# A compound contains 4.07% H, 24.27% C, and the remaining is chlorine. Its molar mass is 98.96 g/mol. What are its empirical formula and molecular formula?

Jun 22, 2018

The empirical formula is $\text{CH"_2"Cl}$. The molecular formula is ${\text{C"_2"H"_4"Cl}}_{2}$.

#### Explanation:

Step 1. Calculate the empirical formula

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 $\text{H}$ to $\text{Cl}$.

Your compound contains 24.27 % $\text{C}$, and 4.07 % $\text{H}$.

Assume that you have 100 g of sample.

Then it contains 24.27 g of $\text{C}$ and 4.07 g of $\text{H}$.

$\text{Mass of O = (100 - 24.27 - 4.07) g = 71.66 g}$

$\text{Moles of C" = 24.27 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)( "g C")))) = "2.021 mol C}$

$\text{Moles of H" = 4.07 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "4.038 mol H}$

$\text{Moles of Cl" = 71.66 color(red)(cancel(color(black)("g Cl"))) × "1 mol Cl"/(35.45 color(red)(cancel(color(black)( "g Cl")))) = "2.021mol Cl}$

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

$\underline{\boldsymbol{\text{Element"color(white)(m) "Mass/g"color(white)(Xl) "Moles"color(white)(Xll) "Ratio"color(white)(m)"Integers}}}$
$\textcolor{w h i t e}{m} \text{C} \textcolor{w h i t e}{X X X m m l} 24.27 \textcolor{w h i t e}{X m l} 2.021 \textcolor{w h i t e}{m m l} 1 \textcolor{w h i t e}{m m m m m l l} 1$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{X X X X m l l} 4.07 \textcolor{w h i t e}{m m l} 4.038 \textcolor{w h i t e}{X m l l} 1.998 \textcolor{w h i t e}{m m m l l} 2$
$\textcolor{w h i t e}{m} \text{Cl} \textcolor{w h i t e}{X X X m m} 71.66 \textcolor{w h i t e}{X m l} 2.021 \textcolor{w h i t e}{m m l} 1.000 \textcolor{w h i t e}{m m m l l} 1$

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

Step 2. Calculate the molecular formula of the compound

The empirical formula mass of $\text{CH"_2"Cl}$ is 49.48 u.

The molecular mass is 98.96 u.

The molecular mass must be an integral multiple of the empirical formula mass.

"MM"/"EFM" = (98.96color(red)(cancel(color(black)("u"))))/(49.48 color(red)(cancel(color(black)("u")))) = 2.000 ≈ 2

The molecular formula must be twice the empirical formula.

${\text{MF" = ("EF")_2 = ("CH"_2"Cl")_2 = "C"_2"H"_4"Cl}}_{2}$