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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?

When I worked the problem I got $\pu{29.9 g/mol}$ is that right?

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No! The answer $\pu{29.9 gm/mol}$ is wrong.

The best way to solve these type of problems is Ideal Gas Equation. You need to do a little modification as provided in answer by Rauru Ferro, i.e. $$P V = n R T$$

Since $ n = m / M $, we can also write above as $$P V = \left( \frac{m}{M} \right) R T.$$

Since density $d = m/V$, $$P = \left( \frac{d}{M} \right) R T.$$

Now you can easily apply this modified equation in your problem. Before applying this, all you need to do is to convert the pressure from torr to atm: $$\pu{1 atm} = \pu{760 torr}.$$

And also you will need to see the units of gas constant and other quantities as these type of problems seems much easier, but not-focusing-on-units can lead you to the wrong answer.

The answer for your problem will be $\pu{31.9 g/mol}$ approximately.

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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 &= \frac{dV}{M}& \implies&& pV &= \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.

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Density is $\pu{0.977 g/l}$ at specified conditions.

What happens to the density if you increase the pressure from $710$ to $760$? This tells you to multiply by $\frac{760}{710}$.

What happens to the density if you cool it from $373$ to $273$? This tells you to multiply by $\frac{373}{273}$.

If $\pu{1 l}$ has this mass, what is the mass of $\pu{22.414 l}$?

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From gas equation: $$ P V = n R T$$

As $n = m / M$ and density $d = m / V$, $$P = d R T / M,$$ so $$M = 0.977 \cdot 0.082 \cdot 373 / (710/760) = \pu{32.1 g/mol}.$$

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