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Unit # 3 Stoichiometry: Calculations with Chemical Formulas and Equations

CHM 1045 : General Chemistry and Qualitative Analysis. Unit # 3 Stoichiometry: Calculations with Chemical Formulas and Equations. Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL. Textbook Reference : Module #3 & 4. Chemical Reaction. H 2 O (g). CO 2 (g).

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Unit # 3 Stoichiometry: Calculations with Chemical Formulas and Equations

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  1. CHM 1045: General Chemistry and Qualitative Analysis Unit # 3Stoichiometry:Calculations with Chemical Formulas and Equations Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL • Textbook Reference: • Module #3 & 4

  2. Chemical Reaction H2O(g) CO2 (g) The actual phenomenon that occurs when chemical interact with each other. flame Methane gas is mixed with air and then it is light-up by a spark. O2 (g) CH4 (g) What is happening here?

  3. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Chemical Equations Reactants Products Symbolic representations of chemical reactions States Coefficients Subscripts Balanced Chemical Equations represent events that occur at the atomic level which we cannot perceive; but they explain the mass ratio of substances involved in a chemical equation (stoichiometry)

  4. Predicting Products: Types of Reactions What happens when substances react? AB  A + B A + B AB (1) Decomposition:(2) Combination (Synthesis): (3) Double Displacement (Replacement) or Metathesis, Exchange (4) Single Displacement (Replacement)(5) Combustion AB + CD  AD + CB where A & C are Metals, B & D Nonmetals MN + M MN + N (More Active Metal) M or N MN + (More Active Non-Metal) • : reactions of oxygen with an organic compounds (hydrocarbons, alcohols) that produce CO2 + H2O and a flame. • C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)

  5. Chemical Equations What happens when you mix (cause a reaction of) the following? Sodium Chloride • Sodium + Chlorine  • Dihydrogen Monoxide  • Magnesium + Hydrochloric Acid  • Hydrochloric Acid + Calcium Hydroxide  • Combustion (burning with oxygen) of: Sucrose (C12H22O11) Octane Hydrogen + oxygen Magnesium Chloride + Hydrogen Hydrogen hydroxide + Calcium chloride + Oxygen  Carbon dioxide + water + Oxygen  Carbon dioxide + water Write balanced chemical equations for each.

  6. * Predicting Products, writing Formulas and Balancing Equations • Alonso’s Rules for BE: • Easy element 1st hard elements last. • One element at a time. • Use fractions when necessary. • Sodium + Chlorine  • Dihydrogen Monoxide  • Magnesium + Hydrochloric Acid  • Hydrochloric Acid + Calcium Hydroxide  • Combustion (burning with oxygen) of: • Sucrose (C12H22O11) • Octane Sodium Chloride 2 Na + Cl2 H2O  Mg + HCl  HCl + Ca(OH)2  C12H22O11 + O2 C8H18 + O2 NaCl H2 + O2 MgCl2 + H2 HOH + CaCl2 CO2+ H2O CO2+ H2O 2 Hydrogen + oxygen 2 2 Magnesium Chloride + Hydrogen 2 Hydrogen hydroxide + Calcium chloride 2 2 Carbon dioxide + water 12 12 11 Carbon dioxide + water 16 18 2 12.5 8 9 25

  7. Decomposition Reactions Simple: Binary compounds break down into their constituent elements 2H2O 2H2 + O2 2NaCl(l) 2Na (l) + Cl2(g) 2NaN3(s) 2Na(s) +3N2(g) electrolysis electrolysis heat {AirBags Movie*} ∆ sodium azide (N31-) Important Exception: Catalyst 2H2O2 2H2O + O2 {Peroxide Movie}

  8. CaCO3 (s) CaO (s) + CO2 (g) Decomposition Reactions ComplexCompounds decompose into simpler compounds Allcarbonatesbreak down to metal oxides and carbon dioxide Chloratesbreak down to metal chlorides and oxygen • 2 KClO3 (s)  2 KCl (s) + 3O2 (g) ∆ Acidsbreak down to nonmetal oxides and water • 2 H3PO4 (aq)  P2O5(g) + 3H2O (l) • 2HNO3 (aq)  N2O5(g) + H2O (l) • 2 NaOH (aq) Na2O (s) + H2O (l) Basesbreak down to metal oxides and water

