# How do you calculate the partial pressure of oxygen, O2, in whose composition as weight percentage is given as: CO2 = 0.04%, O2 = 22.83, N2 = 75.33% and H2O = 1.8%, f the pressure of air is given as 760 mm Hg?

Oct 22, 2016

The partial pressure of ${\text{O}}_{2}$ is 155 mmHg.

#### Explanation:

The partial pressure depends on the number of moles, so our first task is to convert the mass percentages to moles.

Assume that you have 100 g of the gas.

Then you have 0.04 g of ${\text{CO}}_{2}$, 22.83 g of ${\text{O}}_{2}$, 75.33 g of ${\text{N}}_{2}$, and 1.8 g of $\text{H"_2"O}$.

${\text{Moles of CO"_2 = 0.04 color(red)(cancel(color(black)("g CO"_2))) × ("1 mol CO"_2)/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.000 91 mol CO}}_{2}$

${\text{Moles of O"_2 = 22.83 color(red)(cancel(color(black)("g O"_2))) × ("1 mol O"_2)/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.7134 mol O}}_{2}$

${\text{Moles of N"_2 = 75.33 color(red)(cancel(color(black)("g N"_2))) × ("1 mol N"_2)/(28.01 color(red)(cancel(color(black)("g N"_2)))) = "2.689 mol N}}_{2}$

$\text{Moles of H"_2"O" = 1.8 color(red)(cancel(color(black)("g H"_2"O"))) × ("1 mol H"_2"O")/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0999 mol H"_2"O}$

$\text{Total moles" = ("0.000 91 + 0.7134 + 2.689 + 0.0999) mol" = "3.503 mol}$

According to Dalton's Law of Partial Pressures,

${P}_{\text{CO₂" + P_"O₂" + P_"N₂" +P_"H₂O" = P_"Total" = "760 mmHg}}$

Dalton's Law can also be expressed as

color(blue)(bar(ul(|color(white)(a/a) P_i = chi_iP_"Total"color(white)(a/a)|)))" "

where $i$ represents a particular component and ${\chi}_{i}$ is its mole fraction.

For oxygen,

chi_"O₂" = n_"O₂"/n_"Total"= (0.7134 color(red)(cancel(color(black)("mol"))))/(3.503 color(red)(cancel(color(black)("mol")))) = 0.2037

${P}_{\text{O₂" = "0.2037 × 760 mmHg" = "155 mmHg}}$