Chemical Equations. Chemical Equations. Chemical Equations. Combination Reactions. Decomposition Reactions

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1 John W. Moore Conrad L. Stanitski Peter C. Jurs Chemical Equations Reactants yeast Products C 6 H 1 O 6 (aq) C H 5 OH(l) + CO (g) glucose ethanol carbon dioxide Chapter 4 Quantities of Reactants and Products Stephen C. Foster Mississippi State University Conditions may be shown over the arrow. e.g. heat (Δ) reflux catalyst present (yeast) Physical states are often listed: (g) gas (s) solid (l) liquid (aq) aqueous (dissolved in water) Chemical Equations Balanced equations obey the law of conservation of mass (Lavoisier 1789). Mass is neither created nor destroyed in a chemical reaction. C H 6 (g) + 7 O (g) 4 CO (g) + 6 H O(l) Nanoscale molecules 7 molecules 4 molecules 6 molecules Macroscale moles 7 moles 4 moles 6 moles (30.0)= 60.0 g 7(3.0)= 4.0 g 4(44.0)= g 6(18.0)= g C H 6 molar mass 84.0 g 84.0 g O molar mass Chemical Equations Stoichiometry The relationship between the number of reactant and product molecules in a chemical equation. CaCO 3 (s) + HNO 3 (aq) A stoichiometric coefficient Ca(NO 3 ) (aq) + CO (g) + H O(l) Combination Reactions + X Z XZ Element plus halogen or O : Mg(s) + O (g) MgO(s) I (s) + Zn(s) ZnI (s) There are other types: SO (g) + O (g) SO 3 (g) Decomposition Reactions + XZ X Z Often initiated by heat: C CaCO 3 (s) CaO(s) + CO (g) heat KNO 3 (s) KNO (s) + O (g) Occasionally by shock: 4 C 3 H 5 (NO 3 ) 3 (l) 1 CO (g) + 10 H O(l) + 6 N (g) + O (g) nitroglycerin 1

2 Decomposition Reactions Displacement Reactions + + A XZ AZ X Metals may displace another metal from its salt Fe(s) + CuSO 4 (aq) FeSO 4 (aq) + Cu(s) Zn(s) + AgNO 3 (aq) Zn(NO 3 ) (aq) + Ag(s) Some other examples: F (g) + LiCl(s) LiF(s) + Cl (g) Na(s) + H O(l) NaOH(aq) + H (g) Displacement Reactions Exchange Reactions + + AD XZ AZ XD AgNO 3 (aq) + HCl(aq) AgCl(s) + HNO 3 (aq) Pb(NO 3 ) (aq) + K CrO 4 (aq) PbCrO 4 (s) + KNO 3 (aq) Balancing Chemical Equations 1. Write an unbalanced equation with correct formulas for all substances.. Balance the atoms of one of the elements. Start with the most complex molecule. Change the coefficients in front of the molecules. Do NOT alter the chemical formulas. 3. Balance the remaining elements. Balancing Chemical Equations Balance : Al + Fe O 3 Al O 3 + Fe step 1 Al + Fe O 3 Al O 3 + Fe 1 Al (Fe + 3O) (Al + 3O) 1Fe step Al + Fe O 3 Al O 3 + Fe 1 Al (Fe + 3O) (Al + 3O) Fe balanced step 3 Al + Fe O 3 Al O 3 + Fe Al (Fe + 3O) (Al + 3O) Fe step 4 balanced

