Question #8cd77

Apr 19, 2017

The empirical formula is ${\text{C"_5"H}}_{12}$.

Explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of $\text{C}$ to $\text{H}$.

$\text{Mass of C = 12 g}$

$\text{Mass of alkane = mass of C + mass of H}$

$\text{14.4 g = 12 g + mass of H}$

$\text{Mass of H = (14.4 – 12) g = 2.4 g}$

$\text{Moles of C" = 12 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01color(red)(cancel(color(black)( "g C")))) = "0.999 mol C}$

$\text{Moles of H "= 2.4 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "2.38 mol H}$

To get this into an integer ratio, we divide both numerator and denominator by the smaller value.

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

$\boldsymbol{\text{Element"color(white)(Ag) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(m)×5color(white)(m)"Integers}}$
$\textcolor{w h i t e}{m l l} \text{C} \textcolor{w h i t e}{X X X m m} 12 \textcolor{w h i t e}{X m m l} 0.999 \textcolor{w h i t e}{X m m} 1 \textcolor{w h i t e}{m m m m} 5 \textcolor{w h i t e}{m m m l l} 5$
$\textcolor{w h i t e}{m l l} \text{H" color(white)(XXXXmll)2.4 color(white)(mml)} 2.38 \textcolor{w h i t e}{X m m l l} 2.38 \textcolor{w h i t e}{m m} 11.9 \textcolor{w h i t e}{m m} 12$

There are 5 mol of $\text{C}$ for 12 mol of $\text{H}$.

The empirical formula of the alkane is ${\text{C"_5"H}}_{12}$.

Here is a video that illustrates how to determine an empirical formula.