How to determine the molecular formula of this organic compound?

An organic compound containing only carbon, hydrogen and oxygen was analysed gravimetrically. When completely oxidized in air, #0.900 g# of the compound produced #1.80 g# of carbon dioxide and #0.736 g# of water. A separate #2.279 g# sample, when vaporized in a #1.00 dm3# vessel at #100°C#, had a pressure of #84 kPa#. Determine the molecular formula of the compound.

1 Answer
Sep 16, 2016

Answer:

WARNING! Long answer! The molecular formula is #"C"_4"H"_8"O"_2#.

Explanation:

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" = 1.80 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.4912 g C"#

#"Mass of H" = 0.736 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0823 g H"#

#"Mass of O" = "Mass of compound - mass of C - mass of O" = "0.900 g - 0.4912 g - 0.0823 g" = "0.3265 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.

#"Element"color(white)(Xll) "Mass/g"color(white)(Xmll) "Moles"color(white)(m) "Ratio" color(white)(m)"Integers"#
#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.4912 color(white)(mll)"0.040 90" color(white)(Xll)2.004color(white)(mm)2)#
#color(white)(ll)"H" color(white)(XXXXl)0.0823 color(white)(mll)"0.0816" color(white)(mml)4.00 color(white)(Xmll)4#
#color(white)(ll)"O" color(white)(mmmml)0.3265color(white)(mll)"0.020 41"color(white)(mll)1color(white)(mmmll)1#

The empirical formula is #"C"_2"H"_4"O"#.

Use the Ideal Gas Law to calculate the molar mass

#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#

Since #n = "mass"/"molar mass" = m/M#,

we can write the Ideal Gas Law as

#color(blue)(bar(ul(|color(white)(a/a) PV = m/MRTcolor(white)(a/a)|)))" "#

We can rearrange this to get

#M = (m/V)(RT)/P#

or

#color(blue)(bar(ul(|color(white)(a/a)M = (mRT)/(PV)color(white)(a/a)|)))" "#

#m = "2.279 g"#
#R = "8.314 kPa·dm"^3"·K"^"-1""mol"^"-1"#
#T = "(100 + 273.15) K" = "373.15 K"#
#P = "84 kPa"#

#M = ("2.279 g" × 8.314 color(red)(cancel(color(black)("kPa·dm"^3"·K"^"-1")))"mol"^"-1" × 373.15 color(red)(cancel(color(black)("K"))))/(84 color(red)(cancel(color(black)("kPa")))) = "84.2 g/mol"#

Calculate the molecular formula

The empirical formula mass of #"C"_2"H"_4"O"# is 44.05 u.

The molecular mass is 84.2 u.

The molecular mass must be an integral multiple of the empirical formula mass.

#"MM"/"EFM" = (84.2 color(red)(cancel(color(black)("u"))))/(44.05 color(red)(cancel(color(black)("u")))) = 1.91 ≈ 2#

The molecular formula must be twice the empirical formula.

#"Molecular formula" = ("C"_2"H"_4"O")_2 = "C"_4"H"_8"O"_2#