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Chapter 4

Chapter 4. Chemical Equations and Stoichiometry. Chapter goals. Balance equations for simple chemical reactions. Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of a limiting reagent.

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Chapter 4

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  1. Chapter 4 Chemical Equations and Stoichiometry

  2. Chapter goals • Balance equations for simple chemical reactions. • Perform stoichiometry calculations using balanced chemical equations. • Understand the meaning of a limiting reagent. • Calculate the theoretical and percent yields of a chemical reaction. • Use stoichiometry to analyze a mixture of compounds or to determine the formula of a compound.

  3. Chemical Equations • shorthand notation for chemical change • based on the law of conservation of matter (A. Lavoisier, 18th century): valid for ordinary reactions. For extraordinary reactions, such as nuclear reactions, the more general law of conservation of energy holds

  4. Symbolism • starting materials (reactants) on left • products on right • reactants separated from products by  or • reactants separated from each other with +, same for products •  heat • h light • balanced by both mass and charge by using coefficients • use correct molecular formulas  

  5. Symbolism (contd…) • trailing subscripts • (g) gas • (l) liquid • (s) solid • (aq) aqueous solution • catalysts, special solvents, special conditions written above and/or below arrow,  heat, h light

  6. Balancing Chemical Equations • done by inspection • usually best to begin with largest molecule • for reactions of organic compounds, such as combustion reactions, balance C atoms first. Secondly balance H atoms. Then, balance O atoms • convert any fractional coefficient to whole number by multiplying all equation times the denominator of fraction

  7. Example, Balance Al(s) + O2(g) Al2O3(s) Balancing Al 2 Al(s) + O2(g)Al2O3(s) Balancing O 2 Al(s) + 1.5 O2(g)Al2O3(s) fractional coefficients not allowed then, multiply  2 (2  1.5 = 3) 4 Al(s) + 3 O2(g)2Al2O3(s)

  8. Example, Balance NH3(g)+ O2(g)  NO2(g) + H2O(l) Balancing H 2 NH3(g)+ O2(g)  NO2(g) + 3 H2O(l) Balancing N 2 NH3(g)+ O2(g) 2 NO2(g) + 3 H2O(l) 3.5  2 = 7 O  (4 + 3) O Balancing O 2 NH3(g)+ 3.5 O2(g)  2 NO2(g) + 3 H2O(l) 4 NH3(g)+ 7 O2(g)  4 NO2(g) + 6 H2O(l)

  9. Types of Reactions • combination • A + B  C • decomposition • A  B + C • combustion • A + O2 products

  10. Combustion Reactions • reaction with molecular oxygen • of hydrocarbons produces •  carbon dioxide and water • e.g.. CH4(g) + O2(g)  CO2(g) + H2O(l) • (methane) • CH4(g) + 2 O2(g)  CO2(g) + 2H2O(l) • Also: 2 H2(g) + O2(g)  2 H2O (l)

  11. Combustion Reactions • C5H12(g) + O2(g)  CO2(g) + H2O(l) • Balance C atoms • C5H12(g) + O2(g) 5CO2(g) + H2O(l) • Balance H atoms • C5H12(g) + O2(g)  5CO2(g) + 6H2O(l) • Balance O atoms, 5x2 + 6 = 16 O • C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(l)

  12. Combustion Reactions • C6H14(g) + O2(g)  CO2(g) + H2O(l) • Balance C atoms • C6H14(g) + O2(g) 6CO2(g) + H2O(l) • Balance H atoms • C6H14(g) + O2(g)  6CO2(g) + 7H2O(l) • Balance O atoms, 6x2 + 7 = 19 O • C6H14(g) + 9.5O2(g)  6CO2(g) + 7H2O(l) • Multiply both sides of equation by 2 • 2C6H14(g) + 19O2(g)  12CO2(g) + 14H2O(l)

  13. Combustion Reactions • NH3(g) + O2(g)  NO(g) + H2O(g) • N atoms are balanced Balance H atoms 2NH3(g) + O2(g)  NO(g) + 3H2O(g) N atoms are not balanced now 2NH3(g) + O2(g) 2NO(g) + 3H2O(g) • Balance O atoms, 2 + 3 = 5 O • 2NH3(g) + 2.5O2(g)  2NO(g) + 3H2O(g) • Multiply both sides of equation by 2 • 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

