You need to know two things to solve this problem:

- That 1 mole of
*any ideal gas occupies* #"22.4 L"# at **STP**;
- The relationship between moles and molar mass

#"moles" = ("mass of substance")/("molar mass")#

So, you know that the vapor weighs #"0.96 g"# at **STP**. **STP** conditions imply a temperature of #"273.15 K"# and a pressure of #"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"#

**SIde note** - you can use the ideal gas law equation, #PV = nRT#, to double check this result;

Since #n= "m"/"MM"#, you get that

#"n" = "m"/"MM" = "0.01116 moles"#

Since #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"#

Your compound's empirical formula is #(CH_2)_x#, or #C_xH_(2x)#. The value of #x# 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"#

Ideally, #x# should be as close to an integer as possible; in this case, the closest integer would be #6#, which would make the compound's molecular mass equal to #"84.0 g/mol"# and the molecular formula

#C_6H_12#.

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" = |84.0 - 86.02|/84.0 * 100 = "2.4%"#

which is a relatively small percent error #-># the actual molecular formula is #C_6H_12#.