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Chapter 3 Stoichiometry

Chapter 3 Stoichiometry. 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.

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Chapter 3 Stoichiometry

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  1. Chapter 3 Stoichiometry

  2. 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. 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 asN. One mole (1 mol) contains 6.022x1023 entities (to four significant figures)

  4. Counting Objects of Fixed Relative Mass 12 red marbles @ 7g each = 84g 12 yellow marbles @4g each=48g So equal numbers will always have the same 7:4 ratio = 84:48 55.85g Fe = 6.022 x 1023 atoms Fe 32.07g S = 6.022 x 1023 atoms S These values come from the atomic mass values for Fe and S in the Periodic Table

  5. Oxygen, O2 32.00 g One Mole of Common Substances Water, H2O 18.02 g CaCO3 100.09 g Copper 63.55 g

  6. 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 (also called atomic weight) Molecular (or formula) mass Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) amu (also called molecular weight) Molar mass (M) Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units) g/mol (also called gram-molecular weight)

  7. 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.

  8. Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of compound 12 atoms 6 atoms 6 atoms 12 moles of atoms 6 moles of atoms 6 moles of atoms Moles of atoms/ mole of compound 6(6.022 x 1023) atoms 12(6.022 x 1023) atoms 6(6.022 x 1023) atoms Atoms/mole of compound Mass/moleculeof compound 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 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

  9. no. of grams Mass (g) = no. of moles x 1 mol 1 mol No. of moles = mass (g) x no. of grams 6.022x1023 entities No. of entities = no. of moles x 1 mol 1 mol No. of moles = no. of entities x 6.022x1023 entities Interconverting Moles, Mass, and Number of Chemical Entities

  10. PROBLEM: amount(mol) of Ag 107.9g Ag mass(g) of Ag mol Ag mass(g) of Fe amount(mol) of Fe mol Fe 55.85g Fe 6.022x1023atoms Fe atoms of Fe mol Fe Sample Problem 3.1 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. multiply by M of Ag (107.9g/mol) SOLUTION: = 3.69g Ag 0.0342mol Ag x PLAN: (b) To convert g of Fe to atoms we first have to find the #mols of Fe and then convert mols to atoms. divide by M of Fe (55.85g/mol) SOLUTION: 95.8g Fe x = 1.72mol Fe multiply by 6.022x1023 atoms/mol = 1.04x1024 atoms Fe 1.72mol Fe x

  11. mass(g) of (NH4)2CO3 amount(mol) of (NH4)2CO3 number of (NH4)2CO3 formula units mol (NH4)2CO3 6.022x1023 formula units (NH4)2CO3 96.09g (NH4)2CO3 mol (NH4)2CO3 Sample Problem 3.2 Calculating the Moles and Number of Formula Units in a Given Mass of a Compound PROBLEM: 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. divide by M multiply by 6.022x1023 formula units/mol SOLUTION: The formula is (NH4)2CO3. 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 x = 2.61x1023 formula units (NH4)2CO3

  12. Mass % of element X = moles of X in formula x molar mass of X (g) x 100 molecular (or formula) mass of compound (g) Mass % of element X = atoms of X in formula x atomic mass of X (amu) x 100 molecular (or formula) mass of compound(amu)

  13. moles of X in one mol of compound mass (g) of X in one mol of compound Mass fraction of X Mass % of X Flow Chart of Mass Percentage Calculation M= (g/mol) of X Divide by mass(g) of one mol of compound Multiply by 100

  14. 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 %

  15. 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.

  16. mol Na mol O mol Cl 16.00 O 35.45g Cl 22.99g Na 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 = 0.123 mol Na mass(g) of each element divide by M(g/mol) 4.35g Cl = 0.123 mol Cl amount(mol) of each element 7.83g O = 0.489 mol O use # of moles as subscripts preliminary formula Na1 Cl1 O3.98 Na1 Cl1 O3.98 NaClO4 NaClO4 change to integer subscripts empirical formula NaClO4 is sodium perchlorate.

  17. 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 divide mol mass by mass of empirical formula to get a multiplier preliminary formula convert to integer subscripts empirical formula

  18. mol H mol O mol C 16.00g O 1.008g H 12.01g C 3.33 3.33 3.33 90.08g molar mass of lactate 30.03g mass of CH2O 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 6.71g H 53.3g O 3.33mol C 6.66mol H 3.33mol O C3.33 H6.66 O3.33 CH2O empirical formula C3H6O3 is the molecular formula 3

  19. m 4 m 2 CnHm + (n+ ) O2 = n CO2(g) + H2O(g) Combustion Analysis Determining Composition by Looking at amounts of Products of Combustion Combustion Train for the Determination of the Chemical Composition of Organic Compounds.

