How does speed of sound change when you extract air out of the experiment? So I understand that pressure and density cancel each other out so speed of sound is mostly related to the temperature of the gas.
But I was wondering what exactly, experimentally happens with the speed of sound while you are extracting air out of a tube towards a vacuum. Does it stay the same and then suddenly drop to zero near the vacuum?
Thanks in advance
 A: 
So I understand that pressure and density cancel each other out
so speed of sound is mostly related to the temperature of the gas.

You are right. In air the speed of sound is given by
$$v=\sqrt{\gamma\frac{p}{\rho}}$$
where $p$ is the pressure, $\rho$ is the density,
and $\gamma$ is the adiabatic constant of air ($=1.4$).
According to the kinetic theory of gases this expression is a constant
(only depending on temperature and the average mass of the air molecules).

what exactly, experimentally happens with the speed of sound
while you are extracting air out of a tube towards a vacuum.
Does it stay the same and then suddenly drop to zero near the vacuum?

When you reach vacuum, then the equation above equation becomes
$$v=\sqrt{\gamma\frac{p}{\rho}}=\sqrt{\frac{0}{0}}=?$$
which is undefined. It is not $0$, not $\infty$, and nothing
in between. Such a result was to be expected.
In vacuum there is no sound, because there are no molecules
carrying any sound.
When there is no sound, then it doesn't make sense asking
for the speed of this sound.
Even before reaching the absolute vacuum, there are so few
air molecules in your container, that they rarely collide
with each other. And when the mean free path between two
collisions becomes longer than the wavelength of sound,
then the concept of sound waves is not applicable anymore.
