# Question #aa570

Feb 16, 2017

The empirical formula is ${\text{C"_3"H}}_{8}$.

#### Explanation:

We can calculate the mass of $\text{C}$ from the mass of ${\text{CO}}_{2}$.

$\text{Mass of C" = 12 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "3.27 g C}$

$\text{Mass of H" = "4 g - 3.27 g" = "0.73 g}$

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

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

$\boldsymbol{\text{Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(m) "Ratio"color(white)(m)"×2"color(white)(ml)"×3"color(white)(m)"Integers}}$
$\textcolor{w h i t e}{m l} \text{C} \textcolor{w h i t e}{X X X m m} 3.27 \textcolor{w h i t e}{m m l} 0.272 \textcolor{w h i t e}{X m} 1 \textcolor{w h i t e}{X m m l l} 2 \textcolor{w h i t e}{m m l} 3 \textcolor{w h i t e}{m m m l} 3$
$\textcolor{w h i t e}{m l} \text{H} \textcolor{w h i t e}{X X X m m} 0.73 \textcolor{w h i t e}{m m l} 0.72 \textcolor{w h i t e}{m m l l} 2.64 \textcolor{w h i t e}{X m} 5.29 \textcolor{w h i t e}{m} 7.94 \textcolor{w h i t e}{m m} 8$

The empirical formula is ${\text{C"_3"H}}_{8}$.