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Determination of Molar Mass of Propane

Determination of Molar Mass of Propane. or . . . How to correctly use downward displacement of water to collect a gas. Collection of gas by downward displacement of water. What two conditions must a gas meet in order to be amenable to collection by d.d. of water?

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Determination of Molar Mass of Propane

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  1. Determination of Molar Mass of Propane or . . . How to correctly use downward displacement of water to collect a gas

  2. Collection of gas by downward displacement of water

  3. What two conditions must a gas meet in order to be amenable to collection by d.d. of water? 1. Gas cannot react with water; 2. Gas must not be water soluble.

  4. For the above diagram, what gas is collected? Methane, CH4 But the gas collected is NOT pure. What is the contaminant? Water vapour. Why? Water vapour is “swept up” during the collection process.

  5. Some Questions • How can we account for the amount of water vapour present in the gas collected? • What experimental condition might affect the amount of water vapour “swept up”? Temperature. We can account for the amount of water vapour present in a gas collected by d.d. of water . . .

  6. Pressure of Water Vapour (p 558)

  7. How do we use the pressure of water vapour table? Do we add or subtract the pressure of water vapour to the pressure of the gas collected? We subtract it. If P of collected gas is 101.03 kPa at 20oC, we adjust the pressure by subtracting the pressure of water vapour at 20oC. Pdry gas = (101.03 – 2.33) = 98.70 kPa

  8. How can we easily measure the pressure of the gas inside the collection vessel? Can we get a barometer inside the gas collection vessel? No. But look at the following:

  9. Notice that the water level inside and outside the collection vessel are equal. How does the pressure inside the collection vessel compare with Patm? It is the same—as long as water levels inside & outside collection vessel are equal, Pinside = Poutside

  10. Sample Problem 1 A sample of hydrogen is collected by d.d. of water using the following reaction: Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g) What volume of dry hydrogen can be expected when 0.25 g of Mg reacts with excess HCl? What additional info req’d? The water level inside and outside the collection vessel are equal; temperature is 23oC, atmospheric pressure is 102.3 kPa.

  11. 1 mol Mg  1 mol H2 0.25 g ↓ /24.3 g/mol 0.0103 mol → → → → → → → 0.0103 mol n = 0.0103 mol H2 T = 23oC, or = 296 K. Is the PH2 = Patm? (refer to table) PH2= (102.3 – 2.99) = 99.31 kPa R = 8.31 L*kPa/(mol*K) V = (n*R*T)/P = 0.26 L of dry H2 expected.

  12. Now . . . Let’s determine the molar mass of propane by d.d. of water. Equipment: • electronic balance • thermometer • barometer (or dial 416-661-0123) • propane cylinder with delivery hose • d.d. of water set-up • large graduated collection vessel Suggest an experimental procedure. Let’s get busy . . .

  13. Sample Data: mass of propane cylinder before = 725.34 g mass of propane cylinder after = 724.52 g Mass of propane collected = 0.82 g Twater = Troom = 21.7oC Patm = 761.9 mmHg = 101.5 kPa

  14. V of propane collected = 0.45 L (Water level inside and outside collection vessel are equal. Don’t forget to account for Pwater vapour.) Ppropane = (101.5 – 2.81) = 98.7 kPa

  15. PV = nRT n = PV/RT = (98.7 kPa)*(0.45 L) (8.314 LkPamol-1K-1)*(22 + 273)K = 0.01811 mol of dry propane collected Molar mass of propane = 0.82 g/0.01811 mol = 45 g/mol Accepted value = 44 g/mol (< 3% error)

  16. HW P 560 # 40

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