# Question 7c4d8

Mar 5, 2016

WARNING! Long Answer! The empirical formula is $\text{C"_2"H"_2"O"_3"K}$. The molecular formula is ${\text{C"_4"H"_4"O"_6"K}}_{2}$.

#### Explanation:

The equation for the combustion reaction is

$\text{2K"_2"C"_4"H"_4"O"_6 + "5O"_2 → "2K"_2"O" + "8CO"_2 + "4H"_2"O}$

Part 1. Empirical formula.

("C, H, O, K") + "O"_2 → "CO"_2color(white)(l) +color(white)(l) "H"_2"O"color(white)(l) +color(white)(l) "K"_2"O"
$\textcolor{w h i t e}{m l} \text{0.5000 g"color(white)(mmmmm) "0.389 g"color(white)(ll) "0.0796 g"color(white)(ll) "0.208 g}$

We can calculate the masses of $\text{C, H}$, and $\text{K}$ from the masses of their oxides.

$\text{Mass of C" = 0.389 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.1062 g C}$

$\text{Mass of H" = 0.0796 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.008 91 g H}$

$\text{Mass of K" = 0.208 color(red)(cancel(color(black)("g K"_2"O"))) × "78.20 g K"/(94.20 color(red)(cancel(color(black)("g K"_2"O")))) = "0.1727 g K}$

$\text{Mass of O" = "mass of (C, H, O, K) – mass of (C + H + K)" = "0.5000 g – (0.1062 + 0.00891 + 0.1727) g" = "(0.5000 – 0.2878) g" = "0.2122 g O}$

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

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

$\text{Element"color(white)(X) "Mass/g"color(white)(Xm) "mmol"color(white)(mll) "Ratio" color(white)(mll)"Integers}$
stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.1062 color(white)(Xml)"8.843" color(white)(Xll)2.002color(white)(Xmm)2
$\textcolor{w h i t e}{l l} \text{H" color(white)(XXXXl)0.00891 color(white)(mll)"8.839} \textcolor{w h i t e}{X l l} 2.001 \textcolor{w h i t e}{X X X} 2$
$\textcolor{w h i t e}{l l} \text{K" color(white)(XXXml)0.1727color(white)(mml)"4.417} \textcolor{w h i t e}{X l l} 1 \textcolor{w h i t e}{X X X m l l l} 1$
$\textcolor{w h i t e}{l l} \text{O" color(white)(XXXXl)0.2122 color(white)(mll)"13.26} \textcolor{w h i t e}{X m l} 3.002 \textcolor{w h i t e}{X X X} 3$

The empirical formula is $\text{C"_2"H"_2"O"_3"K}$.

Part 2. Molecular formula.

The relative empirical formula mass of $\text{C"_2"H"_2"O"_3"K}$ is 113.13.

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

$\text{MM" = n × "EFM}$, or

n = "MM"/"EFM" = 226.27/113.13 = 1.9992 ≈ 2#

The molecular formula must be twice the empirical formula.

${\text{MF" = "EF"_2 = ("C"_2"H"_2"O"_3"K")_2 = "C"_4"H"_4"O"_6"K}}_{2}$

Part 3. The chemical equation

The first experiment is a combustion reaction.

The equation for the reaction is

$\text{2K"_2"C"_4"H"_4"O"_6 + "5O"_2 → "2K"_2"O" + "8CO"_2 + "4H"_2"O}$