Chapter 3 - Molecules Moles and Chemical Equations PDF

Larry Brown Tom Holme www.cengage.com/chemistry/brown Chapter 3 Molecules, Moles, and Chemical Equations Technological University of the Philippines - Taguig Campus Basic Arts and Sciences Department Chemistry Sec...

Larry Brown Tom Holme www.cengage.com/chemistry/brown Chapter 3 Molecules, Moles, and Chemical Equations Technological University of the Philippines - Taguig Campus Basic Arts and Sciences Department Chemistry Section Chapter Objectives List at least three characteristics of explosive chemical reactions. Explain balancing a chemical reaction as an application of the law of conservation of mass. List at least three quantities that must be conserved in chemical reactions. Write balanced chemical equations for simple reactions, given either an unbalanced equation or a verbal description. 2 Chapter Objectives Explain the concept of a mole in your own words. Interpret chemical equations in terms of both moles and molecules. Interconvert between mass, number of molecules, and number of moles. Determine a chemical formula from elemental analysis (i.e., from % compositions). 3 Chapter Objectives Define the concentration of a solution and calculate the molarity of solutions from appropriate data. Calculate the molarity of solutions prepared by dilution or calculate the quantities needed to carry out a dilution to prepare a solution of a specified concentration. Distinguish between electrolytes and nonelectrolytes and explain how their solutions differ. 4 Chapter Objectives Describe the species expected to be present (ions, molecules, etc.) in various simple solutions. Recognize common strong acids and bases. Write molecular and ionic equations for acid-base neutralization reactions. 5 Explosions 1. Explosions release a large amount of energy when a fairly complex molecule decomposes into smaller, simpler compounds. 2. Explosions occur very quickly. 3. Modern explosives are generally solids. 6 Explosions Dynamite is an explosive made from liquid nitroglycerin and an inert binder to form a solid material. Solids are easier to handle than liquids 7 Explosions The destructive force of explosions is due in part to expansion of gases, which produces a shockwave. 8 Chemical Formulas and Equations Chemical formulas provide a concise way to represent chemical compounds. Nitroglycerin, shown earlier, becomes C3H5N3O9 A chemical equation builds upon chemical formulas to concisely represent a chemical reaction. 9 Writing Chemical Equations Chemical equations represent the transformation of one or more chemical species into new substances. Reactants are the original materials and are written on the left hand side of the equation. Products are the newly formed compounds and are written on the right hand side of the equation. Reactants → Products 10 Writing Chemical Equations Chemical formulas represent reactants and products. Phase labels follow each formula. solid = (s) liquid = (l) gas = (g) aqueous (substance dissolved in water) = (aq) Some reactions require an additional symbol placed over the reaction arrow to specify reaction conditions. Thermal reactions: heat () Photochemical reactions: light (h) 11 Writing Chemical Equations Different representations for the reaction between hydrogen and oxygen to produce water. 12 Balancing Chemical Equations The law of conservation of matter: matter is neither created nor destroyed. Chemical reactions must obey the law of conservation of matter. The same number of atoms for each element must occur on both sides of the chemical equation. A chemical reaction simply rearranges the atoms into new compounds. 13 Balancing Chemical Equations Balanced chemical equation for the combustion of methane. 14 Balancing Chemical Equations Chemical equations may be balanced via inspection, which really means by trial and error. Numbers used to balance chemical equations are called stoichiometric coefficients. The stoichiometric coefficient multiplies the number of atoms of each element in the formula unit of the compound that it precedes. Stoichiometry is the various quantitative relationships between reactants and products. 15 Balancing Chemical Equations Pay attention to the following when balancing chemical equations: Do not change species Do not use fractions (cannot have half a molecule) Make sure you have the same number of atoms of each element on both sides 16 Example Problem 3.1 Write a balanced chemical equation describing the reaction between propane, C3H8, and oxygen, O2, to form carbon dioxide and water. 17 Aqueous Solutions and Net Ionic Equations Reactions that occur in water are said to take place in aqueous solution. Solution: homogeneous mixture of two or more substances. Solvent: solution component present in greatest amount. Solute: solution component present in lesser amount. The preparation of a solution is a common way to enable two solids to make contact with one another. 18 Solutions, Solvents, and Solutes For solutions, the concentration is a key piece of information. Concentration: relative amounts of solute and solvent. Concentrated: many solute particles are present. Dilute: few solute particles are present. 