What is the molecular formula for phosphorus?

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Problem #1: 50.0 cm3 of phosphorus vapor at 526 °C and 101 kPa weighs 0.093 g.

(a) Calculate the relative molecular mass of phosphorus.
(b) Given that the relative atomic mass of phosphorus is 30.974, deduce the number of atoms of phosphorus in one molecule of the element.

Solution:

1) Use PV = nRT to determine the moles of gas present:

(101 kPa) (0.0500 L) = n (8.314 L kPa/mol K) (799 K)

n = 7.60212 x 10¯1 mol

2) Divide grams by moles to get molecular weight:

0.093 g / 7.60212 x 10¯1 mol = 122 g/mol

3) Divide molecular weight by atomic weight:

122 / 30.974 = 3.94

molecular phosphorous is P4


Problem #2: The volume of vapor of 1.043 g of phosphorus is 600.0 mL at 800.0 K temperature and 700.0 mmHg pressure. Calculate the number of P atoms in each molecule of phosphorus.

Solution:

1) Find moles of molecules:

PV = nRT

(700.0 torr) (0.6000 L) = (x) (62.36368 torr-liters/mol-K) (800.0 K)

8.41836 x 10¯3 mol

Note the use of a different value of R from the one the ChemTeam normally uses.

2) Find moles of atoms:

1.043 g / 30.97376 g mol¯1 = 3.367366 x 10¯2 mol

3) Calculate number of P atoms in one P molecule:

3.367366 x 10¯2 mol of atoms / 8.41836 x 10¯3 mol of molecules = 4

molecular phosphorous is P4


Problem #3: At 100.0 °C and 120.0 torr, the mass density of phosphorus vapor is 0.6388 kg/m3. What is the molecular formula of phosphorus under these conditions?

Solution:

1) Convert density to g/L:

0.6388 kg/m3 = 638.8 g/m3

Since 1 m3 = 1 x 103 dm3

d = 638.8 / 1 x 103 = 0.6388 g / dm3

2) Use a rearrangement of PV = nRT to solve for the molar mass:

Molar mass = (dRT) / P

MM = [(0.6388 g L¯1) (0.08206 L atm mol¯11) (373 K)] / 0.158 atm

MM = 123.75 g/mol

A brief discussion of the above rearrangement may be found here.

3) Divide the molar mass of the compound by the atomic mass:

123.75 g mol¯1 / 30.9738 g mol¯1 = 3.995

The molecular formula is P4


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