# Question ce300

Oct 18, 2016

The molecular formula is ${\text{C"_4"H}}_{10}$.

#### Explanation:

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

$\text{Mass of C" = 3.03 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.8269 g C}$

$\text{Mass of H" = 1.55 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.1734 g H}$

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

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

$\text{Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(mm) "Ratio"color(white)(m)×2color(white)(mm)"Integers}$
stackrel(————————————————————)(color(white)(ml)"C" color(white)(XXXl)0.8269 color(white)(m)"0.068 85" color(white)(Xl)1color(white)(Xmml)2color(white)(mmmml)2
$\textcolor{w h i t e}{m l} \text{H} \textcolor{w h i t e}{X X X l} 0.1734 \textcolor{w h i t e}{m} 0.1720 \textcolor{w h i t e}{m l l l} 2.498 \textcolor{w h i t e}{X l} 4.996 \textcolor{w h i t e}{m m l} 5$

The empirical formula is ${\text{C"_2"H}}_{5}$.

The empirical formula mass is $\text{(2×12.01 + 5×1.008) u = 29.06 u}$

The molecular mass is $\text{58 u}$.

"Molecular mass"/"Empirical formula mass" = (58 color(red)(cancel(color(black)("u"))))/(29.06 color(red)(cancel(color(black)("u")))) = 2.0 ≈ 2.

∴ The molecular formula is ("C"_2"H"_5)_2 = "C"_4"H"_10#.