Question

A 4.66 g sample of phosphorus was burned in a large excess of chlorine, and the phosphorus chloride product was found to have a mass of 20.8 g. The vapor of this phosphorus chloride effused at a rate that was 1.77 times slower than that of the same amount of CO2 at the same temperature and pressure. Determine the molar mass and the molecular formula of this phosphorus chloride.

Answer 1

we know that

for effusion

r2/r1 = sqrt( M1/M2)

so in this case

rCO2/r= sqrt(M/MCO2)

1.77 = sqrt( M/44)

M = 137.8476 g/mol

so

the molar mass of the compound is 137.8476 g/mol

now

moles of P taken = 4.66 / 31 = 0.15

now

moles of the PxCly produced = 20.8 / 137.8476 = 0.15

so

x = 0.15/0.15 = 1

so the compound is PCly

now

molar mass of PCly = atomic mass of P + (y * atomic mass of Cl)

137.8476 = 31 + 35.45y

y = 3

so

the molecular formula of this compound is PCl3