Chapter 4 Skills Balance Equations for simple chemical reactions Perform stoichiometry calculations using balanced chemical equations Mass relationships Amounts tables Understand the impact of a limiting reactant on a chemical reaction Calculate the theoretical and percent yields of a chemical reaction Stoichiometric uses Analysis of mixtures Determination of empirical or molecular formulas http:\\now.brookscole.com\kotz6e 11/10/2018 Kull Chem 105 Chapter 2  

Balance Equations for simple chemical reactions 4. Balance the following equations: (a) Cr(s) + Cl2(g)  CrCl3(s) (b) SiO2(s) + C(s)  Si(s) + CO(g) (c) Fe(s) + H2O(g)  Fe3O4(s) + H2(g) 4.4 (a) 2 Cr(s) + 3 Cl2(g)  2 CrCl3(s) (b) SiO2(s) + 2 C(s)  Si(s) + 2 CO(g) (c) 3 Fe(s) + 4 H2O(g)  Fe3O4(s) + 4 H2(g) 11/10/2018

Perform stoichiometry calculations using balanced chemical equations 14. The formation of water-insoluble silver chloride is useful in the analysis of chloride-containing substances. Consider the following unbalanced equation: BaCl2(aq) + AgNO3(aq)  AgCl(s) + Ba(NO3)2(aq) (a) Write the balanced equation. (b) What mass AgNO3, in grams, is required for complete reaction with 0.156 g of BaCl2? What mass of AgCl is produced? 11/10/2018

GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS Mass product Mass reactant Stoichiometric factor Moles reactant Moles product 11/10/2018

Perform stoichiometry calculations using balanced chemical equations 11. Iron metal reacts with oxygen to give iron(III) oxide, Fe2O3. (a) Write a balanced equation for the reaction. (b) If an ordinary iron nail (assumed to be pure iron) has a mass of 2.68 g, what mass of Fe2O3, in grams, is produced if the nail is converted completely to the oxide? (c) What mass of O2, in grams, is required for the reaction? 11/10/2018

Answers 4.4 (a) 2 Cr(s) + 3 Cl2(g)  2 CrCl3(s) (b) SiO2(s) + 2 C(s)  Si(s) + 2 CO(g) (c) 3 Fe(s) + 4 H2O(g)  Fe3O4(s) + 4 H2(g) 4.14 (a) BaCl2(aq) + 2 AgNO3(aq)  2 AgCl(s) + Ba(NO3)2(aq) (b) 0.156 g BaCl2 · · · = 0.255 g AgNO3 0.156 g BaCl2 · · · = 0.215 g AgCl 4.11 (a)4 Fe(s) + 3 O2(g)  2 Fe2O3(s) (b) 2.68 g Fe · · · = 3.83 g Fe2O3 (c) 3.83 g Fe2O3– 2.68 g Fe = 1.15 g O2 11/10/2018

Chemical Equations Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. Because of the principle of the Law of the Conservation of Matter , an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Demo of conservation of matter, See Screen 4.3. 2HgO(s) ---> 2 Hg(liq) + O2(g) 11/10/2018

Chemical Analysis Active Figure 4.8 11/10/2018

Limiting reactant Vinegar is acetic acid: CH3COOH Baking soda is sodium bicarbonate: NaHCO3 Mixing the two is simply and acid base reaction. CH3COOH + NaHCO3 ---> CH3COONa + H2CO3 (60.0 g/mole) (84 g/mole) That last product is carbonic acid which quickly decomposes into carbon dioxide and water: H2CO3 ---> H2O + CO2 The CO2 is what you see foaming and bubbling in this reaction. 11/10/2018

Understand the impact of a limiting reactant on a chemical reaction 24. Aluminum chloride, AlCl3, is made by treating scrap aluminum with chlorine. 2 Al(s) + 3 Cl2(g)  2 AlCl3(s) If you begin with 2.70 g of Al and 4.05 g of Cl2, (a) Which reactant is limiting? (b) What mass of AlCl3 can be produced? (c) What mass of the excess reactant remains when the reaction is completed? 11/10/2018