  14. Stoichiometry • calculations involving chemical equations Steps • write balanced chemical equation • convert given quantities to moles • determine limiting reagent if necessary • calculate number of moles of desired substance • convert to desired units

  15. Stoichiometry Calculations are based on relationships similar to those in the following example CH4(g)(methane) + 2 O2(g)  CO2(g) + 2H2O(l) 1 molecule of CH4reacts with 2 molecules of O2 to produce 1 molecule of CO2 + 2 molecules of H2O 1 mole of CH4reacts with 2 moles of O2 to produce1 mole of CO2 + 2 moles of H2O 16.04 g of CH4reacts with 2x32.00 of O2 to produce 44.01 g of CO2+ 2x18.02 g of H2O

  16. Example: Iron reacts with steam to form Fe3O4 and H2. Calculate mol H2 produced by reaction of 10.0 g iron with excess steam. Fe(s) + H2O(g)  Fe3O4(s) + H2(g) 3 Fe(s) + 4 H2O(g)  Fe3O4(s) + 4 H2(g)

  17. First, we convert g of Fe into moles Fe 1 mol Fe 10.0 g Fe x —————— = 0.179 mol Fe 55.85 g Fe Second, we use Fe moles to calculate H2’s. Water is not used because it is excess 4 mol H2 0.179 mol Fe x ————— = 0.239 mol 3 mol Fe

  18. Example: Tin(IV) oxide reacts with carbon at high temperature to form elemental tin and carbon monoxide. How many kg of carbon are needed to convert 50.0 kg of tin(IV) oxide to elemental tin? • Sn4+ • O2– • SnO2 • SnO2 + C  Sn + CO To balance the equation we need 2 O on the right and so on.

  19. SnO2 + 2 C  Sn + 2 CO 1x 103 g 1 mol SnO2 50.0 kg SnO2 x ———— x —————— 1 kg 150.7 g SnO2 2 moles C 12.01 g C 1 kg C x —————— x ———— x ———— 1 mol SnO2 1 mol C 1x 103 g C 7.97 kg C (we may use kmol too) (332 moles SnO2 664 moles C)

  20. The air in a closed container consists of 1.40 mol O2, 7.00 mol N2, with traces of other gases. A small lamp fueled by methanol, CH3OH, is lighted in the container. How many mL of methanol (d=0.791 g/mL) will be consumed when the lamp goes out? • CH3OH(l) + O2(g) • CH3OH(l) + O2(g) CO2(g) + H2O(l) • 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(l) I recommend you to practice this balancing

  21. 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(l) We can determine the amount of CH3OH out of O2’s 2 mol CH3OH 1.40 mol O2 x —————— = 0.933 mol CH3OH 3 mol O2 32.04 g CH3OH 1 mL CH3OH 0.933 mol x —————— x —————— 1 mol CH3OH 0.791 g CH3OH = 37.8 mL CH3OH

  22. Limiting Reagent (reactant) • the reactant that is present in quantity smaller to completely react other reactant; is consumed completely during the reaction; determines amount of product yielded It can be also seen as the reagent that theoretically produces the smallest amount of product(s). • excess reagent (reactant): the reactant present in quantity greater than needed for the reaction; part of it remains after the reaction is completed

  23. Example: Copper(II) oxide reacts with ammonia to form copper, water, and nitrogen. If 236.1 g copper(II) oxide are treated with 64.38 g ammonia, how much copper (grams) is produced? How many g of excess reagent remain? • CuO • NH3 • CuO(s) + NH3(g)  Cu(s) + H2O(l) + N2(g) • 3CuO + 2NH3(g)  3Cu(s) + 3H2O(l) + N2(g)

