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

##”CH”_4##

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 .

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

• ##”4.5 g carbon, C”##
• ##”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

##”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”##

##”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

##color(green)(bar(ul(|color(white)(a/a)color(black)(“C”_1″H”_4 implies “CH”_4)color(white)(a/a)|)))##

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