Question

When 1.00 g of solid phosphorous is heated to 400 degree C in a 500 mL flask, it is converted to vapor. The pressure inside the flask is 678 torr. What is the molecular formula of the vapor? (i.e. is phosphorous vapor monatomic, diatomic, triatomic, etc.?)

Answer 1

Ans. Part 1: Number of moles of phosphorous atoms in 1.0 g sample = Mass/ Molar mass

                                                                        = 1.0 g / (30.973762 g/ mol)

                                                                        = 0.03228539 mol

Part 2: Calculate number of moles of vapor phase:

It’s assumed that vapor phosphorous behave as ideal gas under the given conditions.

Given,

            Pressure, p = 678 torr = 0.892 atm                                                ; [760 torr= 1 atm ]

            Volume, V = 500 mL = 0.5 L                                                           ; [1 mL = 10^-3 L]

            Temperature, T = 400⁰C = 673.15 K

Ideal gas equation: PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in above equation-     

            1.0 atm x 0.5 L = n x (0.0821 atm L mol^-1K^-1) x 673.15 K

            Or, n = 0.5 atm L / (55.266 atm L mol^-1) = 0.0090472 mol

Therefore, the number of moles in vapor phase = 0.0090472 mol

# Now,

Ratio of moles of P-atoms in two phases = Moles in solid phase / Moles in vapor phase

                                                            = 0.0322 mol / 0.0090 mol

                                                            = 3.578 = 4 (rounded off to nearest whole number)

Thus, molecular formula of vapor-phase phosphorous = P₄

The ratio of “4” indicates that four phosphorous atoms remain linked together in the gaseous phase as a single entity or mole in vapor phase.