  24. 3CuO + 2NH3(g)  3Cu(s) + 3H2O(l) + N2(g) 1 mol CuO 236.1 g CuO x —————— = 2.968 mol CuO 79.55 g CuO 1 mol NH3 64.38 g NH3 x —————— = 3.780 mol NH3 17.03 g NH3 Now, let’s see which one of the two, CuO or NH3, theoretically produces the smallest amount of a product, Cu in this case. 3 mol Cu 2.968 mol CuO————— = 2.968 mol Cu 3 mol CuO And, now, ammonia, NH3, …

  25. 3CuO + 2NH3(g)  3Cu(s) + 3H2O(l) + N2(g) 3 mol Cu 3.780 mol NH3————— = 5.670 mol Cu 2 mol NH3 Then, the amount of Cu we can obtain with CuO, 2.968 mol, is less than what we can get with NH3, i.e., CuO is the LR. Conclusion: we must use the amount of CuO to calculate the amount of Cu yielded and for the amount of NH3 that has reacted. 2.968 mol CuO 3.780 mol NH3 ───────── = 0.989 < ───────── = 1.89 3 (coeff.) 2 (coeff.)

  26. 3CuO + 2NH3(g)  3Cu(s) + 3H2O(l) + N2(g) How many moles of Cu? 2.968 mol Cu 63.55 g Cu 2.968 mol Cu x ————— = 188.6 g Cu 1 mol Cu How many g of NH3 in excess? started with 3.780 mol NH3. The reacted from CuO, 2 mol NH3 2.968 mol CuO x ————— = 1.979 mol NH3 used up 3 mol CuO Then, by subtraction 3.780 – 1.979 = 1.801 mol NH3 17.03 g left over 1.801 mol NH3 x ———— = 30.67 g NH3 1 mol

  27. 4 Li(s) + O2(g) 2 Li2O(s) start (given) 4.0 mol 1.0 mol 0 mol reacted -4.0 mol -1.0 mol +2.0 mol Finish 0 mol 0 mol 2.0 molNoLR Which one is the limit reagent, Li or O2? 2 mol Li2O 4.0 mol Li ───────= 2 molLi2O theoret. produced 4 mol Li 2 mol Li2O 1.0 mol O2───────=2 mol Li2O theoret. produced 1 molO2 Then, there is no L. R. Both react completely

  28. 4 Li(s) + O2(g) 2 Li2O(s) Start (given)4.0 mol 0.5 mol 0 mol reacted -2.0 mol -0.5 mol +1.0 mol Finish 2.0 mol 0 mol 1.0 molO2 LR Which one is the limit reagent, Li or O2? 2 mol Li2O 4.0 molLi ───────= 2 mol Li2O theoret. produced 4 mol Li 2 mol Li2O 0.5 molO2───────= 1mol Li2O theort. produced 1 molO2 Then, O2 is the L. R. 1.0 mole of Li2O formed; 2 mole of Li excess

  29. 4 Li(s) + O2(g) 2 Li2O(s) start 8.0 mol 3.0 mol 0 mol reacted -8.0 mol -2.0 mol +4.0 mol Finish 0 mol 1.0 mol 4.0 mol Li LR Homework…

  30. Example: Dihydrogen sulfide and sulfur dioxide react to form sulfur & water. How much sulfur is formed when 5.00 g dihydrogen sulfide are mixed with 5.00 g of sulfur dioxide? How many g of excess reagent remain after reaction? Equation H2S(g) + SO2(g) S8(s) + H2O(l) balancing (try it) 16 H2S(g) + 8 SO2(g)  3 S8(s) + 16H2O(l)

  31. Firstly: calculate moles (of molecules) of H2S and SO2. Then, determine Limiting Ractant 1 mol H2S 5.00 g H2S x ————— = 0.147 mol H2S 34.1 g H2S 1 mol SO2 5.00 g SO2 x ————— = 0.0780 mol SO2 64.1 g SO2 How much S8? 16 H2S(g) + 8 SO2(g)  3 S8(s) + 16 H2O(l) 3 mol S8 0.147 molH2S x ————— = 0.0276molS8 16 molH2S (smaller) 3 mol S8 0.0780 mol SO2 x ————— = 0.0292 mol S8 8 mol SO2 H2S is the L.R.