Answers 4.24 (a) 2.70 g Al · = 0.100 mol Al 4.06 g Cl2 · = 0.0573 mol Cl2 The required mole ratio is 2 mol Al to 3 mol Cl2. More Al is available than required, so Cl2 is the limiting reactant. (b) 0.0573 mol Cl2 · · = 5.09 g AlCl3 (c) 0.0573 mol Cl2 · · = 1.03 g Al used 2.70 g Al available – 1.03 g Al used = 1.67 g Al remains Equation 2 Al(s) + 3 Cl2(g)  2 AlCl3(s) Initial amount (mol) 0.100 0.0573 0 Change (mol) –0.0382 –0.0573 +0.0382 Amount after rxn (mol) 0.062 0 0.0382 4.34 0.196 g CO2 · · · = 0.748 g NaHCO3 · 100% = 43.5% NaHCO3 11/10/2018

Understand the impact of a limiting reactant on a chemical reaction 20. ■ Ammonia gas can be prepared by the reaction of a metal oxide such as calcium oxide with ammonium chloride. CaO(s) + 2 NH4Cl(s)  2 NH3(g) + H2O(g) +CaCl2(s) If 112 g of CaO and 224 g of NH4Cl are mixed, what mass of NH3 can be produced? 11/10/2018

Answers 4.20 112 g CaO · = 2.00 mol CaO 224 g NH4Cl · = 4.19 mol NH4Cl  CaO has fewer moles and is the limiting reactant 112 g CaO · · · = 68.0 g NH3 Equation CaO(s) + 2 NH4Cl(s)  2 NH3(g) + H2O(g) + CaCl2(s) Initial (mol) 2.00 4.19 0 0 0 Change (mol) –2.00 –4.00 +4.00 +2.00 +2.00 Amt after rxn(mol) 0 0.19 4.00 2.00 2.00 11/10/2018

Calculate the theoretical and percent yields of a chemical reaction 30. A reaction studied by Wächtershäuser and Huber (see “Black Smokers and the Origins of Life”) is 2 CH3SH + CO  CH3COSCH3 + H2S If you begin with 10.0 g of CH3SH, and excess CO, (a) What is the theoretical yield of CH3COSCH ? (b) If 8.65 g of CH3COSCH3 is isolated, what is its percent yield? 11/10/2018

Answers 4.30 (a) 10.0 g CH3SH · · · = 9.37 g CH3COSCH3 (b) · 100% = 92.3% yield 4.34 0.196 g CO2 · · · = 0.748 g NaHCO3 · 100% = 43.5% NaHCO3 4.32 2.634 g mixture – 2.125 g after heating = 0.509 g H2O lost 0.509 g H2O · · · = 2.41 g CuCl2·2 H2O · 100% = 91.4% CuCl2·2 H2O 11/10/2018

Chemical Analysis An impure sample of the mineral thenardite contains Na2SO4. Mass of mineral sample = 0.123 g The Na2SO4 in the sample is converted to insoluble BaSO4. The mass of BaSO4 is 0.177 g What is the mass percent of Na2SO4 in the mineral? 11/10/2018

Chemical Analysis Na2SO4(aq) + BaCl2(aq) --> 2 NaCl(aq) + BaSO4(s) 0.177 g BaSO4 (1 mol/233.4 g) = 7.58 x 10-4 mol BaSO4 7.58 x 10-4 mol BaSO4 (1 mol Na2SO4/1 mol BaSO4) = 7.58 x 10-4 mol Na2SO4 7.58 x 10-4 mol Na2SO4 (142.0 g/1 mol) = 0.108 g Na2SO4 (0.108 g Na2SO4/0.123 g sample)100% = 87.6% Na2SO4 11/10/2018

Chemical Analysis An impure sample of the mineral thenardite contains Na2SO4. Sample mass = 0.123 g The Na2SO4 is converted to insoluble BaSO4. The mass of BaSO4 is 0.177 g What is the mass percent of Na2SO4 in the mineral? 11/10/2018

Determining the Formula of a Hydrocarbon by Combustion Active Figure 4.9 11/10/2018

Using Stoichiometry to Determine a Formula Burn 0.115 g of a hydrocarbon, CxHy, and produce 0.379 g of CO2 and 0.1035 g of H2O. CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O What is the empirical formula of CxHy? 11/10/2018

