Question

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 27.0 psi, and the lab was at room temperature (25.0°C). What mass of vaporized triethylamine in grams was left in the lecture bottle?

Answer 1

About 3.81 grams.

Step-by-step explanation:

We can use the ideal gas law. Recall that:

`\displaystyle PV = nRT`

Note that the universal gas constant R has the value 0.08206 L-atm/mol-K.

Hence, convert the measured pressure to atms (1 atm = 14.7 psi):

`\displaystyle 27.0\text{ psi} \cdot \frac{1\text{ atm}}{14.7\text{ psi}} = 1.84\text{ atm}`

Rearrange the equation to solve for n, the number of moles of vaporized triethylamine and evaluate. The temperature is (25.0 + 273.15 ) K = 298.2 K:

`\displaystyle \begin{aligned} n& = (PV)/(RT) \\ \\ & = \frac{(1.84\text{ atm})(0.500\text{ L})}{\left(\frac{0.08206\text{ L - atm}}{\text{mol-K}}\right)(273.2\text{ K})} \\ \\ &= 0.0376\text{ mol N(CH$_2$CH$_3$)$_3$}\end{aligned}`

Convert from moles to grams:

`\displaystyle 0.0376\text{ mol N(CH$_2$CH$_3$)$_3$} \cdot \frac{101.22\text{ g N(CH$_2$CH$_3$)$_3$}}{1\text{ mol N(CH$_2$CH$_3$)$_3$}} = 3.81\text{ g N(CH$_2$CH$_3$)$_3$}}`

In conclusion, there is about 3.81 grams of vaporized triethylamine in the lecture bottle.