Question

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

Answer 1

`"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"`.