  32. Theoretically, 0.147 mol of H2S produces 0.0276 mol S8, and 0.0780 mol SO2produces 0.0292 molS8. Then, H2S is the Limiting Reactant, and SO2 is the excess reagent. To determine the amount of sulfur we use H2S: 16 H2S(g) + 8 SO2(g)  3 S8(s) + 16 H2O(l) 3 mol S8already done 0.147 mol H2S x ————— = 0.0276 mol S8 16 mol H2S MW (S8) = 8 x 32.066 = 256.6 g/mol 256.5 g S8 0.0276 mol S8 x ————— = 7.08 g S8 1 mol S8

  33. How many g of excess reagent remain after reaction? That is SO2 We need to determine the amount of SO2 that reacts with 0.147 mol of H2S (the LR): 16 H2S(g) + 8 SO2(g)  3 S8(s) + 16 H2O(l) 8 mol SO2 0.147 mol H2S x ————— = 0.0735 mol SO2 16 mol H2S 0.0780 mol SO2 initial – 0.0735 mol reacted = 0.0045 mol remaining after reaction MW (SO2) = 32.066 + 2 x 16.00 = 64.07 g/mol 64.07 g SO2 0.0045 mol SO2 x —————— = 0.29 g SO2 (excess) 1 mol SO2

  34. Percent Yield • Theoretical Yield: the maximum amount of product that can be obtained from a chemical reaction. It is the one we calculate from the chemical equations • Actual Yield: the amount of product that is experimentally obtained from a reaction—it is less than the theoretical yield (due to …) actual yield % yield = ———————— x 100 theoretical yield % yield  100%

  35. Example: Benzene, C6H6, reacts with bromine to form bromobenzene, C6H5Br, and hydrogen bromide. Reaction of 8.00 g benzene with excess bromine yielded 12.85 g bromobenzene. Calculate % yield of bromobenzene. • actual yield = 12.85 g C6H5Br • C6H6 + Br2 C6H5Br + HBr • To calculate %yield we need to know the theoretical yield. We will use 8.0 g C6H6 (LR) because bromine is the excess reagent

  36. C6H6 + Br2 C6H5Br + HBr MW (C6H6) = 6 x 12.011 + 6 x 1.008 = 78.11 g/mol How many moles of C6H6? 1mol C6H6 8.00 g C6H6 x —————— = 0.102 mol C6H6 78. 11 g C6H6 How many moles of C6H5Br are produced? 1mol C6H5Br 0.102 mol C6H6 x —————— = 0.102 mol C6H5Br 1 mol C6H6 Converting moles of C6H5Br to grams… The theoretic. yield 157.0 g C6H5Br 0.102 mol C6H5Br x ——————— = 16.0 g C6H5Br 1 mol C6H5Br

  37. the percent yield of C6H5Br actual yield % yield = ———————— x 100 theoretical yield 12.85 g C6H5Br % yield = ——————— x 100 = 80.3% 16.0 g C6H5Br

  38. 64.0 g of methanol, CH3OH, were expected to be produced through the reaction CO(g) + 2H2(g)  CH3OH(l) One student got 56.0 g of methanol for that reaction in the laboratory. What was the %yield of methanol? 64.0 g of methanol is the theoretical yield (expected) 56.0 g is the actual yield (in the laboratory) 56.0 g % Yield = ————  100 = 87.5 % 64.0 g

  39. Analysis of mineral sample The mineral cerussite is mostly PbCO3, but other substances are present. To analyze for the PbCO3 content, a sample of mineral is first treated with nitric acid to dissolve PbCO3 PbCO3(s) + 2 HNO3(aq)  Pb(NO3)2(aq) + H2O(l) + CO2(g) On adding sulfuric acid, lead(II) sulfate precipitates. Pb(NO3)2(aq) + H2SO4(aq)  PbSO4(s) + 2 HNO3(aq) Solid PbSO4(s) is isolated and weighed. Suppose a 0.583 g sample of mineral produced 0.628 g PbSO4. What is the mass % of PbCO3 in the mineral sample? KEY: 1 mol PbCO3  1 mol Pb(NO3)2  1 mol PbSO4

