First we calculate the empirical formula.
We can calculate the masses of #"C"# and #"H"# from the masses of their oxides (#"CO"_2# and #"H"_2"O"#).
#"Mass of C" = 2.389 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.6519 g C"#
#"Mass of H" = 1.22 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.1365 g H"#
#"Mass of C + Mass of H" = "0.6519 g + 0.1365 g" = "0.7883 g"#
This is less than the mass of the sample.
The missing mass must be caused by #"O"#.
#"Mass of O = 1.00 g - 0.7883 g = 0.212 g"#
Now, we must convert these masses to moles and find their ratios.
From here on, I like to summarize the calculations in a table.
#bb("Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(mm) "Ratio"color(white)(mm)"Integers")#
#color(white)(ml)"C" color(white)(XXXm)0.6519 color(white)(ml)0.05422
color(white)(Xmlll)4.092color(white)(Xmmm)4#
#color(white)(ml)"H" color(white)(XXXm)0.1365 color(white)(ml)0.1354 color(white)(mmm)10.22 color(white)(Xmmm)10#
#color(white)(ml)"O" color(white)(XXXm)0.212 color(white)(mll)0.01325 color(white)(mmml)1 color(white)(Xmmmmm)1#
The empirical formula is #"C"_4"H"_10"O"#.
Now, we use the Ideal Gas Law to determine the molar mass.
#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#
Since #n = m/M#, we can rearrange this equation to get
#PV = (m/M)RT#
And we can solve this equation to get
#M = (mRT)/(PV)#
Thus, in this problem,
#m = "1.00 g"#
#R = "0.082 06 dm"^3·"atm·K"^"-1""mol"^"-1"#
#T = "25.0 °C" = "298.15 K"#
#P = "1 atm"#
#V = "324 cm"^3 = "0.324 dm"^3#
∴ #M = (1.00 color(red)(cancel(color(black)("g"))) × "0.082 06" color(red)(cancel(color(black)("dm"^3·"atm·K"^"-1")))"mol"^"-1" × 298.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("atm"))) × 0.324 color(red)(cancel(color(black)("dm"^3)))) = "75.5 g/mol"#
Finally, we can calculate the molecular formula.
The empirical formula mass of #"C"_4"H"_10"O"# is #"(4×12.01 + 10×1.008 + 1×16.00) u = 74.12 u"#
The molecular mass is #"75.5 u"#.
#"Molecular mass"/"Empirical formula mass" =(75.5 color(red)(cancel(color(black)("u"))))/(74.12 color(red)(cancel(color(black)("u")))) = 1.02 ≈ 1#.
∴ The molecular formula is #("C"_4"H"_10"O")_1 = "C"_4"H"_10"O"#.
