# 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 $1.00 \mathrm{dm} 3$ vessel at 100°C, had a pressure of $84 k P a$. Determine the molecular formula of the compound.

Sep 16, 2016

WARNING! Long answer! The molecular formula is ${\text{C"_4"H"_8"O}}_{2}$.

#### Explanation:

Calculate the empirical formula.

We can calculate the masses of $\text{C}$ and $\text{H}$ from the masses of their oxides (${\text{CO}}_{2}$ and $\text{H"_2"O}$).

$\text{Mass of C" = 1.80 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.4912 g C}$

$\text{Mass of H" = 0.736 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0823 g H}$

$\text{Mass of O" = "Mass of compound - mass of C - mass of O" = "0.900 g - 0.4912 g - 0.0823 g" = "0.3265 g}$

Now, we must convert these masses to moles and find their ratios.

From here on, I like to summarize the calculations in a table.

$\text{Element"color(white)(Xll) "Mass/g"color(white)(Xmll) "Moles"color(white)(m) "Ratio" color(white)(m)"Integers}$
stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.4912 color(white)(mll)"0.040 90" color(white)(Xll)2.004color(white)(mm)2)
$\textcolor{w h i t e}{l l} \text{H" color(white)(XXXXl)0.0823 color(white)(mll)"0.0816} \textcolor{w h i t e}{m m l} 4.00 \textcolor{w h i t e}{X m l l} 4$
$\textcolor{w h i t e}{l l} \text{O" color(white)(mmmml)0.3265color(white)(mll)"0.020 41} \textcolor{w h i t e}{m l l} 1 \textcolor{w h i t e}{m m m l l} 1$

The empirical formula is $\text{C"_2"H"_4"O}$.

Use the Ideal Gas Law to calculate the molar mass

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Since $n = \text{mass"/"molar mass} = \frac{m}{M}$,

we can write the Ideal Gas Law as

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = \frac{m}{M} R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this to get

$M = \left(\frac{m}{V}\right) \frac{R T}{P}$

or

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} M = \frac{m R T}{P V} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

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

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 $\text{C"_2"H"_4"O}$ 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.

${\text{Molecular formula" = ("C"_2"H"_4"O")_2 = "C"_4"H"_8"O}}_{2}$