# Question eac7c

Jan 22, 2016

$\text{73 g/mol}$

#### Explanation:

This is a great example of how you can use a variation of the ideal gas law equation to find the molar mass without finding the number of moles first.

As you know, a compound's molar mass tells you the mass of one mole of that substance. This means that you can find a compound's molar mass by diving the mass of a sample of said compound by the total number of moles present in that sample

color(blue)("molar mass" = M_M = "mass"/"no. of moles" = m/n = ["grams"/"mol"])

This of course implies that you can find the number of moles contained in a sample of a given compound by dividing the mass of the sample by the molar mass of the compound

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

Now, the most common form of the ideal gas law equation looks like this

$\textcolor{b l u e}{P V = n R T} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of that gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas, i.e. the temperature expressed in Kelvin

Notice that you can replace the number of moles by using the mass and the molar mass of the gas

$P V = {\overbrace{\frac{m}{M} _ M}}^{\textcolor{red}{= n}} \cdot R T$

This variation of the ideal gas law equation will allow you to find the molar mass of the gas without calculating the number of moles present in the sample.

So, rearrange to isolate ${M}_{M}$ on one side of the equation and plug in your values - do not forget that you must use the units that match those used in the expression of the universal gas constant!

These conversion factors will thus come in handy

$\text{1 atm " = " 760 torr}$

$\text{1 g" = 10^3"mg}$

$\text{1 L" = 10^3"mL}$

Also, keep in mind that the molar mass is expressed in grams per mole!

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

M_M = (10.1 * 10^(-3) "g" * 0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 23)color(red)(cancel(color(black)("K"))))/(10/760color(red)(cancel(color(black)("atm"))) * 255 * 10^(-3)color(red)(cancel(color(black)("L"))))#

${M}_{M} = \text{73.190 g/mol}$

Now, you should round this off to one sig fig, the number of sig figs you have for the pressure of the gas, but I'll leave it rounded to two sig figs, just for good measure

${M}_{M} = \textcolor{g r e e n}{\text{73 g/mol}}$