How do I find molar mass using gas laws?
A gaseous sample of a compound has a density of $\pu{0.977 g/L}$ at $\pu{710.0 torr}$ and $\pu{100.00 ^\circ C}$. What is the molar mass of this compound? 


The ideal gas law states $$pV = nRT\tag{1}\label{id-gas}$$ where $p$ is defined as the pressure, any units, $V$ is the volume in liters, $n$ is the amount of substance of gas, $R$ is the universal gas constant, and $T$ is temperature in kelvin.

The information given to us is the density of $\pu{0.977 g/L}$, a pressure of $\pu{710.0 torr}$, and temperature of $\pu{100 ^\circ C}$.

Knowing that amount of substance, $$n = \frac{M}{m},$$ where $M$ is molecular mass in $\pu{g/mol}$, and $m$ is the mass of gas in gram, we can see the unit gram cancels out and we are left with the unit mole.

However, we are given a density, $$d = \frac{m}{V} = \pu{0.977 g/L}.$$ 

Plugging into $\eqref{id-gas}$ the molecular mass and mass for $n$ we get:

$$ pV = \frac{M}{m}RT .$$

Isolating $m$ in density, $m = d\cdot V$, and isolating $m$ in $n = m/M\Leftrightarrow m = n\cdot M$, we can combine those two equations for $m$:

$$n\cdot M = d\cdot V$$

Isolate $n$, and plug back into our ideal gas equation $\eqref{id-gas}$:
\begin{align}
 n &amp;= \frac{dV}{M}&amp;
 \implies&amp;&amp;
 pV &amp;= \frac{dV}{M}RT
\end{align}

Volume cancels out, and we isolate for $M$:
$$ M = d \cdot \frac{RT}{p}. $$

Plug in our known values, knowing that $\pu{1 torr} = \pu{1 mm Hg}$ and $\pu{760 mm Hg} = \pu{1 atm}$:

$$ M = \pu{0.977 g//L} \cdot \frac{
  \pu{0.082057 L * atm // mol * K}\cdot\pu{373.15K}}{
  \pu{710 torr}\cdot \pu{mmHg//torr}\cdot\pu{{1atm}//760mmHg}}
     = \pu{32.0 g//mol}$$

And our answer has three significant figures.