Question

Post Lab #1 A student was assigned the task of determining the molar mass of an unknown gas. The student measured the mass of a sealed 843 mL rigid flask that contained dry air. The student then flushed the flask with the unknown gas, resealed it, and measured the mass again. Both the air and the unknown gas were at 23.0°C and 750. torr. The data for the experiment are shown in the table below. Volume of sealed flask 843 mL Mass of sealed Mask and dry air 157,70 % Mass of sealed flask and unknown gas 158.08 g (a) Calculate the mass, in grams of the dry air that was in the sealed flask. (The density of dry air is 1.18 g L-1 at 23.0°C and 750.torr.) (b) Calculate the mass, in grams, of the sealed flask itself (i.e., if it had no air in it). (©) Calculate the mass, in grams, of the unknown gas that was added to the sealed flask. (d) Using the information above, calculate the value of the molar mass of the unknown gas. After the experiment was completed, the instructor informed the student that the unknown gas was carbon dioxide (44.0 g/mol). (e) Calculate the percent error in the value of the molar mass calculated in part (d).

Answer 1

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a) Density = mass of dry air - Volume of dry air = volume of sealed flask 1. 18g l = mass of dry air & 1000ml 843 ml | Mass of dry air = 16 18 g 1 X 843 ml XIL o 1000 ml = 0.9947g 6 Mass of sealed flask and dry air = 157-70 g. Mass of sealed flask = 157.70 g - 0.99479 = 156,70°9 c mass of sealed flask and unknown gas = 158-08 g ] mass of unknown gas = 158.089 - 15 6.70 g d) According to ideal gas equation moles of a gas & PV P-Pressure T-Temperatur RTV-Volume P = 750 torr atm = 760 torr = 0.9868 atm T = 23°C + 273·15= 296. 15k R = 0.0821 Latm kt molt V = 843 ml = 0. 843 L moles of gas = 0.9868 atm x 0.843L 0.0821 latin kalmal x 296.15k = 0.0342 mol Moles - mass & Molar mass - Mass gas. molar mass - moles gas Molar mass = 1.38 g = 40.35 g/mol R 0.0342mel e) Percent error = Actual value-Theolog" theoretical vatue Theoretical value X100 - 44.0-40-359 mal = 44. - 448/mol 44.0 = 8.29% L 40.35 g mol x 100 44.0 x100 og opmol x100 = 3-65x100