3 Balancing Chemical Equations Combustion of rocket fuel: C H 8 N + N O 4 N + H O + CO C + 8H + 4N + 4O 1C + H + N + 3O Balance C and N in C H 8 N first: C H 8 N + N O 4 N + 4 H O + CO C + 8H + 4N + 4O C + 8H + N + 8O Still. Adjust N and O C H 8 N + N O 4 3 N + 4 H O + CO Balanced Balancing Chemical Equations Polyatomic ion on both sides of an equation? Balance as units. NaNO 3 (s) + H SO 4 (aq) Na SO 4 (aq) + HNO 3 (aq) Na + NO 3 + H + SO 4 Na + SO 4 + H + NO 3 Balance Na in Na SO 4 NaNO 3 (s) + H SO 4 (aq) Na SO 4 (aq) + HNO 3 (aq) Na + NO 3 + H + SO 4 Na + SO 4 + H + NO 3 NaNO 3 (s) + H SO 4 (aq) Na SO 4 (aq) + HNO 3 (aq) Na + NO 3 + H + SO 4 Na + SO 4 + H + NO 3 balanced The Mole and Chemical Reactions C H 6 (g) + 7 O (g) 4 CO (g) + 6 H O(l) moles of C H 6 react with 7 moles of O moles of C H 6 produce 4 moles of CO mol C H 6 7 mol O mol C H 6 4 mol CO etc. Mole ratios: mol C H 6 7 mol O =1 7 mol O mol C H 6 =1 The Mole and Chemical Reactions What mass of O and Br is produced by the reaction of 5.0 g of TiO with excess BrF 3? 3 TiO (s) + 4 BrF 3 (l) 3 TiF 4 (s) + Br (l) + 3 O (g) Notes: Check the equation is balanced! Stoichiometric ratios: 3TiO 3O ; 3TiO Br ; and many others Excess BrF 3 = enough BrF 3 to react all the TiO. The Mole and Chemical Reactions What mass of O and Br is produced by the reaction of 5.0g of TiO with excess BrF 3? 3 TiO (s) + 4 BrF 3 (l) 3 TiF 4 (s) + Br (l) + 3 O (g) n TiO = mass TiO / FM TiO = 5.0 g x = mol TiO g The Mole and Chemical Reactions What mass of O and Br is produced by the reaction of 5.0g of TiO with excess BrF 3? 3 TiO (s) + 4 BrF 3 (l) 3 TiF 4 (s) + Br (l) + 3 O (g) Mass of O produced = n O (mol. wt. O ) = mol x 3.00 g/mol = 10.0 g 3 TiO Br 3 mol TiO 3 mol O 3 mol O mol TiO = mol O 3 mol TiO Br n Br = mol TiO = mol Br 3 TiO Mass of Br = mol g = 33.4 g Br mol Br 3

4 Practice Problem 4.8 The purity of Mg can be found using the reaction Mg(s) + HCl(aq) MgCl (aq) + H (g) Calculate the % Mg in a 1.7-g sample that produced 6.46 g of MgCl when reacted with excess HCl. Practice Problem 4.8 FW of MgCl Mg + HCl MgCl + H n MgCl = 6.46 g MgCl = (35.45) = 95.1 g/mol 95.1 g = mol MgCl More difficult What should you calculate? How much pure Mg will make 6.46 g of MgCl? Express as a % of the original mass. Use mole ratio Mg MgCl Mg required: mol MgCl 1 Mg 1 MgCl = mol of pure Mg Practice Problem 4.8 Mg + HCl MgCl + H Calculate mass of pure Mg needed Given 10 slices of cheese and 14 slices of bread. How many sandwiches can you make? mol Mg x 4.31 g = g Mg Given 1.7 g of impure Mg. Purity (as mass %) = g x 100% = 95.9 % 1.7 g Balanced equation 1 cheese + bread 1 sandwich 1 cheese bread 1 cheese 1 sandwich bread 1 sandwich Two methods can be used: Product Method Calculate the product from each starting material. The reactant giving the smallest number is limiting. 10 cheese x 1 sandwich = 10 sandwiches 1 cheese 14 bread x 1 sandwich = 7 sandwiches bread Bread is limiting. It will be used up first Correct answer Reactant Method Pick a reactant; calculate the amount of the other(s) needed. Enough? Yes = Your choice is the limiting reactant. No = Another reactant is limiting. e.g. choose bread cheese needed: 14 bread (1 cheese / bread ) = 7 Available. bread is limiting. e.g. choose cheese bread needed: 10 cheese ( bread/1 cheese) = 0 Not available. bread is limiting. 4

5 Bread is limiting base all other calculations on the limiting reactant. Sandwiches made 14 bread (1 sandwich / bread ) = 7 sandwiches Cheese remaining 14 bread (1 cheese / bread ) = 7 cheese used. Started with 10 cheese. Cheese remaining 10 7 = 3 slices How much water will be produced by the combustion of 5.0 g of H in the presence of 100. g of O? Write a balanced equation: H (g) + O (g) H O(l) H n H = 5.0 g = 1.40 mol H.016 g O n O = 100. g = 3.15 mol O 3.00 g Moles: H + O H O Product Method H O From H n HO = 1.40 mol H = 1.40 mol H H O From O n HO = 3.15 mol O = 6.50 mol 1O O gave less. O is limiting. Use O in all calcs. m HO = 6.50 mol H O 18.0 g = 113. g water H + O H O Moles available: Reactant Method e.g. choose H O needed: 1.40 mol H (1 O / H )= 6.0 mol Not available. O is limiting. You only need one calculation. Had you chosen O H needed: 3.15 mol O ( H /1 O )= 6.50 mol Available. O is limiting. H O formed: 3.15 mol O (H O/1O ) = 6.50 mol. = 113. g Reactions with Limited Reactants Consider : 4 NH 3 (g) + 5 O (g) 4 NO(g) + 6 H O(g) If 374 g of NH 3 and 768 g of O are mixed, what mass of NO will form? Balanced equation? yes n NH3 = 374 g g n O = 768g 3.00 g = 1.96 mol = 4.00 mol Reactions with Limited Reactants 4 NH 3 (g) + 5 O (g) 4 NO(g) + 6 H O(g) Mol available: From NH 3 NO formed: 1.96 mol NH 3 4 NO = 1.96 mol NO 4 NH 3 From O NO formed: 4.00 mol O 4 NO 5 O Smallest amount. O is limiting. = 19.0 mol NO 5

