So my chemistry teacher gave me this packet having to do with The Ideal Gas Law. One of the homework questions says, "An unknown gas at $\pu{20.0 ^\circ C}$ and $\pu{1900 torr}$ has a density of $\pu{3.95 g/L}$. Identify the compound." I have tried this problem many times and have derived many formulas to no avail. Can someone please help me?


closed as off-topic by Mithoron, airhuff, aventurin, Tyberius, M.A.R. ಠ_ಠ May 22 '18 at 21:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Have you clarified what you have tried or where specifically you seem to be getting stuck. One thing you could do is suppose you have one mole of the compound. Using the data you have, you can then calculate the volume. Can you see a way forward from there? @ThomasMcCarthy $\endgroup$ – Tyberius May 22 '18 at 18:56

Suppose the mass of the gas is $m$ and its molar mass and the volume under the given conditions are $M$ and $V$, respectively. Thus, the number of moles ($n$) and the density ($d$) of the gas under the given conditions is: $$n=\frac{m}{M} \space \text {and}$$ $$d=\frac{m}{V} \space \Longrightarrow V=\frac{m}{d}$$ Substitute these values in the Ideal Gas Law, $PV=nRT$. $$P\frac{m}{d}=\frac{m}{M}RT$$ Thus, $$M=\frac{d}{P}RT$$ Since you know $P$, $T$, and $d$, numerically, you can calculate $M$ in $\mathrm{gmol^{-1}}$, using appropriate $R$ value.

  • $\begingroup$ Ok so I did the math for that and my answer was 38.0268 g/mol. The final part to the question is to identify the compound and I was unable to find a gas with the molar mass of 38.0268 g/mol. $\endgroup$ – Thomas McCarthy May 22 '18 at 4:49
  • $\begingroup$ Well it could be C-13 trideutero methyl flouride, that will have an approximate formula mass of 38 $\endgroup$ – Nuclear Chemist May 22 '18 at 5:09
  • 1
    $\begingroup$ @ThomasMcCarthy - 38.0268 has far too many digits. The density was only given to three significant figures so no more should be used. Thus the answer should be 38.0. So I'd guess $\ce{F_2}$ $\endgroup$ – MaxW May 22 '18 at 5:18
  • $\begingroup$ @MaxW: Actually it is 30.00 so MaxW assumption is correct: the gas is $\ce{F2}$. $\endgroup$ – Mathew Mahindaratne May 22 '18 at 5:39

PV = nRT
Assume you have 1 liter volume so V = 1
P = 1900 torr
n = moles = ?
R = 62.63 L-torr/mol-K
T = 20 + 273 = 293 K
n = PV/RT = (1900)(1)/(62.63)(293)
n = 0.104 moles
Molar mass = g/mole = 3.95 g/0.104 mole = 37.98 = 38.0 g/mol
Unknown could be F2 gas


Not the answer you're looking for? Browse other questions tagged or ask your own question.