The volume of the oxygen sample collected was 720. mL at 25.0℃ and a barometric pressure of 755 torr. What would the volume of the oxygen be at STP?

Question

A certain mass of oxygen was collected over water when potassium chlorate was decomposed by heating. The volume of the oxygen sample collected was 720. mL at 25.0℃ and a barometric pressure of 755 torr. What would the volume of the oxygen be at STP?

Answer 1

Volume = 720 ml = 0.720 L

Pressure = 755 torr

Vapor pressure of H₂O at 25ºC = 23.8 torr

Pressure of O₂ gas = 755 torr - 23.8 torr = 731.2 torr = 0.962 atm (you must subtract v.p. of water)

Temperature = 25ºC = 298K

From here, we can do the problem one of two ways. We can find moles of O₂ using PV = nRT and then convert that answer to volume by multiply it by 22.4 L/mole. Or we can use P1V1/T1 = P2V2/T2 using the above values for P1, V1 and T1 and using P2 = 1 atm, V2 =? and T2 = 298 (same as T1)

Since I used the first approach the last time I answered this, I'll now use the second approach:

(0.962 atm)(0.720 L) = (1 atm)(V2)

V2 = 0.693 L (vs. 0.634 L) ... not sure where the discrepancy is. Could be in rounding the conversion to atm and/or rounding the value of R.