Formulas, Equations, and Moles II

Amounts of Reactants and Products Chapter 3 z z Formulas, Equations, and Moles: II z In lab, we want to predict amount of product we can expect from a reaction Have to use stoichiometry: quantitative study of reactants and products in a reaction (mole to mole ratio) Use the mole to relate one substance in a reaction to another (equation coefficients, see Table 3.1)  2H2 (g) + O2 (g) Æ 2H2O (g) Mole Ratios 2 ways to calculate moles of substance z Use molar mass: z Concentration: amount of solute present in a given amount of solution Molarity: moles of solute in 1 L of solution  Mass ÷ molar mass = moles z z We use coefficients in a balanced equation to determine the mole ratio in which substances need to react. MgCl2(aq) +2AgNO3(aq)Æ 2AgCl(s) + Mg(NO3)2(aq)  1 molecule of MgCl2 reacts with 2 molecules of z  moles solute / liters solution; M = mol / L z AgNO3 to produce 2 molecules of AgCl and 1 molecule of Mg(NO3)2  Molecules are not practical! Scale up to moles! If I have 1.50 mL of a 0.100 M solution of NaCl, how many moles are in the sample?  M * V = moles 4 Mole Diagram Stoichiometry Calculations Mole to mole conversions MgCl2(aq) +2AgNO3(aq)Æ 2AgCl(s) + Mg(NO3)2(aq) 1.00 mol xs ? z Mass Mole z z How many moles of solid product do you predict to form? z Cannot compare grams of one to grams of another!! Mole Particles Molarity Courtesy: www.unit5.org/christjs/Stoichiometry/Mole%20Island%20Diagram.ppt 6 1 Mole Ratios z z z Predict products (with phases) and balance each equation. Calculate the moles of precipitate (ppt) that can be made in the following reactions. NaCl (aq) + Hg(NO3)2 (aq) Æ  0.500 mol NaCl z z z FeCl3 (aq) + NaOH (aq) Æ  2.35 mol NaOH z Amounts of Reactants and Products Li2SO4 (aq) + AgCH3COO (aq) Æ z  0.1098 mol AgCH3COO We use the balanced chemical equation to determine relative amounts of reactants and d products. d t How can we predict the amount of product we should get? Worked Ex. 3.5 – 3.7; Problems 3.7, 3.8 8 Stoichiometry Calculations z z z z Mass to mass conversions Given mass of reactant, find mass of precipitate: MgCl2 + 2 AgNO3Æ 2 AgCl + Mg(NO3)2 1.2482 g MgCl g 2 and excess AgNO g 3  Convert grams of MgCl2 to moles using molar mass Group Work z z NaOH + 10.54 g Al(NO3)3 Æ Al(OH)3 + NaNO3 xs ? z How many grams of aluminum hydroxide can be made by starting with 10.54 grams of sodium hydroxide? (Hint: Is the equation balanced?) z The mass calculated is the theoretical yield (the most that can theoretically be made in lab).  Convert moles of MgCl2 to moles of AgCl using mol-mol ratio  Convert moles of AgCl to grams using molar mass Yields of Chemical Reactions Mass to Mass Conversions If you ran that reaction in lab and only obtained 4.9548 g (actual yield) of Al(OH)3, what percent of the maximum did you obtain? z Percent yield = Actual / Theoretical x 100% z How many grams of KClO3 are needed to completely destroy one gummy bear? (Gummy bears have an average mass of 2.4882 88 g g. Roughly oug y 50% o of a gu gummy y bea bear’s s mass is sucrose - C12H22O11.) z _KClO3(l) + _C12H22O11(s) → _KCl(l) + _CO2(g) + _H2O(g) z z z Theoretical will be calculated; actual yield will be given in a problem (or obtained in lab) Worked Ex. 3.8, 3.9; Problems 3.9, 3.10 12 2 Limiting Reagents: 2H2 + O2 Æ 2H2O Limiting Reagents z z z z Analogy: making martinis I have 7 martini glasses, 36 olives, 12 shots of vodka, and 8 stirrers Each martini requires 1 glass glass, 2 olives olives, 2 shots of vodka, and 1 stirrer How many martinis can I make? Limiting Reactant 13 Limiting Reagents 14 Limiting Reagents 2H2 + O2 Æ 2H2O 2H2 + O2 Æ 2H2O What is the limiting reactant? What is the limiting reactant? 15 Limiting Reagents 16 Group Work 2H2 + O2 Æ 2H2O z What is the limiting reactant? z z z z z z Al(s) + Cl2(g) Æ AlCl3(s) Begin with 1.67 g Al (s) and 1.67 g Cl2 (g). Calculate mass of product that can be made from each reactant? 1) Which reactant produces less product? 2) How many grams of AlCl3(s) will be produced, theoretically? 3) What is the percent yield if you obtained 1.76 g AlCl3 in the lab? 4) How many grams of the excess reagent will be left over? 17 3 Group Work z z End-of-chapter problem 3.71: Hydrogen and chlorine react to yield hydrogen chloride. How many grams of HCl are formed from reaction of 3.56 3 56 g o of hydrogen yd oge a and d8 8.94 9 g grams a so of chlorine? Worked Ex. 3.10, Problem 3.11, 3.12 Solution Stoichiometry Example z Consider the reaction:  3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq) z If we mix 25.0 mL of 0.200 M CaCl2 solution with 50.0 mL of 0.250 M Na3PO4 solution, what mass of precipitate is formed? 