6 Reactions with Limited Reactants Mass of NO formed? 4 NH 3 (g) + 5 O (g) 4 NO(g) + 6 H O(g) 1.96 mol 4.00 mol 19.0 mol O is limiting. Base all calculations on O. NO formed = 19.0 mol NO Mass of NO = 19.0 mol NO x 30.01g = 576 g NO Reactions with Limited Reactants What mass of MgI is made by the reaction of 75.0 g of Mg with 75.0 g of I? Mg + I MgI Balanced? Calculate moles YES 75.0 g of Mg = 75.0g/(4.31 g mol -1 ) = mol Mg 75.0 g of I = 75.0g/(53.9 g mol -1 ) = mol I Limiting reactant? 1Mg 1I so I = limiting Since 1MgI 1I produce mol MgI Mass of MgI = mol x 78. g/mol = 8. g Percent Yield Theoretical yield The amount of product predicted by stoichiometry. Actual yield The quantity of desired product actually formed. Percent yield Actual yield % yield = x 100% Theoretical yield Percent Yield Few reactions have 100% yield. Possible reasons Side reactions may produce undesired product(s). Incomplete reaction due to poor mixing or reaching equilibrium Product loss during isolation and purification. Percent Yield.50 g of copper heated with an excess of sulfur made.53 g of copper(i) sulfide 16 Cu(s) + S 8 (s) 8 Cu S(s) What was the percent yield for this reaction? n Cu used:.50 g 63.55g = mol Cu 16 mol Cu used 8 mol Cu S made Theoretical yield: mol Cu 8 Cu S 16 Cu = mol Cu S Percent Yield.50 g Cu + S 8 (excesss) made.53 g Cu S What was the %-yield? Theoretical yield = mol Cu S 159. g = mol Cu S = g Cu S Actual yield =.53 g Cu S (in problem).53 g Percent yield = x 100% = 80.8% g 6

7 Atom Economy Examines the fate of all starting-material atoms. % atom economy = atomic mass of atoms in useful product(s) atomic mass of all reactants used High atom economy = low waste production x 100% Ideal reaction: high % yield and high atom economy. Empirical formula = simplest ratio of atoms in a molecule. Found for organic compounds by combustion analysis O Sample in a furnace H O absorber Mg(ClO 4 ) C and H are converted to CO & H O. CO absorber NaOH Both are trapped and the weight gain measured. Other elements (N, O ) with other traps or by mass difference. Vitamin C (176.1 g/mol) contains C, H & O only. If g is burned in O, 1.50 g CO and g H O form. Find its empirical & molecular formula. Mass of C 1.50 g CO CO C g C = g C g CO CO C or 1.50 g CO g C = g C g CO Vitamin C contains C, H & O only. Combustion of g of vitamin C produced 1.50 g of CO and g of H O. Mass of H g H O.0158 g H g H O = g H mass of O = sample mass (mass of C + mass of H) Mass of O = g ( g g) = g Convert to moles: g C g/mol g H g/mol g O g/mol = mol C = H = mol O Find the mole ratio (divide by smallest ): C / = 1.00 H / = 1.34 O / = 1.00 Close to 1 : 1⅓ : 1 (C : H : O) Multiply by 3 to get an integer ratio (3 : 4 : 3) Empirical formula is C 3 H 4 O 3 7

8 Vitamin C contains C, H & O only molar mass of vitamin C is g/mol, find its empirical and molecular formula. Empirical formula = C 3 H 4 O 3 Empirical mass = 3(1) + 4(1) + 3(16) g = 88 g Molar mass = g Molar mass x (empirical mass) Vitamin C has the molecular formula C 6 H 8 O 6 8