19 Solutions, Solvents, and Solutes Solution preparation: Solid CuSO4, the solute, is transferred to a flask. Water, the solvent, is added. The flask is shaken to speed the dissolution process. Two solutions of CuSO4. Solution on the left is more concentrated, as seen from its darker color. 20 Solutions, Solvents, and Solutes Compounds can be characterized by their solubility. Soluble compounds dissolve readily in water. Insoluble compounds do not readily dissolve in water. Solubility can be predicted using solubility guidelines. 21 Solutions, Solvents, and Solutes Solubility guidelines for soluble salts 22 Solutions, Solvents, and Solutes Solubility guidelines for insoluble salts 23 Example Problem 3.2 Which of the following compounds would you predict are soluble in water at room temperature? a) KClO3 b) CaCO3 c) BaSO4 d) KMnO4 24 Solutions, Solvents, and Solutes Electrolytes are soluble compounds that conduct current when dissolved in water. Weak electrolytes dissociate partially into ions in solution. Strong electrolytes dissociate completely into ions in solution. Nonelectrolytes do not dissociate into ions in solution. 25 Solutions, Solvents, and Solutes a) Sugar, a nonelectrolyte, does not conduct current when dissolved in water. b) Acetic acid, a weak electrolyte, weakly conducts current when dissolved in water. c) Potassium chromate, a strong electrolyte, strongly conducts current when dissolved in water. 26 Chemical Equations for Aqueous Reactions When a covalently bonded material dissolves in water and the molecules remain intact, they do not conduct current. These compounds are nonelectrolytes. C6H12O6(s)⎯→C6H12O6(aq The water molecules are not shown explicitly, although their presence is indicated by the “(aq)” on the product side. 27 Chemical Equations for Aqueous Reactions When an ionic solid dissolves in water, it breaks into its constituent ions. This is called a dissociation reaction. These compounds conduct electric current and are electrolytes. NaCl (s) ⎯⎯ → Na (aq) + Cl (aq) + − 28 Chemical Equations for Aqueous Reactions Aqueous chemical reactions can be written as a molecular equation. The complete formula for each compound is shown. Note, all of the species may not be molecules. HNO3 (aq) + NH3 (g) ⎯⎯→ NH4 NO3 (aq) 29 Chemical Equations for Aqueous Reactions Dissociation of reactants and products is emphasized by writing a total ionic equation. H+ (aq) + NO− (aq)+ NH (g) ⎯⎯→ NH+ (aq) + NO− (aq) 3 3 4 3 Note: HNO3 is a strong acid and thus dissociates completely, while NH3 does not dissociate 30 Chemical Equations for Aqueous Reactions Spectator ions are ions uninvolved in the chemical reaction. When spectator ions are removed, the result is the net ionic equation. Total ionic equation H+ (aq) + NO− (aq)+ NH (g) ⎯⎯→ NH+ (aq) + NO− (aq) 3 3 4 3 Net ionic equation H + (aq) + NH (g) ⎯⎯→ NH+ (aq) 3 4 Spectator ion = NO−3 31 Acid-Base Reactions Acids are substances that dissolve in water to produce H+(or H3O+) ions. Examples: HCl, HNO3, H3PO4, HCN Bases are substances that dissolve in water to produce OH– ions. Examples: NaOH, Ca(OH)2, NH3 32 Acid-Base Reactions Strong acids and bases completely dissociate in water. HCl(g) + H O(l ) ⎯⎯→ H O (aq) + Cl (aq) + − 2 3 NaOH(s) ⎯⎯→ Na (aq) + OH (aq) + − 33 Acid-Base Reactions All common strong acids and bases. 34 Acid-Base Reactions Weak acids and bases partially dissociate in water. Notice the two-way arrows, which emphasize that the reaction does not proceed completely from left to right. ⎯ CH3COOH(aq) + H2O(l )  ⎯⎯→ H 3O + (aq) + CH 3COO − (aq) ⎯ NH3 (aq) + H2O(l )  ⎯⎯→ NH + 4 (aq) + OH − (aq) 35 Acid-Base Reactions Some common weak acids and bases. 36 Acid-Base Reactions Mixing an acid and a base leads to a reaction known as neutralization, in which the resulting solution is neither acidic nor basic. Net ionic equation for neutralization of strong acid and strong base. H O+ (aq) + OH− (aq) ⎯⎯→ 2H O(l) 3 2 37 Example Problem 3.3 When aqueous solutions of acetic acid and potassium hydroxide are combined, a neutralization reaction will occur. Write the following equations: a) molecular b) total ionic c) net ionic 38 Precipitation Reactions A precipitation reaction is an aqueous reaction that produces a solid, called a precipitate. Net ionic reaction for the precipitation of lead(II) iodide. Pb 2+ (aq) + 2I− (aq) ⎯⎯ → PbI (s) 2 39 Precipitation Reactions Precipitation reaction between aqueous solutions of KI and Pb(NO3)2, which are both colorless. The bright yellow solid, PbI2, is produced. PbI2 is insoluble as predicted by the solubility guidelines. 40 Example Problem 3.4 When aqueous sodium carbonate and barium chloride are combined, the solution becomes cloudy white with solid barium carbonate. Write the following equations: a) molecular b) total ionic c) net ionic 41 Interpreting Equations and the Mole Balanced chemical equations are interpreted on the microscopic and macroscopic level. Microscopic interpretation visualizes reactions between molecules. Macroscopic interpretation visualizes reactions between bulk materials. 