# Question #96443

Mar 1, 2016

Empirical formula $C H$

#### Explanation:

Combustion analysis or combustion reaction involving hydrocarbons ( compounds composed only of carbon and hydrogen) have the following equation (unbalanced)

$H y \mathrm{dr} o c a r b o n$ + ${O}_{2}$ $\implies$ $C {O}_{2}$ + ${H}_{2} O$

What we are given:

20g $H y \mathrm{dr} o c a r b o n$ + ${O}_{2}$ $\implies$ 67.6g $C {O}_{2}$ + 13.8g ${H}_{2} O$

The empirical formula of an compound is defined as the lowest ratio of elements within that compound.

For example:

Glucose; ${C}_{6} {H}_{12} {O}_{6}$ , in its lowest ratio, the empirical formula for glucose would be $C {H}_{2} O$

Erythrulose; ${C}_{4} {H}_{8} {O}_{4}$ in its lowest ratio, the empirical formula is the same as glucose: $C {H}_{2} O$.

Step 1

The first step in finding the empirical formula of our hydrocarbon is to convert the grams of $C {O}_{2}$ and ${H}_{2} O$ into mols

$C$ = $12.1 g$
$O$ = 16.00g * 2 = $32.0 g$

$67.6 g$ $C {O}_{2}$ * $\frac{1 m o l}{44.1 g}$ = 1.53 mol $C {O}_{2}$

$H$ = 1.01g * 2 = $2.02 g$
$O$ = $16.00 g$

$13.8 g$ ${H}_{2} O$ $\frac{1 m o l}{18.02 g}$ = .766 mol ${H}_{2} O$

We will now factor the amount of ${O}_{2}$ that was used in the reaction.

With the Law of Conservation of Mass in mind, we are given that 20g of $H y \mathrm{dr} o c a r b o n$ reacts with ${O}_{2}$ to form 67.6g $C {O}_{2}$ and 13.8g of ${H}_{2} O$ or a total of 81.4g.

The equation 81.4g $\left(H {2}_{O} + C {O}_{2}\right)$ - 20g $H y \mathrm{dr} o c a r b o n$ = 61.4g ${O}_{2}$

$O$ = 16.00g, but oxygen is a diatomic, so we multiply by 2 to get 32.00g

$61.4 g$ ${O}_{2}$ * $\frac{1 m o l {O}_{2}}{32.00 g}$ = 1.92 mol ${O}_{2}$

Step 2

Now we multiply our three mol ratios by the smallest to get a whole number ratio between them

$C {O}_{2}$ ; $\frac{1.53 m o l}{.766 m o l}$ = 2.00 mol $C {O}_{2}$

${H}_{2} O$ ; $\frac{.766 m o l}{.766 m o l}$ = 1.00 mol ${H}_{2} O$

${O}_{2}$ ; $\frac{1.92 m o l}{.766 m o l}$ = 2.51 mol ${O}_{2}$

Since ${O}_{2}$ is at 2.51 mol, we multiply everything by 2

2.00 mol $C {O}_{2}$ * 2 = 4.00 mol $C {O}_{2}$

1.00 mol ${H}_{2} O$ * 2 = 2.00 mol ${H}_{2} O$

2.51 mol ${O}_{2}$ * 2 = 5.02 mol ${O}_{2}$ $\implies$ 5.00 mol ${O}_{2}$

Step 3

going back to our equation, we can balance it with the mol ratios

$H y \mathrm{dr} o c a r b o n$ + 5 ${O}_{2}$ $\implies$ 4 $C {O}_{2}$ + 2 ${H}_{2} O$

Left hand side: 10 $O$
Right hand side: 10 $O$, 4$C$, 4$H$

For the equation to be balanced, the $H y \mathrm{dr} o c a r b o n$ must have 4 $C$ and 4 $H$. The empirical formula is the lowest ratio of the elements $\left(C , H\right)$, thus it is ${C}_{1} {H}_{1}$ or just $C H$