# Question fbbce

Jun 17, 2017

The empirical formula is ${\text{C"_6"H}}_{7}$.

#### Explanation:

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

$\text{Mass of C" = 3.02 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.8241 g C}$

$\text{Mass of H" = 0.72 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0806 g H}$

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

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

$\text{Element"color(white)(X) "Mass/g"color(white)(Xm) "Moles"color(white)(mll) "Ratio"color(white)(m)×6 color(white)(mll)"Integers}$
stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.8241 color(white)(mll)"0.06862" color(white)(Xl)1color(white)(mmm)6color(white)(Xmmmm)6#
$\textcolor{w h i t e}{l l} \text{H" color(white)(XXXXl)0.0806 color(white)(mll)"0.0800} \textcolor{w h i t e}{m l l l} 1.165 \textcolor{w h i t e}{m} 6.990 \textcolor{w h i t e}{X X X} 7$

The empirical formula is $\text{C"_6"H"_7}$.