# An air sample contains 0.038% CO_2. If the total pressure is 758 mm Hg, what is the partial pressure of CO_2?

Dec 30, 2017

${P}_{C {O}_{2}}$ = $0.288 \text{mmHg}$ if %CO_2 is mole percent. If %CO_2 is weight percent then one will need the mass values of the other components of the air mixture.
Given 0.038% (mole/mole) => ${X}_{C {O}_{2}}$ = $\left(0.0038 \text{mole" CO_2)/(100"moles"AIR}\right)$ = $0.00038$
${X}_{C {O}_{2}}$ = ${P}_{C {O}_{2}} / {P}_{\text{Total}}$ => ${P}_{C {O}_{2}}$= ${X}_{C {O}_{2}}$dot${P}_{C {O}_{2}}$
= 0.00038(758"mmHg")=$0.288 \text{mmHg}$