# A sample of gas occupies a volume of 272 mL at a pressure of .999 atm and a temp of 357K. If the sample weighs 1.03 g, what is the molecular weight of the gas?

Oct 21, 2015

$\text{111 g/mol}$

#### Explanation:

You know that a gas occupies a volume of $\text{227 mL}$ at those specific conditions for pressure and temperature. Moreover, you know that this sample has a mass of $\text{1.03 g}$.

Now, the molar mass of a substance tells you exactly what the mass of one mole of that substance is. This means that in order to be able to find the molar mass of the gas, you need to know exactly how many moles of gas you have in that sample.

To do that, use the ideal gas law equation

$P V = n R T \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.082 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature of the gas, expressed in Kelvin

Plug in your values and solve for $n$ - keep in mind that the volume must be expressed in liters!

$P V = n R T \implies n = \frac{P V}{R T}$

n = (0.999color(red)(cancel(color(black)("atm"))) * 272 * 10^(-3)color(red)(cancel(color(black)("mL"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 357color(red)(cancel(color(black)("K")))) = "0.009282 moles"

This means that the molar mass of the gas will be

${M}_{\text{M}} = \frac{m}{n}$

${M}_{\text{M" = "1.03 g"/"0.009282 moles" = "110.97 g/mol}}$

Rounded to three sig figs, the answer will be

M_"M" = color(green)("111 g/mol")