# An experiment shows that a 245mL gas sample has a mass of 0.436g at a pressure of 757 mmHg and a temperature of 29C. What is the molar mass of the gas?

##### 1 Answer

#### Explanation:

The idea here is that you need to use the idel gas law equation to determine how many **moles** of gas you have in that sample.

Once you know how many moles of gas have that specific mass, you can solve for its *molar mass*.

So, the ideal gas law equation looks like this

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

*universal gas constant*, usually given as

The important thing to notice now is that the values given to you for pressure, temperature, and volume **do not** use the units required by the universal gas constant.

This means that you're going to have to do some conversions before plugging them into the ideal gas law equation. More specifically, you need to have

the temperature form degrees Celsius to Kelvinthe pressure from mmHg to atmthe volume from mililiters to liters

With this in mind, rearrange the ideal gas law equation and solve for

#PV = nRT implies n = (PV)/(RT)#

#n = (757/760color(red)(cancel(color(black)("atm"))) * 245 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 29)color(red)(cancel(color(black)("K"))))#

#n = "0.00985 moles"#

Now, **molar mass** is defined as *mass* per unit of mole

#color(blue)("molar mass" = "mass"/"no. of moles")#

In your case, the molar mass of the sample will be

#M_"M" = m/n#

#M_"M" = "0.436 g"/"0.00985 moles" = "44.264 g/mol"#

Now, I'll leave this rounded to three sig figs, despite the fact that you only have two sig figs for the temperature

#M_"M" = color(green)("44.3 g/mol")#