A gas sample weighing 3.78 grams occupies a volume of 2.28 L at STP. What is the molecular mass of the sample?

1 Answer
Feb 21, 2017

"37.6 g mol"^(-1)

Explanation:

The trick here is to realize that if you know the volume of a gas at STP, you can use the fact that 1 mole of any ideal gas occupies "22.7 L" under STP conditions to calculate how many moles of gas you have in your sample.

color(red)(ul(color(black)("Under STP conditions: " "1 mole of an ideal gas = 22.7 L")))

In your case, you know that your sample of gas occupies "2.28 L" under STP conditions, which are currently defined as a pressure of "100 kPa" and a temperature of 0^@"C".

This means that your sample will contain

2.28 color(red)(cancel(color(black)("L"))) * overbrace("1 mole gas"/(22.7color(red)(cancel(color(black)("L")))))^(color(blue)("molar volume of a gas at STP")) = "0.10044 moles gas"

Now, the molar mass of the gas is the mass of exactly 1 mole of the gas. In your case, you know that you get "3.78 g" for every 0.10044 moles, which means that you have

1 color(red)(cancel(color(black)("mole"))) * "3.78 g"/(0.10044color(red)(cancel(color(black)("moles")))) = "37.6 g"

Since this is the mass of 1 mole of gas, you can say that the molar mass of the gas is

color(darkgreen)(ul(color(black)("molar mass = 37.6 g mol"^(-1))))

The answer is rounded to three sig figs.

SIDE NOTE A lot of textbooks and online sources still use the old definition of STP conditions, i.e. a pressure of "1 atm" and a temperature of 0^@"C".

Under these conditions, 1 mole of any ideal gas occupies "22.4 L". If this is the value of the molar volume of a gas at STP given to you, simply redo the calculations using "22.4 L" instaed of "22.7 L".