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

Answer:

#"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"#.