  20. PROBLEM: 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? Sample Problem 3.6 Determining a Molecular Formula from Combustion Analysis PLAN: difference (after-before) = mass of oxidized element find the mass of each element in its combustion product preliminary formula empirical formula molecular formula find the mols

  21. CO2 85.35g-83.85g = 1.50g H2O 37.96g-37.55g = 0.41g 2.016g H 12.01g C 44.01g CO2 18.02g H2O 12.01g C 1.008g H 16.00g O 0.409g C 0.046g H 0.545g O 176.12g/mol 88.06g Sample Problem 3.6 Determining a Molecular Formula from Combustion Analysis continued SOLUTION: There are 12.01g C per mol CO2 . 1.50g CO2 = 0.409g C There are 2.016g H per mol H2O. 0.41g H2O = 0.046g H O must be the difference: 1.000g - (0.409 + 0.049) = 0.545 = 0.0341mol C = 0.0461mol H = 0.0341mol O C1H1.3O1 C3H4O3 = 2.000 C6H8O6

  22. 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

  23. H2 + F2 2 HF

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

  25. PLAN: SOLUTION: C8H18 + O2 CO2 + H2O C8H18 + O2 CO2 + H2O 2C8H18 + 25O216CO2 + 18H2O 2C8H18 + 25O216CO2 + 18H2O balance the atoms adjust the coefficients check the atom balance 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g) specify states of matter Sample Problem 3.7 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. PROBLEM: translate the statement 25/2 8 9 The mole ratios are the same as the molecular coefficients in the balanced reaction.

  26. Calculation Amounts of Reactants and Products in a Chemical Reaction

  27. 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

  28. 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) 3mol O2 2mol Cu2S 2mol SO2 64.07g SO2 2mol Cu2S mol SO2 103g Cu2O mol Cu2O kg Cu2O 143.10g Cu2O 3mol O2 32.00g O2 kg O2 2mol Cu2O mol O2 103g O2 Sample Problem 3.8 Calculating Amounts of Reactants and Products continued SOLUTION: (a) 10.0mol Cu2S = 15.0mol O2 (b) 10.0mol Cu2S = 641g SO2 (c) 2.86kg Cu2O = 20.0mol Cu2O 20.0mol Cu2O = 0.960kg O2

  29. 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. 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) 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

  30. mass of N2H4 mass of N2O4 limiting mol N2 multiply by M mol of N2H4 mol of N2O4 g N2 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

  31. An Ice Cream Sundae Analogy for Limiting Reactions

  32. N2H4(l) + N2O4(l) N2(g) + H2O(l) mol N2H4 32.05g N2H4 3 mol N2 28.02g N2 2mol N2H4 mol N2 mol N2O4 92.02g N2O4 3 mol N2 mol N2O4 Sample Problem 3.10 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant continued SOLUTION: 2 3 4 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

  33. 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.

  34. SiO2(s) + 3C(s) SiC(s) + 2CO(g) 103g SiO2 mol SiO2 kg SiO2 60.09g SiO2 40.10g SiC kg mol SiC 103g 51.4kg 66.73kg 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%

  35. 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.

  36. PLAN: Molarity is the number of moles of solute per liter of solution. 1.80mol HBr 1000mL 455 mL soln 1 L 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

  37. 0.460moles 1 L 141.96g Na2HPO4 mol Na2HPO4 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

  38. Converting a Concentrated Solution to a Dilute Solution

  39. 0.15mol NaCl L soln L solnconc 6mol 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 0.80L soln = 0.12mol NaCl divide by M of concentrated soln 0.12mol NaCl = 0.020L soln L of concentrated soln

  40. mass Mg(OH)2 L HCl divide by M divide by M mol Mg(OH)2 mol HCl mol ratio 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.

  41. Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) mol Mg(OH)2 58.33g Mg(OH)2 2 mol HCl 1 mol Mg(OH)2 1L 0.10mol HCl 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

  42. 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.

  43. L of Hg(NO3)2 L of Na2S multiply by M multiply by M mol Hg(NO3)2 x 1mol HgS mol Na2S 1mol Na2S mol ratio x 1mol HgS mol ratio 1mol Hg(NO3)2 mol HgS mol HgS 232.7g HgS 1 mol HgS Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions in Solution continued SOLUTION: Hg(NO3)2(aq) + Na2S(aq) HgS(s) + 2NaNO3(aq) 0.050L Hg(NO3)2 0.020L Na2S 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

  44. End of Chapter 3

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