Question

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.

Answer 1

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`