# A 0.642 g sample of an unknown gas was collected over water at 25.0 degrees C and 1.04 atm. The collection cylinder contained 151.3 mL of gas after the sample was released. How do you find the molar mass of the unknown gas?

Dec 29, 2016

The molar mass of the gas is 103 g/mol.

#### Explanation:

We can use the Ideal Gas Law to solve this problem.

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Since $n = \frac{m}{M}$, we can rearrange this equation to get

$P V = \left(\frac{m}{M}\right) R T$

And we can solve this equation to get

$M = \frac{m R T}{P V}$

Since you are collecting the gas over water,

${P}_{\text{atm" = P_"gas" + P_"water}}$

At 25.0 °C, the partial pressure of water is 0.0313 atm

${P}_{\text{gas" = P_"atm" - P_"water" = "1.04 atm - 0.0313 atm" = "1.009 atm}}$

$m = \text{0.642 g}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{25.0 °C" = "298.15 K}$
$P = \text{1.009 atm}$
$V = \text{151.3 mL" = "0.1513 L}$
M = (0.642 color(red)(cancel(color(black)("g"))) × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 298.15 color(red)(cancel(color(black)("K"))))/(1.009 color(red)(cancel(color(black)("atm"))) × 0.1513 color(red)(cancel(color(black)("L")))) = "103 g/mol"