Formulas, Equations and Moles - General Chemistry - Notes | CHM 121, Study notes of Chemistry

Download Formulas, Equations and Moles - General Chemistry - Notes | CHM 121 and more Study notes Chemistry in PDF only on Docsity! Chapter 3 Formulas, Equations, and Moles General Chemistry John E. McMurry 2 •2 HCl(aq) + CaCO3(s)  CaCl2(aq) + H2O(l) +CO2(g) •HCl and CaCO3 are called reactants • CaCl2, H2O,CO2 are called the products •Reactants are separated from products with “” that means “yields” •States matter: for solids use (s), liquids (l), gases (g), and for substances dissolved in water (aqueous solutions) use (aq). Chemical Equations Balancing Chemical Equations HgI2(s) + 2KNO3(aq)Hg(NO3)2(aq) + 2KI(aq) right side: 1 Hg 2 I 2 K 2 N 6 O left side: 1 Hg 2 N 6 O 2 K 2 I Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 3/6 Balancing Chemical Equations 1.Find suitable coefficients—the numbers placed before formulas to indicate how many formula units of each substance are required to balance the equation. 2H2O(l)2H2(g) + O2(g) 1.Write the unbalanced equation using the correct chemical formula for each reactant and product. H2O(l)H2(g) + O2(g) Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 3/7 Balancing Chemical Equations 1.Reduce the coefficients to their smallest whole- number values, if necessary, by dividing them all by a common denominator. 2H2O(l)2H2(g) + O2(g) 4H2O(l)4H2(g) + 2O2(g) divide all by 2 2 Mg + O2  2 MgO 2 Al + 3 Br2  2 AlBr3 6 HCl + 2 Al  2 AlCl3 + 3 H2 CuSO4 + Na2CO3  CuCO3 + Na2SO4 It’s ok as it is 2 Al + 3 H2SO4  Al2(SO4)3 + 3 H2 P2O5 + 3 H2O  2 H3PO4 B2H4 + 6 H2O  2 H2BO3 + 6 H2 Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 3/15 Avogadro’s Number and the Mole Avogadro’s Number (NA): One mole of any substance contains 6.022 x 1023 formula units. One mole of any substance is equivalent to its molecular or formula mass. 1 mole = 28.0 gC2H4: 1 mole = 36.5 gHCl: How many moles of carbon are there in 36.0 grams of carbon? 1 mole carbon = 12.0 grams carbon carbon moles 3.00 carbon grams 12.0 carbon mole 1 X carbon g 0.36  How many moles of sodium are there in 56.25 grams? 1 mole sodium = 23.0 grams sodium Na moles 2.45 Na g 23.0 Na mole 1 X Na g 6.255  How many nitrogen atoms are there in 5.25 grams of nitrogen atoms? # of atoms # of moles Mass 6.02 X 1023 Atomic mass 23 23 10 X 2.26 N mole 1 atoms 10 X 6.02 X N g 14.0 N mole 1 X N g 25.5  How many moles of CO2 are there in 26.5 grams CO2? •Molar Mass – Mass of one mole of a compound •Molecular weight •Formula weight •Carbon 1 X 12.0 g = 12.0 •Oxygen 2 X 16.0 g = 32.0 Total 44.0 g/mol 2 2 2 2 CO mole 0.602 CO g 44.0 CO mole 1 X CO g 5.26  How many moles of CO2 are there in 125.0 grams of CO2? # of units # of moles Mass 6.02 X 1023 molar mass 2 2 2 2 CO moles 2.84 CO g 44.0 CO mole 1 X CO g 0.125  How many molecules is this? molecules 10 X 1.71 CO mole 1 molecules 10 X 6.02 X CO moles 2.84 24 2 23 2  Using Conversion Factors in Chemical Reactions How many moles of N2 will be needed to make 3 moles of NH3 in this reaction? N2 + 3 H2  2 NH3 2 3 2 3 N moles 1.5 NH moles 2 N mole 1 X NH moles 3  How many moles of O2 will be needed to make 0.750 moles of CO2 in this reaction? C3H8 + 5 O2  3 CO2 + 4 H2O 2 2 2 2 O moles 1.25 CO moles 3 O mole 5 X CO moles 0.750  Putting it All Together grams  moles  moles  grams Grams of substance A Number of moles of A Number of moles of B Molar mass of substance A Balanced Reaction Coefficients Molar mass of substance B Grams of substance B Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 3/30 Stoichiometry: Chemical Arithmetic How many grams of NaOH are needed to react with 25.0 g Cl2? 2NaOH(aq) + Cl2(g) NaOCl(aq) + NaCl(aq) + H2O(l) Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas: 1 mol NaOH 40.0 g NaOH25.0 g Cl2 70.9 g Cl2 1 mol Cl2 1 mol Cl2 2 mol NaOH = 28.2 g NaOH xx x Consider the reaction 2 Al + 3Cl2  2 AlCl3 How many grams of chlorine gas will be needed to react with 5.75 grams of aluminum? Almoles 0.213 Alg 27.0 Almole 1 X Alg 75.5  2 2 Cl mol 320.0 Almol 2 Cl mol 3 X Almol 213.0  2 2 2 2 Cl g 7.22Cl mol 1 Cl g 71.0 X Cl mol 320.0  4 FeS2(s) + 11 O2(g)  2 Fe2O3(s) + 8 SO2(g) What mass of O2 is needed to react with 15.0 grams of FeS2? Strategy: g FeS2  mol FeS2  mol O2  g O2 2 2 2 2 2 2 2 2 O g 0.11O mol 1 O g 32.0 X FeS mol 4 O mol 11 X FeS g 120. FeS mol 1 X FeS g 0.15  Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 3/35 Reactions with Limiting Amounts of Reactants Limiting Reactant: The reactant that is present in limiting amount. The extent to which a chemical reaction takes place depends on the limiting reactant. Excess Reactant: Any of the other reactants still present after determination of the limiting reactant. N2 + 3 H2  2 NH3 5.00 grams of H2 are reacted with excess N2, and 15.5 grams of NH3 are recovered. What is the percent yield? 