  9.  Na2O + CO2 + H2O Ammonium carbonate powder is heated strongly

  10. 2 Mg (s) + O2 (g)  2 MgO (s) Zn (s) + S (s)  ZnS (s) 2 H2 (g) + O2 (g) 2 H2O (l) 2 Al (s) + 3 Br2 (l)  2 AlBr3 (s) Simple: Two or more elements react to form one compound A + B  AB Combination (Synthesis) Reactions Now let’s balance equations {Mg Movie} {ZnS Movie*} {H2O Movie*} {AlBr3 Movie*}

  11. Bromine liquid is poured over aluminum metal Hydrogen chloride and ammonia gas are mixed together. Sulfur dioxide gas is bubbled into water.

  12. Example: what quantity of Baking Soda will react with 100mL of vinegar? NaHCO3 (s) + HC2H3O2 (l) NaC2H3O2 (aq) + HHCO3 (aq) H2CO3 (aq)  H2O (l) + CO2 (g) (2nd Rx decomposition) Involve two Compounds Elements (or polyatiomic groups) in the two compounds exchange partners Metathesis (Double Displacement) AB + CD  AD + CB where A & C are Metals, B & D Nonmetals {Movie: Bicarb + Vineg with Stoichio&LimitReag *}

  13. Examples: HCl (aq) + NaOH(aq) NaCl (aq) + HOH (l) 2 HCl (aq) + Ca(OH)2(aq) CaCl2 (aq) + 2 HOH (l) HCl (aq) + NH4OH(aq) NH4Cl (aq) + 2 HOH (l) Metathesis (Double Displacement): Acid-Base Neutralization Reaction Acid: compound containing hydrogen and a non metal (HN) Bases: a metal hydroxide (MOH) HN + MOH  MN + HOH Acid +Base Salt+ Water {Movie: A-B RxNo Ind} {Movie: A-B Rx Ind+pH meter*}

  14. A more active element displacing a less active elements from a compound. Single Displacement Reactions (Single Replacement Rx.) (More Active Metal) element MN + M MN + N M or N compound MN + (More Active Non-Metal) Activity (Electromotive) Series: Metals: Li > K > Ba > Sr > Ca > Na > Mg > Al > Zn > Cr > Fe > Cd > Co > Ni > Sn > Pb > (H) > Sb > Bi > Cu > Hg > Ag > Pd > Pt > Au Halogens:F >Cl > Br > I

  15. Examples: Cu (s) + 2 AgNO3 (aq) Cu (s) + Zn(NO3)2 (aq) Cl2(g) + 2 NaBr (aq) A more active element displacing a less active elements from a compound. Single Displacement Reactions {Movie:Cu+AgNO3} • 2 Ag +Cu(NO3)2 (aq) • No Reaction • 2 NaCl (aq) + Br2 (aq) Activity (Electromotive) Series: Metals: Li > K > Ba > Sr > Ca > Na > Mg > Al > Zn > Cr > Fe > Cd > Co > Ni > Sn > Pb > (H) > Sb > Bi > Cu > Hg > Ag > Pd > Pt > Au Halogens: F >Cl > Br > I Activity series can also be found in form of Reduction Potential table.

  16. Most Active Nonmetal Most Active Metal

  17. H+OH-

  18. Reactions with Oxygen What is the difference? Oxidation Rx. (1) Oxidation Reactions: are combination reactions involving oxygen. {Movie: Mg, Fe, P, S +conc. O2 {Metal Oxides*} (2) Combustion Reaction: Rapid reactions of oxygen with an organic compounds (hydrocarbons, alcohols) that produce CO2 + H2O and a flame. Examples: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g) Combustion Rx {Movie: CH3OH + O2*}

  19. When oxygen is scarce…. *

  20. gram-Molar Mass(g-MM) =Atomic Weigh, Formula Weigh or Molecular Weight Mole = 6.022 x 1023 particles Mass : Weight (in grams)

  21. gram-Molar Mass (g-MM): AW, FW, MW * the mass in grams of 1 mole of a substance (units= g/mol) For an element we find it on the periodic table. • For compounds the same as the formula & molecular weight (but in g/mol) • Example: the g-MM of Al2(SO4)3, would be • 2 Al: 2x(26.98 amu) = 53.96 • + 3 S: 3x(32.06 amu) = 96.18 • +3x4 O: 12x(16.00 amu) =192.00 • 342.14 amu (g/mol)

