Question

A fluoride of phosphorus in gaseous state was found to diffuse 2.12 times more slowly than nitrogen under similar conditions. Calculate the molecular mass and molecular formula of the fluoride which contains one atom of phosphorus per molecule.

Answer 1

To solve this we must know the Graham’s law of diffusion. Graham’s law of diffusion states that at constant temperature and pressure, the atoms with high molecular mass effuse or diffuse slower than the atoms with low molecular mass. Graham’s law thus states that the square root of the molecular mass is inversely proportional to the rate of diffusion.

Formula Used:

$\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{\sqrt {{M_2}} }}{{\sqrt {{M_1}} }}$

Complete step-by-step solution: We know that Graham’s law states that the square root of the molecular mass is inversely proportional to the rate of diffusion. The equation for Graham’s law is as follows:
$\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{\sqrt {{M_2}} }}{{\sqrt {{M_1}} }}$
Where, ${r_1}$ and ${r_2}$ are the rates of diffusion,
${M_1}$ and ${M_2}$ are the molecular masses.
Thus,
$\dfrac{{{r_{{\text{phosphorus fluoride}}}}}}{{{r_{{\text{nitrogen}}}}}} = \dfrac{{\sqrt {{M_{{\text{nitrogen}}}}} }}{{\sqrt {{M_{{\text{phosphorus fluoride}}}}} }}$
We are given that a fluoride of phosphorus in gaseous state was found to diffuse 2.12 times more slowly than nitrogen under similar conditions. Thus,
$\dfrac{1}{{2.12}} = \dfrac{{\sqrt {{M_{{\text{nitrogen}}}}} }}{{\sqrt {{M_{{\text{phosphorus fluoride}}}}} }}$
We know that the molecular mass of nitrogen is $28{\text{ g/mol}}$. Thus,
$\dfrac{1}{{2.12}} = \dfrac{{\sqrt {28{\text{ g/mol}}} }}{{\sqrt {{M_{{\text{phosphorus fluoride}}}}} }}$
$\Rightarrow {\left( {\dfrac{1}{{2.12}}} \right)^2} = \dfrac{{28{\text{ g/mol}}}}{{{M_{{\text{phosphorus fluoride}}}}}}$
$\Rightarrow {M_{{\text{phosphorus fluoride}}}} = \dfrac{{28{\text{ g/mol}} \times {{\left( {2.12} \right)}^2}}}{{{{\left( 1 \right)}^2}}}$
$\therefore {M_{{\text{phosphorus fluoride}}}} = 125.84{\text{ g/mol}}$
Thus, the molecular mass of a fluoride of phosphorus is $125.84{\text{ g/mol}}$.
We are given that the fluoride of phosphorus contains one atom of phosphorus per molecule. Thus, assume that the molecular formula is ${\text{P}}{{\text{F}}_{\text{x}}}$.
We know that the molecular mass of a fluoride of phosphorous is $125.84{\text{ g/mol}}$, the molecular mass of phosphorus is $31{\text{ g/mol}}$ and the molecular mass of fluorine is $19{\text{ g/mol}}$. Thus,
${\text{P}}{{\text{F}}_{\text{x}}} = {\text{P}} + {\text{x}} \times {\text{F}}$
$\Rightarrow 125.84{\text{ g/mol}} = 31{\text{ g/mol}} + {\text{x}} \times 19{\text{ g/mol}}$
$\Rightarrow {\text{x}} \times 19{\text{ g/mol}} = \left( {125.84 - 31} \right){\text{ g/mol}}$
$\Rightarrow {\text{x}} = \dfrac{{\left( {125.84 - 31} \right){\text{ g/mo}}l}}{{19{\text{ g/mol}}}}$
$\therefore {\text{x}} = 4.99 \approx 5$

Thus, the molecular formula for the fluoride of phosphorus is ${\text{P}}{{\text{F}}_{\text{5}}}$.

Note: Graham’s law is more accurate for the molecular effusion that involves the movement of one gas at a time. For diffusion, Graham’s law is approximate. This is because the process of diffusion involves movement of more than one gas at a time.