# What is the empirical formula of methane given methane can be decomposed into 4.5 g of carbon and 1.5 g of hydrogen?

Oct 30, 2016

${\text{CH}}_{4}$

#### Explanation:

Your goal when trying to figure out a compound's empirical formula is to find the smallest whole number ratio that exists between its constituent elements.

The problem tells you that a sample of methane underwent combustion and produced

• $\text{4.5 g carbon, C}$
• $\text{1.5 g hydrogen, H}$

The first thing to do here is to convert these masses to moles by using the molar masses of the two elements. You will have

$\text{For C: " 4.5 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.3747 moles C}$

$\text{For H: " 1.5 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.008color(red)(cancel(color(black)("g")))) = "1.4881 moles H}$

Now, in order to find the mole ratio that existed between these two elements in the sample that underwent combustion, you must divide both values by the smallest one.

You will have

"For C: " (0.3747 color(red)(cancel(color(black)("moles"))))/(0.3747color(red)(cancel(color(black)("moles")))) = 1

"For H: " (1.4881 color(red)(cancel(color(black)("moles"))))/(0.3747color(red)(cancel(color(black)("moles")))) = 3.97 ~~ 4

Since $1 : 4$ is already the smallest whole number ratio that can exist between these two elements, you can say that the empirical formula for methane is

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{C"_1"H"_4 implies "CH}}_{4}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$