Chapter 3: Stoichiometry

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1 Chapter 3: Stoichiometry Chem 6A Michael J. Sailor, UC San Diego 1 Announcements: Thursday (Sep 29) quiz: Bring student ID or we cannot accept your quiz! No notes, no calculators Covers chapters 1 and 2 Need to know your name, PID, and section # Chem 6A Michael J. Sailor, UC San Diego 2

2 Cover page of Thursday s quiz Chem 6A 2010 (Sailor) Name: Student ID Number: Section Number: QUIZ H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Some useful constants and relationships: Ideal gas constant: L. atm. mol -1. K -1 = J. mol -1. K -1 Avogadro constant: x mole -1 Planck's constant = h = x J. s speed of light: 3.00 x 10 8 m/s J = 1 L. atm 1J = 1kg. m 2 /s 2 1 atm = 760 Torr 1 ev = x J E = -RHh/n 2 RH = 3.29 x Hz C2 = second radiation constant = 1.44 x 10-2 K. m Emitted power (W) = (constant)t 4 Surface area (m 2 ) max = 1 5 C 2 E = hc " Chem 6A Michael J. Sailor, UC San Diego 3 The Periodic Table of the Elements QUIZ THURS Oct 20 (front page) Chem 6A Michael J. Sailor, UC San Diego 4

3 The Periodic Table of the Elements QUIZ THURS Oct 20 (back page) Provide the names of the following elements: Element H Li Fr Pt Ag Sb Etc 20 elements total Name hydrogen lithium francium platinum silver antimony Chem 6A Michael J. Sailor, UC San Diego 5 Problem: Mass-to-mass calculations Since the bronze age (4000 b.c.), copper metal has been produced by smelting, in which Cu 2 O ore is reduced with excess carbon (charcoal). How much copper can be produced from smelting of 1.00 kg of pure Cu 2 O? a) 222 g b) 888 g c) 444 g d) 2252 g e) none of the above Chem 6A Michael J. Sailor, UC San Diego 6

4 Solution: Mass-to-mass calculations (1) The balanced equation for the reaction is: 2Cu 2 O + C 4Cu + CO 2 (2) Find out how many moles of Cu are there, then convert to grams Cu: 1000 g Cu 2 O Mol Cu 2 O 4 mol Cu g Cu g Cu 2 O 2 mol Cu 2 O 1 mol Cu = 888 g Cu Chem 6A Michael J. Sailor, UC San Diego 7 Problem: Limiting reactant In the Haber process, hydrogen (H 2 ) reacts with nitrogen (N 2 ) in a reactor to make ammonia. The reactor is initially charged with 2 mol of H 2 and 3 mol of N 2 and the reaction is allowed to go to completion. Fill out the table below indicating the amounts of reactants and products present after the reaction is complete. 3 H 2 + N 2 2 NH 3 Moles H 2 Moles N 2 Moles NH 3 Chem 6A Michael J. Sailor, UC San Diego 8

5 Solution: Limiting reactant The reactor is initially charged with 2 mol of H 2 and 3 mol of N 2 Balanced equation: 3 H 2 + N 2 2 NH 3 If all 2 mol of H 2 reacts, how many moles of N 2 are needed? Before reaction Moles H 2 2 Moles N 2 3 Moles NH 3 0 After reaction Chem 6A Michael J. Sailor, UC San Diego 9 Solution: Limiting reactant The reactor is initially charged with 2 mol of H 2 and 3 mol of N 2 Balanced equation: 3 H 2 + N 2 2 NH 3 If all 2 mol of H 2 reacts, how many moles of N 2 are needed? = 2/3 mol N 2 needed We have 3 mole of N 2, so N 2 is in excess and H 2 is limiting Chem 6A Michael J. Sailor, UC San Diego 10

