Molar mass of acetone

  • Thread starter Hammertong
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  • #1
Hammertong
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A 125mL flask contains 0.239g of acetone vapor at 766mm Hg and 100*C. Calculate the molar mass of acetone.
766x125/373Kx273K/760=92.210mL
.239g/92.210/1000=2.592g/mL
Where am I going wrong? I have no idea if I'm anywhere close.
 

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  • #2
Redbelly98
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Those equations aren't really what we want.

Hint: molar mass is the number of grams per moles.
 
  • #3
Hammertong
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I know what molar mass is, but I don't understand how to get it from what I'm given. My teacher goes over the equations once then moves on. Then he expects us to be able to do it backwards, forwards, and sideways.
I figured it out, I just needed to multiply my end answer by 22.4L to get weight of one mole at STP.
 
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  • #4
symbolipoint
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Not sure anymore how all the units work; "R" is the complicated part, but the question seems like an ideal gas type. PV=nRT. You can determine the number of moles from that, "n". Next, you compare the number of moles to the number of grams of mass, which you are given.
 
  • #5
epenguin
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These things are basically simple proportions.

A mole as you say occupies 22.4 l under standard conditions (still called STP or NTP?) . Got a texbook to check out how those are defined?
The excercise is not under standard conditions, but a higher temperature.
Again it is simple proportions - you have to know what absolute temperature means.
You do not need to use explicitly that equation - if you use the Charles temperature proportionality law you are using it implicitly anyway .
You can actually work out R from the above information.
Hope these thing become clear and simple. :smile:
 

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