# Question dbc04

Jan 2, 2017

WARNING! Very long answer! (a) The balanced equation is $\text{C"_2"H"_6"(g)" + "4Cl"_2"(g)" → "C"_2"H"_2"Cl"_4"(l)" + "4HCl(g)}$. (b) You need to use 7.01 g of ethane.

#### Explanation:

(a) Write the balanced equation

This is really a three-part question:

i. Find the empirical formula.
ii. Find the molecular formula.
iii. Write the equation.

i. Find 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 14.3 % $\text{C}$ and 84.5 % $\text{Cl}$.

Then "% H = 100 % - % C - % Cl" = "100 % - 14.3 % - 84.5 %" = 1.2 %

Assume that we have 100 g of sample.

Then it contains 14.3 g of $\text{C}$, 1.2 g of $\text{H}$, and 84.5 g of $\text{Cl}$.

$\text{Moles of C" = 14.3 color(red)(cancel(color(black)("g c"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "1.191 mol H}$

$\text{Moles of H" = 1.2 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "1.19 mol H}$

$\text{Moles of Cl" = 84.5 color(red)(cancel(color(black)("g O"))) × "1 mol Cl"/(35.45color(red)(cancel(color(black)( "g Cl")))) = "2.384 mol Cl}$

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

$m a t h b f \text{Element"color(white)(m) mathbf"Mass/g"color(white)(X) mathbf"Moles"color(white)(Xll)mathbf "Ratio"color(white)(m)mathbf"Integers}$
color(white)(m)"C" color(white)(XXXmll)14.3color(white)(Xmm)1.191 color(white)(mml)1.001color(white)(mmml)1
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m m m l} 1.2 \textcolor{w h i t e}{m m l l} 1.19 \textcolor{w h i t e}{m m m} 1 \textcolor{w h i t e}{m m m m m l} 1$
$\textcolor{w h i t e}{m} \text{Cl} \textcolor{w h i t e}{m m m m l} 84.59 \textcolor{w h i t e}{m m l} 2.384 \textcolor{w h i t e}{m m l} 2.003 \textcolor{w h i t e}{m m m l} 2$

The empirical formula of $\text{Y}$ is ${\text{CHCl}}_{2}$.

ii. Calculate the molecular formula

The empirical formula mass of ${\text{CHCl}}_{2}$ is 83.92 u.

The molecular mass is 168 u.

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

"MM"/"EFM" = (168 color(red)(cancel(color(black)("u"))))/(83.92 color(red)(cancel(color(black)("u")))) = 2.00 ≈ 2#

∴ The molecular formula is twice the empirical formula.

${\text{Molecular formula" = ("CHCl"_2)_2 = "C"_2"H"_2"Cl}}_{4}$

Compound $\text{Y}$ is ${\text{C"_2"H"_2"Cl}}_{4}$

iii. Write the balanced equation

The unbalanced equation must be

$\text{C"_2"H"_6"(g)" + "Cl"_2"(g)" → "C"_2"H"_2"Cl"_4"(l)" + "HCl(g)}$

The balanced equation is

$\text{C"_2"H"_6"(g)" + "4Cl"_2"(g)" → "C"_2"H"_2"Cl"_4"(l)" + "4HCl(g)}$

(b) Calculate the mass of ethane required

$\text{57.0 % yield" = "22.3 g Y}$

$\text{Theoretical yield" = 100 color(red)(cancel(color(black)("% yield"))) × "22.3 g Y"/(57.0 color(red)(cancel(color(black)("% yield")))) = "39.12 g Y}$

${\text{Mass of C"_2"H"_6 = 39.12 color(red)(cancel(color(black)("g Y"))) × (1 color(red)(cancel(color(black)("mol Y"))))/(167.85 color(red)(cancel(color(black)("g Y")))) × (1 color(red)(cancel(color(black)("mol Y"))))/(1 color(red)(cancel(color(black)("mol C"_2"H"_6)))) × ("30.07 g C"_2"H"_6)/(1 color(red)(cancel(color(black)("mol C"_2"H"_6)))) = 7.01"g C"_2"H}}_{6}$

The mass of ethane required is 7.01 g.