# Question e7e77

Apr 30, 2016

$\text{CH}$

#### Explanation:

The idea here is that when you're dealing with gases kept under the same conditions for pressure and temperature, the mole ratio that exists between them in the balanced chemical equation is equivalent to a volume ratio.

You can prove this by using the ideal gas law equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

If you take $P$ to be the pressure and $T$ to be the absolute temperature at which your combustion reaction takes place, you can say that you have

$P \cdot {V}_{C {O}_{2}} = {n}_{C {O}_{2}} \cdot R T \to$ for carbon dioxide

$P \cdot {V}_{{H}_{2} O} = {n}_{{H}_{2} O} \cdot R T \to$ for water

If you divide these two equations, you will be left with

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot {V}_{C {O}_{2}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot {V}_{{H}_{2} O}} = \frac{{n}_{C {O}_{2}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}{{n}_{{H}_{2} O} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}$

This if equivalent to

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{V}_{C {O}_{2}} / {V}_{{H}_{2} O} = {n}_{C {O}_{2}} / {n}_{{H}_{2} O}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ the mole ratio is equal to the volume ratio

So, you know that your reaction produces carbon dioxide and water in a $2 : 1$ volume ratio, which tells you that the reaction produces carbon dioxide and water in a $2 : 1$ mole ratio.

If you take $\textcolor{b l u e}{n}$ to be the number of moles of water produced by the reaction, you can say that

${\text{hydrocarbon" + color(blue)(5n)/2"O"_ (2(g)) -> color(blue)(2n)"CO"_ (2(g)) + color(blue)(n)"H"_ 2"O}}_{\left(g\right)}$

Now, notice that all the moles of carbon that were initially present in the hydrocarbon are now a part of the carbon dioxide and all the moles of hydrogen that were present in the hydrocarbon are now a part of the water.

Since you know that one mole of carbon dioxide contains one mole of carbon, and one mole of water cotnains two moles of hydrogen, you can say that the hydrocarbon contained

color(blue)(2n)color(red)(cancel(color(black)("moles CO"_2))) * "1 mole C"/(1color(red)(cancel(color(black)("mole CO"_2)))) = (2n)color(white)(a)"moles C"

color(blue)(n)color(red)(cancel(color(black)("moles H"_2"O"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = (2n)color(white)(a)"moles H"

This means that the hydrocarbon's empirical formula, which tells you the smallest whole number ratio that exists between a compound's constituent elements, will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{C"_ (2n)"H"_(2n) implies "C"_1"H"_1 implies "CH}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

May 1, 2016

The empirical formula of the Hydrocarbon is $C H$

#### Explanation:

Let the molecular formula of the hydrocarbon whose empirical formula is to be determined , be ${C}_{x} {H}_{y}$. The empirical formula represents the least ratio of number of atoms of constituent elements present in the molecule of the compound. So we are to find out the least value of the ratio $x : y$ to write the empirical formula of the compound.

Let us try. The balanced equation of the combustion of hydrocarbon,
${C}_{x} {H}_{y}$ may be written as below

${C}_{x} {H}_{y} \left(\text{*}\right) + \left(x + \frac{y}{4}\right) {O}_{2} \left(g\right) \to x C {O}_{2} \left(g\right) + \frac{y}{2} {H}_{2} O \left(g\right)$

$\text{ * physical state unknown}$

Now on the basis of this balanced equation we see that the number of moles of $C {O}_{2}$ and ${H}_{2} O$ produced on cobustion of 1 mole of ${C}_{x} {H}_{y}$ are
$C {O}_{2} \left(g\right) \to$ x moles and ${H}_{2} O \left(g\right) \to \frac{y}{2}$ moles

So whatever be the amount of hydrocarbon taken the ratio of number of moles of $C {O}_{2}$ and ${H}_{2} O$ produced on complete combustion of ${C}_{x} {H}_{y}$ will be $x : \frac{y}{2} = 2 x : y$

Now the given condition in our problem is that under similar condition of temperature and pressure the ratio of the volumes of
$C {O}_{2} \left(g\right)$ and ${H}_{2} O \left(g\right)$ produced on combustion is $2 : 1$

Now by Avogadro's law we know that under similar condition of temperature and pressure same volume of gas always contains same number of molecules i.e. same number of moles of gas

So when Temperature and Pressure of gas remaining constant

Volume(V) is proportional to no. of moles (n) Or vice versa

So in our case
The ratio of number of moles of $C {O}_{2}$ and ${H}_{2} O$ produced on combustion of ${C}_{x} {H}_{y}$ i.e. $x : \frac{y}{2} \mathmr{and} 2 x : y$ will be equal to ratio of their volumes( $2 : 1$)

I.e. 2x:y=2;1 =>x:y=1:1# (least value of the ratio)

Hence the empirical formula of the Hydrocarbon is $C H$