Suppose the O₂ gas evolved by a certain reaction taking place at 40 °C is collected over water, and the final volume of gas in the collection tube is measured to be 45.9 mL.
Calculate the mass of O₂ that is in the collection tube.
The key to solve the problem is to find the volume that occupied the O₂. Since the sample was collected over water, the volume measured (45.9 mL) also includes some water vapor.
According to Dalton's Law:
Total Pressure = Poxygen + Pwater
If we search we will find that the Vapor Pressure of water at 40 °C is 55.32 mmHg. So:
Pwater = 55.32 mmHg.
If we convert that into atm:
Pwater = 55.32 mmHg / (760 mmHg/1 atm)
Pwater = 0.073 atm
We are not given the atmospheric pressure but we will assume that it is 1 atm.
Total Pressure = 1 atm
With those values we can find the partial pressure of our gas:
Total Pressure = Poxygen + Pwater
Poxygen = Total Pressure - Pwater = 1 atm - 0.073 atm
Poxygen = 0.927 atm
Remember the definition of the partial pressure of a gas. The partial pressure of a gas (in a mixture of gases) is the pressure of that gas if it alone occupied the entire volume. Now that we found the partial pressure of oxygen, we can apply the ideal gas Law using the total volume that we measured in the experiment to find the number of moles.
Poxygen * V = n * R * T
We already know these values:
T = 40 °C = (40 + 273.15) K
T = 313.15 K
R = 0.082 atm * L /(mol *K)
n = ?
V = 45.9 mL / (1000 mL/1L) = 0.0459 L
Using those values and solving the equation for the number of moles:
Poxygen * V = n * R * T
n = Poxygen * V / (R *T)
n = 0.927 atm * 0.0459 L / (0.082 atm*L/(mol*K) * 313.15 K)
n = 0.001657 moles
And finally using the molar mass of Oxygen:
molar mass of O₂ = 32 g/mol
mass of O₂ = 0.001657 moles * 32 g/mol
mass of O₂ = 0.053 gr
So the answer to our problem using two SF is 0.053 g