Trimethylphosphine is a common liquid compound that vaporizes readily. Determine the molar mass of trimethylphosphine?

Trimethylphosphine gas :
Pressure, P = (161/760) atm
Volume, V = 125 mL = 0.125 L
Mass, m = 0.0899 g
Molar mass, M = ? g/mol
Gas constant, R = 0.0821 atm L / (mol K)
Absolute temperature, T = (273.2 + 0.0) K = 273.2 K

PV = nRT
PV = (m/M)RT
M = mRT/(PV)

Molar mass, M = 0.0899 × 0.0821 × 273.2 / [(161/760) × 0.125] g/mol = 76.1 g/mol

We will use PV = nRT to determine the number of moles of gas present.

(161 mmHg / 760 mmHg/atm) (0.125 mL) = (n) (0.08206 L atm / mol K) (273 K)

n = 0.00118203 mol

Note that I changed mmHg to atm. You don't have to do that but then you would need to use a value of R with the units L mmHg / mol K.

Now, determine the molar mass:

0.0899 g / 0.00118203 mol = 76.0 g/mol

This is a common test question. Be prepared!

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