Question

112cm³ of a gaseous fluoride of phosphorus has a mass of 0.63 g at STP. Calculate the relative molecular mass of the fluoride. If the molecule of the fluoride contains only one atom of phosphorus, then determine the formula of the phosphorus fluoride (F=19, P=31).

Answer 1

Mole concept: Finding molecular formula:

Elements:

• Phosphorous $\mathrm{P}$
• Fluorine $\mathrm{F}$

Step 1:Finding molar mass of the compound:

• $112c{m}^3$ of a gaseous fluoride of phosphorus that has a mass $=0.63g$
• At STP, the $22400{\mathrm{cm}}^3$ of gas = 1 mole gas.
• Therefore, will have mass $=\frac{0.63\mathrm{g}x22400\frac{c{m}^3}{mole}}{112c{m}^3}$

$=126g/mol$

Step 2: Determine the formula of the fluoride of phosphorous.

• The molecular mass = atomic mass of $\mathrm{P}$ + Atomic mass of $\mathrm{F}$
• Plug in the values, 126 g/mol = 31 g/mol + Total atomic mass of $\mathrm{F}$
• So, the total mass of $\mathrm{F}$ = 95 g
• But, the Atomic mass of $\mathrm{F}$ = 19 g/mol
• So, plug in the values $=\frac{95\mathrm{g}}{19\frac{\mathrm{g}}{\mathrm{mol}}}$

No. gram atoms of $\mathrm{F}$ = 5 moles.

Step 3: The molecular formula and its name

• We have given 1 g atom phosphorous. It combines with 5 g atoms to form ${\mathrm{PF}}_5$.
• Simply, 1 atom of phosphorous combines with 5 atoms to form ${\mathrm{PF}}_5$.

Hence, there are 5 atoms of F combined with 1 atom of $\mathrm{P}$ so the molecular formula is ${\mathrm{PF}}_5$, called Phosphorous pentafluoride.