dr. Wolf’s CHM 101 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 3-30 Dr. Wolf’s CHM 101 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) 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 3-31 Dr. Wolf’s CHM 101 limiting mol N2 mol of N2 g N2 An Ice Cream Sundae Analogy for Limiting Reactions 3-32 Dr. Wolf’s CHM 101 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 3-33 4.68mol N2 mol N2O4 N O 2 4 92.02g N2O4 3 mol N2 Dr. Wolf’s CHM 101 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 Yield Calculations Actual Yield - Amount of product actually produced in the reaction. It can be expressed in grams or moles. Theoretical Yield - Amount of product if the reaction proceeded as complete as possible determined by the limiting reagent. Again, it can be expressed in grams or moles. Percent Yield - The ratio of Actual Yield to Theoretical Yield (both expressed in the same units) times 100. 3-34 Dr. Wolf’s CHM 101 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 3-35 Dr. Wolf’s CHM 101 40.10g SiC kg 103g x100 =77.0% = 66.73kg Stoichiometry of Solutions Concentration in Terms of Molarity A solution consists of a smaller amount of a substance, the solute, dissolved in a larger amount of another substance, the solvent. The concentration of the solution is expressed as the amount of solute dissolved in a given amount of solution. The term most commonly used is Molarity (M), defined as moles of solute per liter of solution. 3-36 Dr. Wolf’s CHM 101 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 3-37 Dr. Wolf’s CHM 101 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 3-38 Dr. Wolf’s CHM 101 Converting a Concentrated Solution to a Dilute Solution 3-39 Dr. Wolf’s CHM 101 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 isotonic 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 3-40 Dr. Wolf’s CHM 101 0.80L soln 0.15mol NaCl = 0.12mol NaCl L soln L solnconc 0.12mol NaCl = 0.020L soln 6mol 3-41 Dr. Wolf’s CHM 101 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 3-42 divide by M Dr. Wolf’s CHM 101 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 1L mol HCl Dr. Wolf’s CHM 101 = 3.4x10-3 mol HCl 1 mol Mg(OH)2 0.10mol HCl 3-43 MgCl2(aq) + 2H2O(l) = 3.4x10-2 L HCl Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions in Solution PROBLEM: PLAN: 3-44 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. Dr. Wolf’s CHM 101 Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions in Solution continued SOLUTION: L of Hg(NO3)2 Hg(NO3)2(aq) + Na2S(aq) 2NaNO3(aq) 0.050L Hg(NO3)2 multiply by M mol Hg(NO3)2 mol ratio x 0.010 mol/L x 0. 10 mol/L x 1mol HgS 1mol Hg(NO3)2 1mol Na2S 3-45 mol HgS Dr. Wolf’s CHM 101 232.7g HgS 1 mol HgS multiply by M = 2.0x10-3 mol HgS Hg(NO3)2 is the limiting reagent. 5.0x10-4 L of Na2S 0.020L Na2S x 1mol HgS = 5.0x10-4 mol HgS mol HgS HgS(s) + = 0.12g HgS mol Na2S mol ratio mol HgS End of Chapter 3 3-46 Dr. Wolf’s CHM 101