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?

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 #"CO"_2#, 22.83 g of #"O"_2#, 75.33 g of #"N"_2#, and 1.8 g of #"H"_2"O"#.

#"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#

#"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#

#"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#

#"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"#

#"Total moles" = ("0.000 91 + 0.7134 + 2.689 + 0.0999) mol" = "3.503 mol"#

According to Dalton's Law of Partial Pressures,

#P_"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_"O₂" = "0.2037 × 760 mmHg" = "155 mmHg"#