Question

how? :(
an unknown gas contains 85.6% c and 14.4% h. at 0.455 atm and 425 k, the gas has a density of 0.549 g/l. what is the molecular formula for the gas?

Answer 1

@thomaster :(

Answer 2

\[PV=NRT\]
N=mass/molar mass=M/Mm
\[PV=M/Mm*R*T\]
Density=mass/volume=M/V
\[P=D*R*T/Mm\]
Solve for Mm
\[Mm=DRT/P\]=42.08g
grams of C=Grams total*percent carbon
\[42.8*0.856=36.02\]
moles carbon=grams carbon/molar mass
36.02/12=3
Do the same thing for H
\[42.8*0.144=6.06\]
6.06/1.01=6
The formula is C3H6

Answer 3

n / V = P / RT = (0.455 atm) / ((0.08205746 L atm/K mol) x (425 K)) =
0.01305 mol/L

(0.549 g/L) / (0.01305 mol/L) = 42.069 g/mol

Supposing the given percentages are percents by mass:
(42.069 g/mol) x (0.856) / (12.01078 g C/mol) = 2.998
(42.069 g/mol) x (0.144) / (1.007947 g H/mol) = 6.010
Round to the nearest while numbers to find the molecular formula:
C3H6