Question

Determine the density of NH3 gas at 435 K and 1.00 atm... I know that the ideal gas law is PV=nRT. I tried plugging in (1)V=(17)(0.082)(435). I got the volume (I think), but what do I get the density?

Answer 1

Density is grams/volume (m/V)
We can rewrite PV = nRT to find an exprssion that includes density.
We know that moles * molecular weight = mass,
or n * M = m
so n = m/M
Plug that into PV=nRT
PV = nRT/M
Now we notice that m/V is present if you rearrange:
PM/(RT) =n/V = D (D for density)
So PM/(RT) = D
You can then plug in values to solve for density :)

Answer 2

thank you very much! I think I just got a little bit confused about n and m/M. I'm still a little fuzzy on molarity.

Answer 3

ohh okay. So M when we are talkin about PV=nRT and stoichiometry actually refers to molecular weight, not molarity.
the units for molecular weight are g/mol
That's why
n * M = m
mol *g/mol = g
M, indeed, does mean molarity, when talking about concentration but that is a different area of chemistry :)
Does that make any more sense now?

Answer 4

ah yeah. That clears it up... so when I am calculating the molecular weight, those are the values I get when (for example) I convert 1 mol O2 to molecular weight (16*2)? Thank you :)

Answer 5

yesiree! so in this case, you will calculate the moleular weight of NH3, so you do 14.01g/mol + 3(1.0079g)/mol which is approximately 17 grams/mol. This is M of NH3

Answer 6

14.01 g/mol being the weight of the N and 1.0079 of H

Answer 7

thank you so much! Now I understand my chemistry hw :p. You have a new fan :D

Answer 8

awwww yay! I am glad that you understand your homework better now! that is my goal :)