20 Titrations - Solution Stoichiometry z Titration: method of determining the concentration (or volume) of an unknown solution by using a solution with a known concentration (standard)  moles = M * V z z Equivalence point: point at which acid has completely reacted with and been neutralized by base (moles acid = moles base) End point: point at which the indicator changes color (slight change in solution’s pH) Titrations z 23.78 mL of 0.2500 M NaOH neutralized 20.00 mL of HCl. What is the concentration of HCl? Start with a balanced equation. Titration set-up z z z z z Titrant in buret (usually) Unknown in flask (usually) Can’t use dilution calculation! N d mol-mol Need l l ratio i off acid-base id b Solution stoichiometry: use molarity to convert between volume and moles (instead of molar mass)  M = mol / L Group Work z How many mL of 1.018 M H2SO4 are needed to neutralize 20.00 mL of 0.9989 M NaOH?  Write balanced equation  Determine number of moles of known  Calculate mole ratio to determine moles of unknown  Divide by concentration to find volume z Worked Ex. 3.14, Problems 3.20, 3.21, 3.22 4 Concentration of Solutions: Molarity z z Concentration of Solutions: Molarity Concentration: amount of solute present in a given amount of solution Molarity: moles of solute in 1 L of solution z z z  moles solute / liters solution; M = mol / L z z z 1.00 M NaCl = 1.00 mol NaCl in 1 L solution Knowing concentration, volume, and/or mass allows us to calculate: z  1) mass of solid needed to make a solution of a certain concentration, or z  2) concentration of solution if mass and volume are 1.00 M NaCl = 1.00 mol NaCl in 1 L solution 0.500 M NaCl = 0.500 mol NaCl in 1 L sol’n. To find moles of solute: M * V = mol How many moles in 5.67 mL of 0.500 M NaCl? What volume is needed to make a 0.15 M solution using 5.67 mol of solute? Worked Ex. 3.11, 3.12, Problems 3.14 – 3.17 known. 25 Concentration of ions z z What is the concentration of Cl- ions in a 0.150 M FeCl3 solution? 1 mol FeCl3 Æ 1 mol Fe3+ + 3 mol Cl- 26 Concentration of Solutions z We can calculate concentration (molarity) if we know mass and volume.  Mass solute ÷ molar mass = moles solute  Moles solute ÷ volume solution = Molarity z Or we can calculate the mass of solute needed to make a solution of certain concentration (molarity).  Molarity * volume solution = moles solute  Moles solute * molar mass = mass solute needed Concentration Calculations z z z What is the concentration of a 1.00 L solution made by dissolving 5.00 g NaCl in water? 5.00 grams NaCl in a 500.0 mL solution? What mass of NaCl do we need to make 1.00 L of a 0.500 M solution? Concentration of Solutions – Mad Libs z z z z z z 1) 2) 3) 4) 5) 6) Water-soluble ionic compound Number greater than 500 Number between 0 and 1 N Number b b between 1 and d5 Any water-soluble compound Number between 500 and 1000 5 Concentration of Solutions - Mad Libs z Dilution How many grams of (1) _______ are required to prepare a (2) _____ mL solution whose concentration is (3) _____? z What is the concentration of a solution made by adding (4) ______ grams of (5) ______ to _______ mL of water? z Dilution: lowering the concentration of solution (by adding water)  number of moles of solute stays the same, amount of solvent increases z The number of moles of solute stays the same; volume increases (lower molarity)  moles = M * V  Mconc * Vconc = Mdil * Vdil z Dilution z How would you prepare 500.0 mL of 0.500 M H2SO4 starting with concentrated (18.0 M) solution? (Hint: What variable are you solving for?) o ) z How much water do you need to add to 25.0 mL of a 4.50 M NaOH solution to make a 1.00 M NaOH solution? Worked Ex. 3.13; Problems 3.18, 3.19 Percent Composition of Compounds z z We can reverse this calculation. Know percentages, need to calculate empirical formulas. M1V1 = M2V2 32 Percent Composition of Compounds z z Percent by mass: percent by mass of each element in a compound H2O2  2 moles H H, 2 moles O  Remember to add acid to water. z or Example: What is the final concentration of solution made by adding 100.0 mL of 3.00 M HCl to 300.0 mL of DI water?  H = 2 mol x 1.008 g/mol H = 2.016 g H  O = 2 mol x 16.00 g/mol O = 32.00 g O  H2O2 = 2.016 g H + 32.00 g O = 34.02 g  %H = 2.016 g H / 34.02 g H2O2 x 100% = 5.926% z What is %O? Solve 2 ways….. Empirical Formulas z Calculate the empirical formulas:  50% S, 50% O  43.64% P, 56.36% O z Assume 100 g of the material:  50 g O x 1 mol / 16.0 g = 3.125 mol O  50 g S x 1 mol / 32.066 g = 1.559 mol S z Assume 100 g of material:  43.64 g P Æ mol P (1.4089 mol P)  56.36 g O Æ mol O (3.5225 mol O)  Ratio of two moles (2.5 O / 1 P); need whole numbers 36 6 Molecular Formulas z z z We just found the empirical formula to be P2O5. This compound has a molecular mass of 141.9 g/mol. We know the molecular formula has a molar mass of 283.8 g/mol. How many multiples of P2O5 do we need to reach a mass of 283.8? Practice Problem z A compound contains 30.4 % nitrogen and 69.6 % oxygen. The molecular mass of the compound is 92 g/mol.  What is the empirical formula of the compound?  What is the molecular formula of the compound? Worked Ex. 3.15 – 3.17; Problems 3.24 – 3.29 38 Group Work z A compound contains 40.00% C, 6.71% H, and the rest is O. Determine the empirical formula of this compound. z If the molar mass of the substance is 180.16 g/mol, determine the molecular formula of this substance.  CH2O  C6H12O6 7