An old 0.500 L lecture bottle of triethylamine (N(CH₂CH₃)₃) was found in a lab and needed for a synthesis reaction. A pressure regulator indicated a pressure of 18.5 psi, and the lab was at room temperature (25.0°C). What mass of vaporized triethylamine in grams was left in the lecture bottle?

Mass of vaporized triethylamine : 2.606 g

Given

0.5 L triethylamine

P = 18.5 psi

T = 25 °C

Required

mass of vaporized triethylamine

Solution

Conversion :

P 18.5 psi = 1,26 atm

T = 25 +273 = 298 K

Ideal gas law :

PV=nRT

n = PV/RT

Input the value :

n = (1.26 atm x 0.5 L) /(0.08205 x 298)

n = 0.0258

MW triethylamine = 101 g/mol

Mass  triethylamine :

= n x MW

= 0.0258 x 101 g/mol

= 2.606 g

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