A gas sample weighing 3.78 grams occupies a volume of 2.28 L at STP. What is the molecular mass of the sample?
The trick here is to realize that if you know the volume of a gas at STP, you can use the fact that
#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
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 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
#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
Under these conditions,