Question

If the empirical formula is `C_2H_5`, and the molecular mass is `58*g*mol^-1`, what is the molecular formula?

Answer 1

If I am reading your question correctly, it is `H_3C-CH_2CH_2CH_3` or `C_4H_10`.

Explanation:

Butane is a `C_4` alkane. The isomer depicted is the straight chain, n -butane. It has 1 other isomer.

Answer 2

`"C"_4"H"_10`

Explanation:

I assume that you have to determine the molecular formula of butane starting from its empirical formula and its molar mass, which is `~= "58 g"`.

If you started from butane's percent composition of carbon and hydrogen, and provided that your calculations were correct, then the empirical formula should have come out as `"C"_2"H"_5`.

Now, you know that the empirical formula tells you the simplest ratio of atoms between the elements that make up a compound.

In your case, you know that a butane molecule contains 5 atoms of hydrogen for every 2 atoms of carbon.

Since you know what the compound's molar mass is, you can say that

`(2 xx M_"M carbon" + 5 xx M_"M hydrogen") * color(blue)(n) = "58 g"`

You know that the molecule contains at least two atoms of carbon and five atoms of hydrogen, but you don't exactly how many of each you have `->` this is what `color(blue)(n)` represents.

So, use the molar masses of carbon and hydrogen to get the value of `color(blue)(n)`

`(2 xx "12.01 g/mol" + 5 xx "1.01 g/mol") * color(blue)(n) = "58 g"`

`color(blue)(n) = (58color(red)(cancel(color(black)("g"))))/(29.07color(red)(cancel(color(black)("g")))) = "1.995 ~= 2`

This means tha tthe molecular formula of butane is

`("C"_2"H"_5)_2 = color(green)("C"_4"H"_10)`