Using Stoichiometry to Determine a Formula CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O First, recognize that all C in CO2 and all H in H2O is from CxHy. 0.379 g CO2 +O2 Puddle of CxHy 0.115 g 1 CO2 molecule forms for each C atom in CxHy 0.1035 g H2O 1 H2O molecule forms for each 2 H atoms in CxHy 11/10/2018

Using Stoichiometry to Determine a Formula CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O First, recognize that all C in CO2 and all H in H2O is from CxHy. 1. Calculate amount of C in CO2 8.61 x 10-3 mol CO2 --> 8.61 x 10-3 mol C 2. Calculate amount of H in H2O 5.744 x 10-3 mol H2O -- >1.149 x 10-2 mol H 11/10/2018

Using Stoichiometry to Determine a Formula CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O Now find ratio of mol H/mol C to find values of x and y in CxHy. 1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C3H4 11/10/2018

Analysis of mixtures 34. At higher temperatures NaHCO3 is converted quantitatively to Na2CO3. 2 NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g) Heating a 1.7184-g sample of impure NaHCO3 gives 0.196 g of CO2. What was the mass percent of NaHCO3 in the original 1.7184-g sample? 11/10/2018

Answers 4.34 0.196 g CO2 · · · = 0.748 g NaHCO3 · 100% = 43.5% NaHCO3 11/10/2018

Analysis of mixtures 32. A 2.634-g sample containing CuCl22H2O and other materials was heated. The sample mass after heating to drive off the water was 2.125 g. What was the mass percent of CuCl22H2O in the original sample? 11/10/2018

Answers 4.30 (a) 10.0 g CH3SH · · · = 9.37 g CH3COSCH3 (b) · 100% = 92.3% yield 4.34 0.196 g CO2 · · · = 0.748 g NaHCO3 · 100% = 43.5% NaHCO3 4.32 2.634 g mixture – 2.125 g after heating = 0.509 g H2O lost 0.509 g H2O · · · = 2.41 g CuCl2·2 H2O · 100% = 91.4% CuCl2·2 H2O 11/10/2018

Stoichiometric uses – Determining a formula 44. To find the formula of a compound composed of iron and carbon monoxide, Fex(CO)y, the compound is burned in pure oxygen to give Fe2O3 and CO2. If you burn 1.959 g of Fex(CO)y and obtain 0.799 g of Fe2O3 and 2.200 g of CO2, what is the empirical formula of Fex(CO)y? 11/10/2018

Stoichiometric uses – Determining a formula 42. ■ An unknown compound has the formula CxHyOz. You burn 0.1523 g of the compound and isolate 0.3718 g of CO2 and 0.1522 g of H2O. What is the empirical formula of the compound? If the molar mass is 72.1 g/mol, what is the molecular formula? (See Exercise 4.9.) 11/10/2018

Answers 4.44 0.799 g Fe2O3 · · = 0.0100 mol Fe 2.200 g CO2 · · = 0.04999 mol CO The empirical formula is Fe(CO)5 4.42 0.3718 g CO2 · · = 0.008448 mol C 0.008448 mol C · = 0.1015 g C 0.1522 g H2O · · = 0.01690 mol H 0.01690 mol H · = 0.01703 g H mass of O = sample mass – mass of C – mass of H = 0.1523 g – 0.1015 g C – 0.01703 g H = 0.0338 g O 0.0338 g O · = 0.00211 mol O The empirical formula is C4H8O The empirical formula mass is equal to the molar mass. The molecular formula is also C4H8O. 11/10/2018

Amounts tables and Chemical stoichiometry 18. Ethane, C2H6, burns in oxygen. (a) What are the products of the reaction? (b) Write the balanced equation for the reaction. (c) What mass of O2, in grams, is required for complete combustion of 13.6 of ethane? (d) What is the total mass of products expected from the combustion of 13.6 g of ethane? 11/10/2018

Answers 4.18 (a) CO2, carbon dioxide, and H2O, water (b) 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) (c) 13.6 g C2H6 · · · = 50.7 g O2 (d) 13.6 g C2H6+50.7 g O2 = 64.3g rxts= 64.3g products Equation 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) I (mol) 0.452 1.58 0 0 C (mol) –0.452 –1.58 +0.904 +1.36 E (mol) 0 0 0.904 1.36 11/10/2018