# What is the mass of a 22.41*L volume of ethane that has a pressure of 780*mm*Hg?

Apr 26, 2017

We get about $16 \cdot g$, i.e. a molar quantity.

#### Explanation:

$1 \cdot a t m$ pressure will support a column of mercury that is $760 \cdot m m \cdot H g$ high. Measurement of pressure in $m m \cdot H g$ is thus useful (because a mercury column is easy to set up!), because it can measure a pressure of about an atmosphere, or very LOW pressures, such as when you do a vacuum distillation, and evacuate the system out to $0.01 \cdot m m \cdot H g$.

If you attempt to measure pressures much HIGHER than $1 \cdot a t m$ with a mercury column, YOU WILL GET metallic mercury all over the laboratory, and this is a major clean-up job. Nevertheless, mercury barometers sometimes read a pressure of slightly greater than $760 \cdot m m$.

So let's convert the pressure to $\text{atmospheres}$.

$\text{Pressure} = \frac{780 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 1.026 \cdot a t m$.

The Ideal Gas Equation tells us that $n = \frac{P V}{R T}$

i.e. $\text{Mass"/"Molar mass} = \frac{P V}{R T}$

OR $\text{Mass"="Molar Mass} \times \frac{P V}{R T}$

And now we fill in the $\text{noombers}$:

$\text{Mass} = \frac{16.33 \cdot g \cdot \cancel{m o {l}^{-} 1} \times 1.026 \cdot \cancel{a t m} \times 22.410 \cdot \cancel{L}}{0.0821 \cdot \cancel{L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1} \times 273.13 \cdot \cancel{K}}$

=??*g

We get an answer in $\text{grams}$ so we must be doing something right.

We could have also used the fact that under standard conditions of $273 \cdot K$, and $1 \cdot a t m$, ONE mol of Ideal Gas will occupy $22.4 \cdot L$. And we know the molar mass of methane.