# Formaldehyde consists of 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. What is its empirical formula?

Feb 2, 2016

The empirical formula is $\text{CH"_2"O}$.

#### Explanation:

Assume that you have 100 g of formaldehyde.

Then you have 40.0 g of $\text{C}$, 6.7 g of $\text{H}$, and 53.3 g of $\text{O}$.

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

$\text{Moles of C" =40.0 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "3.331 mol C}$

$\text{Moles of H" = 6.7 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008color(red)(cancel(color(black)("g H")))) = "6.65 mol H}$

$\text{Moles of O" = 53.3 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "3.331 mol O}$

To get the molar ratio, we divide each number of moles by the smallest
number ($3.331$).

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

$\text{Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(X) "Ratio" color(white)(X)"Integers}$
stackrel(—————————————————-—)(color(white)(l)"C" color(white)(XXXX)40.0 color(white)(Xlll)3.331 color(white)(Xll)1 color(white)(XXXlll)1)
$\textcolor{w h i t e}{X} \text{H} \textcolor{w h i t e}{X X X X l} 6.7 \textcolor{w h i t e}{X X l} 6.65 \textcolor{w h i t e}{X X l} 2.00 \textcolor{w h i t e}{X X l} 2$
$\textcolor{w h i t e}{X} \text{O} \textcolor{w h i t e}{X X X l l} 53.3 \textcolor{w h i t e}{X X l} 3.331 \textcolor{w h i t e}{X l l} 1.00 \textcolor{w h i t e}{X X l} 1$

The ratio comes out as $\text{C:H:O} = 1 : 2 : 1$.

Thus, the empirical formula is $\text{CH"_2"O}$.