# Question #eac7c

##### 1 Answer

#### 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 = m/n implies n = m/M_M#

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

#color(blue)(PV = nRT)" "# , where

*universal gas constant*, usually given as

**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

#PV = overbrace(m/M_M)^(color(red)(=n)) * RT#

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 **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

#"1 atm " = " 760 torr"#

#"1 g" = 10^3"mg"#

#"1 L" = 10^3"mL"#

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

#PV = m/M_M * RT implies M_m = (m * RT)/(PV)#

#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 = "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 = color(green)("73 g/mol")#