Collection of Gas Over Water

Problem:

193 mL of O₂ was collected over water on a day when the atmospheric pressure was 762 mmHg. The temperature of the water was 23.0 ^o C. How many grams of oxygen were collected?

Strategy:

1. Use Dalton's law and the vapor pressure of water at 23.0 ^o C to correct the pressure to units of atmoshperes.
   P_T = P_oxygen +P_water
   At 23.0 ^o C the vapor pressure of water is 21.1 mmHg. (This can be found on a vapor pressure table.)
   762 mmHg = P_oxygen + 21.1 mmHg
   P_oxygen = 762 mmHg - 21.1 mmHg
   P_oxygen =741 mmHg
   
   
   
2. Convert the corrected pressure to atmospheres.
   (741 mmHg) (1 atm / 760 mmHg) = 0.975 atm
   
   
3. Use the ideal gas law to find out how many moles of gas were produced:
   PV = nRT (remember to put volume in liters and temperature in Kelvin)
   (0.975 atm) (.193 L) = n (.0821 L atm / mol K) (298 K)
   n = (0.975 atm) (.193 L) / (.0821 L atm / mol K) (298 K)
   n = 7.69 X 10^-4 mol
   
   
4. Use the number of moles and the molecular weight of oxygen to find out how many grams of oxygen were collected.
   
   (7.69 X 10^-4 mol) (32.0 g / 1 mol) = 2.46 X 10^-2 g