# 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.
PT = Poxygen +Pwater
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 = Poxygen + 21.1 mmHg
Poxygen = 762 mmHg - 21.1 mmHg
Poxygen =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.