What is the empirical formula, if given the following percent composition: 62.1% C, 5.21%H, 12.1% N, and 20.7% O?

Mar 5, 2016

The empirical formula is ${\text{C"_12"H"_12"NO}}_{3}$.

Explanation:

Your percentages don't add up to exactly 100 %, but they are within the limits of experimental error.

Assume that you have 100 g of the compound.

Then you have 62.1 g of $\text{C}$, 5.21 g of $\text{H}$, 12.1 g of $\text{N}$, and 20.7 g of $\text{O}$.

Our job is to calculate the ratio of the moles of each element.

$\text{Moles of C" = 62.1color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "5.171 mol C}$

$\text{Moles of H" = 5.21 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "5.169 mol H}$

$\text{Moles of N" = 12.1 color(red)(cancel(color(black)("g N"))) × "1 mol N"/(14.01 color(red)(cancel(color(black)("g N")))) = "0.8637 mol N}$

$\text{Moles of O" = 20.7 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "1.294 mol O}$

To get the molar ratio, we divide each number of moles by the smallest number
($\text{0.8637}$).

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

$\text{Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio" color(white)(m)×2color(white)(mll)"Integers}$
stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXX)62.1 color(white)(Xmll)"5.171" color(white)(Xll)5.987color(white)(m) 11.97color(white)(Xmm)12
$\textcolor{w h i t e}{l l} \text{H" color(white)(XXXXll)5.21 color(white)(mm)"5.169} \textcolor{w h i t e}{X l l} 5.985 \textcolor{w h i t e}{m} 11.97 \textcolor{w h i t e}{X X X} 12$
$\textcolor{w h i t e}{l l} \text{N" color(white)(XXXml)12.1 color(white)(mml)"0.8637} \textcolor{w h i t e}{X l} 1 \textcolor{w h i t e}{m m m l} 2 \textcolor{w h i t e}{X X X m l l l} 2$
$\textcolor{w h i t e}{l l} \text{O" color(white)(XXXXl)20.7 color(white)(mml)"1.294} \textcolor{w h i t e}{X l l} 1.498 \textcolor{w h i t e}{m l l} 2.998 \textcolor{w h i t e}{X X X} 3$

The empirical formula is ${\text{C"_12"H"_12"NO}}_{3}$.