4. Hemoglobin, a protein in red blood cells, carries O 2, from the lungs to the body s cells. Iron (as Fe 2+ ) makes up 0.33 mass % of hemoglobin. If the molar mass of hemoglobin is 6.8x10 4 g/mol, how many Fe 2+ ions are in one molecule? Assume 1 mol hemoglobin: mass Fe 2+ = 0.0033 (6.8x10 4 g mol -1 ) = 224.4 g Looking at one molecule of hemoglobin: mass Fe 2+ in one molecule = 224.4 amu # of Fe 2+ 224.4 amu ions = = 4.02 55.85 amu 4 Fe 2+ per molecule of hemoglobin 1
Solution Stoichiometry A solution is a homogenous mixture of a solvent plus a solute. 9 July 2013 2
Concentration of Solutions Amount of solute present in a given quantity of solvent or solution Expressed as molarity grams solute molar mass solute M = molarity = moles of solute total liters of solution Density = mass volume 9 July 2013 3
Example Calculate the molarity of a solution that contains 12.5 g of pure sulfuric acid (H 2 SO 4 ) in 1.75 L of solution (Molar mass H 2 SO 4 = 98.1 g) Molarity = 12.5 g H 2 SO 4 1 mol H 2 SO 4 98.1 g H 2 SO 4 1.75 L = 0.0728 M H 2 SO 4 9 July 2013 4
Concentration An intensive (not extensive) property Independent of the volume of solution (like density or temperature) 4 July 2013 5
Preparing Solutions in the Laboratory 9 July 2013 6
Practice Problem How many grams of KI is required to make 500. ml of a 2.80 M KI solution? Reworded: Suppose I have 500. ml of a 2.80 M KI solution. How many grams of KI solute does it contain? ( It meaning the entire 500. ml volume.) 9 July 2013 7
Practice Problem How many grams of KI is required to make 500. ml of a 2.80 M KI solution? grams KI = 500. ml 1 L 1000 ml 2.80 mol KI L soln 166 g KI mol KI = 232 g KI 9 July 2013 8
Practice Problem Sucrose has a molar mass of 342.29 g/mol. It is a fine, white, crystalline powder with a pleasing, sweet taste. What is the molarity of an aqueous solution made by placing 75.5 g of 85% pure sugar in 500.00 ml of water? 9 July 2013 9
What does 3.5 M FeCl 3 mean? 1. 3.5 mol of dry, 100% pure FeCl 3 dissolved in 1.00 L total solution volume 2. [Fe 3+ ] = 3.5 M and [Cl - ] = 3 x 3.5 M 9 July 2013 10
Example In 2 L of 0.20 M Al 2 (SO 4 ) 3, (a) what is the molarity of Al 2 (SO 4 ) 3? (b) How many moles of Al are there? (c) What is the molarity of [Al 3+ ] and [SO 4 2- ] ions? a. [Al 2 (SO 4 ) 3 ] = 0.20 M b. mol Al = 2 L 0.20 mol L = 0.4 mol Al c. Al 2 + = 2 0.20M and SO 4 2 = 3 0.20M 9 July 2013 11
Example Hydrochloric acid is sold commercially as a 12.0 M solution. How many moles of HCl are in 300.0 ml of 12.0 M solution? mol HCl 1 L = 300.0 ml 1000 ml = 3.60 mol HCl 12 mol HCl 1 L soln 9 July 2013 12
Dilution 9 July 2013 13
Dilution Add solvent to a concentrated solution From higher molarity lower molarity The solution volume increases while the number of moles of solute remains the same M dil V dil = no of moles = M conc V conc 9 July 2013 14
Example Isotonic saline is a 0.15 M aqueous solution of NaCl. How would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution? M dil V dil = no of moles = M conc V conc 0.15 M 0.80 L = 6.0 M V conc V conc = 0.02 L = 20. ml A 0.020 L portion of the concentrated solution must be diluted to a final volume of 0.80 L. 9 July 2013 15
Practice Problem 1. Calculate the maximum number of moles and grams of H 2 S that can form when 158 g of aluminum sulfide reacts with 131 g of water: Al 2 S 3 + H 2 O Al(OH) 3 + H 2 S [unbalanced] What mass of the excess reactant remains? 9 July 2013 16
Practice Problem 2. Sodium borohydride (NaBH 4 ) is used industrially in many organic syntheses. One way to prepare it is by reacting sodium hydride with gaseous diborane (B 2 H 6 ). Assuming an 88.5% yield, how many grams of NaBH 4 can be prepared by reacting 7.98 g of sodium hydride and 8.16 g of diborane? 9 July 2013 17
Practice Problem 3. Calculate each of the following quantities: a. Volume in milliliters of 2.26 M potassium hydroxide that contains 8.42 g of solute b. Number of Cu 2+ ions in 52 L of 2.3 M copper(ii) chloride c. Molarity of 275 ml of solution containing 135 mmol of glucose 9 July 2013 18
Practice Problem 4. Calculate each of the following quantities: a. Volume of 2.050 M copper(ii) nitrate that must be diluted with water to prepare 750.0 ml of a 0.8543 M solution b. Volume of 1.63 M calcium chloride that must be diluted with water to prepare 350. ml of a 2.86x10-2 M chloride ion solution c. Final volume of a 0.0700 M solution prepared by diluting 18.0 ml of 0.155 M lithium carbonate with water 9 July 2013 19
Announcements Quiz # 3 on July 16 Tuesday Coverage: Chapter 3 Stoichiometry of Formulas and Equations Long Test # 1 on July 18, Escaler Hall (tentative) Coverage: Chapters 1-3 9 July 2013 20
Problem Sets Chapter Practice problems 1 6, 7, 8, 22, 24, 26, 30, 34, 47, 49, 53, 55 2 11, 13, 17, 33, 61, 65, 67, 69, 71, 73, 75 3 6, 10, 12, 14, 16, 18, 20, 23, 26, 28, 30, 32, 33, 38, 40, 41, 43, 45, 49, 51, 65, 66, 69, 71, 73, 75, 83, 85, 89, 93 Practice on some MC questions: http://www.mhhe.com/physsci/chemistry/silberberg/student/ olc/quiz.mhtml 9 July 2013 21