42 Interpreting Chemical Equations Balanced chemical reactions provide stoichiometric ratios between reactants and products. Ratios relate relative numbers of particles. 2H2 (g) + O2 (g) ⎯⎯→ 2H2O(g) 2 molecules H2: 1 molecule O2: 2 molecules H2O 100 molecules H2: 50 molecule O2: 100 molecules H2O 43 Avogadro’s Number and the Mole A mole is a means of counting the large number of particles in samples. One mole is the number of atoms in exactly 12 grams of 12C (carbon-12). 1 mole contains Avogadro’s number (6.022 x 1023 particles/mole) of particles. The mass of 6.022 x 1023atoms of any element is the molar mass of that element. 44 Avogadro’s Number and the Mole One mole samples of various elements. All have the same number of particles. 45 Avogadro’s Number and the Mole Balanced chemical reactions also provide mole ratios between reactants and products. 2H 2 (g) + O 2 (g) ⎯⎯ → 2H 2O(g) 2 moles H2: 1 mole O2: 2 moles H2O 46 Determining Molar Mass The molar mass of a compound is the sum of the molar masses of all the atoms in a compound.  2 mol H  1.0 g H   16.0 g    +  1 mol O   1 mol H   1 mol O  = 18.0 g/mol H2O 47 Example Problem 3.5 Determine the molar mass of each of the following compounds: a) PbN6 b) C3H5N3O9 c) Hg(ONC)2 48 Calculations Using Moles and Molar Mass Molar mass allows conversion from mass to number of moles, much like a unit conversion. 1 mol C7H5N3O6= 227.133 g C7H5N3O6 1 mol C 7 H 5 N 3 O 6 300.0 g C 7 H 5 N 3O 6  227.133 g C 7 H 5 N 3O 6 = 1.320 mol C 7 H 5N 3O 6 49 Calculations Using Moles and Molar Mass Avogadro’s number functions much like a unit conversion between moles to number of particles. 1 mol C7H5N3O6= 6.022  1023C7H5N3O6 molecules How many molecules are in 1.320 moles of nitroglycerin? 6.022  1023 molecules C 7 H 5N O 1.320 mol C 7 H 5N O 3 6  3 6 1 mol C 7 H 5N O 3 6 = 7.949  1023 molecules C H N O 7 5 3 6 50 Example Problem 3.6 A sample of the explosive TNT (C7H5N3O6) has a mass of 5. g. How many moles of TNT are in this sample? How many molecules are in this sample? 51 Example Problem 3.7 How many pounds of halite (C2H6N4O5) correspond to 315 moles? (1 pound = 454 g) 52 Elemental Analysis: Determining Empirical and Molecular Formulas Empirical formulas can be determined from an elemental analysis. An elemental analysis measures the mass percentage of each element in a compound. The formula describes the composition in terms of the number of atoms of each element. The molar masses of the elements provide the connection between the elemental analysis and the formula. 53 Elemental Analysis: Determining Empirical and Molecular Formulas Assume a 100 gram sample size Percentage element  sample size = mass element in compound. (e.g., 16% carbon = 16 g carbon) Convert mass of each element to moles using the molar mass. Divide by smallest number of moles to get mole to mole ratio for empirical formula. When division by smallest number of moles results in small rational fractions, multiply all ratios by an appropriate integer to give whole numbers. 2.5  2 = 5, 1.33  3 = 4, etc. 54 Example Problem 3.8 The explosive known as RDX contains 16.22% carbon, 2.72% hydrogen, 37.84% nitrogen, and 43.22% oxygen by mass. Determine its empirical formula. 55 Elemental Analysis: Determining Empirical and Molecular Formulas A molecular formula is a whole number multiple of the empirical formula. Molar mass for the molecular formula is a whole number multiple of the molar mass for the empirical formula. If the empirical formula of a compound is CH2 and its molar mass is 42 g/mol, what is its molecular formula? 56 Example Problem 3.9 An alloy contains 70.8 mol % palladium and 29.2 mol % nickel. Express the composition of this alloy as weight percentage (wt %). 57 Molarity Molarity, or molar concentration, M, is the number of moles of solute per liter of solution. Provides relationship among molarity, moles solute, and liters solution. moles of solute Molarity (M ) = liter of solution If we know any two of these quantities, we can determine the third. 58 Example Problem 3.10 A solution is prepared by dissolving 45.0 g of NaClO in enough water to produce exactly 750 mL of solution. What is the molarity of this solution? 59 Dilution Dilution is the process in which solvent is added to a solution to decrease the concentration of the solution. The number of moles of solute is the same before and after dilution. Since the number of moles of solute equals the product of molarity and volume (M  V), we can write the following equation, where the subscripts denote initial and final values. M i  Vi = M f  Vf 60 Example Problem 3.11 A chemist requires 1.5 M hydrochloric acid, HCl, for a series of reactions. The only solution available is 6.0 M HCl. What volume of 6.0 M HCl must be diluted to obtain 5.0 L of 1.5 M HCl? 61 Explosive and Green Chemistry Green chemistry: the philosophy that chemical processes and products should be designed with the goal of reducing environmental impacts Firing of guns involves detonating a primer, which then induces a larger explosion. Traditional primers are lead- based, e.g., Pb(N3)2 Research is underway to find less toxic primers 62