# Question a782a

Mar 7, 2016

The empirical formula is ${\text{CH}}_{2}$.

#### Explanation:

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

$\text{Mass of C" = 2.64 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.7204 g C}$

$\text{Mass of H" = 1.08 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.1208 g H}$

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

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

$\text{Element"color(white)(X) "Mass/g"color(white)(Xm) "Moles"color(white)(mll) "Ratio" color(white)(mll)"Integers}$
stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.7204 color(white)(mll)"0.05998" color(white)(Xl)1color(white)(Xmmmm)1#
$\textcolor{w h i t e}{l l} \text{H" color(white)(XXXXl)0.1208 color(white)(mll)"0.1198} \textcolor{w h i t e}{m l l l} 1.998 \textcolor{w h i t e}{X X X} 2$

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