# Burning then sample of an organic compound with the weight of 0.00480 g it is afforded: 0.00703 g of the CO2 and 0.00283 g of the H2O, Determine the empirical formula of the compound which consists of hydrogen, carbon and oxygen?

Sep 22, 2016

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

#### Explanation:

We can calculate the masses of $\text{C}$ and $\text{H}$ from the masses of their oxides (${\text{CO}}_{2}$ and $\text{H"_2"O}$).

$\text{Mass of C" = 7.03 color(red)(cancel(color(black)("mg CO"_2))) × "12.01 mg C"/(44.01 color(red)(cancel(color(black)("mg CO"_2)))) = "1.918 g C}$

$\text{Mass of H" = 0.283 color(red)(cancel(color(black)("mg H"_2"O"))) × "2.016 mg H"/(18.02 color(red)(cancel(color(black)("mg H"_2"O")))) = "0.3166 mg H}$

$\text{Mass of O" = "Mass of compound - mass of C - mass of O" = "4.80 mg - 1.918 mg - 0.3166 mg" = "2.565 mg}$

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

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

$\text{Element"color(white)(X) "Mass/mg"color(white)(X) "Millimoles"color(white)(m) "Ratio"color(white)(m)"Integers}$
stackrel(—————————————————-———)(color(white)(ll)"C" color(white)(XXXmm)1.918 color(white)(mmm)0.1597 color(white)(Xlll)1color(white)(Xmmml)1
$\textcolor{w h i t e}{l l} \text{H} \textcolor{w h i t e}{X X X X m} 0.3166 \textcolor{w h i t e}{m m l} 0.3145 \textcolor{w h i t e}{m l l l} 1.969 \textcolor{w h i t e}{X X l} 2$
$\textcolor{w h i t e}{l l} \text{O} \textcolor{w h i t e}{m m m m l} \textcolor{w h i t e}{l l} 2.565 \textcolor{w h i t e}{m m l l} 0.1603 \textcolor{w h i t e}{m m} 1.004 \textcolor{w h i t e}{m m l} 1$

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