6 Solution: Limiting reactant The reactor is initially charged with 2 mol of H 2 and 3 mol of N 2 and the reaction is allowed to go to completion. Balanced equation: 3 H 2 + N 2 2 NH 3 Before reaction What s used? What s left? After reaction, Moles H 2 = 2-2 = 0 Moles H Moles N 2 3 2/ Moles NH Moles N 2 = 3-2/3 = 7/3, or 2.33 Moles NH 3 = mol H 2 2 mol NH 3 = 4/3, or mol H 2 Chem 6A Michael J. Sailor, UC San Diego 11 Solution: Limiting reactant What if we chose to calculate it assuming the N 2 is limiting? Balanced equation: 3 H 2 + N 2 2 NH 3 If all 3 mol of N 2 reacts, how many moles of H 2 are needed? Before reaction Moles H 2 2 Moles N 2 3 Moles NH 3 0 After reaction Chem 6A Michael J. Sailor, UC San Diego 12

7 Solution: Limiting reactant If all 3 mol of N 2 reacts, how many moles of H 2 are needed? Balanced equation: 3 H 2 + N 2 2 NH 3 = 9 mol H 2 needed We only have 2 mole of H 2, so H 2 is limiting Chem 6A Michael J. Sailor, UC San Diego 13 Chapter 3: Stoichiometry (cont) Chem 6A, Section D Sept 29, 2011 Chem 6A Michael J. Sailor, UC San Diego 14

8 Problem: Limiting reactant Since the bronze age (4000 b.c.), copper metal has been produced by smelting, in which Cu 2 O ore is reduced with excess carbon (charcoal). How much copper can be produced from smelting of 1.00 kg of pure Cu 2 O with 25 g of charcoal? Assume the byproduct of the reaction is CO2. a) 222 g b) 888 g c) 444 g d) 2252 g e) none of the above Chem 6A Michael J. Sailor, UC San Diego 15 Solution: Limiting reactant The balanced equation for the reaction is: 2Cu 2 O + C 4Cu + CO 2 Find out how many moles of Cu 2 O we have: 1000 g Cu 2 O Mol Cu 2 O g Cu 2 O Find out how many moles of C we have: 25 g C Mol C g C how many moles of C are needed: = 6.99 mol Cu 2 O = 2.08 mol C 1000 g Cu 2 O Mol Cu 2 O 1 mol C g Cu 2 O 2 mol Cu 2 O = 3.49 mol C needed Chem 6A Michael J. Sailor, UC San Diego 16

9 Solution: Limiting reactant We have 2.08 mol C. We need 3.49, so all of the C will be used up before all of the Cu 2 O has reacted. So C (carbon) is the limiting reactant. how many moles of Cu will be produced using this much C? 25 g C Mol C 4 mol Cu g Cu g C 1 mol C Mol Cu = 529 g Cu Chem 6A Michael J. Sailor, UC San Diego 17 Problem: Combustion analysis Combustion analysis is carried out on g of a compound that contains only carbon, hydrogen, and oxygen. The masses of water and carbon dioxide produced are g and g, respectively. What is the empirical formula of the compound? See problem 3.34 from the book Chem 6A Michael J. Sailor, UC San Diego 18

10 Solution: Combustion analysis Like excess-limiting reagent problem except the oxygen is never limiting: C x H y O z + xs O 2 x CO 2 + y/2 H 2 O Figure out number of moles and mass of carbon, hydrogen: g CO 2 1mol CO 2 1 mole C g CO 2 1 mol CO 2 = moles C moles C g C 1 mol C = g C g H 2 O 1mol H 2 O 2 mole H g H 2 O 1 mol H 2 O = moles H moles H g H 1 mol H = g H Chem 6A Michael J. Sailor, UC San Diego 19 Solution: Combustion analysis C x H y O z + xs O 2 x CO 2 + y/2 H 2 O Mass left over = = g mass C x H y O z mass H mass C So g is the mass of O (NOT O 2 ) in the sample. The number of moles of O is: g O 1mol O g O = mol O Chem 6A Michael J. Sailor, UC San Diego 20

