how many grams of oxygen are contained in 10.5 liters of oxygen measured over water at 25 C and 740 mm Hg? Vapor pressure of water at 25 C is 24mm Hg

Answer 1

When a gas is collected over water, you must deduct the water pressure from the total pressure to obtain the pressure of the gas.

Pressure of gas = 740 mm - 24 mm = 716 mm Hg

To find the moles, use PV = nRT and solve for n....

n = PV/RT = (716)(10.5)/(62.36)(298) = 0.405 moles

mass = 0.405 moles x 32g/mole = 12.9 g

Answer 2

How many grams of O₂ are contained in 10.5 L O₂ measured over water at 25°C and 740 mm Hg? The vapor pressure of water at 25°C is 24 mm Hg.

Given: V = 10.5 L

T = 25°C = 298 K (25+273) We measure in °C and calculate in K

This is a gas laws problem. The 10.5 L contains O₂ and H₂O vapor. So you must first calculate the pressure of the O₂ alone by subtracting the vapor pressure of the water from the total pressure:

Pressure of O₂ = total pressure – water pressure = 740 mm Hg - 24 mm Hg = 716 mm Hg.

So now you have P, V, T and you can calculate n, the number of mol O₂

PV = nRT and R is the universal gas constant, 62.4 (L•mm Hg)/(mol•K)

rearrange to solve for n: PV/RT = n

n = (716 mm Hg • 10.5 L) / (62.4(L•mm Hg)/(mol•K) • 298K)

n = 0.404 mol O₂

Convert mol O₂ to mass O₂ using the molar mass in dimensional analysis:

0.404 mol O₂ = 32.00 g O₂ / 1 mol O₂ = 12.9 g O₂