  22. 2 x C: 2x(12.0 amu) 6 x H: 6x(1.0 amu) 30.0 amu Formula Weight (FW) • Sum of the atomic weights for the atoms in a chemical formula unit(ionic compound) • So, the formula weight of calcium chloride, CaCl2, would be Ca: 1x(40.1 amu) 2 x Cl: 2x(35.5 amu) 111.1 amu Molecular Weight (MW) • Sum of the atomic weights of the atoms in a molecule (covalent compound) • For the molecule ethane, C2H6, the molecular weight would be

  23. (1) ( 197) (# atoms of element) (atomic weight of Au) x 100 % element = % Au = x 100 ( 2,301) (MW of Pb5Au(TeSb)4S5) (Mass) Percent Composition Percentage mass of a element (Na) in compound (NaCl): g-MM = 2,301g/ Nagyagite Gold Ore: Pb5Au(TeSb)4S5 Problem: (1) calculate mass % Au in Nagyagite. (2) If you buy 1 kg of the ore, how much gold does it have? = 8.56 %

  24. Using mass % to determine mass of one particular element in a sample What is the mass of carbon in a 25g sample of carbon dioxide? There are two parts to this problem: (1) What is the percentage mass of carbon in carbon dioxide? % C = (2) What is the mass of carbon in a 25 g sample of carbon dioxide? x 27% ? g C = 25 g CO2

  25. Dermatological Chemical Biological The Mole Concept Avogadro's Number: 6.022,141,410,704,090,840,990,72 x 1023 602,214,141,070,409,084,099,072 . pentillion trillion sextillion quadrillion billion million thousand 602 sextillion

  26. Using Equivalences as Mole Ratios: From Equivalences we obtain useful Ratios or Conversion factors: g-MM = 1 Mole () = 6.022 x 10 23 particles NaCl = 58g/η(Atoms or molecules) or or or

  27. * Mole Calculations: g-MM Moles # of Particles Which ratios will you need? g-MM Moles: ? g = 3.20 mol of NaCl or ? mol = 3.20 g of NaCl = 1.9 x 10 24 f.u. Moles # of Particles: ? f.u. = 3.2 mol NaCl or = 5.3 x 10 28 mol ? mol = 3.2 x 10 52 f.u. NaCl

  28. Mole Calculations: g-MM Moles # of Particles g-MM # of Particles: ? g = 4.2 x 10 34 f. u. of NaCl 4.1 x 1012 g 4.1 x 1012g ? f. u. of NaCl = 3.2 g of NaCl 3.3 x 1022 f.u. 3.3 x 1022f.u.

  29. * Mole Calculations: g-MM Moles # of Particles (atoms or molecules) How many molecules of H2O in 29g of water? How many atoms? ? atoms = 0.50 mole Fe(CO)3(PH3)2 = 3.0 X 1023 f.u. 2+ 15 1+ 3+ 3+ 6 = = 4.5 x 1024 atoms

  30. The Determination of Empirical Formulas of Compounds by Elemental Analysis CxHy Combustion furnace

  31. Types of Formulas • Structural formulas(skeletal or space-filling) show the order in which atoms are bonded and their three-dimensional shape. • Empirical formulas give the lowest whole-number ratioof atoms of each element in a compound. • Molecular formulas give the exact number of atoms of each element in a compound. H2O HO CH Why are empirical formulas needed? Benzene, C6H6

  32. Elemental Analyses How do we determine the formula of a compound? Compounds are broken down and the massesof their constituent elements are measured. From these masses the empirical formulas can be determined. HxOy moles Mole Ratio Expt. Data: (68g) 4g H 64g O EmpF HO

  33. * Calculating Empirical Formulas Problem: The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%),nitrogen (10.21%),and oxygen (23.33%). Find the empirical formula of PABA. carbon (61.31g), hydrogen (5.14g),nitrogen (10.21g),oxygen (23.33g) carbon (61.31%), hydrogen (5.14%),nitrogen (10.21%),oxygen (23.33%). Percent means out of 100, so assume a 100g sample of the compound, then…. Calculate the empirical formula (mole ratio) from the percent composition (% mass).

