Question

Consider a sample of a hydrocarbon at$\mathbf{\text{0.959atm}}$and$\mathbf{\text{298K}}$. Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapour at$\mathbf{\text{1.51atmand375K}}$. This mixture has a density of$\mathbf{\text{1.391g/L}}$and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon and name it.

Short answer

The identity of the unknown hydrocarbon is ethane, and its formula is ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$.

Step-by-step solution

Step 1:Volume of the hydrocarbon

Sample of a hydrocarbon at 0.959 atm and 298 K produces a mixture of carbon dioxide and water vapor at 1.51atm and 375Kupon combustion.

Products occupy 4 times the volume of the hydrocarbon

Mass of the products

From the given information, we get the reaction:

${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}{\text{+O}}_{\text{2}}$→${\text{xCO}}_{\text{2}}\text{+}\frac{\text{y}}{\text{2}}{\text{H}}_{\text{2}}\text{O}$

To solve for x and y, we must firstly calculate the mass of the products using the p formula:$\frac{\text{mass}}{\text{volume}}\text{.}$

To do so, we must assume that the volume of the hydrocarbon is$\text{1.000L}$which makes the volume of the products $\text{4.000L.}$

$\begin{array}{l}\text{massofproducts=ρ×volume}\\ \text{=1.391}\frac{\text{g}}{\text{L}}\text{×4.000L}\\ \text{=5.564gofproducts}\end{array}$

Step 3:Number of moles for the hydrocarbon and its products

Next, with the given information, the ideal gas law will give us the number of moles for the hydrocarbon and its products:

$\text{PV=nRT}$

Where R is the gas constant $\text{=0.08206}\frac{\text{L×atm}}{\text{K×mol}}$

$\begin{array}{c}{\text{n}}_{\text{hydrocarbon}}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{0.959atm×1.000L}}{\text{0.08206}\frac{\text{L×atm}}{\text{K×mol}}\text{×298K}}\\ {\text{=0.0392mol}}_{\text{x}}{\text{H}}_{\text{y}}\\ {\text{n}}_{\text{products}}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{1.51atm×4.000L}}{\text{0.08206}\frac{\text{L×atm}}{\text{K×mol}}\text{×375K}}\\ \text{=0.1963molproducts}\end{array}$

Step 4: Molecular formula of the hydrocarbon

With this, we can create equations,

$\begin{array}{l}\text{0.0392x+0.0392}\left(\frac{\text{y}}{\text{2}}\right)\text{=0.1963molproducts}\\ \text{0.0392x×}\left({\text{44.01g/molCO}}_{\text{2}}\right)\text{+0.0392}\left(\frac{\text{y}}{\text{2}}\right)\text{×}\left({\text{18.02g/molH}}_{\text{2}}\text{O}\right)\\ \text{=5.564gofproducts}\end{array}$

and use them to solve for x and y. Solving for the two values, we get$\text{x=2andy=6}$giving us the formula of the hydrocarbon${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ or ethane.

Therefore,the identity of the unknown hydrocarbon is ethane, and its formula is.${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$