# Question 00913

Nov 27, 2015

${\text{C"_text(n)"H}}_{\textrm{2 n}}$

#### Explanation:

As you know, a hydrocarbon is an organic compound that only contains carbon and hydrogen.

This means that if you know how much carbon it contains, you can automatically say how much hydrogen it contains. IN this case, the sample contains 86% carbon and 14% hydrogen, since

86% + 14% = 100%

To make calculations easier, asu\sume that you have a $\text{100.0-g}$ sample of this hydrocarbon. This means that the sample will contain

• 86% carbon $\to \text{ 86 g C}$
• 14% hydrogen $\to \text{ 14 g H}$

Your strategy will now be to determine how many moles of each element you have in your sample.

To do that, use the molar masses of the two elements

$\text{For C: " 86color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "7.160 moles C}$

$\text{For H: " 14color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "13.89 moles H}$

Next, divide both values by the smallest one to get the empirical formula of the compound

"For C: " (7.160 color(red)(cancel(color(black)("moles"))))/(7.160 color(red)(cancel(color(black)("moles")))) = 1

"For H: " (13.89 color(red)(cancel(color(black)("moles"))))/(7.160 color(red)(cancel(color(black)("moles")))) = 1.9399 ~~ 2#

The empirical formula of the compound will be

${\text{C"_1"H"_2 implies "CH}}_{2}$

To get the actual molecular formula of the compound, you would need to know its molar mass. However, possible molecular formulas for $\text{X}$ will take the form

${\left({\text{CH}}_{2}\right)}_{\textcolor{b l u e}{n}}$ $\implies {\text{C"_text(n)"H}}_{\textrm{2 n}}$