# Question 15384

Jan 24, 2017

WARNING! Long answer! The molecular formula is $\text{C"_4"H"_10"O}$.

#### Explanation:

First we 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 $\text{H"_2"O}$).

$\text{Mass of C" = 2.389 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.6519 g C}$

$\text{Mass of H" = 1.22 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.1365 g H}$

$\text{Mass of C + Mass of H" = "0.6519 g + 0.1365 g" = "0.7883 g}$

This is less than the mass of the sample.

The missing mass must be caused by $\text{O}$.

$\text{Mass of O = 1.00 g - 0.7883 g = 0.212 g}$

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

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

$\boldsymbol{\text{Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(mm) "Ratio"color(white)(mm)"Integers}}$
color(white)(ml)"C" color(white)(XXXm)0.6519 color(white)(ml)0.05422 color(white)(Xmlll)4.092color(white)(Xmmm)4
$\textcolor{w h i t e}{m l} \text{H} \textcolor{w h i t e}{X X X m} 0.1365 \textcolor{w h i t e}{m l} 0.1354 \textcolor{w h i t e}{m m m} 10.22 \textcolor{w h i t e}{X m m m} 10$
$\textcolor{w h i t e}{m l} \text{O} \textcolor{w h i t e}{X X X m} 0.212 \textcolor{w h i t e}{m l l} 0.01325 \textcolor{w h i t e}{m m m l} 1 \textcolor{w h i t e}{X m m m m m} 1$

The empirical formula is $\text{C"_4"H"_10"O}$.

Now, we use the Ideal Gas Law to determine 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 $n = \frac{m}{M}$, we can rearrange this equation to get

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

And we can solve this equation to get

$M = \frac{m R T}{P V}$

Thus, in this problem,

$m = \text{1.00 g}$
$R = \text{0.082 06 dm"^3·"atm·K"^"-1""mol"^"-1}$
$T = \text{25.0 °C" = "298.15 K}$
$P = \text{1 atm}$
$V = {\text{324 cm"^3 = "0.324 dm}}^{3}$

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 $\text{C"_4"H"_10"O}$ is $\text{(4×12.01 + 10×1.008 + 1×16.00) u = 74.12 u}$

The molecular mass is $\text{75.5 u}$.

"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"#.