3 3 3 2 3 2 2 2 NH g 4.28NH mol 1 NH g 17.0 X H mol 3 NH mol 2 X H g 2.00 H mol 1 X H g 5.00  Theoretical Yield = 28.4 g %6.54%100X NH g 28.4 NH g 15.5 yieldpercent 3 3  Limiting Reagent Problems •How many bicycles can be made using 30 tires and 20 frames? •When two pieces of starting information are given, the first task is to determine which will run out first, and stop the reaction. •In chemistry start each as a separate problem, and continue to moles of product. •Then pick the limiting reagent which produces the smallest amount of product Chapter 3/40 Reactions with Limiting Amounts of Reactants If 3 moles of ethylene oxide react with 5 moles of water, which reactant is limiting and which reactant is present in excess? C2H4O(aq) + H2O(l) C2H6O2(l) Tippetts’ Method for Limiting Reagent A1 + A2  B Start as two separate problems A1 and A2 1.Grams A1  moles A1  moles B Grams A2  moles A2  moles B 1.Choose the smallest number of moles of B This will be from the limiting reagent 2.Finish that problem be converting it to grams of B Pick smallest = Limiting 2 HCl + Mg  MgCl2 + H2 How many grams of MgCl2 can be made from 50.0 grams of HCl and 30.0 grams of Mg? 2 2 MgCl mol 685.0 HCl mol 2 MgCl mol 1 X HCl g 36.5 HCl mol 1 X HCl g 0.50  2 2 MgCl mol 23.1 Mg mol 1 MgCl mol 1 X Mg g 24.3 Mg mol 1 X Mg g 0.30  2 2 2 2 MgCl g 65.3 MgCl mol 1 MgCl g 95.3 X MgCl mol 685.  Preparation of Solutions Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 3/45 Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 3/46 Concentrations of Reactants in Solution: Molarity Molarity converts between mole of solute and liters of solution: molarity = moles of solute liters of solution L mol or 1.00 M 1.00 L 1.00 mol = 1.00 1.00 mol of sodium chloride placed in enough water to make 1.00 L of solution would have a concentration equal to: A 500.0-g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? 47 43 43 43 43 POK mol 355.2 POK g 212.27 POK mol 1 POK g .0500        M 57.1 L 50.1 POK mol 355.2 43  Preparation By Dilution of Stock Solution Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 3/50 Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 3/51 Diluting Concentrated Solutions Add 6.94 mL 18.0 M sulfuric acid to enough water to make 250.0 mL of 0.500 M solution. Mi = 18.0 M Mf = 0.500 M Vi = ? mL Vf = 250.0 mL = 6.94 mL 18.0 M 250.0 mL Vi = Mi Mf Vf 0.500 M = Sulfuric acid is normally purchased at a concentration of 18.0 M. How would you prepare 250.0 mL of 0.500 M aqueous H2SO4? x What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution? M1V1 = M2V2 (2.00 M)(V1) = (0.800 M)(150.0 mL) V1 = 60.0 mL •Consider the reaction: 2 NaOH + H2SO4  Na2SO4 + 2 H2O How many grams of NaOH would be needed to react with 125 mL of 0.400 M H2SO4? mL H2SO4  mol H2SO4  mol NaOH  g NaOH 42 42 42 42 SOH mol 0500.0 SOH L. 1 SOH mol 0.400 X SOH L. 0.125  NaOH mol 100.0 SOH mol 1 NaOH mol 2 X SOH mol 0500.0 42 42  NaOH g 00.4 NaOH mol 1 NaOH g 40.0 X NaOH mol 100.0  •Consider the reaction: 2 HCl(aq) + Mg(s)  MgCl2 + H2 How many mL of 4.50 M HCl will be needed to completely react with 15.0 g of Mg? g Mg  mol Mg  mol HCl  mL HCl HCl mol 22.1 Mg mol 1 HCl mol 2 Mg g 24.3 Mg mol 1 X Mg g 0.15 X HCl mL 271 HCl mol 4.50 HCl mL 1000 X HCl mol 22.1  Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 3/57 Titration How can you tell when the reaction is complete? HCl(aq) + NaOH(aq) NaCl(aq) + 2H2O(l) Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known. Once the reaction is complete you can calculate the concentration of the unknown solution. Copyright © Cengage Learning. All rights reserved 60 •Mass percent of an element: •For iron in iron(III) oxide, (Fe2O3): mass of element in compound mass % = × 100% mass of compound 2( 55.85 g) 111.70 g mass % Fe = = × 100% = 69.94% 2( 55.85 g)+3( 16.00 g) 159.70 g Copyright © Cengage Learning. All rights reserved 61 •Empirical formula = CH Simplest whole-number ratio •Molecular formula = (empirical formula)n [n = integer] •Molecular formula = C6H6 = (CH)6 Actual formula of the compound Formulas Percent Composition and Empirical Formula 1.Convert % composition to actual mass 2.Convert mass to moles of each 3.Divide each # of moles by the smallest 1.Results will be subscripts in the formula 2.Convert to whole number if needed 1 to 1.33, 1 to 1.50, 1 to 1.67