  34. Assuming 100.00 g of para-aminobenzoic acid, ? mol C = 61.31 g x = 5.105 mol C ? mol H = 5.14 g x = 5.09 mol H ? mol N = 10.21 g x = 0.7288 mol N ? mol O = 23.33 g x = 1.456 mol O 1 mol 12.01 g 1 mol 1.01 g 1 mol 14.01 g 1 mol 16.00 g Calculating Empirical Formulas 5.105 mol 0.7288 mol 61% C = 7.005  7 = 6.984  7 = 1.000 = 2.001  2 5.09 mol 0.7288 mol 5% H 0.7288 mol 0.7288 mol 10% N 1.458 mol 0.7288 mol 23% O What is the smallest mole ratio of the elements in this compound? Calculate the mole ratioby dividing by the smallest number of moles. These are the subscripts for the empirical formula: C7H7NO2

  35. Combustion Analysis: is amethod of experimentally determining empirical formulas CxHy + O2 CO2 + H2O mass of CO2 mass of H2O mass mass of C? mass of H? How do you calculate the mass of C in CO2 and that of H in H2O? mass of H2O mass of CO2 Magnesium perchlorate Sodium hydroxide {Movie} CxHy • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this • C is determined from the mass of CO2 produced • H is determined from the mass of H2O produced • O is determined by difference after the C and H have been determined Combustion furnace

  36. Calculating Empirical Formulas • CxHy (g) + O2 (g) CO2 (g) + H2O (g) (12g:32g=44g) 14.6 g 5.00 g A 5.00 g sample of an unknown hydrocarbon was burned and produced 14.6 g of CO2. What is the empirical formula of the unknown compound? Empirical Formula CH3 ? g C = 14.6 g CO2 = 3.98 g C g H = 5.00 CxHy – 3.98 g C = 1.02 g H ? mol C = 3.98 g C = 0.332 mol C / 0.332 = 1.00 ? mol H = 1.02 g H = 1.01 mol H / 0.332 = 3.04

  37. 2006 A ?g C = ?g N =

  38. 2003 B

  39. Stoichiometry (mass relationships within chemical equations)

  40. Stoichiometric Calculations 2 2 The coefficients in the balanced equation can also be interpreted as mole ratios of reactants and products Mole Ratios from Balanced Equation:

  41. * Stoichiometric Calculations How many grams of O2 are required to form 2.35 g of MgO? 2 Mg (s) + O2(g) 2 MgO (s) No direct calculation grams grams Change: • grams of MgO  mol MgO • mol of MgO  mol O2 • mol of O2  grams of O2 2 Mg (s) + O2(g)  2 MgO (s) mole ratio from balanced equation

  42. Stoichiometric Calculations Balanced Eq. uses MOLE language C6H12O6 + 6 O2 6 CO2 + 6 H2O ? g H2O Grams H2O Grams C6H12O6 (3) (1) (2) Moles C6H12O6 Moles H2O (1) Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… (2) use the coefficients to find the moles of H2O… (3) and then turn the moles of water to grams

  43. Stoichiometry: Limiting Reactants (or, too much of one reactant andnot enough of the other) Make cookies until you run out of one of the ingredients In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make. { MovieLimitingReactants: Zn + 2 HCl  ZnCl2 + H2 }

  44. Limiting & Excess Reactants • The limiting reactant is the reactant present in the smallest stoichiometric amount • In other words, it’s the reactant you’ll run out of first • Which is Limiting which is Excess? • Limiting H2; ExcessO2

  45. Limiting & Excess Reactants Problem 2 Mg (s) + O2(g)  2 MgO (s) If 5.0g of both Mg and O2 are used: • Which is the limiting and the excess reactants? • How much of the excess will be left unreacted ? • How much MgO will be produced ?

  46. Limiting & Excess Reactants Problem 5 g 5 g 2 Mg (s) + O2(g)  2 MgO (s) ? g O2 = 5.0 g Mg ? g Mg = 5.0 g O2 = 3.29 g O2 = 7.59 g Mg Mg is limiting reactant and O2 is excess reactant Which one do I use to determine the MgO produced by rx? ?g MgO = 5.0 g Mg

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