Questions: 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 33.0 psi, and the lab was at room temperature (25°C). What mass of vaporized triethylamine was left in the lecture bottle?

Solution Steps

First, we need to convert the pressure from psi to atmospheres (atm) because the ideal gas law uses pressure in atm. The conversion factor is \(1 \, \text{atm} = 14.696 \, \text{psi}\).

\[ P = \frac{33.0 \, \text{psi}}{14.696 \, \text{psi/atm}} = 2.246 \, \text{atm} \]

The ideal gas law requires temperature in Kelvin. Convert the given temperature from Celsius to Kelvin.

\[ T = 25^{\circ} \text{C} + 273.15 = 298.15 \, \text{K} \]

The ideal gas law is given by:

\[ PV = nRT \]

Where:

• \(P\) is the pressure in atm,
• \(V\) is the volume in liters,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant, \(0.0821 \, \text{L atm/mol K}\),
• \(T\) is the temperature in Kelvin.

Rearrange the equation to solve for \(n\):

\[ n = \frac{PV}{RT} \]

Substitute the known values:

\[ n = \frac{(2.246 \, \text{atm})(0.500 \, \text{L})}{(0.0821 \, \text{L atm/mol K})(298.15 \, \text{K})} = 0.0459 \, \text{mol} \]

The molar mass of triethylamine \((\text{N(CH}_2\text{CH}_3)_3)\) is calculated as follows:

• Nitrogen (N): \(1 \times 14.01 \, \text{g/mol} = 14.01 \, \text{g/mol}\)
• Carbon (C): \(6 \times 12.01 \, \text{g/mol} = 72.06 \, \text{g/mol}\)
• Hydrogen (H): \(15 \times 1.008 \, \text{g/mol} = 15.12 \, \text{g/mol}\)

Total molar mass = \(14.01 + 72.06 + 15.12 = 101.19 \, \text{g/mol}\)

Now, calculate the mass:

\[ \text{mass} = n \times \text{molar mass} = 0.0459 \, \text{mol} \times 101.19 \, \text{g/mol} = 4.6445 \, \text{g} \]

Final Answer