11 Solution: Combustion analysis C x H y O z + xs O 2 x CO 2 + y/2 H 2 O Dividing by the least common denominator (O in this case): Element Moles Fraction C H O So the empirical formula is C 2 H 6 O Chem 6A Michael J. Sailor, UC San Diego 21 Water: H!+ 2!- O 105 H!+ Chem 6A Michael J. Sailor, UC San Diego 22

12 Sodium Chloride (NaCl) Crystal Cl - Na + Unit Cell Chem 6A Michael J. Sailor, UC San Diego 23 Dissolution of NaCl in water ion-dipole interactions replace ion-ion interactions Solvent Solute = H 2 O = Na + = Cl - Chem 6A Michael J. Sailor, UC San Diego 24

13 Ethanol in water dipole-dipole interactions: Like Dissolves Like Hydrogen Bonds Chem 6A Michael J. Sailor, UC San Diego 25 Problem: Determine molar concentration What is the molarity of a solution made by adding enough water to 20.0 g of NaCl to make 50.0 ml of solution? MW of NaCl is g/mol. Set up but do not solve. Chem 6A Michael J. Sailor, UC San Diego 26

14 Solution: Determine molar concentration Molarity is defined as moles of solute/liter of solution: 20.0 g NaCl mol NaCl g NaCl L solution (= 6.84 M) Chem 6A Michael J. Sailor, UC San Diego 27 Problem: Dilution calculations What volume of M sucrose solution is needed to make 2.00 L of a 0.05 M sucrose solution? Set up but do not solve. Chem 6A Michael J. Sailor, UC San Diego 28

15 Solution: Dilution calculations 1. How many moles of sucrose are needed? 0.05 mol sucrose 2 Liters Liter = 0.1 mol sucrose 2. What volume of solution do I need? 0.1 mol sucrose Liter mol sucrose = L of solution Chem 6A Michael J. Sailor, UC San Diego 29 Problem: Dilution calculations Calculate the volume (in liters) of a 0.1 M KMnO 4(aq) stock solution that is needed to prepare 250 ml of 1.50 x 10-3 M KMnO 4(aq) (set up but do not solve). Chem 6A Michael J. Sailor, UC San Diego 30

16 Solution: Dilution calculations Volume = 1.5 x 10-3 mol KMnO ml L L L 1000 ml 0.1 mol KMnO 4(aq) (= 3.75 x 10-3 L) Chem 6A Michael J. Sailor, UC San Diego 31 Problem: Titration calculations A g sample of iron ore (mixture of Fe 2 O 3 and SiO 2 ) is dissolved in hydrochloric acid and 22.3 ml of M KMnO 4 are required to reach the stoichiometric point. What is the percent by mass of iron present in the sample? The balanced net ionic equation for the titration reaction is given below: 5Fe 2+ + MnO H + 5Fe 3+ + Mn H 2 O a) 63.6 % b) 7.28 % c) 36.4 % d) 100 % e) none of the above Chem 6A Michael J. Sailor, UC San Diego 32

17 Solution: Titration calculations Find out how many moles of Fe are there: 1.18 x 10-2 mol KMnO ml L 5 mol Fe 2+ L 1000 ml - 1 mol MnO 4(aq) = x 10-3 mol Fe 2+ = x 10-3 mol Fe. How many grams of Fe were there? x 10-3 mol Fe g Fe Mol Fe = grams Fe. So the % by mass Fe is: g/0.202g x 100% = 36.4% Chem 6A Michael J. Sailor, UC San Diego 33 Sections SECTION TIME LOCATION D01 M 2-2:50 pm WLH 2115 D02 M 3-3:50 pm WLH 2115 D03 M 4-4:50 pm WLH 2115 D04 W 2-2:50 pm WLH 2115 D05 W 3-3:50 pm WLH 2115 D06 W 4-4:50 pm WLH 2115 D07 F 2-2:50 pm WLH 2115 D08 F 3-3:50 pm WLH 2115 D09 F 4-4:50 pm WLH Chem 6A Michael J. Sailor, UC San Diego

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