You need to know two things to solve this problem:
- That 1 mole of any ideal gas occupies "22.4 L"22.4 L at STP;
- The relationship between moles and molar mass
"moles" = ("mass of substance")/("molar mass")moles=mass of substancemolar mass
So, you know that the vapor weighs "0.96 g"0.96 g at STP. STP conditions imply a temperature of "273.15 K"273.15 K and a pressure of "1.0 atm"1.0 atm. This means that the number of moles you have is
n=V/V_("molar") = (250 * 10^(-3)"L")/("22.4 L") = "0.01116 moles"n=VVmolar=250⋅10−3L22.4 L=0.01116 moles
SIde note - you can use the ideal gas law equation, PV = nRTPV=nRT, to double check this result;
Since n= "m"/"MM"n=mMM, you get that
"n" = "m"/"MM" = "0.01116 moles"n=mMM=0.01116 moles
Since m = "0.96 g"m=0.96 g, the value of the molar mass will be
"molar mass" = m/n = ("0.96 g")/("0.01116 moles") = "86.02 g/mol"molar mass=mn=0.96 g0.01116 moles=86.02 g/mol
Your compound's empirical formula is (CH_2)_x(CH2)x, or C_xH_(2x)CxH2x. The value of xx is determined by dividing the molar mass of the compound by the molar mass of the empirical formula
x = ("86.02 g/mol")/("14.0 g/mol") = "6.14"x=86.02 g/mol14.0 g/mol=6.14
Ideally, xx should be as close to an integer as possible; in this case, the closest integer would be 66, which would make the compound's molecular mass equal to "84.0 g/mol"84.0 g/mol and the molecular formula
C_6H_12C6H12.
Since this problem describes the Dumas method of molecular weight determination, an experimental method used to determine molar mass, you could calculate the percent error for the result
"%error" = |"accepted value - experimental value"|/("accepted value") * 100%error=|accepted value - experimental value|accepted value⋅100
"%error" = |84.0 - 86.02|/84.0 * 100 = "2.4%"%error=|84.0−86.02|84.0⋅100=2.4%
which is a relatively small percent error ->→ the actual molecular formula is C_6H_12C6H12.
