# Question 79408

Feb 23, 2017

The empirical formula is ${\text{C"_6"H"_13"O}}_{3}$.

#### Explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of $\text{C}$ to $\text{H}$ to $\text{O}$.

Your compound contains 54.53 % $\text{C}$, 9.15 % $\text{H}$, and 36.32 % $\text{O}$.

Assume that you have 100 g of sample.

Then it contains 54.53 g of $\text{C}$, 9.15 g of $\text{H}$, and 36.32 g of $\text{O}$.

$\text{Moles of C" = 54.53 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)( "g C")))) = "4.540 mol B}$

$\text{Moles of H" = 9.15 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "9.077 mol H}$

$\text{Moles of O" = 36.32 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)( "g O")))) = "2.270 mol O}$

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

$\boldsymbol{\text{Element"color(white)(m) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mm)"×3"color(white)(mm)"Integers}}$
color(white)(m)"C" color(white)(XXXmm)54.53 color(white)(Xm)4.540 color(white)(mmll)2.000color(white)(mml)6.000color(white)(mmml)6#
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{X X X X m l l} 9.15 \textcolor{w h i t e}{m m} 9.077 \textcolor{w h i t e}{X m l l} 4.304 \textcolor{w h i t e}{m l l} 12.91 \textcolor{w h i t e}{m m m l} 13$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{X X X m m} 36.32 \textcolor{w h i t e}{X m} 2.270 \textcolor{w h i t e}{m m l l} 1 \textcolor{w h i t e}{m m m m l} 3 \textcolor{w h i t e}{m m m m m l} 3$

The empirical formula is ${\text{C"_6"H"_13"O}}_{3}$.