Question

The partial pressures of CH4, N2, and O2 in a sample are 0.19atm, 0.62 atm, and 7.4 atm. If there are 0.031g of methane, what mass of oxygen is present (in grams)?

Answer 1

1st find the mole fraction of CH4

p(CH4) = Ptotal*X(CH4)

0.19 = (0.19 + 0.62 + 7.4)*X(CH4)

X(CH4) = 0.0231

Molar mass of CH4 = 1*MM(C) + 4*MM(H)

= 1*12.01 + 4*1.008

= 16.042 g/mol

mass of CH4 = 0.031 g

we have below equation to be used:

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(0.031 g)/(16.042 g/mol)

= 1.932*10^-3 mol

X(CH4) = mol of CH4 / total moles

0.0231 = 1.932*10^-3 mol/total moles

total moles = 0.0835 moles

now find the mole fraction of O2

p(O2) = Ptotal*X(O2)

7.4 = (0.19 + 0.62 + 7.4)*X(O2)

X(CH4) = 0.901

X(O2) = mol of O2 / total moles

0.901 = mol of O2/0.0835 moles

mol of O2 = 0.0753 mol

Molar mass of O2 = 32 g/mol

we have below equation to be used:

mass of O2,

m = number of mol * molar mass

= 7.53*10^-2 mol * 32 g/mol

= 2.41 g

Answer: 2.41 g