Question

How do you calculate the partial pressure of oxygen, ${O_2}$, in whose composition as weight percentage is given as: $C{O_2} = 0.04\%$, ${O_2}$ = 22.83, ${N_2}$ = 75.33% and ${H_2}O$ = 1.8%. If the pressure of the air is given as 760mmHg.

Answer 1

The partial pressure of oxygen is calculated by multiplying the total vapour pressure with the mole fraction. The mole fraction is determined by dividing the number of mole of components by the total moles.

Complete step by step answer:
For $C{O_2}$ the weight percentage 0.04 % means that 0.04 grams is present in 100 grams.
For ${O_2}$ the weight percentage 22.83 % means that 22.83 grams is present in 100 grams.
For ${N_2}$ the weight percentage 75.33% means that 75.33 grams is present in 100 grams.
For ${H_2}O$ the weight percentage 1.8% means 1.8 grams is present in 100 grams.
To calculate the partial pressure of oxygen first determine the number of moles.
The molar mass of carbon dioxide is 44.01 g/mol.
The molar mass of oxygen is 32.00 g/mol.
The molar mass of nitrogen is 28.01 g/mol.
The molar mass of water is 18.02 g/mol.
The formula to calculate the number of moles is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles
m is the mass
M is the molar mass
To calculate the number of moles of carbon dioxide substitute the value in the above equation.
$\Rightarrow n = \dfrac{{0.04}}{{44.01g/mol}}$
$\Rightarrow n = 0.0009$
To calculate the number of moles of oxygen substitute the value in the above equation.
$\Rightarrow n = \dfrac{{22.83}}{{32g/mol}}$
$\Rightarrow n = 0.713$
To calculate the number of moles of nitrogen substitute the value in the above equation.
$\Rightarrow n = \dfrac{{75.33}}{{28.01g/mol}}$
$\Rightarrow n = 2.68$
To calculate the number of moles of water substitute the value in the above equation.
$\Rightarrow n = \dfrac{{1.8}}{{18.02g/mol}}$
$\Rightarrow n = 0.099$
The vapour pressure of the oxygen is calculated by the equation as shown below.
${P_{{O_2}}} = {X_{{O_2}}} \times {P_{Total}}$
Where,
${P_{{O_2}}}$ is the vapour pressure of the oxygen
${X_{{O_2}}}$ is the mole fraction of oxygen
${P_{total}}$ is the total vapour pressure
The mole fraction of oxygen is calculated by the formula as shown below.
${X_{{O_2}}} = \dfrac{{{n_{{O_2}}}}}{{{n_{total}}}}$
Where,
${X_{{O_2}}}$ is the mole fraction of oxygen
${n_{{O_2}}}$ is the number of moles of oxygen
${n_{total}}$ is the total moles
To calculate the mole fraction of oxygen, substitute the values in the above equation.
$\Rightarrow {X_{Oxygen}} = \dfrac{{0.7134mol}}{{3.503mol}}$
$\Rightarrow {X_{{O_2}}} = 0.2037$
To calculate the vapour pressure of the oxygen, substitute the values in the formula.
$\Rightarrow {P_{{O_2}}} = 0.2037 \times 760mmHg$
$\Rightarrow {P_{{O_2}}} = 155mmHg$
Therefore, the partial pressure of oxygen, ${O_2}$ is 155mmHg.

Note:
According to Dalton’s law of partial pressure states the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressure exerted by individual gases.
${P_{Total}} = {P_{C{O_2}}} + {P_{{O_2}}} + {P_{{N_2}}} + {P_{{H_2}O}} = 760mmHg$