WA 6 Dearing

7. A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0°C. (a) Calculate the density of the gas in grams per liter. D = m/V, 4.65g / 2.10L = D 2.21 g/L (b) What is the molar mass of the gas? (10 points) n = PV/RT, P 1.00atm x V 2.10L / R 0.08206 x 27C + 273 = 300K = 0.085mol 4.65g / 0.085mol = 54.7 g/mol 8. A certain anesthetic contains 64.9 percent carbon, 13.5 percent hydrogen, and 21.6 percent oxygen by mass. At 120°C and 750 mmHg, 1.00 L of the gaseous compound weighs 6.90 g. What is the molecular formula of the compound? (10 points) Assume 100g, 64.9g C / 12amu = 5.4mol, 13.5g H / 1amu = 13.5mol, 21.6g O / 16amu = 1.35mol, Divide by smallest, 5.4 / 1.35 = 4, 13.5 / 1.35 = 10, C 4 H 10 O aw = 74 PV = nRT, Solve for n, n= PV/RT, P 750mmHG / 760 = .98atm, T 120C + 273 = 393K (P .98atm x V 1.0L) / (R 0.0821 x 393K) = n 0.0307mol, m 6.90g / 0.0307mol = 224mw 224 / 74 = 3, C 4 H 10 O x 3 = Molecular formula C 12 H 30 O 3 9. What is the mass of the solid NH 4 Cl formed when 73.0 g of NH 3 ( g ) are mixed with an equal mass of gaseous HCl? What is the volume and identity of the gas remaining, measured at 14.0°C and 752 mmHg? (8 points) NH 3 ( g ) + HCl( g ) → NH 4 Cl (Reference: Chang 5.59) 73.0g NH 3 / 17amu = 4.3 mol 73.0g HCl / 36.5amu = 2.0 mol HCl = limiting reactant w/2.0mol, 2.0mol NH 3 produced w/ 2.3mol excess remaining 2.0mol NH 4 Cl x 53.5amu = 107.0g NH 4 Cl (s) produced PV = nRT = V= nRT/P, 752mmHG/760 = 0.98atm, 14C + 273 = 287K V = (n 2.3mol x R 0.0821 x T 287K) / P 0.98atm = 55.3L NH 3 remaining 10. A mixture of gases contains 0.31 mol CH 4 , 0.25 mol C 2 H 6, and 0.29 mol C 3 H 8 . The total pressure is 1.50 atm. Calculate the partial pressures of the gases. (8 points) (Reference: Chang 5.67) P1 + P2 + P3 = P T P1 0.31 mol CH 4 + P2 0.25 mol C 2 H 6 + P3 0.29 mol C 3 H 8 = P T 0.85 total mol P i of CH 4 = 0.31mol / 0.85 x 1.5atm = 0.55 atm P i of C 2 H 6 = 0.25mol / 0.85 x 1.5atm = 0.44 atm P i of C 3 H 8 = 0.29mol / 0.85 x 1.5atm = 0.51 atm 2