I believe the correct vapour density is 61.5. See below.
Our first task is to calculate the empirical formula of the compound.
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
The compound contains 58.55 % #"C"#, 4.05 % #"H"#, and 11.36 % #"N"#.
That adds up to 73.96 %.
The remainder, 26.04 % must be #"O"#.
So, our job is to calculate the molar ratio of #"C"# to #"H"# to #"N"# to #"O"#.
Assume that you have 100 g of sample.
Then it contains 58.55 g of #"C"#, 4.05 g of #"H"#, 11.36 g of #"N"#, and 26.04 g of #"O"#.
#"Moles of C" = 58.55 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)( "g C")))) = "4.8751 mol C"#
#"Moles of H" = 4.05 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "4.018 mol H"#
#"Moles of N" = 11.36 color(red)(cancel(color(black)("g N"))) × "1 mol N"/(14.01 color(red)(cancel(color(black)( "g O")))) = "0.810 85 mol N"#
#"Moles of O" = 26.04 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)( "g O")))) = "1.6275 mol O"#
From this point on, I like to summarize the calculations in a table.
#bbul("Element"color(white)(m) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(m)"Integers")#
#color(white)(m)"C" color(white)(XXXmm)58.55 color(white)(Xm)4.8751
color(white)(mm)6.0120color(white)(mmm)6#
#color(white)(m)"H" color(white)(XXXXmll)4.05 color(white)(mm)4.018 color(white)(Xml)4.955 color(white)(mmmll)5#
#color(white)(m)"N"color(white)(XXXmm)11.36 color(white)(Xm)"0.810 85"color(white)(m)1color(white)(mmmmmll)1#
#color(white)(m)"O"color(white)(XXXmm)26.04 color(white)(Xm)1.6275color(white)(mll)2.0072color(white)(mmml)2#
The empirical formula is #"C"_6"H"_5"NO"_2#.
Now, we find the molecular formula.
#"Empirical formula mass" = "(72.06 + 5.04 + 14.01 + 32.00) u" = "123.10 u"#
#"Vapour density" = 6.15 = M_text(r gas)/M_text(r H₂) = M_text(r gas)/2.016#
∴ #M_text(r gas) = 6.15 × 2.016 = 12.40#
This is impossible!
However, if the vapour density is 61.5, we get
#M_text(r gas) = 61.5 × 2.016 = 124.0#
The molecular mass must be an integral multiple of the empirical formula mass.
#"MM"/"EFM" = 124.0/123.10 = 1.007 ≈ 1#
∴ The molecular formula must be the same as the empirical formula: #"C"_6"H"_5"NO"_2#.