  40. Calculating moles and grams First we calculate moles of PbSO4 (FW = 303.3) 1 mol PbSO4 0.628 g PbSO4x —————— = 0.00207 mol PbSO4 303.3 g PbSO4 Now, we determine moles of PbCO3 1 mol Pb(NO3)2 1 mol PbCO3 0.00207 mol PbSO4x——————— x ——————— 1 mol PbSO4 1 mol Pb(NO3)2 = 0.00207 mol PbCO3 that must be converted to g

  41. grams and % of PbCO3 (FW = 267.2 g/mol) g PbCO3 out of mol PbCO3 267.2 g PbCO3 0.00207 mol PbCO3x —————— = 0.553 g PbCO3 1 mol PbCO3 Now the % PbCO3 in the mineral sample (0.583 g) 0.553 g PbCO3 Mass percent PbCO3= ——————— x 100 = 94.9% 0.583 g sample

  42. Combustion Analysis • sample is burned in oxygen • C  CO2 • H  H2O

  43. Example: • compound containing only C, H, & O • combustion of 28.64 mg sample of compound  • 88.02 mg CO2 • 27.03 mg H2O • MW = 286.5 (it is a given data) • determine molecular formula

  44. Example: • CxHyOz + O2  CO2 + H2O • C  CO2, H  H2O, O  CO2 & H2O • det. mass of C and of H • det. mass of O • calc. mol C, H, & O • det. empirical formula • det. molecular formula

  45. Example: 1 mmol CO2 1 mmol C 88.02 mg CO2 x —————— x —————— x 44.01 mg CO2 1 mmol CO2 12.01 mg C x —————— = 24.02 mg C = 2.000 mmol of C 1 mmol C … now H

  46. Example: 1 mmol H2O 2 mmol H 27.03 mg H2O x —————— x —————— x 18.02 mg H2O 1 mmol H2O 1.008 mg H x —————— = 3.02 mg H = 3.000 mmol of H 1 mmol H … now mg of oxygen mg sample = mg C + mg H + mg O 28.64 mg = 24.02 mg C + 3.024 mg H + mg O mg O = 28.64 – 24.02 – 3.024 = 1.60 mg O

  47. Example: 1 mmol C 24.02 mg C x ————— = 2.000 mmol of C 12.01 mg 1 mmol H 3.024 mg H x ————— = 3.000 mmol of H 1.008 mg H 1 mmol O 1.60 mg O x ————— = 0.100 mmol of O 16.00 mg O Now, we will divide all by the smallest…

  48. Example: C2.0H3.0O0.1 0.10.1 0.1 C20H30O is the empirical formula weight of E.F. = 2012.01 + 301.008 + 16.00 = 286.4 MW 286.5 n = —————— = ————— = 1 W of EF 286.4 molecular formula = C20H30O (same as emp)

  49. Problem: Methane, CH4, burns in oxygen. (a) What are the products of the reaction? CO2 + H2O (b) Write the balance equation for the reaction. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) (c) What mass (g) of O2 is required for complete combustion of 25.5 g of methane? Firstly, we need to calculate moles of methane 1 mol CH4 25.5 g CH4 x ─────── = 1.59 mol CH4Then, use 16.04 g CH4coeffients and molar mass O2 2 mol O2 32.00 g O2 1.59 mol CH4 —————  ————— = 102 g O2 1 mol CH4 1 mol O2

  50. Problem: Hexane, C6H14, burns in oxygen to give CO2 and H2O. (a) Write a balance equation for the reaction. 2 C6H14(l) + 19 O2(g)  12 CO2(g) + 14 H2O(l) (b) If 215 g C6H14 are mixed with 215 g of O2, what masses of CO2 and H2O are produced? Firstly, we need to calculate moles of C6H14 and O2 1 mol C6H142.50 215 g C6H14x ─────── = 2.50 mol C6H14───= 1.25 86.17 g C6H142 1 mol O26.72 215 g O2————— = 6.72 mol O2─── = 0.354 32.00 g O219 Now, the Limiting Reagent (It seems to be O2)

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